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https://sbnwiki.astro.umd.edu/wiki/index.php?title=Example_Python_Reader_for_PDS4_Images&diff=prev&oldid=1427	[ "# Difference between revisions of \"Example Python Reader for PDS4 Images\"\n\n## Introduction\n\nThis document describes how to use Python to read in an image from a PDS4 data product. The code will not validate the label, but will read the data based on the label keywords. The code below is limited to reading BOPPS BIRC images, but can be used as an example for other arrays.\n\n## Requirements\n\nThis example assumes the user is running Python 2.7, with a recent NumPy package installed. The visualization example uses matplotlib.\n\n## Goal and Method\n\nThe goal is to read in an image from a BOPPS BIRC data product into a Numpy array. We will provide the script with the name of the label, the script will then\n\n1. Open the label.\n2. Find the data product file name.\n3. Determine the Array_2D_Image data type and shape.\n4. Read in the data array.\n5. Return the array.\n\n## Implementation\n\nFor this basic example, we designed the reader as a function in a module named birc. The user calls a single function, birc.read_image(), passing the name of the label as the first argument. The function will load the label using the ElementTree module and find the first Array_2D_Image element to read in. A second function, read_pds4_array(), determines the correct data type and shape, then reads the data from the file. A class specifically designed for PDS4 Array_2D_Image objects, aptly named PDS4_Array_2D_Image, is initialized with the data, the label describing the data, and the local_identifier of the array. The local_identifier is not required, but is present in the BIRC labels, so our class assumes it is included. The class then determines the image orientation. The image is stored as a class attribute data. The class attribute display_data is also provided, which can be used for displaying with the origin in the lower left corner. The code is listed below.\n\n## Example\n\nUsing the birc.py module listed below, this example reads in an image, zeros out the first 10 rows of the array, then displays the data with matplotlib.\n\n```import birc\nimport matplotlib.pyplot as plt\n\n# the image is im.data\n# the image for displaying purposes is im.display_data\n\n# zero out the first 10 rows\nim[:10] = 0\n\n# display the result\nplt.clf()\nplt.imshow(im.display_data, origin='lower')\nplt.draw()\n```\n\n## birc.py\n\n```\"\"\"\nbirc --- Example PDS4 Array_2D_Image reader for BOPPS/BIRC data\n===============================================================\n\nExample\n-------\n\nimport birc\nimport matplotlib.pyplot as plt\n\n# array is im.data\n# array for displaying is im.display_data\n\nplt.clf()\nplt.imshow(im.display_data, origin='lower')\nplt.draw()\n\nClasses\n-------\nPDS4_Array_2D_Image - A PDS4 2D image.\n\nFunctions\n---------\nread_image - Read a BIRC image described by a PDS4 label file.\n\n\"\"\"\n\nimport numpy as np\nimport xml.etree.ElementTree as ET\n\nclass PDS4_Array_2D_Image(object):\n\"\"\"A PDS4 array for 2D images, with limited functionality.\n\nParameters\n----------\ndata : ndarray\nThe data.\nlocal_identifier : string\nThe label's local_identifier for this array.\nlabel : ElementTree\nThe PDS4 label that contains the description of the array.\n\nAttributes\n----------\ndata : ndarray\nSee Parameters.\nlabel : ElementTree\nSee Parameters.\nlocal_identifier : string\nSee Parameters.\n\nhorizontal_axis : int\nThe index of the horizontal axis for display.\nvertical_axis : int\nThe index of the vertical axis for display.\n\ndisplay_data : ndarray\nThe data array rotated into display orientation, assuming the\ndisplay will draw the image with the origin in the lower left\ncorner. The vertical axis will be axis 0, the horizontal axis\nwill be axis 1.\n\n\"\"\"\n\ndef __init__(self, data, local_identifier, label):\nself.data = data\nself.local_identifier = local_identifier\nself.label = label\nself._orient()\n\ndef _orient(self):\n\"\"\"Set object image orientation attributes.\"\"\"\n\n# namespace definitions\nns = {'pds4': 'http://pds.nasa.gov/pds4/pds/v1',\n'disp': 'http://pds.nasa.gov/pds4/disp/v1'}\n\n# find local_identifier in File_Area_Observational\narray = None\nxpath = ('./pds4:File_Area_Observational/pds4:Array_2D_Image/'\n'[pds4:local_identifier]')\nfor e in self.label.findall(xpath, ns):\nthis_local_id = e.find('./pds4:local_identifier', ns).text.strip()\nif this_local_id == self.local_identifier:\narray = e\nbreak\n\nassert array is not None, \"Array_2D_Image with local_identifier == {} not found.\".format(self.local_identifier)\n\n# find display_settings_to_array for local_identifier in\n# Display_Settings\ndisplay_settings = None\nxpath = ('./pds4:Observation_Area/pds4:Discipline_Area/'\n'disp:Display_Settings')\nfor e in self.label.findall(xpath, ns):\nlir = e.find('./disp:Local_Internal_Reference', ns)\nreference = lir.find('./disp:local_identifier_reference', ns).text.strip()\nif reference == self.local_identifier:\ndisplay_settings = e\nbreak\n\nassert display_settings is not None, \"Display_Settings for local_identifier == {} not found.\".format(self.local_identifier)\n\n# determine display directions\ndisplay_dir = e.find('./disp:Display_Direction', ns)\nh = display_dir.find('./disp:horizontal_display_direction', ns)\nv = display_dir.find('./disp:vertical_display_direction', ns)\nself.display_directions = (h.text.strip(), v.text.strip())\n\n# determine horizonal and vertical axes\nhaxis = display_dir.find('./disp:horizontal_display_axis', ns).text.strip()\nfor axis in array.findall('./pds4:Axis_Array', ns):\nif axis.find('./pds4:axis_name', ns).text.strip() == haxis:\nsn = int(axis.find('./pds4:sequence_number', ns).text.strip())\nself.horizontal_axis = sn - 1\n\nvaxis = display_dir.find('./disp:vertical_display_axis', ns).text.strip()\nfor axis in array.findall('./pds4:Axis_Array', ns):\nif axis.find('./pds4:axis_name', ns).text.strip() == vaxis:\nsn = int(axis.find('./pds4:sequence_number', ns).text.strip())\nself.vertical_axis = sn - 1\n\n@property\ndef display_data(self):\n# only need to roll one axis for a 2D image\nim = np.rollaxis(self.data, self.vertical_axis)\nif 'Right to Left' in self.display_directions:\nim = im[:, ::-1]\nif 'Top to Bottom' in self.display_directions:\nim = im[::-1]\nreturn im\n\n\"\"\"Read a BIRC image described by a PDS4 label file.\n\nOnly the first Array_2D_Image is returned, based on BIRC PDS4\nsample data files.\n\nParameters\n----------\nfile_name : string\nThe name of the PDS4 label file describing the BIRC image.\n\nReturns\n-------\nim : PDS4_Array_2D_Image\nThe image\n\nRaises\n------\nNotImplementedError\n\n\"\"\"\nimport os\n\n# namespace definitions\nns = {'pds4': 'http://pds.nasa.gov/pds4/pds/v1'}\n\nlabel = ET.parse(file_name)\n\n# Find the first File_Area_Observational element with an Array_2D_Image\nfind = label.findall(\n'pds4:File_Area_Observational/[pds4:Array_2D_Image]', ns)\n\nif len(find) > 1:\nraise NotImplementedError(\"Multiple Array_2D_Image elements found.\")\nelse:\nfile_area = find\n\nfile_area, './pds4:Array_2D_Image', ns,\ndirname=os.path.dirname(file_name))\n\nreturn PDS4_Array_2D_Image(data, local_identifier, label)\n\nParameters\n----------\nfile_area : ElementTree Element\nThe File_Area_Observational element from the PDS4 label.\nxpath : string\nThe array is described by the element `file_area.find(xpath)`.\nns : dictionary\nNamespace definitions for `file_area.find()`.\ndirname : string\nThe original data label's directory name, used to find the array\nfile.\n\nReturns\n-------\ndata : ndarray\nThe array.\nlocal_identifier : string\nThe local_identifier of the array, or `None` if not present.\n\nRaises\n------\nNotImplementedError\n\n\"\"\"\n\nimport os\n\nfile_name = file_area.find('pds4:File/pds4:file_name', ns).text.strip()\nfile_name = os.path.join(dirname, file_name)\n\narray = file_area.find(xpath, ns)\n\nlocal_identifier = array.find('./pds4:local_identifier', ns).text.strip()\n\n# Examine the data type, and translate it into a numpy dtype\npds4_to_numpy_dtypes = {\n\"IEEE754MSBSingle\": '>f4'\n}\ntry:\nk = array.find('pds4:Element_Array/pds4:data_type', ns).text.strip()\ndtype = np.dtype(pds4_to_numpy_dtypes[k])\nexcept KeyError:\nraise NotImplementedError(\"PDS4 data_type {} not implemented.\".format(k))\n\n# determine the shape\nndim = int(array.find('pds4:axes', ns).text.strip())\nshape = ()\nfor i in range(ndim):\nk = ('./pds4:Axis_Array/[pds4:sequence_number=\"{}\"]/pds4:elements').format(i + 1)\nshape += (int(array.find(k, ns).text.strip()), )\n\n# verify axis order\naxis_index_order = array.find('./pds4:axis_index_order', ns).text.strip()\nassert axis_index_order == 'Last Index Fastest', \"Invalid axis order: {}\".format(axis_index_order)\n\noffset = array.find('./pds4:offset', ns)\nassert offset.attrib['unit'].lower() == 'byte', \"Invalid file offset unit\"\nwith open(file_name, 'r') as inf:\ninf.seek(int(offset.text.strip()))\ndata = np.fromfile(inf, dtype, count=np.prod(shape)).reshape(shape)\n\nreturn data, local_identifier\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5231015,"math_prob":0.6176084,"size":11646,"snap":"2021-43-2021-49","text_gpt3_token_len":2873,"char_repetition_ratio":0.15856382,"word_repetition_ratio":0.34528553,"special_character_ratio":0.2726258,"punctuation_ratio":0.20426123,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9566692,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-01T22:39:25Z\",\"WARC-Record-ID\":\"<urn:uuid:83a78354-2655-4dc1-8167-64ba1273831b>\",\"Content-Length\":\"34379\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f094125c-6d5f-4399-b716-425757ba6d53>\",\"WARC-Concurrent-To\":\"<urn:uuid:a53279d3-95e8-41a4-b6f1-7138771ce6cd>\",\"WARC-IP-Address\":\"129.2.14.91\",\"WARC-Target-URI\":\"https://sbnwiki.astro.umd.edu/wiki/index.php?title=Example_Python_Reader_for_PDS4_Images&diff=prev&oldid=1427\",\"WARC-Payload-Digest\":\"sha1:RRWYTNYM75TBDJ4F27MMFZFE4WE5G3QR\",\"WARC-Block-Digest\":\"sha1:37W5FC7LJY6M7NVJEWSB6OM5JZOT2MB7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964360951.9_warc_CC-MAIN-20211201203843-20211201233843-00148.warc.gz\"}"}
https://gmatclub.com/forum/the-greatest-common-factor-of-16-and-the-positive-integer-n-109870.html?sort_by_oldest=true	[ "GMAT Question of the Day: Daily via email \| Daily via Instagram New to GMAT Club? Watch this Video\n\n It is currently 13 Jul 2020, 07:24", null, "### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we’ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.", null, "", null, "# The greatest common factor of 16 and the positive integer n\n\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager", null, "", null, "Joined: 28 Aug 2010\nPosts: 139\nThe greatest common factor of 16 and the positive integer n [#permalink]\n\n### Show Tags\n\n9\n59", null, "00:00\n\nDifficulty:", null, "", null, "", null, "75% (hard)\n\nQuestion Stats:", null, "60% (02:20) correct", null, "40% (02:25) wrong", null, "based on 691 sessions\n\n### HideShow timer Statistics\n\nThe greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3. Which of the following could be the greatest common factor of n and 210?\n\nA. 3\nB. 14\nC. 30\nD. 42\nE. 70\nVeritas Prep GMAT Instructor", null, "V\nJoined: 16 Oct 2010\nPosts: 10666\nLocation: Pune, India\nRe: The greatest common factor of 16 and the positive integer n [#permalink]\n\n### Show Tags\n\n14\n5\najit257 wrote:\nThe greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3. Which of the following could be the greatest common factor of n and 210?\n\n3\n\n14\n\n30\n\n42\n\n70\n\ni am not so sure about the oa.\n\nGCF (n, 16) = 4\nThis means 4 is a factor of n but 8 and 16 are not. (If 8 were a factor of n too, the GCF would have been 8. Similarly for 16)\n\nGCF (n, 45) = 3\nThis means 3 is a factor of n but 9 and 5 are not. Same logic as above.\n\n210 = 2357\nn has 4 and 3 as factors and it doesn't have 5 as a factor.\nso GCF of n and 210 could be 6 (if 7 is not a factor of n) or 42 (if 7 is a factor of n)\n\nNote: 3 is definitely not the GCF of n and 210 because they definitely have 32 in common. So GCF has to be at least 6.\n_________________\nKarishma\nVeritas Prep GMAT Instructor\n\nDirector", null, "", null, "Status: -=Given to Fly=-\nJoined: 04 Jan 2011\nPosts: 776\nLocation: India\nSchools: Haas '18, Kelley '18\nGMAT 1: 650 Q44 V37\nGMAT 2: 710 Q48 V40\nGMAT 3: 750 Q51 V40\nGPA: 3.5\nWE: Education (Education)\nRe: The greatest common factor of 16 and the positive integer n [#permalink]\n\n### Show Tags\n\n5\n4\nEDIT.... My explanation was wrong :p [Correcting it]\n\nCorrected Version:\n\nLet's try the Prime Box Approach\n\nPrime Box is simply a collection of all prime factors of a given number!\n\n(1) Prime Box of 16 = \|2, 2, 2, 2\|\n(2) Prime Box of 45 = \|3, 3, 5\|\n\n(3) Prime Box of n = \|2, 2, 3....\|\n(4) Prime Box of 210 = \|2, 5, 3, 7\|\n\nFrom 3 and 4:\nThe GCF of n and 210 must be a multiple of 6.\n\nSo we can eliminate A, B and E!\n\nFrom 2 and 3:\nn is not a multiple of 5. If it were, the GCF of n and 45 would have been 15!\nSo we can eliminate C\n\nThe only remaining choice is 'D'\n_________________\n\nOriginally posted by AmrithS on 22 Feb 2011, 18:36.\nLast edited by AmrithS on 22 Feb 2011, 18:51, edited 2 times in total.\n##### General Discussion\nManager", null, "", null, "Joined: 28 Aug 2010\nPosts: 139\nRe: The greatest common factor of 16 and the positive integer n [#permalink]\n\n### Show Tags\n\nthanks Karishma ...my doubt is exactly the same thing but for 16 ..how come 42 is correct if n and 16 have a gcf of 4\nVeritas Prep GMAT Instructor", null, "V\nJoined: 16 Oct 2010\nPosts: 10666\nLocation: Pune, India\nRe: The greatest common factor of 16 and the positive integer n [#permalink]\n\n### Show Tags\n\n1\najit257 wrote:\nthanks Karishma ...my doubt is exactly the same thing but for 16 ..how come 42 is correct if n and 16 have a gcf of 4\n\nThat is because 210 has only one 2. Even though n has a 4 as factor, 210 does not. Therefore GCF of n and 210 does not have 4 as a factor. Does it make sense now?\n_________________\nKarishma\nVeritas Prep GMAT Instructor\n\nManager", null, "", null, "Joined: 28 Aug 2010\nPosts: 139\nRe: The greatest common factor of 16 and the positive integer n [#permalink]\n\n### Show Tags\n\nso lets say if another choice was given as 6 then what could have been the ans or it would not be that close.\nVeritas Prep GMAT Instructor", null, "V\nJoined: 16 Oct 2010\nPosts: 10666\nLocation: Pune, India\nRe: The greatest common factor of 16 and the positive integer n [#permalink]\n\n### Show Tags\n\najit257 wrote:\nso lets say if another choice was given as 6 then what could have been the ans or it would not be that close.\n\nThen there would have been two correct options: 6 and 42. Either can be the GCF of n and 210 depending on what exactly n is.\nGMAT never has 2 correct options and hence such a scenario is not possible. Only one of 6 and 42 would be in the answer choices.\n_________________\nKarishma\nVeritas Prep GMAT Instructor\n\nMath Expert", null, "V\nJoined: 02 Sep 2009\nPosts: 65238\n\n### Show Tags\n\n4\nrosgmat wrote:\nThe greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3. Which of the following could be the greatest common factor of n and 210?\n\na) 3\nb) 14\nc) 30\nd) 42\ne) 70\n\nThe greatest common factor of 2^4=16 and n is 4 --> n is a multiple of 2^2=4 but not the higher powers of 2, for example 2^3=8 or 2^4=16, because if it were then the greatest common factor of 16 and n would be more than 4;\n\nThe greatest common factor of 3^25=45 and n is 3 --> n is a multiple of 3 but not the higher powers of 3 and not 5, because if it were then the greatest common factor of 3^25=45 and n would be more than 3;\n\nSo, n is a multiple of 2^23=12, not a multiple of higher powers of 2 or 3, and not a multiple of 5 (so n=12x where x could be any positive integer but 2, 3, or 5). Now, as 210=2357 then the greatest common factor of n and 210 could be 6 or 67=42 (if 7 is a factor of n).\n\n_________________\nManager", null, "", null, "Joined: 17 Feb 2011\nPosts: 139\nConcentration: Real Estate, Finance\nSchools: MIT (Sloan) - Class of 2014\nGMAT 1: 760 Q50 V44\n\n### Show Tags\n\n3\nYou can do the prime boxes.\n\nPrime box of 16: 2, 2, 2, 2\nPrime box of 45: 3, 3, 5\n\nPrime box of 210: 2, 5, 3, 7\n\nSo, n has at least two 2's and one 3, but n hasn't got any 5. Now, checking alternatives:\nA) wrong, as n and 210 share at least one 2 and one 3.\nB) wrong again, no 3 in 14.\nC) wrong, as 30 has a 5\nD) correct. 42 prime box is 2, 3, 7, so it meets all requirements.\nE) wrong, 70 prime box has 2, 7 and 5\nIntern", null, "", null, "Joined: 09 Jul 2012\nPosts: 22\nLocation: India\nConcentration: Strategy, Sustainability\nGMAT 1: 700 Q50 V34\n\n### Show Tags\n\n1\nThe greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3. Which of the following could be the greatest common factor of n and 210?\nSenior Manager", null, "", null, "Joined: 23 Mar 2011\nPosts: 392\nLocation: India\nGPA: 2.5\nWE: Operations (Hospitality and Tourism)\nRe: The greatest common factor of 16 and the positive integer n [#permalink]\n\n### Show Tags\n\n2\nmun23 wrote:\nThe greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3. Which of the following could be the greatest common factor of n and 210?\n\n(A)3\n(B)14\n(C)30\n(D)42\n(E)70\n\nNeed easy explanation to solve it quickly\n\nHi, let me try to explain in simpler way:\n\nGCF = 4 = 2^2\n16 = 2^4\nthat means prime box of n = 2^2 , ? ? ?\n\nGCF = 3\n45 = 3^25\nthat means prime box of n = 3^1, ? ? ?\n\nOverall prime box of n = 2^2, 3^1, ???\n\nNow, 210 = 3752\nfrom above we know the prime factor and powers of n (not complete ??)\ntherefore GCF = 3^1 2^1 * 7^1 (not 5 - we have seen above, but at least a 7 is possible)\nThus at least a GCF of 42 is possible here\n\nHope this helps\nManager", null, "", null, "Joined: 09 Apr 2013\nPosts: 185\nLocation: United States\nConcentration: Finance, Economics\nGMAT 1: 710 Q44 V44\nGMAT 2: 740 Q48 V44\nGPA: 3.1\nWE: Sales (Mutual Funds and Brokerage)\nRe: The greatest common factor of 16 and the positive integer n [#permalink]\n\n### Show Tags\n\n1\nSimple explanation?\n\nFirst, do a factor tree for each number.\n\nYou'll see that 5 can't be a factor of N (otherwise it would have been the highest factor between N and 45).\n\nThe highest factor that could theoretically exist between N and 210 is therefore all of the factors of 210 besides those we've ruled out. 210 is factored to 2 * 3 * 5 * 7. we've ruled out 5, so 2 * 3 * 7 = 42. Answer is D.\n\nIf the question asked \"the highest factor that we KNOW exists\" rather than \"COULD\" exist, the answer would be six since 2 and 3 are both factors of N, as well as of 210.\nManager", null, "", null, "B\nJoined: 10 Mar 2014\nPosts: 177\n\n### Show Tags\n\nBunuel wrote:\nrosgmat wrote:\nThe greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3. Which of the following could be the greatest common factor of n and 210?\n\na) 3\nb) 14\nc) 30\nd) 42\ne) 70\n\nThe greatest common factor of 2^4=16 and n is 4 --> n is a multiple of 2^2=4 but not the higher powers of 2, for example 2^3=8 or 2^4=16, because if it were then the greatest common factor of 16 and n would be more than 4;\n\nThe greatest common factor of 3^25=45 and n is 3 --> n is a multiple of 3 but not the higher powers of 3 and not 5, because if it were then the greatest common factor of 3^25=45 and n would be more than 3;\n\nSo, n is a multiple of 2^23=12, not a multiple of higher powers of 2 or 3, and not a multiple of 5 (so n=12x where x could be any positive integer but 2, 3, or 5). Now, as 210=2357 then the greatest common factor of n and 210 could be 6 or 67=42 (if 7 is a factor of n).\n\nHi Bunnel,\n\nI have resolved this till n is a multiple of 2^23=12\n\nwhy we are not considering 5 but we are considering 7.\n\nThanks\nMath Expert", null, "V\nJoined: 02 Sep 2009\nPosts: 65238\n\n### Show Tags\n\nPathFinder007 wrote:\nBunuel wrote:\nrosgmat wrote:\nThe greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3. Which of the following could be the greatest common factor of n and 210?\n\na) 3\nb) 14\nc) 30\nd) 42\ne) 70\n\nThe greatest common factor of 2^4=16 and n is 4 --> n is a multiple of 2^2=4 but not the higher powers of 2, for example 2^3=8 or 2^4=16, because if it were then the greatest common factor of 16 and n would be more than 4;\n\nThe greatest common factor of 3^25=45 and n is 3 --> n is a multiple of 3 but not the higher powers of 3 and not 5, because if it were then the greatest common factor of 3^25=45 and n would be more than 3;\n\nSo, n is a multiple of 2^23=12, not a multiple of higher powers of 2 or 3, and not a multiple of 5 (so n=12x where x could be any positive integer but 2, 3, or 5). Now, as 210=2357 then the greatest common factor of n and 210 could be 6 or 67=42 (if 7 is a factor of n).\n\nHi Bunnel,\n\nI have resolved this till n is a multiple of 2^23=12\n\nwhy we are not considering 5 but we are considering 7.\n\nThanks\n\nNotice that the GCF of 45 (a multiple of 5) and n is 3 (not a multiple of 5). This means that n itself cannot be a multiple of 5.\n\nAs for 7: we know for sure that 2 and 3 are factors of n and 5 is not a factor of n. We know nothing about its other primes, so any prime greater than 5 theoretically can be a factor of n.\n_________________\nSVP", null, "", null, "Status: The Best Or Nothing\nJoined: 27 Dec 2012\nPosts: 1706\nLocation: India\nConcentration: General Management, Technology\nWE: Information Technology (Computer Software)\nRe: The greatest common factor of 16 and the positive integer n [#permalink]\n\n### Show Tags\n\n1\n16........ n .......................... n ....... 45\n\nGCF = 4 ................................. GCF = 3\n\nSo n may be either 4 * 3 = 12 or a multiple of 12 to satisfy the GCF values given.\n\n210 = 7 * 2 * 5 * 3\n\n2, 3 & 5 are associated with 16 & 45. Including them in calculations will disturb the already provided GCF values\n\nOnly 7 stands out.\n\nSo 12 * 7 = 84\n\nGCF of 84 & 210 = 42\n\nDirector", null, "", null, "S\nJoined: 17 Dec 2012\nPosts: 635\nLocation: India\nThe greatest common factor of 16 and the positive integer n [#permalink]\n\n### Show Tags\n\n1\nn has to be a multiple of (22)3 = 12\nA common factor between 210= (2357) and multiple of 12 is 23=6\nSo the G.C.F of n and 210 has to be a multiple of 6\nThe two choices that are multiples of 6 are 30 and 42.\nBur n is not a multiple of 5 .So 30 can be ruled out and the answer is 42.\n_________________\nSrinivasan Vaidyaraman\nMagical Logicians\nHolistic and Holy Approach\nCurrent Student", null, "B\nJoined: 23 Dec 2013\nPosts: 132\nLocation: United States (CA)\nGMAT 1: 710 Q45 V41\nGMAT 2: 760 Q49 V44\nGPA: 3.76\nRe: The greatest common factor of 16 and the positive integer n [#permalink]\n\n### Show Tags\n\n1\n1\najit257 wrote:\nThe greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3. Which of the following could be the greatest common factor of n and 210?\n\nA. 3\nB. 14\nC. 30\nD. 42\nE. 70\n\nThis problem works best if you break it down into pieces with Venn diagrams.\n\n16 and n have a GCF of 4 = 2^2. That means n has at least two 2's as factors.\n\n45 and n share one factor: 3. That means n has at least one 3 and two 2's (32^2).\n\nThe next step is to break down 210 into 2375. 5 is not a possible factor of n because it would've been a GCF with 45 and n. But 7 could be a common divisor because it's not expressly forbidden anywhere. So the GCF in this case is 237 = 42.\nGMATH Teacher", null, "P\nStatus: GMATH founder\nJoined: 12 Oct 2010\nPosts: 938\nRe: The greatest common factor of 16 and the positive integer n [#permalink]\n\n### Show Tags\n\najit257 wrote:\nThe greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3. Which of the following could be the greatest common factor of n and 210?\n\nA. 3\nB. 14\nC. 30\nD. 42\nE. 70\n\n$$?\\,\\,\\,:\\,\\,\\,GCF\\left( {n\\,,2 \\cdot 3 \\cdot 5 \\cdot 7} \\right)\\,\\,\\underline {{\\rm{could}}\\,\\,{\\rm{be}}}$$\n\n$$n \\ge 1\\,\\,\\,{\\mathop{\\rm int}}$$\n\n$$GCF\\left( {{2^4},n} \\right) = {2^2}\\,\\,\\,\\, \\Rightarrow \\,\\,\\,\\,\\,\\left\\{ \\matrix{ {n \\over {{2^2}}} = {\\mathop{\\rm int}} \\hfill \\cr {n \\over {{2^{\\, \\ge \\,3}}}} \\ne {\\mathop{\\rm int}} \\hfill \\cr} \\right.$$\n\n$$GCF\\left( {{3^2} \\cdot 5,n} \\right) = 3\\,\\,\\,\\, \\Rightarrow \\,\\,\\,\\,\\,\\left\\{ \\matrix{ {n \\over 3} = {\\mathop{\\rm int}} \\hfill \\cr {n \\over {{3^{\\, \\ge \\,2}}}} \\ne {\\mathop{\\rm int}} \\,\\,\\,\\,\\,;\\,\\,\\,{n \\over 5} \\ne {\\mathop{\\rm int}} \\,\\, \\hfill \\cr} \\right.$$\n\n$$? = {2^1} \\cdot {3^1} \\cdot {7^{0\\,{\\rm{or}}\\,1}}\\,\\,\\,\\,\\,\\mathop \\Rightarrow \\limits^{{\\rm{alternatives}}\\,!} \\,\\,\\,\\,42\\,\\,\\,\\,\\,\\,\\left( D \\right)$$\n\nThis solution follows the notations and rationale taught in the GMATH method.\n\nRegards,\nFabio.\n_________________\nFabio Skilnik :: GMATH method creator (Math for the GMAT)\nOur high-level \"quant\" preparation starts here: https://gmath.net\nTarget Test Prep Representative", null, "V\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 11096\nLocation: United States (CA)\nRe: The greatest common factor of 16 and the positive integer n [#permalink]\n\n### Show Tags\n\najit257 wrote:\nThe greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3. Which of the following could be the greatest common factor of n and 210?\n\nA. 3\nB. 14\nC. 30\nD. 42\nE. 70\n\nIf the greatest common factor (GCF) of 16 and n is 4, n could be 4, 12, 20, 28, 36, etc. In other words, n is an odd multiple of 4.\n\nSince the GCF of 45 and n is 3, and since 45 and 4 have no common factor other than 1, n must be a multiple of 3 x 4 = 12. However, since n is an odd multiple of 4, n actually has to be an odd multiple of 12 also.\n\nIf n = 12, we see that GCF(45, 12) = 3 and GCF(12, 210) = 6. However, 6 is not one of the choices.\n\nIf n = 36, we see that GCF(45, 36) = 9, but GCF(45, n) is supposed to be 3. So n can’t be 36.\n\nIf n = 60, we see that GCF(45, 60) = 15, but GCF(45, n) is supposed to be 3. So n can’t be 60.\n\nIf n = 84, we see that GCF(45, 84) = 3 and GCF(84, 210) = 42. We see that 42 is one of the choices.\n\n_________________\n\n# Scott Woodbury-Stewart\n\nFounder and CEO\n\[email protected]\n\n214 REVIEWS\n\n5-STARS RATED ONLINE GMAT QUANT SELF STUDY COURSE\n\nNOW WITH GMAT VERBAL (BETA)\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nCEO", null, "", null, "V\nJoined: 03 Jun 2019\nPosts: 3243\nLocation: India\nGMAT 1: 690 Q50 V34", null, "WE: Engineering (Transportation)\nThe greatest common factor of 16 and the positive integer n [#permalink]\n\n### Show Tags\n\najit257 wrote:\nThe greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3. Which of the following could be the greatest common factor of n and 210?\n\nA. 3\nB. 14\nC. 30\nD. 42\nE. 70\n\nGiven: The greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3.\n\nAsked: Which of the following could be the greatest common factor of n and 210?\n\nn = 4k ; where k is not a multiple of 2\n45 = 3^25\nn = 3m; where m is not a multiple of 3 or 5\n210 = 2357\nn = 43k ; where k is any prime other than 2,3,or 5\n\ngcd (n,210) = 23 = 6 or 23*7 = 42\n\nIMO D\n_________________\nKinshook Chaturvedi\nEmail: [email protected]", null, "The greatest common factor of 16 and the positive integer n [#permalink] 29 Mar 2020, 05:56\n\n# The greatest common factor of 16 and the positive integer n", null, "", null, "" ]	[ null, "https://cdn.gmatclub.com/cdn/files/forum/styles/gmatclub_light/theme/images/profile/close.png", null, "https://cdn.gmatclub.com/cdn/files/forum/styles/gmatclub_light/theme/images/profile/close.png", null, 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https://www.origami-resource-center.com/triangle-from-a-rectangle.html	[ "# Make a Triangle from a Rectangle", null, "Some origami models start with an equilateral triangle instead of a square sheet of paper. In this case, you need to make the triangular sheets of paper before you start. This page will show you how to get a triangle from a rectangular sheet of paper. If you are starting with a square sheet of paper, or a rectangle that is not quite long enough, then click here or here.", null, "#### Make a Triangle from a Rectangle\n\n1. Fold and unfold the rectangular sheet of paper in half lengthwise.\n\n2. Fold the top-right corner of the paper down so the corner meets the crease made in step 1 (join the dots). The crease made should intersect with the bottom-right corner.\n\n3. Fold left side of the paper over the flap made in step 2. Use the raw edge of the flap as a guide to position this fold.\n\n4. Unfold and cut along the creases to get an equilateral triangle. Easy." ]	[ null, "https://www.origami-resource-center.com/images/150xNxshape-Triangle-From-Rectangle-icon.jpg.pagespeed.ic.MM9UaeB2tO.jpg", null, "https://www.origami-resource-center.com/images/shape-Triangle-From-Rectangle.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8622032,"math_prob":0.8829964,"size":875,"snap":"2019-51-2020-05","text_gpt3_token_len":197,"char_repetition_ratio":0.1423651,"word_repetition_ratio":0.024691358,"special_character_ratio":0.216,"punctuation_ratio":0.07865169,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9522092,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-12T15:12:51Z\",\"WARC-Record-ID\":\"<urn:uuid:cb094df7-98e5-4ee7-8d54-c73613b97c19>\",\"Content-Length\":\"19022\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:63045cb7-8083-4092-aeab-0f506b703aad>\",\"WARC-Concurrent-To\":\"<urn:uuid:b73d49d7-4f5f-4ea8-8635-9998f202a73a>\",\"WARC-IP-Address\":\"173.247.218.162\",\"WARC-Target-URI\":\"https://www.origami-resource-center.com/triangle-from-a-rectangle.html\",\"WARC-Payload-Digest\":\"sha1:NJIWGIH47L2KJFQDTJC6FQHGJLMTNAEZ\",\"WARC-Block-Digest\":\"sha1:UELMESRB7JW7YTEYQ7WN3WJZMCAYPIMT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540543850.90_warc_CC-MAIN-20191212130009-20191212154009-00484.warc.gz\"}"}
https://in.mathworks.com/matlabcentral/profile/authors/14644008?detail=all	[ "Community Profile", null, "# Dev Gupta\n\nLast seen: 2 months ago Active since 2019\n\nLearner\n\n#### Statistics\n\n•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "#### Content Feed\n\nView by\n\nSolved\n\nDecimation\nWhen dealing to the Roman Army, the term decimate meant that the entire unit would be broken up into groups of ten soldiers, and...\n\n1 year ago\n\nSolved\n\nThere are 10 types of people in the world\nThose who know binary, and those who don't. The number 2015 is a palindrome in binary (11111011111 to be exact) Given a year...\n\n1 year ago\n\nSolved\n\nBinary numbers\nGiven a positive, scalar integer n, create a (2^n)-by-n double-precision matrix containing the binary numbers from 0 through 2^n...\n\n1 year ago\n\nSolved\n\nImplement a bubble sort technique and output the number of swaps required\nA bubble sort technique compares adjacent items and swaps them if they are in the wrong order. This is done recursively until al...\n\n1 year ago\n\nSolved\n\nDot Product\n\n1 year ago\n\nProblem\n\nDot Product\n\n1 year ago \| 0 \| 40 solvers\n\nSolved\n\nFinding peaks\nFind the peak values in the signal. The peak value is defined as the local maxima. For example, x= [1 12 3 2 7 0 3 1 19 7]; ...\n\n1 year ago\n\nSolved\n\nMax index of 3D array\nGiven a three dimensional array M(m,n,p) write a code that finds the three coordinates x,y,z of the Maximum value. 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Example: Input x = 10 Output y is 12 ...\n\n1 year ago\n\nSolved\n\nBit Reversal\nGiven an unsigned integer _x_, convert it to binary with _n_ bits, reverse the order of the bits, and convert it back to an inte...\n\n1 year ago\n\nSolved\n\nConverting binary to decimals\nConvert binary to decimals. Example: 010111 = 23. 110000 = 48.\n\n1 year ago\n\nSolved\n\nFind out sum and carry of Binary adder\nFind out sum and carry of a binary adder if previous carry is given with two bits (x and y) for addition. Examples Previo...\n\n1 year ago\n\nSolved\n\n~~~~~~~ WAVE ~~~~~~~~~\n\|The WAVE generator\| Once upon a time there was a river. 'Sum' was passing by the river. He saw the water of the river that w...\n\n1 year ago\n\nSolved\n\nRemove the air bubbles\nGiven a matrix a, return a matrix b in which all the zeros have \"bubbled\" to the top. That is, any zeros in a given column shoul...\n\n1 year ago\n\nSolved\n\nFind Logic 32\n\n1 year ago\n\nProblem\n\nFind Logic 32\n\n1 year ago \| 1 \| 30 solvers\n\nSolved\n\nBack to basics 15 - classes\nCovering some basic topics I haven't seen elsewhere on Cody. Return the class of the input variable.\n\n1 year ago\n\nSolved\n\nFind Logic 31\n\n1 year ago\n\nProblem\n\nFind Logic 31\n\n1 year ago \| 1 \| 23 solvers\n\nSolved\n\nFind Logic 30\n\n1 year ago\n\nProblem\n\nFind Logic 30\n\n1 year ago \| 1 \| 24 solvers\n\nSolved\n\nPattern matching\nGiven a matrix, m-by-n, find all the rows that have the same \"increase, decrease, or stay same\" pattern going across the columns...\n\n1 year ago" ]	[ null, "https://in.mathworks.com/responsive_image/150/150/0/0/0/cache/matlabcentral/profiles/14644008_1548563151426_DEF.jpg", null, "https://in.mathworks.com/matlabcentral/profile/hunt/MLC_Treasure_Hunt_badge.png", null, "https://in.mathworks.com/images/responsive/supporting/matlabcentral/cody/badges/famous.png", null, "https://in.mathworks.com/images/responsive/supporting/matlabcentral/cody/badges/quiz_master.png", null, "https://in.mathworks.com/images/responsive/supporting/matlabcentral/cody/badges/curator.png", null, "https://in.mathworks.com/images/responsive/supporting/matlabcentral/cody/badges/mathworks_generic_group.png", null, "https://in.mathworks.com/images/responsive/supporting/matlabcentral/cody/badges/community_authored_group.png", null, "https://in.mathworks.com/images/responsive/supporting/matlabcentral/cody/badges/puzzler.png", null, "https://in.mathworks.com/images/responsive/supporting/matlabcentral/cody/badges/promoter.png", null, "https://in.mathworks.com/images/responsive/supporting/matlabcentral/cody/badges/speed_demon.png", null, "https://in.mathworks.com/images/responsive/supporting/matlabcentral/cody/badges/creator.png", null, "https://in.mathworks.com/images/responsive/supporting/matlabcentral/cody/badges/solver.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79330736,"math_prob":0.9373284,"size":4556,"snap":"2021-43-2021-49","text_gpt3_token_len":1267,"char_repetition_ratio":0.1761863,"word_repetition_ratio":0.075,"special_character_ratio":0.27238807,"punctuation_ratio":0.12345679,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99282527,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24],"im_url_duplicate_count":[null,1,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-03T17:14:51Z\",\"WARC-Record-ID\":\"<urn:uuid:387b782c-9418-4e60-b578-8866f67b63d9>\",\"Content-Length\":\"116930\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cf2ab53c-666e-476e-89a0-e0ba7592b2c8>\",\"WARC-Concurrent-To\":\"<urn:uuid:9b5ba5ce-12d9-4f5e-a459-0320ff77598a>\",\"WARC-IP-Address\":\"104.96.217.125\",\"WARC-Target-URI\":\"https://in.mathworks.com/matlabcentral/profile/authors/14644008?detail=all\",\"WARC-Payload-Digest\":\"sha1:6GOCNA23TCDCYK4C5B7V72C5VK5O4S67\",\"WARC-Block-Digest\":\"sha1:CVDBQHRXPTWRV2PAJ35WP7M47QUVHQGI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362891.54_warc_CC-MAIN-20211203151849-20211203181849-00064.warc.gz\"}"}
https://metanumbers.com/107107	[ "# 107107 (number)\n\n107,107 (one hundred seven thousand one hundred seven) is an odd six-digits composite number following 107106 and preceding 107108. In scientific notation, it is written as 1.07107 × 105. The sum of its digits is 16. It has a total of 4 prime factors and 16 positive divisors. There are 76,320 positive integers (up to 107107) that are relatively prime to 107107.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 6\n• Sum of Digits 16\n• Digital Root 7\n\n## Name\n\nShort name 107 thousand 107 one hundred seven thousand one hundred seven\n\n## Notation\n\nScientific notation 1.07107 × 105 107.107 × 103\n\n## Prime Factorization of 107107\n\nPrime Factorization 7 × 11 × 13 × 107\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 107107 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 107,107 is 7 × 11 × 13 × 107. Since it has a total of 4 prime factors, 107,107 is a composite number.\n\n## Divisors of 107107\n\n16 divisors\n\n Even divisors 0 16 8 8\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 145152 Sum of all the positive divisors of n s(n) 38045 Sum of the proper positive divisors of n A(n) 9072 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 327.272 Returns the nth root of the product of n divisors H(n) 11.8063 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 107,107 can be divided by 16 positive divisors (out of which 0 are even, and 16 are odd). The sum of these divisors (counting 107,107) is 145,152, the average is 9,072.\n\n## Other Arithmetic Functions (n = 107107)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 76320 Total number of positive integers not greater than n that are coprime to n λ(n) 3180 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 10176 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 76,320 positive integers (less than 107,107) that are coprime with 107,107. And there are approximately 10,176 prime numbers less than or equal to 107,107.\n\n## Divisibility of 107107\n\n m n mod m 2 3 4 5 6 7 8 9 1 1 3 2 1 0 3 7\n\nThe number 107,107 is divisible by 7.\n\n## Classification of 107107\n\n• Arithmetic\n• Deficient\n\n• Polite\n\n• Square Free\n\n### Other numbers\n\n• LucasCarmichael\n\n## Base conversion (107107)\n\nBase System Value\n2 Binary 11010001001100011\n3 Ternary 12102220221\n4 Quaternary 122021203\n5 Quinary 11411412\n6 Senary 2143511\n8 Octal 321143\n10 Decimal 107107\n12 Duodecimal 51b97\n20 Vigesimal d7f7\n36 Base36 2an7\n\n## Basic calculations (n = 107107)\n\n### Multiplication\n\nn×y\n n×2 214214 321321 428428 535535\n\n### Division\n\nn÷y\n n÷2 53553.5 35702.3 26776.8 21421.4\n\n### Exponentiation\n\nny\n n2 11471909449 1228721805354043 131604706406055483601 14095785289033384682052307\n\n### Nth Root\n\ny√n\n 2√n 327.272 47.4904 18.0907 10.1383\n\n## 107107 as geometric shapes\n\n### Circle\n\n Diameter 214214 672973 3.60401e+10\n\n### Sphere\n\n Volume 5.14686e+15 1.4416e+11 672973\n\n### Square\n\nLength = n\n Perimeter 428428 1.14719e+10 151472\n\n### Cube\n\nLength = n\n Surface area 6.88315e+10 1.22872e+15 185515\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 321321 4.96748e+09 92757.4\n\n### Triangular Pyramid\n\nLength = n\n Surface area 1.98699e+10 1.44806e+14 87452.5\n\n## Cryptographic Hash Functions\n\nmd5 c39287d312d814d143e0a4d1051ffc91 784dbca81c93bb7cd57749a764d4be9e281f5a53 6d56d1763079f9ffbd233f182a18ac63d1bb433ca53b2fe4ea0d078c272f6e9a be3b747ce58a3410196682fa025b2d5af2360d9e73bb58ecade2f40049c2e3e11541bb04a3108b88eb6e67e49951d5dca6dedf2bd0c3f43e756179bda1e72976 03820e0ebdcb85358809d2feda139e83cba4ffa1" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.60157067,"math_prob":0.97097677,"size":4616,"snap":"2021-43-2021-49","text_gpt3_token_len":1626,"char_repetition_ratio":0.120338246,"word_repetition_ratio":0.03660322,"special_character_ratio":0.45732236,"punctuation_ratio":0.07544757,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.996066,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-23T04:36:08Z\",\"WARC-Record-ID\":\"<urn:uuid:b3409f20-54bf-4adc-8df8-db53d1e141ce>\",\"Content-Length\":\"40227\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f45f9702-8a03-4d3f-aa5b-86b6c48ed706>\",\"WARC-Concurrent-To\":\"<urn:uuid:0de75a44-d490-4526-86e1-adccc47afe09>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/107107\",\"WARC-Payload-Digest\":\"sha1:HOQYONYO5QCKHSB7R7CHARNZ7DGK2DNA\",\"WARC-Block-Digest\":\"sha1:7PELVYX7DY35Z6CPJSSX2PYTUGWK2VFM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585561.4_warc_CC-MAIN-20211023033857-20211023063857-00071.warc.gz\"}"}
https://www.pcibex.net/wiki/penncontroller-gettable-filter/	[ "# PennController.GetTable().filterGlobal Commands\n\n`PennController.GetTable( tablename ).filter( \"column\" , \"match\" )`\n\nor `PennController.GetTable( tablename ).filter( \"column\" , /match/ )`\n\nor `PennController.GetTable( tablename ).filter( function )`\n\nReturns a filtered version of the table, containing only rows whose specified column’s value is a match.\n\nIf you use a string then the column’s value must match the text exactly. Alternatively, you can use a regular expression to test the column’s value. You can also use a function that will take each row as an argument and should return `true` to keep the row or `false` to exclude it.\n\nYou can use several `filter`s in chain.\n\nExample:\n\n```Template(\n.filter( row => row.Item > 0 ) // 'Item' should be greater than 0, and\n.filter( \"ButtonText\" , /second/ ) // 'ButtonText' should contain 'second'\n,\nrow => newTrial( \"button trial\" ,\nnewButton(\"test button\", row.ButtonText)\n.print()\n.wait()\n)\n.log( \"Item\" , row.Item )\n.log( \"Text\" , row.ButtonText )\n);\n```\n\nGenerates only one trial from a subset of the table spreadsheet.csv: first we only consider rows where the value in the Item column is a number greater than 0 (this is practically ineffective, for both rows in spreadsheet.csv already satisfy this condition) and we further consider only rows among those rows where the value of the ButtonText column is a text containing the string second (only the second row satisfies this condition)." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.67396504,"math_prob":0.7046261,"size":1405,"snap":"2021-04-2021-17","text_gpt3_token_len":327,"char_repetition_ratio":0.11848679,"word_repetition_ratio":0.018099548,"special_character_ratio":0.24412811,"punctuation_ratio":0.15564202,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96432245,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-24T03:18:13Z\",\"WARC-Record-ID\":\"<urn:uuid:7d87aa76-3198-47f0-8c57-db0482aa3baf>\",\"Content-Length\":\"46457\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:950addd0-62f0-42f9-add4-b7d6697c40c0>\",\"WARC-Concurrent-To\":\"<urn:uuid:627a116a-ab29-469d-98d7-19ea66cdd1ec>\",\"WARC-IP-Address\":\"45.33.65.35\",\"WARC-Target-URI\":\"https://www.pcibex.net/wiki/penncontroller-gettable-filter/\",\"WARC-Payload-Digest\":\"sha1:7WCGLCAMVFGGB3EXFRZA6U7HQUQH2Y4E\",\"WARC-Block-Digest\":\"sha1:S4LFKQCXRYFYKZQFXXDDF2FVOTYTIM6N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703544403.51_warc_CC-MAIN-20210124013637-20210124043637-00537.warc.gz\"}"}
https://www.downtoscuba.com/boyles-law-in-scuba-diving/	[ "", null, "One of the main concepts of physics explained in scuba diving is Boyle’s Law. At a glance, it is a simple expression of a pressure and volume relationship of gases. It can, however, get complex. For the purpose of this article, we are going to keep it simple. What exactly is the relationship between Boyle’s Law and scuba diving? And how does it affect us as scuba divers?\n\n## Boyle’s Law and Scuba Diving\n\nEven though Richard Towneley and Henry Power first noted the relationship between pressure and volume in the 17th century, it was Robert Boyle that confirmed their findings by conducting experiments. He reduced the volume of a closed gas container using mercury and noted the proportional relationship or pressure and volume.\n\nWe don’t have Mercury available. So what about a balloon? We have a closed container of gas. Now if we were to pull this balloon underwater, the water pressure increases linearly with depth. That means our flexible, closed system of gas (balloon) will decrease in volume the deeper we pull it.\n\nThis looks like this:\n\nDepth\n\nSurface\n\n10m\n\n20m\n\n30m\n\n40m\n\nPressure\n\n1 bar\n\n2 bar\n\n3 bar\n\n4 bar\n\n5 bar\n\nVolume\n\n1\n\n1/2\n\n1/3\n\n1/4\n\n1/5\n\nDensity\n\n1x\n\n2x\n\n3x\n\n4x\n\n5x\n\n## What is Boyle’s Law Formula\n\nNow that we know that we are talking about a law that describes the volume of a fixed amount of gas in a closed container when the pressure changes, we know we need pressure and volume as parameters in our formula.\n\nSo that leaves us with starting volume & starting pressure and leaves us with ending volume & ending pressure. Therefore this is what the Boyle’s Law formula is:\n\nP1V1 = P2V2\n\nWhere P1 is first pressure, V1 first volume and after the change we have P2 for the second pressure and V2 for the second volume.\n\nLet’s apply this formula to a dive where we take a balloon with us.\n\nThe balloon at the surface is 2.3 litres in volume and we are at the surface, so 1 bar of absolute pressure.\n\nWe want to dive to 32 meters with this balloon. Therefore, according to the Boyle’s law formula, we have:\n\nP1V1 = P2V2 and can therefore use V2 = P1V1 / P2\n\n(1*2.3) / 4.2 = 0.5476\n\nThis means that the new volume of the balloon at 32 meters depth would be about 0.55 liters.\n\nSince we don’t need to know exact volumes in scuba diving this is overkill. What we do need to understand is that the relationship of gas volume exposed to pressure is proportional. What does this mean for us as divers? For example, we need to add more physical air in order to achieve neutral buoyancy deeper. Furthermore, we cannot hold our breath and ascend. Knowing these facts is why you need to be certified in order to scuba dive.\n\nAs a result of gas volume decreasing with increased pressure, we also notice a gas’s density increase. Increased density in breathing gases leads to faster air consumption rates at depth." ]	[ null, "https://www.downtoscuba.com/wp-content/uploads/2020/03/what-is-boyles-law-696x464.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9388005,"math_prob":0.9515393,"size":2856,"snap":"2020-45-2020-50","text_gpt3_token_len":688,"char_repetition_ratio":0.13113604,"word_repetition_ratio":0.0,"special_character_ratio":0.23284313,"punctuation_ratio":0.087030716,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97440827,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-28T20:53:56Z\",\"WARC-Record-ID\":\"<urn:uuid:f79dda8f-c8cd-4f86-8730-47f17c23c187>\",\"Content-Length\":\"177682\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8f809ad0-067e-40fa-9bf9-bc65ea9b5a56>\",\"WARC-Concurrent-To\":\"<urn:uuid:b3bd76c8-8a4d-46e0-bbcf-be3de3e4d15a>\",\"WARC-IP-Address\":\"173.236.190.126\",\"WARC-Target-URI\":\"https://www.downtoscuba.com/boyles-law-in-scuba-diving/\",\"WARC-Payload-Digest\":\"sha1:EJ2GUMWEBU7SRY2NPDR3SFQGXU4I7R7J\",\"WARC-Block-Digest\":\"sha1:OQNMFCIN5IEB5JHCVOMF3Z3X367I5V7J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107900860.51_warc_CC-MAIN-20201028191655-20201028221655-00427.warc.gz\"}"}
https://ww2.mathworks.cn/matlabcentral/answers/352695-about-decimation-in-dwt	[ "16 views (last 30 days)\nAlexander Voznesensky on 14 Aug 2017\nEdited: Wayne King on 15 Aug 2017\nHi! I'm studying DWT by this link http://slideplayer.com/slide/4998644/. And I have 2 question (slide on 18.07):\n1. We should do decimation anyway: after LP-filter and after HP-filter. After LP-filter it's ok, but after HP-filter we will have an aliasing? Am I right? For ex: We have a signal with f1=100 Hz and f2=400 Hz. Fs=1000 Hz. After LP (0-250 Hz) we have subband with f1=100 Hz, Fs= 500. It's ok (f1<Fs/2). After HP (250-500 Hz) we have subband with f2=400 Hz, Fs= 500. It's not ok. Aliasing (f2>Fs/2)?\n2. How we can use only 2 filters in DWT? I don't understand, why the filters change their cut-off frequency when we use decimated signal? For ex: \"Therefore, if you calculated, say, a low-pass filter with a cutoff frequency of 10 MHz at a sampling frequency of 100 MHz, and then samples taken at a frequency of 10 GHz were applied to the input of this filter, the filter cutoff frequency would be 100 times larger, i.e., 1 GHz .\"\n\nWayne King on 14 Aug 2017\nEdited: Wayne King on 14 Aug 2017\nHi Alexander,\n1. Yes, you are correct that the downsampling causes aliasing (or can cause aliasing). However, the analysis filters and synthesis filters in the wavelet transform are designed in such a way that any aliasing is canceled upon reconstruction.\n2. You are only using two filters because you are using the so-called pyramid algorithm. You can think of the equivalent filters in terms of successive convolutions, but you should keep in the mind the so called Noble identities from multirate signal processing. That allows you to see the effect of interchanging downsampling with filtering (or on the synthesis, the effect of interchanging upsampling with filtering). Read up on the Noble identities, for one example, if you lowpass filter with the scaling filter, G(f), and then downsample by two followed by filtering with the wavelet filter, H(f) that is equivalent to the second-level wavelet coefficients (prior to decimating those by two). Schematically, that is X(f) -> G(f) -> downsample by two -> H(f). The noble identities will tell you that is equivalent to X(f) -> G(f)H(2f) -> downsample by two so the equivalent filter for the second-level wavelet coefficients is G(f)H(2f). Now let's look at that filter in MATLAB.\nN = 64;\n[G,H] = wfilters('sym4');\nGdft = fft(G,N);\nHdft = fft(H,N);\nH2 = Hdft(1+mod(2(0:N-1),N));\nH2 = Gdft.H2;\nNow let's plot Hdft (1st-level wavelet filter response) and H2 (second-level wavelet filter response)\nf = 0:1/N:1/2;\nplot(f,abs(Hdft(1:N/2+1)));\nhold on;\nplot(f,abs(H2(1:N/2+1)),'r');\nlegend('First Level Wavelet','Second Level Wavelet');\nxlim([0 1/2]);\ngrid on;\nxlabel('Cycles/Sample');\n\n#### 1 Comment\n\nAlexander Voznesensky on 14 Aug 2017\nThank you! I was trying to understand what you said. Here is the plot. I see, that we have a scaling effect. For ex, f=100 Hz (after deimation) corresponds to f=200 Hz (before decimation). I supose, this is the reason of using only 2 filters. We should multiply new frequences with factor Mlevel_number and then match them to original f-axis where we have filter's amplitude-frequency response. Formally this is understandable, but... Looks like a trick. Could you recommend some stuff about it?", null, "Wayne King on 15 Aug 2017\nEdited: Wayne King on 15 Aug 2017\nHi Alexander,\nI don't think there is a trick here. The filter responses scale as you go down in resolution but that is because the discrete wavelet transform uses L2 normalization so that the L2 norm is preserved. That is good for lots of applications, but not for others. For example, in the CWT we (and many others) believe that L1 normalization is better and that is why the CWT in MATLAB from 16b on uses L1 normalization.\nOne thing to keep in mind is that the downsampling at each level turns essentially half-band signals into full band signals, so we can say that (this is an approximation) the scaling coefficients at level 1 capture aspects of the data in the frequency interval (-1/4,1/4) while the wavelet coefficients capture information in (-1/2 -1/4) union (1/4,1/2). After downsampling both of these become full-band signals so they both have content over (-1/2, 1/2) but while the scaling coefficients maintain the frequency order, the wavelet coefficients actually map the frequencies in (1/4, 1/2) in reverse order. So a frequency in (1/4,1/2) gets mapped essentially to 2(1/2-f) where f is the original signal. For example:\n% for the classic orthonormal DWT use a power of two and dwtmode 'per'\ndwtmode('per');\nn = 0:255;\nx = cos(2pi7/16n); % frequency is 7/16 cycles/sample almost at Nyquist\nf = -1/2:1/256:1/2-1/256;\nplot(f,fftshift(abs(fft(x))))\nIf you zoom in, you see the frequency is 7/16 as expected. Now we expect that the wavelet coefficients at level one will map this to approximately 2(1/2-7/16)\n[A,D] = dwt(x,'sym4');\nfnew = -1/2:1/128:1/2-1/128;\nplot(fnew,fftshift(abs(fft(D))))\nNow the frequency is 1/8 as expected.\nI think this is cover in a lot detail in Don Percival's and Andrew Walden's textbook on wavelets in time series analysis\n\"Wavelet Methods for Time Series Analysis\"." ]	[ null, "https://www.mathworks.com/matlabcentral/answers/uploaded_files/186358/image.jpeg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92826533,"math_prob":0.9464847,"size":1727,"snap":"2020-24-2020-29","text_gpt3_token_len":474,"char_repetition_ratio":0.11201393,"word_repetition_ratio":0.0,"special_character_ratio":0.2866242,"punctuation_ratio":0.10242587,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9916923,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-14T08:05:03Z\",\"WARC-Record-ID\":\"<urn:uuid:471ce88b-0032-4d36-a7cd-08b00e7c9f5b>\",\"Content-Length\":\"127368\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5e7af662-9789-4219-89b0-7f1808b969b0>\",\"WARC-Concurrent-To\":\"<urn:uuid:2496175e-b55b-474a-8fdb-eb93441ec79b>\",\"WARC-IP-Address\":\"104.96.236.188\",\"WARC-Target-URI\":\"https://ww2.mathworks.cn/matlabcentral/answers/352695-about-decimation-in-dwt\",\"WARC-Payload-Digest\":\"sha1:Q73ODIY3TPAOQ4VPD4WTNWQHWIJCI37T\",\"WARC-Block-Digest\":\"sha1:QMOKFHDP5GSHWRMXNHOUZC3U4YUZYVQW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593657149205.56_warc_CC-MAIN-20200714051924-20200714081924-00273.warc.gz\"}"}
https://catalog.lib.kyushu-u.ac.jp/opac_detail_md/?reqCode=frombib&lang=0&amode=MD100000&opkey=&bibid=11792&start=	[ "## ＜紀要論文＞ON THE SIEGEL-TATUZAWA THEOREM FOR A CLASS OF L-FUNCTIONS\n\n作成者作成者名所属機関所属機関名 Graduate School of Information Engineering, Hiroshima University 広島大学大学院工学研究科情報工学専攻作成者名所属機関所属機関名 Graduate School of Mathematics, Nagoya University 名古屋大学大学院多元数理科学研究科英語 Faculty of Mathematics, Kyushu University 九州大学大学院数理学研究院 2008-03 62 1 201 215 Version of Record restricted access Kyushu Journal of Mathematics \|\| 62(1) \|\| p201-215 http://www2.math.kyushu-u.ac.jp/~kjm/ Kyushu Journal of Mathematics \|\| 62(1) \|\| p201-215 http://www2.math.kyushu-u.ac.jp/~kjm/ Kyushu Journal of Mathematics \|\| 62(1) \|\| p201-215 http://www2.math.kyushu-u.ac.jp/~kjm/ We consider an effective lower bound of the Siegel-Tatuzawa type for general $L$-functions with three standard assumptions. We further assume three hypotheses in this paper that are essential in dev...eloping our argument.Under these assumptions and hypotheses, we prove a theorem of Siegel-Tatuzawa type for general $L$-functions. In particular, we prove such a theorem for symmetric power $L$-functions under certain assumptions.続きを見る\n\n### 詳細\n\nレコードID 11792 査読有 Siegel's zero Tatuzawa's theorem $L$-function 1340-6116 10.2206/kyushujm.62.201 紀要論文 2009.09.25 2018.02.01" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.63171995,"math_prob":0.6316889,"size":575,"snap":"2023-40-2023-50","text_gpt3_token_len":212,"char_repetition_ratio":0.12434326,"word_repetition_ratio":0.040816326,"special_character_ratio":0.27826086,"punctuation_ratio":0.09677419,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95254207,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-24T11:32:17Z\",\"WARC-Record-ID\":\"<urn:uuid:ed8ab9e3-79a6-4600-ba7c-3253b0d3acf0>\",\"Content-Length\":\"88715\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a2b14016-f4ab-4743-8ebb-a1a8ebab7bcb>\",\"WARC-Concurrent-To\":\"<urn:uuid:f8ce8ddd-3dec-42b6-a523-a32a79000675>\",\"WARC-IP-Address\":\"133.5.128.237\",\"WARC-Target-URI\":\"https://catalog.lib.kyushu-u.ac.jp/opac_detail_md/?reqCode=frombib&lang=0&amode=MD100000&opkey=&bibid=11792&start=\",\"WARC-Payload-Digest\":\"sha1:ZIB4QMX4UTBIFGPDW45XOEPIWRZNAJYJ\",\"WARC-Block-Digest\":\"sha1:AE5IISVXK4AD2OCYYBUB45QNV22USCS6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506632.31_warc_CC-MAIN-20230924091344-20230924121344-00129.warc.gz\"}"}
http://www.tomhsiung.com/wordpress/tag/regression-equation/	[ "", null, "The Regression Equation\n\nWhen analyzing data, it is essential to first construct a graph of the data. A scatterplot is a graph of data from two quantitative variables of a population. In a scatterplot, we use horizontal axis for the observations of one variable and a vertical axis for the observations of the other variable. Each pair of observations is then plottted as a point. Note: Data from two quantitative variables of a population are called bivariate quantitative data.\n\nTo measure quantitatively how well a line fits teh data, we first consider the errors, e, made in using the line to predict the y-values of the data points. In general, an error, e, is the signed vertical distance from the line to a data point. To decide which line fits the data better, we first compute the sum of the squared errors. Among all lines, the least-squares criterion is that the line having the smallest sum of squared errors is the one that fits the data best. Or, the least-squares criterion is that the line best fits a set of data points is the one having the smallest possible sum of squared errors.\n\nAlthough the least-squares criterion states the property that the regression line for a set of data points must satify, it does not tell us how to find that line. This task is accomplished by Formula 14.1. In preparation, we introduce some notation that will be used throughout our study of regression and correlation.", null, "Note although we have not used Syy in Formula 14.1, we will use it later.\n\nFor a linear regression y = b0 + b1x, y is the depdendent variable and x is the independent variable. However, in the context of regression analysis, we usually call y the response variable and x the predictor variable or explanatory variable (because it is used to predict or explain the values of the response variable).\n\nExtrapolation\n\nSuppose that a scatterplot indicates a linear relationship between two variables. Then, within the range of the observed values of the predictor variable, we can reasonably use the regression equation to make predictions for the response variable. However, to do so outside the range, which is called extrapolation, may not be reasonable because the linear relationship between the predictor and response variables may not hold there. To help avoid extrapolation, some researchers include the range of the observed values of the predictor variable with the regression equation.\n\nOutliers and Influential Observations\n\nRecall that an outlier is an observation that lies outside the overall pattern of the data. In the context of regression, an outlier is a data point that lies far from the regression line, relative to the other data points. An outlier can sometimes have a significant effect on a regression analysis. Thus, as usual, we need to identify outliers and remove them from the analysis when appropriate – for example, if we find that an outlier is a measurement or recording error.\n\nWe must also watch for influential observations. In regression analysis, an influential observation is a data point whose removal causes the regression equation (and line) to change considerably. A data point separated in the x-direction from the other data points is often an influential observation because the regression line is \"pulled\" toward such a data point without counteraction by other data points. If an influential observation is due to a measurement or recording error, or if for some other reason it clearly does not belong in the data set, it can be removed without further consideration. However, if no explanation for the influential observation is apparent, the decision whether to retain it is often difficult and calls for a judgment by the researcher.\n\nA Warning on the Use of Linear Regression\n\nThe idea behind finding a regression line is based on the assumption that the data points are scattered about a line. Frequently, however, the data points are scattered about a curve instead of a line. One can still compute the values of b0 and b1 to obtain a regression line for these data points. The result, however, will yeild an inappropriate fit by a line, when in fact a curve should be used. Therefore, before finding a regression line for a set of data points, draw a scatterplot. If the data points do not appear to be scattered about a line, do not determine a regression line.\n\nThe Coefficient of Determination\n\nIn general, several methods exist for evaluating the utility of a regression equation for making predictions. One method is to determine the percentage of variation in the observed values of the response variable that is explained by the regression (or predictor variable), as discussed below. To find this percentage, we need to define two measures of variation: 1) the total variation in the observed values of the response variable and 2) the amount of variation in the observed values of the response variable that is explained by the regression.\n\nTo measure the total variation in the observed values of the response variable, we use the sum of squared deviations of the observed values of the response variable from the mean of those values. This measure of variation is called the total sum of squares, SST. Thus, SST = 𝛴(yiy[bar])2. If we divide SST by n – 1, we get the sample variance of the observed values of the response variable. So, SST really is a measure of total variation.\n\nTo measure the amount of variation in the observed values of the response variable that is explained by the regression, we first look at a particular observed value of the response variable, say, corresponding to the data point (xi, yi). The total variation in the observed values of the response variable is based on the deviation of each observed value from the mean value, yiy[bar]. Each such deviation can be decomposed into two parts: the deviation explained by the regression line, y^y[bar], and the remaining unexplained deviation, yiy^. Hence the amount of variation (squared deviation) in observed values of the response variable that is explained by the regression is 𝛴(yi^y[bar])2. This measure of variation is called the regression sum of squares, SSR. Thus, SSR = 𝛴(yi^y[bar])2.\n\nUsing the total sum of squares and the regression sum of squares, we can determine the percentage of variation in the observed values of the response variable that is explained by the regression, namely, SSR / SST. This quantity is called the coefficient of determination and is denoted r2. Thus, r2 = SSR/SST. In a same defintion, the deviation not explained by the regression, yiyi^. The amount of variation (squared deviation) in the observed values of the response variable that is not explained by the regression is 𝛴(yi – yi^)2. This measure of variation is called the error sum of squares, SSE. Thus, SSE = 𝛴(yi – yi^)2.\n\nIn summary, check Definition 14.6", null, "And the coefficient of detrmination, r2, is the proportion of variation in the observed values of the response variable explained by the regression. The coefficient of determination always lies between 0 and 1. A vlaue of r2 near 0 suggests that the regression equation is not very useful for making predictions, whereas a value of r2 near 1 suggests that the regression equation is quite useful for making predictions.\n\nRegression Identity\n\nThe total sum of squares equals the regression sum of squares plus the error sum of squares: SST = SSR + SSE. Because of the regression identity, we can also express the coefficient of determination in terms of the total sum of squares and the error sum of squares: r2 = SSR / SST = (SSTSSE) / SST = 1 – SSE / SST. This formula shows that, when expressed as a percentage, we can also interpret the cofficient of determination as the percentage reduction obtained in the total squared error by using the regression equation instead of the mean, y(bar), to predict the observed values of the response variable.\n\nCorrelation and Causation\n\nTwo variables may have a high correlation without being causally related. On the contrary, we can only infer that the two variables have a strong tendency to increase (or decrease) simultaneously and that one variable is a good predictor of another. Two variables may be strongly correlated because they are both associated with other variables, called lurking variables, that cause the changes in the two variables under consideration.\n\nThe Regression Model; Analysis of Residuals\n\nThe terminology of conditional distributions, means, and standard deviations is used in general for any predictor variable and response variable. In other words, we have the following definitions.", null, "Using the terminology presented in Definition 15.1, we can now state the conditions required for applying inferential methods in regression analuysis.", null, "Note: We refer to the line y = 𝛽0 + 𝛽1x – on which the conditional means of the response variable lie – as the population regression line and to its equation as the population regression equation. Observed that 𝛽0 is the y-intercept of the population regression line and 𝛽1 is its slop. The inferential procedure in regression are robust to moderate violations of Assumptions 1-3 for regression inferences. In other words, the inferential procedures work reasonably well provided the variables under consideration don't violate any of those assumptions too badly.\n\nEstimating the Regression Parameters\n\nSuppose that we are considering two variables, x and y, for which the assumptions for regression inferences are met. Then there are constants 𝛽0, 𝛽1, and 𝜎 so that, for each value x of the predictor variable, the conditional distribution fo the response variable is a normal distribution with mean 𝛽0 + 𝛽1x and standard deviation 𝜎.\n\nBecause the parameters 𝛽0, 𝛽1, and 𝜎 are usually unknown, we must estimate them from sample data. We use the y-intercept and slop of a sample regression line as point estimates of the y-intercept and slop, respectively, of the population regression line; that is, we use b0 to estimate 𝛽0 and we use b1 to estimate 𝛽1. We note that b0 is an unbiased estimator of 𝛽0 and that b1 is an unbiased estimator of 𝛽1.\n\nEquivalently, we use a sample regression line to estimate the unknown population regression line. Of course, a sample regression line ordinarily will not be the same as the population regression line, just as a sample mean generally will not equal the population mean.\n\nThe statistic used to obtain a point estimate for the common conditional standard deviation 𝜎 is called the standard error of the estimate. The standard error of the estimate could be compute by", null, "Analysis of Residuals\n\nNow we discuss how to use sample data to decicde whether we can reasonably presume that the assumptions for regression inferences are met. We concentrate on Assumptions 1-3. The method for checking Assumption 1-3 relies on an analysis of the errors made by using the regression equation to predict the observed values of the response variable, that is, on the differences between the observed and predicted values of the response variable. Each such difference is called a residual, generically denoted e. Thus,\n\nResidual = ei = yiyi^\n\nWe can show that the sum of the residuals is always 0, which, in turn, implies that e(bar) = 0. Consequently, the standard error of the estimate is essentially the same as the standard deviation of the residuals (however, the exact standard deviation of the residuals is obtained by dividing by n – 1 instead of n – 2). Thus, the standard error of the estimate is sometimes called the residual standard deviation.", null, "We can analyze the residuals to decide whether Assumptions 1-3 for regression inferences are met because those assumptions can be translated into conditions on the residuals. To show how, let's consider a sample of data points obtained from two variables that satisfy the assumptions for regression inferences.\n\nIn light of Assumption 1, the data points should be scattered about the (sample) regression line, which means that the residuals should be scattererd about the x-aixs. In light of Assumption 2, the variation of the observed values of the response variable should remain approximately constant from one value of the predictor variable to the next, which means the residuals should fall roughly in a horizontal band. In light of Assumption 3, for each value of the predictor variable, the distribution of the corresponding observed values of the response variable should be approximately bell shaped, which implies that the horizontal band should be centered and symmetric about the x-axis.\n\nFurthermore, considering all four regression assumptions simultaneously, we can regard the residuals as independent observations of a variable having a normal distribution with mean 0 and standard deviation 𝜎. Thus a normal probability plot of the residuals should be roughly linear.", null, "A plot of the residuals against the observed values of the predictor variable, which for brevity we call a residual plot, provides approximately the same information as does a scatterplot of the data points. However, a residual plot makes spotting patterns such as curvature and nonconstant standard deviation easier.\n\nTo illustrate the use of residual plots for regression diagnostics, let's consider the three plots in Figure 15.6. In Figure 15.6 (a), the residuals are scattered about the x-axis (residuals = 0) and fall roughly in a horizontal band, so Assumption 1 and 2 appear to be met. In Figure 15.6 (b) it is suggested that the relation between the variable is curved indicating that Assumption 1 may be violated. In Figure 15.6 (c) it is suggested that the conditional standard deviations increase as x increases, indicating that Assumption 2 may be violated.", null, "Inferences for the Slope of the Population Regression Line\n\nSuppose that the variables x and y satisfy the assumptions for regression inferences. Then, for each value x of the predictor variable, the conditional distribution of the response variable is a normal distribution with mean 𝛽0 + 𝛽1x and standard deviation 𝜎. Of particular interest is whether the slope, 𝛽1, of the population regression line equals 0. If 𝛽1 = 0, then, for each value x of the predictor variable, the conditional distribution of the response variable is a normal distribution having mean 𝛽0 and standard deviation 𝜎. Because x does not appear in either of those two parameters, it is useless as a predictor of y.\n\nOf note, although x alone may not be useful for predicting y, it may be useful in conjunction with another variable or variables. Thus, in this section, when we say that x is not useful for predicting y, we really mean that the regression equation with x as the only predictor variable is not useful for predicting y. Conversely, although x alone may be useful for predicting y, it may not be useful in conjunction with another variable or variables. Thus, in this section, when we say that x is useful for predicting y, we really mean that the regression equation with x as the only predictor variable is useful for predicting y.\n\nWe can decide whether x is useful as a (linear) predictor of y – that is, whether the regression equation has utility – by performing the hypothesis test", null, "We base hypothesis test for 𝛽1 on the statistic b1. From the assumptions for regression inferences, we can show that the sampling distribution of the slop of the regression line is a normal distribution whose mean is the slope, 𝛽1, of the population regression line. More generally, we have Key Fact 15.3.", null, "As a consequence of Key Fact 15.3, the standard variable", null, "has the standard normal distribution. But this variable cannot be used as a basis for the required test statistic because the common conditional standard deviation, 𝜎, is unknown. We therefore replace 𝜎 with its sample estimate Se, the standard error of the estimate. As you might be suspect, the resulting variable has a t-distribution.", null, "In light of Key Fact 15.4, for a hypothesis test with the null hypothesis H0: 𝛽1 = 0, we can use the variable t as the test statistic and obtain the critical values or P-value from the t-table. We call this hypothesis-testing procedure the regression t-test.\n\nConfidence Intervals for the Slop of the Population Regression Line\n\nObtaining an estimate for the slop of the population regression line is worthwhile. We know that a point estimate for 𝛽1 is provided by b1. To determine a confidence-interval estimate for 𝛽1, we apply Key Fact 15.4 to obtain Procedure 15.2, called the regression t-interval procedure.", null, "Estimating and Prediction\n\nIn this section, we examine how a sample regression equation can be used to make two important inferences: 1) Estimate the conditional mean of the response variable corresponding to a particular value of the predictor variable; 2) predict the value of the response variable for a particular value of the predictor variable.", null, "In light of Key Fact 15.5, if we standardize the variable yp^, the resulting variable has the standard normal distribution. However, because the standardized variable contains the unknown parameter 𝜎, it cannot be used as a basis for a confidence-interval formula. Therefore, we replace 𝜎 by its estimate se, the standard error of the estimate. The resulting variable has a t-distribution.", null, "Recalling that 𝛽0 + 𝛽1x is the conditional mean of the response variable corresponding to the value xp of the predictor variable, we can apply Key Fact 15.6 to derivea confidence-interval procedure for means in regression. We call that procedure the conditional mean t-interval procedure.", null, "Prediction Intervals\n\nA primary use of a sample regression equation is to make predictions. Prediction intervals are similar to confidence intervals. The term confidence is usually reserved for interval estimates of parameters. The term prediction is used for interval estimate of variables.", null, "In light of Key Fact 15.7, if we standardize the variable yp – yp^, the resulting variable has the standard normal distribution. However, because the standardized variable contains the unknown parameter 𝜎, it cannot be used as a basis for prediction-interval formula. So we replace 𝜎 by its estimate se, the standard error of the estimate. The resulting variable has a t-distribution.", null, "Using Key Fact 15.8, we can derive a prediction-interval procedure, called the predicted value t-interval procedure.", null, "Inferences in Correlation\n\nFrequently, we want to decide whether two variables are linearly correlated, that is, whether there is a linear relationship between two cariables. In the context of regression, we can make that decision by performing a hypothesis test for the slope of the population regression line. Alternatively, we can perform a hypothesis test for the population linear correlation coefficient, 𝜌. This parameter measures the linear correlation of all possible pairs of observations of two variables in the same way that a sample linear correlation coefficient, r, measures the linear correlation of a sample of pairs. Thus, 𝜌 actually describes the strength of the linear relationship between two variables; r is only an estimate of 𝜌 obtained from sample data.\n\nThe population linear correlation coefficient of two variables x and y always lies between -1 and 1. Values of 𝜌 near -1 or 1 indicate a strong linear relationship between the variables, whereas values of 𝜌 near 0 indicate a weak linear relationship between the variables. As we mentioned, a sample linear correlation coefficient, r, is an estimate of the population linear correlation coefficient, 𝜌. Consequently, we can use r as a basis for performing a hypothesis test for 𝜌.", null, "In light of Key Fact 15.9, for a hypothesis test with the null hypothesis H0: 𝜌 = 0, we use the t-score as the test statistic and obtain the critical values or P-value from the t-table. We call this hypothesis-testing procedure the correlation t-test." ]	[ null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/statistics-denial-statistics-debacles-Malfeasance.jpg", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-10-at-9.18.05-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-14-at-4.40.53-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-15-at-3.49.51-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-15-at-3.51.45-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-15-at-5.45.29-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-15-at-9.25.34-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-15-at-9.57.10-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-15-at-9.58.02-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-15-at-10.15.26-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-15-at-11.06.18-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-15-at-11.07.53-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-15-at-11.15.15-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-15-at-11.24.57-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-16-at-8.06.33-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-16-at-8.30.41-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-16-at-8.46.44-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-16-at-9.31.56-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-16-at-9.45.37-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-16-at-9.47.24-PM.png", null, "http://www.tomhsiung.com/wordpress/wp-content/uploads/2017/10/Screen-Shot-2017-10-16-at-10.18.09-PM.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8928168,"math_prob":0.9923708,"size":19784,"snap":"2020-10-2020-16","text_gpt3_token_len":4139,"char_repetition_ratio":0.2019717,"word_repetition_ratio":0.16422018,"special_character_ratio":0.19935301,"punctuation_ratio":0.10065466,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99940646,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42],"im_url_duplicate_count":[null,7,null,7,null,7,null,7,null,7,null,7,null,7,null,7,null,7,null,7,null,7,null,7,null,7,null,7,null,7,null,7,null,7,null,7,null,7,null,7,null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-25T03:53:00Z\",\"WARC-Record-ID\":\"<urn:uuid:b9650fc3-86a0-4499-89f1-0102711f8b82>\",\"Content-Length\":\"111687\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5190f8e2-4f82-4773-a6c2-65b9bae01056>\",\"WARC-Concurrent-To\":\"<urn:uuid:5a96c69f-1fc8-44eb-b129-fbd6e1f50e6d>\",\"WARC-IP-Address\":\"100.42.56.12\",\"WARC-Target-URI\":\"http://www.tomhsiung.com/wordpress/tag/regression-equation/\",\"WARC-Payload-Digest\":\"sha1:23SUI7ZSMCTMJZ7MLMKE4RIRRKEL7FWH\",\"WARC-Block-Digest\":\"sha1:VGASWKA4JGEZGLSPSAFIWEUL3XIUVK2O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146004.9_warc_CC-MAIN-20200225014941-20200225044941-00308.warc.gz\"}"}
https://www.brighthubengineering.com/cad-autocad-reviews-tips/19291-applications-of-the-cad-software-what-is-geometric-modeling/	[ "Applications of the CAD Software – What is Geometric Modeling?\n\nWhat is Geometric Modeling?\n\nThere are number of applications of the CAD software, one of the most popular applications being geometric modeling. First of all let us see what is geometric modeling? The computer compatible mathematical description of the geometry of the object is called as geometric modeling. The CAD software allows the mathematical description of the object to be displayed as the image on the monitor of the computer.\n\nSteps for Creating the Geometric Model\n\nThere are three steps in which the designer can create geometric models by using CAD software, these are:\n\n1) Creation of basic geometric objects: In the first step the designer creates basic geometric elements by using commands like points, lines, and circles.\n\n2) Transformations of the elements: In the second step the designer uses commands like achieve scaling, rotation and other related transformations of the geometric elements.\n\n3) Creation of the geometric model: During the final step the designer uses various commands to that cause integration of the objects or elements of the geometric model to form the desired shape.\n\nDuring the process of geometric modeling the computer converts various commands given from within the CAD software into mathematical models, stores them as the files and finally displays them as the image. The geometric models created by the designer can open at any time for reviewing, editing or analysis.\n\nRepresentation of the Geometric Models\n\nOf the various forms of representing the objects in geometric models, the most basic is wire frames. In this form the object is displayed by interconnected lines as shown in the figure below (source: Wikipedia). There are three types of wire frame geometric modeling, these are: 2D, 2.1/2D and 3D. They have been described below:\n\n1) 2D: It stands of two dimensional view and is useful for flat objects.\n\n2) 2.1/2D: It gives views beyond the 2D view and permits viewing of 3D object that has no sidewall details.\n\n3) 3D: The three dimension representation allows complete three-dimensional viewing of the model with highly complex geometry. Solid modeling is the most advanced method of geometric modeling in three dimensions.", null, "Reference\n\nBook: CAD/CAM: Computer Aided Design and Manufacturing by Mikell P. Groover and Emory W. Zimmers\n\nThis post is part of the series: Applications of CAD\n\nThis is the series of articles describing various applications of CAD software like geometric modeling, engineering analysis, FEA, design review and evaluation, drafting etc." ]	[ null, "https://img.bhs4.com/CA/6/CA63CD9E220E488788EA94AEE6FA83704586D596_large.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89341444,"math_prob":0.96995556,"size":2682,"snap":"2019-43-2019-47","text_gpt3_token_len":535,"char_repetition_ratio":0.18595967,"word_repetition_ratio":0.009433962,"special_character_ratio":0.19276659,"punctuation_ratio":0.106694564,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9635453,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-21T03:03:18Z\",\"WARC-Record-ID\":\"<urn:uuid:e27bd727-eae0-436d-b996-2458287cc121>\",\"Content-Length\":\"112273\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e745d345-4cfc-405c-9a85-08be34c815b2>\",\"WARC-Concurrent-To\":\"<urn:uuid:99b29a94-a733-45c0-9b47-090550e0c72b>\",\"WARC-IP-Address\":\"104.28.6.60\",\"WARC-Target-URI\":\"https://www.brighthubengineering.com/cad-autocad-reviews-tips/19291-applications-of-the-cad-software-what-is-geometric-modeling/\",\"WARC-Payload-Digest\":\"sha1:PR2W6NAGL2AIPHTZJR4KY27SLHSIJ3W6\",\"WARC-Block-Digest\":\"sha1:GAYFY3Q64EEFMXKT6E7UH6Z734P4MWXY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987751039.81_warc_CC-MAIN-20191021020335-20191021043835-00063.warc.gz\"}"}
https://ricerca.sns.it/handle/11384/125502	[ "We shall consider sections of a complex elliptic scheme E over an affine base curve B, and study the points of B where the section takes a torsion value. In particular, we shall relate the distribution in B of these points with the canonical height of the section, proving an integral formula involving a measure on B coming from the so-called Betti map of the section. We shall show that this measure is the same one which appears in dynamical issues related to the section. This analysis will also involve the multiplicity with which a torsion value is attained, which is an independent problem. We shall prove finiteness theorems for the points where the multiplicity is higher than expected. Such multiplicity has also a relation with Diophantine Approximation and quasi-integral points on E (over the affine ring of B), and in Sections 5 and 6 of the paper we shall exploit this viewpoint, proving an effective result in the spirit of Siegel’s theorem on integral points.\n\n### On the torsion values for sections of an elliptic scheme\n\n#### Abstract\n\nWe shall consider sections of a complex elliptic scheme E over an affine base curve B, and study the points of B where the section takes a torsion value. In particular, we shall relate the distribution in B of these points with the canonical height of the section, proving an integral formula involving a measure on B coming from the so-called Betti map of the section. We shall show that this measure is the same one which appears in dynamical issues related to the section. This analysis will also involve the multiplicity with which a torsion value is attained, which is an independent problem. We shall prove finiteness theorems for the points where the multiplicity is higher than expected. Such multiplicity has also a relation with Diophantine Approximation and quasi-integral points on E (over the affine ring of B), and in Sections 5 and 6 of the paper we shall exploit this viewpoint, proving an effective result in the spirit of Siegel’s theorem on integral points.\n##### Scheda breve Scheda completa Scheda completa (DC)\n2021\nSettore MAT/03 - Geometria\nFile in questo prodotto:\nFile\n1909.01253.pdf\n\nAccesso chiuso\n\nTipologia: Accepted version (post-print)\nLicenza: Non pubblico\nDimensione 361.06 kB\n10.1515_crelle-2021-0056.pdf\n\nOpen Access dal 29/10/2022\n\nTipologia: Published version\nLicenza: Solo Lettura\nDimensione 534.61 kB\nUtilizza questo identificativo per citare o creare un link a questo documento: `https://hdl.handle.net/11384/125502`\n•", null, "ND\n•", null, "5\n•", null, "4" ]	[ null, "https://ricerca.sns.it/sr/cineca/images/thirdparty/pmc_small.png", null, "https://ricerca.sns.it/sr/cineca/images/thirdparty/scopus_small.png", null, "https://ricerca.sns.it/sr/cineca/images/thirdparty/isi_small.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8574664,"math_prob":0.9473476,"size":2520,"snap":"2023-40-2023-50","text_gpt3_token_len":590,"char_repetition_ratio":0.12321144,"word_repetition_ratio":0.7911548,"special_character_ratio":0.21230158,"punctuation_ratio":0.08154506,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9669456,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-05T05:57:56Z\",\"WARC-Record-ID\":\"<urn:uuid:7c0ba5a4-9ca0-4b35-8b29-a16e3a227d66>\",\"Content-Length\":\"50516\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7298b260-32df-4dfd-9814-707e79a5081f>\",\"WARC-Concurrent-To\":\"<urn:uuid:ac424203-b93a-4426-9730-b5d6d295c65d>\",\"WARC-IP-Address\":\"130.186.29.4\",\"WARC-Target-URI\":\"https://ricerca.sns.it/handle/11384/125502\",\"WARC-Payload-Digest\":\"sha1:QB62H6WE4ETQS4COPLLBYDGIB6RJCQWT\",\"WARC-Block-Digest\":\"sha1:2UAYH4ECQFKAAMR7ZLAQGHGP4VTUCD2H\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100545.7_warc_CC-MAIN-20231205041842-20231205071842-00163.warc.gz\"}"}
https://sites.google.com/site/geometry4sage20112012/home/resources/star-lessons/lesson-1-mathematical-thinking	[ "SAGE Academy‎ > ‎Resources‎ > ‎STARS® Lessons‎ > ‎\n\n### Lesson 1: Mathematical Thinking and Logic\n\nLesson 1: Mathematical Thinking and Logic\n\nObjectives\n\nBy the end of this lesson, students will be able to:\n\n• decide whether information drawn from a passage is true, false, or cannot be determined\n• recognize the hypothesis and conclusion of a conditional statement\n• state conditionals in the if … then form\n• recognize and state the converse, inverse, and contrapositive of a statement\n• draw correct conclusions from given statements\n\nIntroduction\n\nGeometry is a word taken from the Greeks with \"geo\" meaning earth and \"metron\" meaning measure. In about 300 BC, a Greek mathematician by the name of Euclid wrote a book called The Elements. It was a collection of all the known geometric facts organized in a logical manner. He included basic terms, like point, line, and plane, along with accepted principles called axioms or postulates. Euclid is known as the \"Father of Geometry.\"", null, "In this lesson, you will apply critical thinking and decide whether information drawn from a passage is true, false, or cannot be determined. You will then learn how to recognize the hypothesis and conclusion of a conditional statement and how to state conditionals in the if… then form. Finally, you will learn to recognize and/or state the converse, inverse, and contrapositive of a statement, as well as draw correct conclusions from given statements.\n\nCritical Thinking\n\nThe study of geometry requires one to be a critical thinker. Approaching and solving geometry problems is like a good detective analyzing a crime scene and looking for clues. You must be able to put your information in a clear, logical sequence to show how you arrived at your conclusion.\n\nExample 1\n\nRead the sentence below and determine if the statements that follow are true, false, or cannot be determined.\n\nThe Wilsons boarded a train to evacuate New Orleans two days prior to Hurricane Gustav coming ashore on Monday, September 1, 2008.\n\n Statement Explanation Gustav was a hurricane. True. The hurricane is named in the sentence. The Wilsons boarded the train on Friday. False. Two days before it came ashore would be Saturday, not Friday. The Wilsons' home is in New Orleans. Cannot be determined. The Wilsons could have been visiting the city, so they do not necessarily live there. The Wilsons have at least one child. Cannot be determined. The sentence does not mention how many people are in the Wilson family. The Wilsons left their car in New Orleans. Cannot be determined. Nothing was mentioned about a car.\n\nAs you can see from the example above, a valid conclusion cannot be drawn when you are missing a number of facts.\n\nExample 2\n\nRead the following passage and determine if the statements that follow are true, false, or cannot be determined.\n\nThe Wilsons arrived safely in Baltimore after boarding a train in New Orleans to escape Hurricane Gustav. Aunt Betty was waiting at the station two hours before they arrived. They hugged each other and headed for the parking garage. Soon after, they were in a car on the highway heading towards a good home-cooked meal.\n\n Statement Explanation The Wilsons left New Orleans to escape Hurricane Gustav. True. It is stated in the sentence. Aunt Betty lives in Baltimore. Cannot be determined. Aunt Betty could have driven from another city to Baltimore to pick up the Wilsons. The Wilsons arrived on time. Cannot be determined. The passage does not mention if the train was on time or not. The Wilsons boarded a train in Baltimore. False. Aunt Betty picked them up in her car. The train's final destination was Baltimore. Cannot be determined. The passage does not mention if the train was only making a stop in Baltimore.\n\nPractice\n\nRead the passage below and determine if the statements that follow are true, false, or cannot be determined.\n\nThe second before she opened her mouth, all Thomas could think was, \"If I hear another horrible actor, I'm going to walk out that door.\" She wasn't, thankfully, but it didn't really matter, because he never would've walked out that door, the door in question being the one in the tiny rehearsal room at Harlequin Studios. It was a grungy-looking rehearsal studio, where everything smelled a little funny, like some undiscovered form of mildew. He wouldn't have walked out that door because auditions weren't over yet, and he was a professional, even though he wouldn't be getting paid. Neither would the actors--the more than 160 of them. The woman who was talking was actually pretty good, good enough to get called back. Her name, Skye Benrexi, was interesting enough alone to merit a callback.\n\nUsed with permission from David L. Williams\n\n1. The rehearsal studio is located in New York City.\n\nCannot be determined", null, "Check Your Answer\n\n2. There were more than 160 actors that came to the audition.\n\nTrue", null, "Check Your Answer\n\n3. Thomas was auditioning for a part.\n\nFalse", null, "Check Your Answer\n\n4. Each actor selected would receive a paycheck.\n\nFalse", null, "Check Your Answer\n\n5. Skye Benrexi will be selected for the part.\n\nCannot be determined", null, "Check Your Answer\n\n6. The rehearsal room is located at Harlequin Studios.\n\nTrue", null, "Check Your Answer\n\n7. The rehearsal room was clean and fresh.\n\nFalse", null, "Check Your Answer\n\n8. Skye auditioned well.\n\nTrue", null, "Check Your Answer\n\n9. Thomas was a professional.\n\nTrue", null, "Check Your Answer\n\n10. Skye Benrexi is a stage name.\n\nCannot be determined", null, "Check Your Answer\n\nConditional Statements\n\nLook at the following conditional statement :\n\nThe first clause begins with the word if - \"if the Sun shines.\" This part of the statement is called the hypothesis. The second clause begins with the word then - \"then you will see your shadow.\" This part of the statement is called the conclusion. The if statement is the given information, and the then statement is the conclusion or result of the given information. Conditional statements are used to establish certain conclusions.\n\nIn mathematics, we can label the if statement with the letter p and the then statement with a q. We can show the conditional statement using symbols in a few different ways:\n\nif p, then q\n\np implies q\n\npq\n\nExample 3\n\nIdentify the hypothesis and conclusion in the following statement:\n\nIf I score a 93% on the test, then I will have an A average.\n\nhypothesis: I score a 93% on the test\n\nconclusion: I will have an A average\n\nExample 4\n\nIdentify the hypothesis and conclusion in the following statement:\n\nIf the cows are lying down in the field, then it will rain.\n\nhypothesis: the cows are lying down in the field\n\nconclusion: it will rain\n\nExample 5\n\nIdentify the hypothesis and conclusion in the following statement:\n\nIf x and y are integers, then their product will be an integer.\n\nhypothesis: x and y are integers\n\nconclusion: their product will be an integer\n\nConditional statements are not always written in the if… then form.\n\nExample 6\n\nIdentify the hypothesis and conclusion in the statement below. Rewrite the statement so it is in the form if… then.\n\nWhen it rains, it pours.\n\nhypothesis: it rains\n\nconclusion: it pours\n\nif… then form: If it rains, then it pours.\n\nExample 7\n\nIdentify the hypothesis and conclusion in the statement below. Rewrite the statement so it is in the form if… then.\n\nA banana is soft if it is spoiled.\n\nhypothesis: a banana is spoiled\n\nconclusion: it is soft\n\nif… then form: If a banana is spoiled, then it is soft.\n\nPractice\n\n1-5. Identify the hypothesis and conclusion in the statements below. If a statement is not in the form if… then, rewrite the statement so it is in the form if… then.\n\n1. If the lights are on in the house, then someone is home.\n\nhypothesis: the lights are on in the house\n\nconclusion: someone is home", null, "Check Your Answer\n\n2. You should come inside if you get cold.\n\nif… then form: If you get cold, then you should come inside.\n\nhypothesis: you get cold\n\nconclusion: you should come inside", null, "Check Your Answer\n\n3. If a triangle has two equal sides, then it is an isosceles triangle.\n\nhypothesis: a triangle has two equal sides\n\nconclusion: it is an isosceles triangle", null, "Check Your Answer\n\n4. The early bird gets the worm.\n\nif… then form: If the bird is early, then it gets the worm.\n\nhypothesis: the bird is early\n\nconclusion: it gets the worm", null, "Check Your Answer\n\n5. A straight angle measures 180 degrees.\n\nif… then form: If an angle is a straight angle, then it measures 180 degrees.\n\nhypothesis: an angle is a straight angle\n\nconclusion: it has a measure of 180 degrees", null, "Check Your Answer\n\nEquivalent Statements\n\nThe statement \"A right angle has a measure of 90 degrees\" can be written as a conditional statement:\n\nIf an angle is a right angle, then it has a measure of 90 degrees.\n\nIt is now in the form of \"if p then q.\" If the p and q are reversed, the statement becomes:\n\nIf an angle has a measure of 90 degrees, then it is a right angle.\n\nWhen the p and q are reversed, the converse is formed. The converse of pq is written qp. In this example, both the conditional statement and its converse are true. This is called a biconditional statement. A statement is biconditional if both it and its converse are true. This is not always the case. Look at another conditional statement:\n\nIf you live in Miami, then you live in Florida.\n\nReversing the two parts of the statement gives us the converse:\n\nIf you live in Florida, then you live in Miami.\n\nIn this case the converse is not necessarily true. A person can live in other cities besides Miami and still be in Florida.\n\nIf we want to negate or say the negative of a statement, we use the word not. Once again, look at the conditional statement \"If an angle is a right angle, then it measures 90 degrees.\" To make the negative of this statement, add not to each side:\n\nIf an angle is not a right angle, then it does not measure 90 degrees.\n\nWe use the symbol ~ for the word not. In this case, we can use symbols to describe this statement as ~p → ~q.\n\nWhen you negate both the hypothesis and the conclusion, you have written the inverse. In the example above, both the conditional and its inverse are true, but this is not always the case. Look at the statement about Florida again:\n\nIf you live in Miami, then you live in Florida.\n\nThe inverse of this statement is not necessarily true:\n\nIf you do not live in Miami, then you do not live in Florida.\n\nWe can also take the inverse of the converse of a statement.\n\nConditional statement: If an angle is a right angle, then it measures 90 degrees.\n\nConverse: If an angle measures 90 degrees, then it is a right angle.\n\nTake the inverse of the converse: If an angle does not measure 90 degrees, then it is not a right angle.\n\nThe inverse of a converse is called the contrapositive. For example, the contrapositive of the statement \"If you live in Miami, then you live in Florida\" is the following:\n\nIf you do not live in Florida, then you do not live in Miami.\n\nIn both cases above, note that the contrapositive is true. The contrapositive is always logically equivalent to the conditional statement (in other words, they both have the same meaning).\n\nHere is a way to remember conditionals:", null, "Example 8\n\nAccept the following conditional statement as true:\n\nIf you live in Winchester, then you attend St. Augustus High School.\n\nWrite the converse, inverse, and contrapositive of each statement. Say whether each is true, false, or not necessarily true (cannot be determined), and which if any are logically equivalent.\n\n Converse If you attend St. Augustus High School, then you live in Winchester. Explanation Not necessarily true. The statement does not indicate whether all students at St. Augustus High School live in Winchester. Inverse If you do not live in Winchester, then you do not attend St. Augustus High School. Explanation Not necessarily true. The statement does not indicate whether all students who do not attend St. Augustus High School do not live in Winchester. Contrapositive If you do not attend St. Augustus High School, then you do not live in Winchester. Explanation True. According to the converse, if you attend St. Augustus High School, then you live in Winchester. The conditional and the contrapositive are logically equivalent. If you live in Winchester, then you attend St Augustus High School, so if you do not attend St Augustus High school, then you do not live in Winchester.\n\nPractice\n\n1. Accept the following conditional statement as true:\n\nIf you are on Flight #107, then you are going to Atlanta.\n\nWrite the converse, inverse, and contrapositive of the statement, and say which statements, if any, are logically equivalent.\n\nConverse: If you are going to Atlanta, then you are on Flight #107. (not necessarily true)\nInverse: If you are not on Flight #107, then you are not going to Atlanta. (not necessarily true)\nContrapositive: If you are not going to Atlanta, then you are not on Flight #107. (true)\nThe conditional and the contrapositive are logically equivalent. The converse and inverse are also logically equivalent.", null, "Check Your Answer\n\n2. Identify the relation of each of the following statements to the original statement below:\n\nIf you use a boogie board, then you live along the coast.\n\na) If you do not live along the coast, then you do not use a boogie board.\n\nb) If you live along the coast, then you use a boogie board.\n\nc) If you do not use a boogie board, then you do not live along the coast.\n\na) contrapositive\n\nb) converse\n\nc) inverse", null, "Check Your Answer\n\nDeductive Reasoning\n\nIn presenting a logical argument, one must combine at least three conditional statements in the following form (the symbol for \"therefore\" is ∴ ).\n\npq\n\nqr\n\npr\n\nAbove, if the first two statements (premises) are true, then the third statement (conclusion) must also be true. Notice the hypothesis of one premise is the same as the conclusion of the other premise. This is called a syllogism. If the first and second premises are true, then the final conclusion is also true.\n\nExample 9\n\nIf a student walks to school, he is late for school. If a student is late for school, he misses part of his first class. What conclusion can you reach based on these premises?\n\n p → q If a student walks to school (p), he is late for school (q). q → r If a student is late for school (q), he misses part of his first class (r). ∴ p → r Therefore, a student who walks to school misses part of his first class.\n\nExample 10\n\nIf you live in Tampa, then you live in Hillsborough County. If you live in Hillsborough County, then you live in Florida. What conclusion can you reach based on these premises?\n\n p → q If you live in Tampa (p), then you live in Hillsborough County (q). q → r If you live in Hillsborough County (q), then you live in Florida (r). ∴ p → r Therefore, if you live in Tampa you live in Florida.\n\nPractice\n\n1. Louis lives in the Ninth Ward. The Ninth Ward is in New Orleans. What conclusion can be reached from these premises?\n\nTherefore, Louis lives in New Orleans.", null, "Check Your Answer\n\n2. Determine if the following pairs of premises form a syllogism. If they do, state the conclusion.\n\npq\n\npr\n\nNo (The hypothesis of one is not the conclusion of the other.)", null, "Check Your Answer\n\n3. Determine if the following pairs of premises form a syllogism. If they do, state the conclusion.\n\nqr\n\npq\n\nYes, pr (The order of the premises can be changed, so pq, qr, pr.)", null, "Check Your Answer\n\nMultiple Premises\n\nSometimes there are more than two premises to consider.\n\npq\n\nqr\n\nrs\n\nst\n\npt\n\nThis is what a good detective does when solving a crime. One clue will lead to another. Eventually, when all the pieces fit together in the proper order, the crime is solved. Look at the premises below that lead to a conclusion. Notice the conclusion of one statement is the hypothesis of the next statement.\n\nIf Alvin has \\$80, then he will go shopping.\n\nIf Alvin goes shopping, then he will buy baseball shoes.\n\nIf Alvin buys baseball shoes, then he will try out for the team\n\nIf Alvin tries out for the team, then he will make the team.\n\nTherefore, if Alvin has \\$80, then he will make the baseball team.\n\nSometimes the steps to prove a syllogism are not as obvious as those in the preceding problems. An equivalent statement may need to be written in order to have the conclusion of one premise be the hypothesis of the other. Equivalent statements can be substituted for one another. Recall that the conditional and its contrapositive are equivalent. Look at the examples below of equivalent statements:\n\npq is equivalent to ~q → ~p\n\n~pq is equivalent to ~qp\n\nq → ~p is equivalent to p → ~q\n\nEach of these could be substituted to produce a syllogism that leads to a conclusion.\n\nExample 11\n\nDetermine if the following pairs of premises form a syllogism. If they do, state the conclusion.\n\nc → ~d\n\ned\n\nUse the fact that the contrapositive of c → ~d is d → ~c\n\nRewrite the pair of premises, changing the order and substituting the contrapositive, to find the conclusion.\n\ned\n\nd → ~c\n\ne → ~c\n\nPractice\n\n1. If Peter makes an A in his geometry class, he will spend the summer in the Caribbean. If he goes to the Caribbean, he gets to go to the summer festival. If he goes to the summer festival, he will perform with the band. What conclusion can be drawn from these premises?\n\nTherefore, if Peter makes an A in his geometry class, he will perform with the band at the summer festival in the Caribbean.", null, "Check Your Answer\n\nTruth Tables\n\nEarlier in the lesson you learned how conditional, converse, inverse, and contrapositive statements can be equivalent. Now we are going to verify which ones are equivalent by using a truth table. A truth table shows all the possibilities in a given situation.\n\nIt is raining.\n\nYou get wet.\n\nBoth of these statements can either be true or false. Assign the letter p to the first statement: \"it is raining.\" The second statement, \"you get wet,\" will be the letter q. Now we will put these two statements together as a conditional statement.\n\nIf it is raining, then you get wet.\n\nIn symbolic form this is written:\n\npq\n\nA truth table is a way to display all the possibilities about whether either of those statements are true. It shows the original, converse, inverse, and contrapositive. First look at a truth table for the two parts of the conditional statement above:\n\n p q True True True False False True False False\n\nNotice that there are four possible combinations of true and false for p and q. Now look at the possibilities for the whole statement, pq. (From now on we will use T for true and F for false.) The third column gives what is called the truth value of a statement. The only time that pq is false is when p is true and q is false. In other words, a true first term, or premise, cannot imply a false conclusion.\n\n p q p → q T T T T F F F T T F F T\n\nIn the first case, T → T. If both of the statements--that it is raining and that you get wet --are true statements, then the conditional statement is true. So, the truth value of this is T.\n\nThe second case is T → F. This means that the first statement, it is raining, is true, but the second statement, that you get wet, is not. If this is the case, the conditional statement cannot be true. The truth value here is F, false.\n\nIn the third case, F → T. Here, the first statement is not true, so it is not raining. The second statement, you get wet, is true. In this case, the truth value is T, because we do not have enough information to know that it is false.\n\nIn the fourth case, F → F. In this case, the first and second statements are both false. As with the example above, since we do not have any information about what happens when it is not raining, the statement cannot be proven false. Here, the truth value is T.\n\nNegation\n\nThe symbol for negation (or \"not\") is a tilde (~). If we want to show that a statement p has been negated, we write ~p. The truth table for negation is the following:\n\n p ~p T F F T\n\nNegation causes a true statement to become false and a false one to become true. Say the original statement is \"10 plus 5 equals 15.\" This is true. The negation of this statement is \"10 plus 5 does not equal 15.\" This is false. In another example, the original statement is \"A pig is a horse.\" This is false. The negation of this statement is \"A pig is not a horse.\" This is true.\n\nConjunction\n\nAnother symbol that we use in truth tables is the conjunction or \"and\" symbol. If we want to connect the statements p and q with the word and, we write it as p q.\n\nLook back at the example above. We will connect the statements with a conjunction:\n\nIt is raining and you get wet.\n\np q\n\nThe truth table for conjunctions is the following:\n\n p q p ∧ q T T T T F F F T F F F F\n\nLook at an example. Check to see if the conjunction statement is true based on the truth table:\n\nGreen is a color and 4 + 5 = 9.\n\nLook at the table. Both of these statements are true. Based on the first row of the table, the entire sentence is true.\n\nNow look at a similar statement:\n\nGreen is a color and 4 + 5 = 10.\n\nSince the first statement is true and the second is false, the entire sentence is false.\n\nExample 12\n\nUse the truth table above to state the truth value of the sentence below.\n\nAn apple is a vegetable and 18 - 5 = 13.\n\nThe first statement is false--an apple is not a vegetable.\n\nThe second statement is true.\n\nSince the first statement is false and the second is true, the entire sentence is false.\n\nExample 13\n\nUse the truth table above to state the truth value of the sentence below.\n\nAn apple is a vegetable and 18 - 5 = 10.\n\nThe first statement is false.\n\nThe second statement is also false.\n\nSince the first statement is false and the second is false, the entire sentence is false.\n\nNotice that the only time p q is true is when both p and q are true.\n\nPractice\n\n1 - 4. Determine the truth value of the following sentences.\n\n1. 5 + 7 = 12 and 5 is an even number.\n\nThe first statement is true, but second statement is false. Therefore, the whole sentence is false.", null, "Check Your Answer\n\n2. London is the capital of England and Paris is the capital of France.\n\nThe first statement is true, and the second statement is true. Therefore, the whole sentence is true.", null, "Check Your Answer\n\n3. Eleven is not a prime number and there are five sides in a pentagon.\n\nThe first statement is false, though the second statement is true. Therefore, the whole statement is false.", null, "Check Your Answer\n\n4. The opposite sides of a trapezoid are always equal and there are 350° in a circle.\n\nThe first statement is false and the second statement is false, so the whole statement is false.", null, "Check Your Answer\n\nEquivalent Statements\n\nLook again at the statements we have been using:\n\np: It is raining.\n\nq: You get wet.\n\nWe can use this statement and truth tables to show which statements are equivalent.\n\nA. Conditional statement:\n\nIf it is raining, then you get wet.\n\npq\n\nWe have seen this truth table before.\n\n p q p → q T T T T F F F T T F F T\n\nB. Converse statement:\n\nIf you get wet, then it is raining.\n\nqp\n\nNotice that the p and q have been switched. The truth table looks like the following:\n\n I II III p q q → p T T T T F T F T F F F T\n\nWe have labeled the columns with Roman numerals. In order to get column III, you must do column II → column I. Notice that the results are different from A.\n\nC. Inverse statement:\n\nIf it is not raining, then you do not get wet.\n\n~p → ~q\n\nBecause p and q have been negated, you have to put the negation columns in the truth table. Columns III and IV are used to form column V.\n\n I II III IV V p q ~p ~q ~p → ~q T T F F T T F F T T F T T F F F F T T T\n\nD. Contrapositive statement:\n\nIf you do not get wet, then it is not raining.\n\n~q → ~p\n\nNotice that the p and q have been switched and negated. We must put in the negation columns. In order to get column V, you must do column IV → column III.\n\n I II III IV V p q ~p ~q ~q → ~p T T F F T T F F T F F T T F T F F T T T\n\nNotice that the last column in the conditional (A) and the contrapositive (D) tables are the same. Therefore, we say that they are equivalent statements. Also, the last column in the converse (B) and the inverse (C) tables are the same. Therefore, they are equivalent.\n\nValidity of an Argument\n\nTruth tables can also be used to determine the validity of an argument. An argument consists of premises (hypothesis) and a conclusion. The premises are connected with the word \"and\" (shown with the symbol ).\n\nLook at an example of an argument.\n\n Premises: If the rain stops, I go outside. The rain stops Conclusion: I go outside.\n\nWrite the statements with symbols.\n\np: The rain stops.\nq: I go outside.\n\nPut premises together with the symbol for a conjunction ().\n\n(pq) p\n\nThe conclusion is q. The premises and conclusion are stated as an implied (→) statement.\n\n[(pq) p] → q\n\nLet's build a truth table to look at this statement.\n\n p q p → q p → q ∧ p [p → q ∧ p] → q\n\nNotice that we start with p and q and build from there. The last column is the final statement. Now we will fill in the truth table, using the four possibilities.\n\n I II III IV V p q p → q p → q ∧ p [p → q ∧ p] → q T T T T T T F F F T F T T F T F F T F T\n\nColumn III is formed by using the rules for → (implied).\n\nIII = I → II\n\nColumn IV is formed by using the rules for (and).\n\nIV = III I\n\nColumn V is formed by using the rules for → (implied).\n\nV = IV → II\n\nIn the last column (V), the results are all true. When this happens it is called a tautology, and the argument is valid.\n\nLook at another argument.\n\nIf it is raining, then I am reading.\n\nTherefore, it is raining.\n\np: It is raining.\n\n Premises: p → q q Conclusion: p Argument: [p → q ∧ q] → p\n\nNow set up the truth table:\n\n I II III IV V p q p → q p → q ∧ q [p → q ∧ q] → p T T T T T T F F F T F T T T F F F T F T\n\nColumn III is formed by using the rules for → (implied).\n\nIII = I → II\n\nColumn IV is formed by using the rules for (and).\n\nIV = III II\n\nColumn V is formed by using the rules for → (implied).\n\nV = IV → I\n\nIn the last column (V), the results are not all true. Therefore, the argument is not valid. Here is another argument.\n\nExample 14\n\nConstruct and describe a truth table about the following argument.\n\nIf wool is expensive, then sweaters are expensive.\n\nSweaters are not expensive.\n\nTherefore, wool is not expensive.\n\np: Wool is expensive.\n\nq: Sweaters are expensive\n\n Premises: p → q ~q Conclusion: ~p Argument: [p → q ∧ ~q] → ~p\n\nNotice that there are negations needed here. Add negation columns to the truth table.\n\n I II III IV V VI VII p q ~p ~q p → q p → q ∧ ~q [p → q ∧ ~q] → ~p T T F F T F T T F F T F F T F T T F T F T F F T T T T T\n\nColumns III and IV are formed by negating Columns I and II.\n\nColumn V is formed by using the rules for → (implied).\n\nV = I → II\n\nColumn VI is formed by using the rules for (and).\n\nVI = V IV\n\nColumn VII is formed by using the rules for → (implied).\n\nVII = VI → III\n\nThe last column is all true. Therefore, the argument is valid.\n\nPractice\n\n1. Construct and explain a truth table about the following argument.\n\nIf John does not go to town, then Marty stays home.\n\nJohn goes to town.\n\nTherefore, Marty does not stay home.\n\np: John goes to town.\n\nq: Marty stays home.\n\n Premises: ~p → q p Argument: [(p → q) ∧ ~p] → ~q\n\n I II III IV V VI VII p q ~p ~q ~p → q (~p → q) ∧ p [(~p → q) ∧ p] → ~q T T F F T T F T F F T T T T F T T F T F T F F T T F F T\n\nColumns III and IV are formed by negating Columns I and II\n\nColumn V is formed by using the rules for → (implied).\n\nV = III → II\n\nColumn VI is formed by using the rules for (and).\n\nVI = V I\n\nColumn VII is formed by using the rules for → (implied).\n\nVII = VI → IV\n\nThe last column is not all true. Therefore, the argument is not valid.", null, "Check Your Answer\n\nEnrichment Activity\n\nResearch the history of geometry. There are many resources that contain information on how geometry was first used by the Babylonians, Egyptians, Hindus, Chinese and Greeks. Famous mathematicians such as Euclid, Archimedes, Thales, and Pythagoras were instrumental in the development of geometry. Do some reading on these ancient times and mathematicians, and write a summary of what you have learned. One possible resource is a book titled Great Moments in Mathematics Before 1650 by Howard Eves, published by the Mathematical Association of America.\n\nLesson Review & Homework\n\nWe started this lesson by defining conditional statements, also known as if … then statements. A conditional statement is made up of a hypothesis and a conclusion. Then, we defined converse, inverse, and contrapositive statements. The lesson also covered which of these statements are logically equivalent--that is, they mean the same thing. The converse and inverse statements are equivalent. The conditional and contrapositive are equivalent.\n\nYou used this information when applying deductive reasoning and finding syllogisms, which are used to form a true conclusion from given premises. You learned about truth tables. You also learned how to use them to understand negation, conjunctions, and the validity of arguments. This kind of logical thinking will be the basis for much of the work you will do as you explore geometry.\n\nHomework\n\n1.\n\nIdentify the hypothesis and conclusion of each of the following conditional statements. Rewrite any statement that is not already in the if… then form.\n\na. If the geese are flying south, then winter is on its way.\n\nb. If you study hard, then you will achieve success.\n\nc. The diameter of a circle is 14 inches if the radius of the circle is 7 inches.\n\nd. Those that live in Florida do not have to pay a state income tax.\n\ne. A scalene triangle has no sides of equal length.\n\n2.\n\nIdentify the relation of the following statements to the original statement below:\n\nThe grass will grow if it rains.\n\na. If it doesn't rain, the grass will not grow.\n\nb. If the grass grows, then it rained.\n\n3.\n\nWrite the converse of the statement below. Is it a good definition of what it means for an animal to be extinct? Why or why not?\n\nIf a type of animal is extinct, then there are no animals of that type left on Earth.\n\n4.\n\nDetermine if the following pairs of premises form a syllogism. If they do, state the conclusion.\n\npq\n\nrq\n\n5.\n\nDetermine if the following pairs of premises form a syllogism. If they do, state the conclusion.\n\npq\n\nrp\n\n6.\n\nDetermine if the following pairs of premises form a syllogism. If they do, state the conclusion.\n\nxy\n\nz → ~y\n\n7.\n\nDetermine if the following pairs of premises form a syllogism. If they do, state the conclusion\n\n~r → ~s\n\nst\n\n8.\n\nForm the negation of each statement.\n\na. The sky is cloudy.\n\nb. The Pittsburgh Steelers football team is not the team with the most Super Bowl wins.\n\nc. Josie is the only person with red hair.\n\n9.\n\nLet p and q represent the following simple statements. Write each sentence below in symbolic form.\n\np: This is a cow.\n\nq: This is a mammal.\n\na. If this is a cow, then this is a mammal.\n\nb. If this is a mammal, then this is a cow.\n\nc. If this is not a mammal, then this is not a cow.\n\n10.\n\nLet q and r represent the following simple statements. Write each symbolic statement below in words.\n\nq: It is Memorial Day.\n\nr: There will be fireworks.\n\na. q ~r\n\nb. r → ~q\n\nc. ~r ~q\n\nd. ~q → ~r\n\n11.\n\n11. Use truth tables to show that qp and ~ p → ~ q are equivalent.\n\n12.\n\nBy using a truth table, determine the validity of the following argument:\n\nIf we turn off the TV, there will be less noise.\n\nThere is less noise.\n\nTherefore, we turned off the TV." ]	[ null, "http://secure.starssuite.com/files/geo2010/Geom_L01_fig01.jpg", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/Geom_01L_fig02.jpg", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null, "http://secure.starssuite.com/files/geo2010/checkanswer_arrow.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91502285,"math_prob":0.8609669,"size":26828,"snap":"2019-51-2020-05","text_gpt3_token_len":6329,"char_repetition_ratio":0.17517895,"word_repetition_ratio":0.15171026,"special_character_ratio":0.22987178,"punctuation_ratio":0.1309801,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9744388,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56],"im_url_duplicate_count":[null,1,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,1,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-26T03:53:33Z\",\"WARC-Record-ID\":\"<urn:uuid:ed76a02e-a1f2-4688-8bc0-90477bb618f8>\",\"Content-Length\":\"83829\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:19118dc1-0054-45c7-b5e7-4636d87ceb11>\",\"WARC-Concurrent-To\":\"<urn:uuid:d43d51bd-7d3c-4af3-9c8a-9179f48302e6>\",\"WARC-IP-Address\":\"172.217.5.238\",\"WARC-Target-URI\":\"https://sites.google.com/site/geometry4sage20112012/home/resources/star-lessons/lesson-1-mathematical-thinking\",\"WARC-Payload-Digest\":\"sha1:V24GWYDVSZEDFLX35M3RXEWP7XJ7LZID\",\"WARC-Block-Digest\":\"sha1:3AUGSBI3TU4UOOFMHA6YYFHVZVFJOHN2\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251684146.65_warc_CC-MAIN-20200126013015-20200126043015-00420.warc.gz\"}"}
https://dsp.stackexchange.com/questions/9590/how-to-determine-the-window-size-and-weights-in-weighted-moving-average-wma-g	[ "# How to determine the window size and weights in Weighted Moving Average (WMA), given desired cut-off frequency?\n\nI am trying to smooth my discrete-time data points using the method of WMA.\n\nCurrently, I am using n as the window size and the weight array, {n/(n(n+1)/2), (n-1)/(n(n+1)/2), ... , 1/(n(n+1)/2)}.\n\nIf the y-value of each point is irrelevant, I can just simply randomly choose my size n.\n\nHowever in my case, I hope to reserve the original values of the data points to the best extent. Thus, I cannot choose a big window that averages everything to flat.\n\nMy cut-off freq. is 3Hz and the sampling rate is 50Hz.\n\nHow may I choose the size of the window n?\n\nYour normalized window is given by\n\n$$w(n)=\\frac{2}{N+1}\\frac{N-n}{N},\\quad n=0,1,\\ldots,N-1$$\n\nThe window satisfies\n\n$$\\sum_{n=0}^{N-1}w(n)=1$$\n\nwhich means that the gain of the corresponding moving average filter is 1 at DC.\n\nFor determining the cut-off frequency, we need to compute the frequency response of the window:\n\n$$W(e^{j\\theta})=\\sum_{n=0}^{N-1}w(n)e^{-jn\\theta}$$\n\nAfter some algebra you get\n\n$$W(e^{j\\theta})=\\frac{2}{N+1}\\frac{1-\\frac{N+1}{N}e^{-j\\theta}+\\frac{1}{N}e^{-j(N+1)\\theta}}{(1-e^{-j\\theta})^2}\\tag{1}$$\n\nNow you need to find the value of $N$ for which the magnitude of (1) at the cut-off frequency $\\theta_c=2\\pi\\frac{3Hz}{50Hz}$ becomes $1/\\sqrt{2}$ (-3dB). Since $N$ must be integer you cannot get any desired cut-off frequency, but the given cut-off is approximately achieved by $N=9$, for which $\|W(e^{j\\theta_c})\|=0.698$ (-3.13dB).\n\n• Thank you so much for the detailed calculation. I made a mistake just now. I have modified my weight array. Could you pls kindly edit your answer according to my updated weight array? I am really new to filter design and thus cannot duplicate ur working process... Thank you! – Sibbs Gambling Jun 14 '13 at 8:43\n• This is exactly the window I assumed you would be using. It now also includes the normalization factor I suggested. So the answer is correct as it stands. If you're happy with it, please accept the answer (by hitting the check mark button) to show that your question has been answered satisfactorily. – Matt L. Jun 14 '13 at 8:48\n• How does the expression above (1) come about, please? Also, can I say that my values within 3Hz are preserved (remain exactly the same)? because I need to use the values of the max and min value of the series, so if the values are distorted after the filter, then I have to do sth. else... – Sibbs Gambling Jun 17 '13 at 6:00\n• Another question: if I find that a bigger value for N actually gives a better performance by experiment, what are the effects when I raise the value of N. Say, now the calculation shows that I need to take N as 9. If I use it as 30, then will the cutoff freq. vary? Will the frequency response within 3Hz remain 1? Actually I do NOT need the frequency response to be 1. As long as the max and min are scaled by a same constant, that will be sufficient. Thanks! – Sibbs Gambling Jun 17 '13 at 6:12\n• If you increase $N$, the cutoff frequency will decrease, because you apply more averaging. You can plot the magnitude of (1) for different values of N to see the effect. Take a grid of frequencies $\\theta\\in [0,\\pi]$, where $\\pi$ corresponds to half the sampling rate, and compute $\|W(e^{j\\theta})\|$ for different values of $N$. – Matt L. Jun 17 '13 at 6:50" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88370687,"math_prob":0.9880989,"size":567,"snap":"2019-51-2020-05","text_gpt3_token_len":155,"char_repetition_ratio":0.124333926,"word_repetition_ratio":0.0,"special_character_ratio":0.2892416,"punctuation_ratio":0.14492753,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9993079,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-28T00:47:17Z\",\"WARC-Record-ID\":\"<urn:uuid:9db3e6c0-74d3-4eb4-b301-aede3ac05d87>\",\"Content-Length\":\"140439\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:15228e77-ecee-41e5-9c5b-597c56a56426>\",\"WARC-Concurrent-To\":\"<urn:uuid:9c3e7b58-fdcd-4d60-8629-329ff4b7cabe>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://dsp.stackexchange.com/questions/9590/how-to-determine-the-window-size-and-weights-in-weighted-moving-average-wma-g\",\"WARC-Payload-Digest\":\"sha1:QQLEBPQ2LCUQSM4RZYMDVEWKVKFWSA5I\",\"WARC-Block-Digest\":\"sha1:C5NHDYRRH3QTMVMLQ4MHRY5PW4QV5KOJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251737572.61_warc_CC-MAIN-20200127235617-20200128025617-00410.warc.gz\"}"}
http://www.puritys.me/docs-blog/article-141-Teajs-GD-library-tutorial.html	[ "2012\nAug\n02\n\n# Teajs GD library tutorial\n\nTags:\n\nGD Extension 主要功能是可以對圖片做處理，簡單的有圖片縮小、放大、切割，也可以用來製作圖片，這篇文章將會實做圖片的縮小，旋轉與銳利化。\n\n## Teajs GD Extension 安裝\n\ng++ -shared -lgd -lpthread -lv8 -lrt -ldl -o gd.so ~/gd.o ../binary/bytestorage.os ../../common.os ../../app.os ../../path.os ../../cache.os\n\n## Javascript GD 畫圖!\n\nTrueColor 是一種由 32 bits 存儲的圖片格式，在三原色 RGB 中各用了 8個 bits ，每一個顏色可以存的數值是 0~255 ，這樣就有 2563 種組合，另外再配上「透明度 alpha」8 bits。\n\n『jpeg』使用的格式是 TrueColor , 『gif』使用的格式則是 Palette ，因此 gif 只有 256 種顏色可以使用，『png』則支援 TrueColor 與 Palette。\n\n## 縮圖前的阿蕯斯", null, "## 縮圖後的阿蕯斯", null, "## Resize 程式範例\n\nResize.sjs\n`var Image = require('gd').Image;var source = \"Arthas-H.jpg\";var img = resize(source , 100,150);img.save(Image.PNG , \"tmp.png\"); function resize(img ,to_w,to_h) { var pos = img.lastIndexOf('.'); var ext = img.substring(pos+1, img.length); var type = Image.JPEG; switch(ext.toLowerCase()) { case 'jpeg':case 'jpg': type = Image.JPEG; break; case 'png': type = Image.PNG; break; case 'gif': type = Image.GIF; break; } var source_img = new Image(type , img); var w = source_img.sx(); var h = source_img.sy(); var image = new Image(Image.TRUECOLOR, to_w, to_h); var destx=0,desty=0; var srcx=0,srcy=0; image = image.copyResized(source_img, destx, desty, srcx, srcy, to_w, to_h , w , h ); return image;} `\n\n## 旋轉前的阿蕯斯", null, "## 旋轉後的阿蕯斯", null, "## Rotate 旋轉原理\n\n# 數學上定義的旋轉，是指用座標軸逆時針旋轉 => 座標點為順時針旋轉，所以算出的結果有一丁點不同。", null, "• 圖中可以算半徑 z = x12+y12 的根號\n• x2 = z × cos(θ1+θ2) = z×(cosθ1×cosθ2 - sinθ1×sinθ2)\n• 因為 cosθ1 = x1/z , sinθ1 = y1/z\n• 得 x2 = x1×cosθ2 - y1×sinθ2\n• 一樣的算法可以求得 y2 = y1×cosθ2 + x1×sinθ2\n\n## Rotate 圖片旋轉程式範例\n\nRotate.sjs\n`var source = \"Arthas-H.jpg\";var img = rotate(source , 10);img.save(Image.PNG , \"tmp.png\"); function rotate(img , angle) { var PI = Math.PI; var radian = PIangle/180; var pos = img.lastIndexOf('.'); var ext = img.substring(pos+1, img.length); var type = Image.JPEG; switch(ext.toLowerCase()) { case 'jpeg':case 'jpg': type = Image.JPEG; break; case 'png': type = Image.PNG; break; case 'gif': type = Image.GIF; break; } var source_img = new Image(type , img); var w = source_img.sx(); var h = source_img.sy(); var p2x = w/2, p2y = h/2; var diffw = Math.abs((p2xMath.cos(radian) + p2yMath.sin(radian))) - Math.abs(p2x); diffw = 2 Math.abs(diffw); var diffh = Math.abs(( -p2xMath.sin(radian) + p2yMath.cos(radian))) - Math.abs(p2y); diffh = 2* Math.abs(diffh); var centerx = Math.round(w/2),centery = Math.round(h/2); var newH = h + diffh; var newW = w + diffw; var image = new Image(Image.TRUECOLOR, newW, newH); var destx=newW/2,desty=newH/2;//rotate in center var srcx=0,srcy=0; image = image.copyRotated(source_img, destx, desty, srcx, srcy, w, h , angle ); return image;} `\n\n## 正常的阿蕯斯", null, "## 銳利化後的阿蕯斯", null, "## 銳利化程式範例\n\nsharpen.sjs\n`var img = sharpen(source, 150);img.save(Image.PNG , \"tmp.png\");function sharpen(img , percent) { var pos = img.lastIndexOf('.'); var ext = img.substring(pos+1, img.length); var type = Image.JPEG; switch(ext.toLowerCase()) { case 'jpeg':case 'jpg': type = Image.JPEG; break; case 'png': type = Image.PNG; break; case 'gif': type = Image.GIF; break; } var source_img = new Image(type , img); var data = source_img.sharpen(percent); return data;} `", null, "" ]	[ null, "http://www.puritys.me/filemanage/blog_files/file_36.jpg", null, "http://www.puritys.me/filemanage/blog_files/file_37.png", null, "http://www.puritys.me/filemanage/blog_files/file_36.jpg", null, "http://www.puritys.me/filemanage/blog_files/file_38.png", null, "http://www.puritys.me/filemanage/blog_files/file_39.gif", null, "http://www.puritys.me/filemanage/blog_files/file_36.jpg", null, "http://www.puritys.me/filemanage/blog_files/file_40.png", null, "http://www.puritys.me/templates/static/images/thumb_nobody.png", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.5311286,"math_prob":0.9905353,"size":4036,"snap":"2019-51-2020-05","text_gpt3_token_len":2144,"char_repetition_ratio":0.15500993,"word_repetition_ratio":0.34262297,"special_character_ratio":0.34340933,"punctuation_ratio":0.23580247,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96880716,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,6,null,2,null,6,null,2,null,2,null,6,null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-20T04:27:03Z\",\"WARC-Record-ID\":\"<urn:uuid:5ef73793-5f62-4f6e-b3f0-48efabb36ff3>\",\"Content-Length\":\"49992\",\"Content-Type\":\"application/http; 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https://www.amcs.uz.zgora.pl/?action=paper&paper=1565	[ "# International Journal of applied mathematics and computer science\n\nPaper details\n\nNumber 3 - September 2020\nVolume 30 - 2020\n\nApproximate state-space and transfer function models for 2x2 linear hyperbolic systems with collocated boundary inputs\n\nKrzysztof Bartecki\n\nAbstract\nTwo approximate representations are proposed for distributed parameter systems described by two linear hyperbolic PDEs with two time- and space-dependent state variables and two collocated boundary inputs. Using the method of lines with the backward difference scheme, the original PDEs are transformed into a set of ODEs and expressed in the form of a finite number of dynamical subsystems (sections). Each section of the approximation model is described by state-space equations with matrix-valued state, input and output operators, or, equivalently, by a rational transfer function matrix. The cascade interconnection of a number of sections results in the overall approximation model expressed in finite-dimensional state-space or rational transfer function domains, respectively. The discussion is illustrated with a practical example of a parallel-flow double-pipe heat exchanger. Its steady-state, frequency and impulse responses obtained from the original infinite-dimensional representation are compared with those resulting from its approximate models of different orders. The results show better approximation quality for the “crossover” input–output channels where the in-domain effects prevail as compared with the “straightforward” channels, where the time-delay phenomena are dominating.\n\nKeywords\ndistributed parameter system, hyperbolic equations, approximation model, state space, transfer function" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8599519,"math_prob":0.9088802,"size":1638,"snap":"2021-04-2021-17","text_gpt3_token_len":320,"char_repetition_ratio":0.11933905,"word_repetition_ratio":0.0,"special_character_ratio":0.16727717,"punctuation_ratio":0.07569721,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98967016,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-23T07:49:54Z\",\"WARC-Record-ID\":\"<urn:uuid:d373435b-ac97-4571-8d98-cd20b292db68>\",\"Content-Length\":\"5884\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f7583f60-ba10-4206-8156-aff59a31a01f>\",\"WARC-Concurrent-To\":\"<urn:uuid:57a862aa-cde1-4ad0-be33-4649efdf4751>\",\"WARC-IP-Address\":\"212.109.142.50\",\"WARC-Target-URI\":\"https://www.amcs.uz.zgora.pl/?action=paper&paper=1565\",\"WARC-Payload-Digest\":\"sha1:Y2GF5VJPUWWLEXZVYXLFZLLTMS4FCHQY\",\"WARC-Block-Digest\":\"sha1:VO35BGXNAFEVGLRPITJPRNKXVGHGTLLK\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703536556.58_warc_CC-MAIN-20210123063713-20210123093713-00782.warc.gz\"}"}
https://collegenursingassignments.com/2020/11/26/m2-a2-finance-2/	[ "# m2 a2 finance\n\nAssignment 2: Time Value of Money\n\nWhen the Genesis Energy and Sensible Essential teams held their weekly meeting, the time value of money and its applicability yielded an extremely stimulating discussion. However, most of the team members from Genesis Energy were very perplexed. Sensible Essentials decided the most expedient way to demonstrate how interest rates as well as time impact the value of money was to use examples. You have been asked to prepare a report analyzing your findings of the three example calculations listed below.\n\nIn this assignment, you will do the following:\n\n1. Calculate the future value of \\$100,000 ten years from now based on the following annual interest rates:\n1. 2%\n2. 5%\n3. 8%\n4. 10%\n1. Calculate the present value of a stream of cash flows based on a discount rate of 8%. Annual cash flow is as follows:\n1. Year 1 = \\$100,000\n2. Year 2 = \\$150,000\n3. Year 3 = \\$200,000\n4. Year 4 = \\$200,000\n5. Year 5 = \\$150,000\n6. Years 6-10 = \\$100,000\n1. Calculate the present value of the cash flow stream in problem 2 with the following interest rates:\n1. Year 1 = 8%\n2. Year 2 = 6%\n3. Year 3 = 10%\n4. Year 4 = 4%\n5. Year 5 = 6%\n6. Years 6-10 = 4\n\nPerform your calculations in an Excel spreadsheet. Copy the calculations in a Word document. In addition, write a 2- to 3-page executive summary in Word format. Your summary should reflect a proper analysis of your findings, including a comparison and contrast of data. Apply APA standards to citation of sources. Use the following file naming convention: LastnameFirstInitial_M2_A2.doc.\n\nBy Wednesday, September 21, 2016, deliver your assignment to the M2: Assignment 2 Dropbox.\n\n Assignment 2 Grading Criteria Maximum Points Calculated the future value of \\$100,000 ten years from now based on an annual interest rate of a) 2%, b) 5%, c) 8%, and d) 10%. 24 Calculated the present value of a stream of cash flows based on a discount rate of 8%. 24 Calculated the present value of the cash flow stream in problem 2 with the interest rates listed in the directions. 24 Summarized the findings of the analysis, including the comparison and contrast of data. 8 Wrote in a clear, concise, and organized manner; demonstrated ethical scholarship in accurate representation and attribution of sources; displayed accurate spelling, grammar, and punctuation. 20 Total: 100\n\n## \"Looking for a Similar Assignment? Get Expert Help at an Amazing Discount!\"", null, "" ]	[ null, "https://mbanursingpapers.com/wp-content/uploads/2019/09/5862ee927d90850fc3ce290d.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9238179,"math_prob":0.9543897,"size":2342,"snap":"2023-40-2023-50","text_gpt3_token_len":588,"char_repetition_ratio":0.1257485,"word_repetition_ratio":0.16504854,"special_character_ratio":0.28095645,"punctuation_ratio":0.11764706,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9834819,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-05T00:10:22Z\",\"WARC-Record-ID\":\"<urn:uuid:df837633-63df-4b1d-bebb-d6ef4bbeba2b>\",\"Content-Length\":\"54455\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e3475b82-bec1-443c-9818-9fbbb302b5eb>\",\"WARC-Concurrent-To\":\"<urn:uuid:eaa1f05b-c32d-4336-bbd2-e6432eae310d>\",\"WARC-IP-Address\":\"68.66.226.126\",\"WARC-Target-URI\":\"https://collegenursingassignments.com/2020/11/26/m2-a2-finance-2/\",\"WARC-Payload-Digest\":\"sha1:2ZIVRZEH5EBNH3CORLTXWJ7HTCTPDSL7\",\"WARC-Block-Digest\":\"sha1:WWHBR5UCZTPIGJI6YCQZTDYAGQOHDJSM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511424.48_warc_CC-MAIN-20231004220037-20231005010037-00786.warc.gz\"}"}
http://qs1969.pair.com/~perl2/?node_id=78994;displaytype=xml	[ "note perlmonkey To implement it using your syntax you could do something like: <CODE>sub operate (&@) { my \\$sub = shift; local \\$~; \\$~ = &\\$sub(\\$_) for @_; return \\$~ }</CODE><BR><BR> So for your examples you would get: <CODE>use strict; sub operate (&@) { my \\$sub = shift; local \\$~; \\$~ = &\\$sub(\\$_) for @_; return \\$~ } my @list = (4, 6, 2, 67, 123, 645, 23, 54 ); my \\$max = operate { \\$~ > \\$_ ? \\$~ : \\$_ } @list; print \\$max, \"\\n\"; my \\$string = operate { \\$~.\\$_ } @list; print \\$string, \"\\n\"; open FILE, \\$0; my \\$count = operate { \\$~ + /operate/ } <FILE>; print \"Count = \\$count\\n\"; </CODE><BR><BR> And the results are <CODE> max = 645 string = 462671236452354 count = 4</CODE> 78977 78977" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.62727,"math_prob":0.96186745,"size":679,"snap":"2022-40-2023-06","text_gpt3_token_len":244,"char_repetition_ratio":0.12148148,"word_repetition_ratio":0.1875,"special_character_ratio":0.48600882,"punctuation_ratio":0.22222222,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97411656,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-02T22:41:14Z\",\"WARC-Record-ID\":\"<urn:uuid:b9b211b7-6445-4c6b-bd0c-93bfa5187938>\",\"Content-Length\":\"1386\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:71a24ee6-1fed-4773-850d-5d0363c4f377>\",\"WARC-Concurrent-To\":\"<urn:uuid:23a6e9fc-862d-4c59-aba8-3908ee858c27>\",\"WARC-IP-Address\":\"216.92.237.59\",\"WARC-Target-URI\":\"http://qs1969.pair.com/~perl2/?node_id=78994;displaytype=xml\",\"WARC-Payload-Digest\":\"sha1:LIQ56CXDSFBZ3K5VMTY72LEZTAJF6DYD\",\"WARC-Block-Digest\":\"sha1:6GTPFV4VDMZBYCB4PULDOIU3IK25PQML\",\"WARC-Identified-Payload-Type\":\"application/xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337360.41_warc_CC-MAIN-20221002212623-20221003002623-00578.warc.gz\"}"}
https://esef2018.com/facebook-nachrichten-ohne-messenger/flixbus-dortmund-hamburg.php	[ "# Algebra equations to solve\n\nAlgebra equations are used to solve larger mathematical problems. These equations allow you to solve equations in a variety of ways, including substitution, addition, subtraction and multiplication. Algebra equations are also used to work out percentages and rates.\n\n## The Best Algebra equations to solve", null, "Once you understand how algebra equations work, you can apply this knowledge to different situations. For example, you can use algebra equations when calculating the price of an item or making a budget for food and other expenses. By working with algebra equations on a regular basis, you'll build your skills and knowledge. You'll also be able to see how these equations work in real-world examples. A good place to start is by learning the basics of algebra. You'll learn how to perform simple operations like addition and subtraction as well as how to solve algebra equations. Once you have these skills down, you'll be able to apply them in different situations.\n\nAlgebra is used to solve equations. Algebra equations can be written in the following ways: The three main types of algebra equations are linear, quadratic, and exponential. Linear equations involve one or two numbers. For example, 1x + 3 = 10. Quadratic equations have two unknown numbers and involve a squared number. For example, 4x2 + 2x + 5 = 25. Exponential equations have one number and involve an exponent (e) sign with a base number. For example for 4e-2x = 6. Algebra can be used to solve equations like the following: To solve the equation 5x - 8 = 7, we must first find the value of \"a\". To do this we use the formula: a = x - (5/8) br> br>Entering this in the formula above, we get: a = 7 - (1/8) br> br>Now that we know how to find \"a\", we can use it to find \"b\". To do this we use: b = a * x br> br>This gives us b = 1 * 7 br> br>The final result is that b = 9 br> br>To solve the equation y - 2 = 3, we must first find\n\nAlgebra is a mathematical field that focuses on solving problems using formulas. Algebra equations are expressions that can be used to solve problems. Algebra equations can be written on paper, in equations, or as word problems. One common type of algebra equation is an equation with two unknowns (also called variables). This type of equation could be used to solve the following types of problems: • Finding the length of a string • Finding the volume of a box • Finding the number of steps it takes to climb a certain number of stairs Algebra equations can also be written in other ways, such as as word problems, by using variables and different symbols. For example, “Bill climbed four flights of stairs to get to his apartment” could represent an algebraic equation such as \"4x + 3 = 16.\" In this case, the letter “x” represents one unknown, and “+” represents addition. The letter “=” represents equality, which means that you need to find the value that makes 4x equal to 16. This could be any number from 1 to 4; for example, 1 would make 4x equal 16 and 2 would make 4x equal 8. However, if you were asked to find out how many steps it took Bill to climb four flights of stairs (meaning you didn't know how many steps there", null, "Not only that it resolves your math problems, it also teaches you solve math! If you're stuck at an algebra problem you can just take a photo or write it and it shows you a step-by-step renovation plus instructions. Would 100% recommend to everyone!\n\nStella Perry", null, "Great app for solving math. Everyone would should be try it, If they weak in math. Excluding it all are good. You must be trying this app. Thank you for giving me your time. It does not have ads it better than a normal calendar it free it a 100/10 go for school to\n\nGabriella Powell\n\nDoing Key Input Solve by substitution method solver How to solve for x in a triangle Math type online App to check math homework System of three equations solver" ]	[ null, "https://esef2018.com/PSY6320a40ff2880/author-profile.jpg", null, "https://esef2018.com/PSY6320a40ff2880/testimonial-one.jpg", null, "https://esef2018.com/PSY6320a40ff2880/testimonial-two.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9563745,"math_prob":0.9971707,"size":3710,"snap":"2022-40-2023-06","text_gpt3_token_len":860,"char_repetition_ratio":0.16405828,"word_repetition_ratio":0.005925926,"special_character_ratio":0.23557952,"punctuation_ratio":0.09646739,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99982256,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-06T10:25:01Z\",\"WARC-Record-ID\":\"<urn:uuid:ccc44639-b74b-431f-9486-003e4e306141>\",\"Content-Length\":\"25498\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:72f97711-471c-4813-8181-85b8400b6b53>\",\"WARC-Concurrent-To\":\"<urn:uuid:05ad13c1-b247-4285-8f4d-5ee682386813>\",\"WARC-IP-Address\":\"67.21.87.14\",\"WARC-Target-URI\":\"https://esef2018.com/facebook-nachrichten-ohne-messenger/flixbus-dortmund-hamburg.php\",\"WARC-Payload-Digest\":\"sha1:RN6IUKTWIVIYRIQ47DFL32OE6BJVQLU4\",\"WARC-Block-Digest\":\"sha1:QBKL7OWMNF43XA6SNOPIFZKUEOTJFB5H\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337803.86_warc_CC-MAIN-20221006092601-20221006122601-00044.warc.gz\"}"}
https://optiekmarcbrabanders.be/Mar-21/668.html	[ "# how to calculate mill critcal speed\n\n• ### How to determine your critical speed ACTIVE\n\nIf you want to race to your full potential it is helpful to determine your critical speed. On paper the athlete with the fastest overall critical speed in three events will most likely be the victor. Critical speed (CS) represents your race pace for any endurance event lasting between 30 and 60 minutes.\n\nChat Online\n• ### critical speed ball mill calculation\n\nTECHNICAL NOTES 8 GRINDING RP KingMineral Technologies complicated in practice and it is not possible to calculate The critical speed of the mill c is defined as the speed\n\nChat Online\n• ### (PDF) Planetary Ball Mill Process in Aspect of Milling Energy\n\nA read is counted each time someone views a publication summary (such as the title abstract and list of authors) clicks on a figure or views or downloads the full-text.\n\nChat Online\n• ### Critical Speed Calculator Nook Industries\n\nCRITICAL SPEED Enter any TWO of the following and click on the \"Calculate\" button\n\nChat Online\n• ### critical speed of ball mill calculation pdf\n\nformula critical speed ball mill . calculate critical speed of ball mill ugcnetnic. Ball Mill Critical Speed Mineral Processing . 10 A Ball Mill Critical Speed (actually ball rod AG or SAG) is the speed at which the centrifugal forces equal gravitational forces at the mill shells inside get price.\n\nChat Online\n• ### What it is the optimun speed for a ball mill\n\nOct 19 2006 · What it is the optimun speed for a ball mill posted in Pyrotechnics I have done a ball mill recenly finished but the motor has too rpms is too fast for use in a ball mill (the pvc cylinder that i use left of the shafts). With the motor i will use a 40 mm pulley because i have a 50 mm driven pulley in one of my two shafts.\n\nChat Online\n• ### How to calculate critical speed in circular motion\n\nApr 14 2012 · In one textbook it says that the critical speed is the minimum speed at which an object can complete the circular motion. It gives the formula v = square root of (g r) However in another textbook it says that the formula is v = square root of (2 g r) How can there be two different\n\nChat Online\n• ### Critical speedWikipedia\n\nThere are two main methods used to calculate critical speed—the Rayleigh–Ritz method and Dunkerley s method. Both calculate an approximation of the first natural frequency of vibration which is assumed to be nearly equal to the critical speed of rotation. The Rayleigh–Ritz method is\n\nChat Online\n• ### Critical Speed Of Mill\n\nhow to calculate critical speed of ball millSBM Machine. Mill Critical Speed Calculation deflection of the lifters is smaller for lower critical speeds. Read more. critical speed of a mill. The Influence of Ball Mill Critical Speed on Production Efficiency. When the rotating speed of ball mill grinder is high the height the ball\n\nChat Online\n• ### calculate critical speed of ball millPopular Education\n\ncomplicated in practice and it is not possible to calculate The critical speed of the mill c is defined as the speed at which a single ball will just . Rod and ball mills in Mular AL and Bhappu R B Editors Mineral Processing Plant Design. Read More.\n\nChat Online\n• ### How to calculate critical speed of ball millHenan\n\nCritical speed of ball mill calculation grinder. Calculations for mill motor power mill speed and media charge Advantages Considering the weight of mill lining and grinding media work out the motor power required in consultation To calculate the motor power required for a cylindrical type ball mill the following formula can be applied Where Nc Critical speed\n\nChat Online\n• ### ball mill calculations critical speedJack Higgins\n\ncalculating critical speed in a ball mill uae Ball Mill Critical Speed Mineral Processing Metallurgy A Ball Mill Critical Speed actually ball rod AG or SAG is the speed at which the centrifugal forces equal gravitational forces at the mill shells inside surface and no balls will fall from its position onto the shell.\n\nChat Online\n• ### mill critical speed formulaJack Higgins\n\nMill Critical Speed Determination. The \"Critical Speed\" for a grinding mill is defined as the rotational speed where centrifugal forces equal gravitational forces at the mill shell s inside surface. This is the rotational speed where balls will not fall away from the mill s shell. Read More\n\nChat Online\n• ### how to calculate aluminum poweder ball mill critical speed\n\nhow to calculate the ball need in ball mill . Critical Speed Calculation Of Ball Mill. ball mill calculation critical speed wildpeppersf. 7 Nov 2013 The critical speed of the mill c is defined as the speed at which a single ball will just remain ball mill critical speed calculation where society/organization is ball mill critical speed\n\nChat Online\n• ### how to calculate critical speed of ball mill\n\nformula for critical speed of rotation of a ball mill. formula for critical speed of a rotating mill dieboldbau. ball mill critical speed calculation Ball mill critical speed Ball mill efficiency ball mill and now only the calculation formula on critical speed Critical rotation speed for ballmilling where D is the inner diameter of a jar and r is the radius of balls.\n\nChat Online\n• ### Mill Critical Speed DeterminationHitlers Hollywood\n\ncritical speed formula for ball mill The formula to calculate critical speed is given below N c 42305 sqtDd N c critical speed of the mill D mill diameter specified in meters d diameter of the ball In practice Ball Mills are driven at a speed of 5090 of the critical speed the\n\nChat Online\n• ### critical speed of grinding mill calculation\n\nThe formula to calculate critical speed is given below. N c = 42.305 /sqt(D-d) N c = critical speed of the mill. D = mill diameter specified in meters. d = diameter of the ball. In practice Ball Mills are driven at a speed of 50-90 of the critical speed the factor being influenced by economic consideration.\n\nChat Online\n• ### ball mill critical speed calculation xls\n\nball mill critical speed calculation xls. how to calculate critical speed of ball mill More information of critical speed of ballmill calculation ball mill critical speed calculation xls. OBTENIR UN PRIX. ball mill calculations excelXinHaigreekhotelguide.\n\nChat Online\n• ### critical speed of ball mill calculation\n\nSAGMILLING . . Mill Critical Speed Determination. The \"Critical Speed\" for a grinding mill is defined as the rotational speed where This is the rotational speed where balls will not fall away from the mill s shell. You may use the Mill Liner Effective\n\nChat Online\n• ### (PDF) Planetary Ball Mill Process in Aspect of Milling Energy\n\nA read is counted each time someone views a publication summary (such as the title abstract and list of authors) clicks on a figure or views or downloads the full-text.\n\nChat Online\n• ### Ball Mill Critical Speed\n\nA Ball Mill Critical Speed (actually ball rod AG or SAG) is the speed at which the centrifugal forces equal gravitational forces at the mill shell s inside surface and no balls will fall from its position onto the shell. The imagery below helps explain what goes on inside a mill as speed varies.\n\nChat Online\n• ### calculate critical speed of ball millPopular Education\n\ncomplicated in practice and it is not possible to calculate The critical speed of the mill c is defined as the speed at which a single ball will just . Rod and ball mills in Mular AL and Bhappu R B Editors Mineral Processing Plant Design. Read More.\n\nChat Online\n• ### Rolling mill speed calculation formula pdf\n\nrolling mill speed calculation formula 2 plays an important role in calculating the rolling.shapes are calculated. Six pass designs used in Swedish mills are analysed. The thesis also includes high-speed rolling of. Calculate those costs.mill roll speeds are usually given in feetminute we. We now have to calculate the peripheral speed of the mill.\n\nChat Online\n• ### calculate critical speed of ball mill\n\ncalculate critical speed of ball mill calculate critical speed of ball mill. Ball mill . A ball mill is a type of grinder used to grind and blend materials for use in mineral dressing The grinding works on the principle of critical speed. Size Reduction and Mill SpeedCritical Calculating how fast a Jar needs to spin is a little\n\nChat Online\n• ### What it is the optimun speed for a ball mill\n\nOct 19 2006 · What it is the optimun speed for a ball mill posted in Pyrotechnics I have done a ball mill recenly finished but the motor has too rpms is too fast for use in a ball mill (the pvc cylinder that i use left of the shafts). With the motor i will use a 40 mm pulley because i have a 50 mm driven pulley in one of my two shafts.\n\nChat Online\n• ### sag mill critical speed calculation\n\nsag mill critical speed calculation. Ball Mill Critical SpeedMineral Processing Metallurgy. A Ball Mill Critical Speed (actually ball rod AG or SAG) is the speed at which the centrifugal forces equal gravitational forces at the mill shell s inside surface and no balls will fall from its position onto the shell.\n\nChat Online" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8320014,"math_prob":0.9566801,"size":8784,"snap":"2021-43-2021-49","text_gpt3_token_len":1819,"char_repetition_ratio":0.26617312,"word_repetition_ratio":0.4261745,"special_character_ratio":0.19000456,"punctuation_ratio":0.042976268,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98210806,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-27T20:27:03Z\",\"WARC-Record-ID\":\"<urn:uuid:7921fa19-05f5-45c9-a09b-e5a3c6cb0173>\",\"Content-Length\":\"17009\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b0cb1bfa-653f-42cd-80d5-44d85d5b6fd7>\",\"WARC-Concurrent-To\":\"<urn:uuid:8791110c-ce79-4393-8a8f-8208195b4e1e>\",\"WARC-IP-Address\":\"104.21.71.35\",\"WARC-Target-URI\":\"https://optiekmarcbrabanders.be/Mar-21/668.html\",\"WARC-Payload-Digest\":\"sha1:XIZOBFDPADI2MM7S72TR3EIPAZ47SIZC\",\"WARC-Block-Digest\":\"sha1:HDOZYXOAOZ7HX7UROT52KPN6GS6YD7EP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323588242.22_warc_CC-MAIN-20211027181907-20211027211907-00515.warc.gz\"}"}
https://eng.kakprosto.ru/kommunalnye-uslugi?page=3	[ "Utilities\n• How to remove readings from a three-phase meter\n• How to calculate the cost of water\n• How to make water in a private house\n• How to fill out a receipt of payment for electricity\n• Where to call if there is no electricity\n• Where to send readings of water meter\n• How to refuse a radio receiving station in Moscow\n• How to install garbage bins in the yard in accordance with the regulations\n• Any gas meter it is better to install\n• How to reprogram a key from the intercom\n• How to apply for benefits utilities\n• Where to call if turned off the hot water\n• If the electricity shelf life\n• Where to go if there is no water\n• How to calculate the rent\n• How to hold water on the land\n• How not to pay for gas at the counter\n• How to pay utility bills\n• How to save gas\n• Do I have to put water meters\n• How to charge for cold water\n• What is the multitariff meter\n• How to choose gas meter\n• How to calculate utilities\n• How to take readings with a mercury 200\n• How to call an electrician\n• What documents are needed for connecting electricity to non-residential plot\n• How to calculate the tariff for hot water\n• How to turn off the radio\n• How to get free electricity", null, "" ]	[ null, "https://tms.dmp.wi-fi.ru/", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8908961,"math_prob":0.7743178,"size":1181,"snap":"2021-31-2021-39","text_gpt3_token_len":284,"char_repetition_ratio":0.23194562,"word_repetition_ratio":0.029166667,"special_character_ratio":0.23539373,"punctuation_ratio":0.0,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9537971,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-01T01:26:12Z\",\"WARC-Record-ID\":\"<urn:uuid:6f681556-651c-4dca-80ff-16b82462a0ea>\",\"Content-Length\":\"72319\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a22d3373-dd26-41a5-a19f-4b676bfc9be6>\",\"WARC-Concurrent-To\":\"<urn:uuid:7c3c549d-bf17-4e63-9977-8c1e0d711fe3>\",\"WARC-IP-Address\":\"178.154.246.3\",\"WARC-Target-URI\":\"https://eng.kakprosto.ru/kommunalnye-uslugi?page=3\",\"WARC-Payload-Digest\":\"sha1:ATB5NFUKWGUWC5COLKKZX4DN7JYC3LJE\",\"WARC-Block-Digest\":\"sha1:RBJSAGMINWVZFQTNAHUMODHLDA3L6KCX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154127.53_warc_CC-MAIN-20210731234924-20210801024924-00684.warc.gz\"}"}
https://www.jpost.com/middle-east/baghdad-2-car-bombs-explode-as-rice-straw-leave	[ "(function (a, d, o, r, i, c, u, p, w, m) { m = d.getElementsByTagName(o), a[c] = a[c] \|\| {}, a[c].trigger = a[c].trigger \|\| function () { (a[c].trigger.arg = a[c].trigger.arg \|\| []).push(arguments)}, a[c].on = a[c].on \|\| function () {(a[c].on.arg = a[c].on.arg \|\| []).push(arguments)}, a[c].off = a[c].off \|\| function () {(a[c].off.arg = a[c].off.arg \|\| []).push(arguments) }, w = d.createElement(o), w.id = i, w.src = r, w.async = 1, w.setAttribute(p, u), m.parentNode.insertBefore(w, m), w = null} )(window, document, \"script\", \"https://95662602.adoric-om.com/adoric.js\", \"Adoric_Script\", \"adoric\",\"9cc40a7455aa779b8031bd738f77ccf1\", \"data-key\");\nvar domain=window.location.hostname; var params_totm = \"\"; (new URLSearchParams(window.location.search)).forEach(function(value, key) {if (key.startsWith('totm')) { params_totm = params_totm +\"&\"+key.replace('totm','')+\"=\"+value}}); var rand=Math.floor(10*Math.random()); var script=document.createElement(\"script\"); script.src=`https://stag-core.tfla.xyz/pre_onetag?pub_id=34&domain=\\${domain}&rand=\\${rand}&min_ugl=0\\${params_totm}`; document.head.append(script);" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9678098,"math_prob":0.96919554,"size":802,"snap":"2023-14-2023-23","text_gpt3_token_len":164,"char_repetition_ratio":0.0802005,"word_repetition_ratio":0.0,"special_character_ratio":0.18952619,"punctuation_ratio":0.10067114,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9789482,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-01T22:26:49Z\",\"WARC-Record-ID\":\"<urn:uuid:255bdca9-a5e9-4f84-b9df-c710aa0a7841>\",\"Content-Length\":\"78653\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:db676f7d-2d37-4a38-9061-ee0da09e24fa>\",\"WARC-Concurrent-To\":\"<urn:uuid:c0ff80d5-b04c-4371-8b70-1e2e83aa45dc>\",\"WARC-IP-Address\":\"159.60.130.79\",\"WARC-Target-URI\":\"https://www.jpost.com/middle-east/baghdad-2-car-bombs-explode-as-rice-straw-leave\",\"WARC-Payload-Digest\":\"sha1:UXQWPVXRYHEP2VKPESIJ56PCRMI2TUC4\",\"WARC-Block-Digest\":\"sha1:CZT3KAZK2E7P53PEQFU5W7VNBOG2VGR2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224648209.30_warc_CC-MAIN-20230601211701-20230602001701-00387.warc.gz\"}"}
https://zbmath.org/?q=an:0747.93071	[ "# zbMATH — the first resource for mathematics\n\nControl of a chaotic system. (English) Zbl 0747.93071\nGiven a Lorenz system subject to control $\\dot x_ 1=-sx_ 1+sx_ 2, \\dot x_ 2=rx_ 1-x_ 2-x_ 1 x_ 3+u, \\dot x_ 3=x_ 1 x_ 2- bx_ 3,$ the authors suggest two different controllers to stabilize unstable equilibrium point of the uncontrolled system. They analyze a particular case $$s=10$$, $$r=28$$, $$b=8/3$$, which has three unstable equilibrium points.\nThe first controller is given by $$u=-k(x_ 1-x_{10})$$. Then a sufficiently large $$k>0$$ guarantees the stability, but the motion may contain chaotic transients of different time lengths depending on the magnitude of $$k$$.\nIn the second case, the controllability minimum principle produces a stabilizing bang-bang control $$u=-10\\text{ sgn}(x^ 2_ 1-(8/3)x_ 3)$$.\nReviewer: J.Kucera (Pullman)\n\n##### MSC:\n 93D15 Stabilization of systems by feedback 93C10 Nonlinear systems in control theory 93C15 Control/observation systems governed by ordinary differential equations\nFull Text:\n##### References:\n J. Brindley and I. M. Motoz, ”Lorenz attractor behavior in a continuously stratified baroclinic fluid,”Phys. Lett., vol. 77A, pp. 441–444, 1980. P. Ehrhard and U. Müller, ”Dynamical behavior of natural convection in a single-phase loop,”J. Fluid Mech. (in press). J.E. Gayek and T.L. Vincent, ”On the asymptotic stability of boundary trajectories,”Int. J. Control, vol. 41, pp. 1077–1086, 1985. · Zbl 0562.93078 J.D. Gibbon and M.J. McGuinness, ”A derivation of the Lorenz equation for some unstable dispersive physical systems,”Phys. Lett., vol. 77A, pp 295–299, 1980. W.J. Grantham and T.L. Vincent, ”A controllability minimum principle.”J. Optimization Theory Applications, vol. 17, pp. 93–114, 1975. · Zbl 0292.93002 H. Haken, ”Analogy between higher instabilities in fluids and lasers,”Phys. Lett., vol. 53A, pp. 77–78, 1975. J.L. Kaplan and J.A. Yorke, ”Preturbulence: a regime observed in a fluid flow model of Lorenz,”Commun. Math. Phys., vol. 67, pp. 93–108, 1979. · Zbl 0443.76059 E. Knobloch, ”Chaos in a segmented disc dynamo,”Phys Lett., vol. 82A, pp. 439–440, 1981. E.N. Lorenz, ”Deterministic non-periodic flow,”J. Atmos. Sci., vol. 20, pp. 130–141, 1963. · Zbl 1417.37129 M.V.R. Malkus, ”Non-peroidic convection at high and low Prandtl number,”Mémoires Société Royale des Sciences de Liége, Series 6, vol. 4, pp. 125–128, 1972. J. Pedlosky, ”Limit cycles and unstable baroclinic waves,”J. Atmos. Sci., vol. 29, p. 53, 1972. J. Pedlosky and C. Frenzen, ”Chaotic and periodic behavior of finite amplitude baroclinic waves,”J. Atmos. Sci., vol. 37, pp. 1177–1196, 1980. C. Sparrow, ”The Lorenz equations: bifurcations, chaos, and strange attractor,”Appl. Math. Sci., vol. 41, 1982. · Zbl 0504.58001 E.D. Yorke and J.A. Yorke, ”Metastable chaos: transition to sustained chaotic behavior in the Lorenz model,”J. Stat. Phys., vol. 21, pp. 263–277, 1979.\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.69030535,"math_prob":0.93608135,"size":3617,"snap":"2021-43-2021-49","text_gpt3_token_len":1180,"char_repetition_ratio":0.109880984,"word_repetition_ratio":0.00754717,"special_character_ratio":0.34089023,"punctuation_ratio":0.27272728,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9891615,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-06T00:01:44Z\",\"WARC-Record-ID\":\"<urn:uuid:4ec19106-8c83-488f-ace0-81963d31267e>\",\"Content-Length\":\"52840\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:974a4d0f-5220-4c44-9910-1b6cb74e3cf5>\",\"WARC-Concurrent-To\":\"<urn:uuid:eb6be642-6d14-49a5-a33d-eb41ac113076>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/?q=an:0747.93071\",\"WARC-Payload-Digest\":\"sha1:GKS7LAOBHY225RTMUU2GYDBS3OKVP2Y5\",\"WARC-Block-Digest\":\"sha1:NHAIEGQ3MKON22UUWAQUVNIZMFOJ5O5T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363226.68_warc_CC-MAIN-20211205221915-20211206011915-00487.warc.gz\"}"}
http://www.next.gr/circuits/Circuit-diagram-for-generating-time-delay-with-555-IC-l43623.html	[ "# Circuit diagram for generating time delay with 555 IC\n\nPosted on Feb 6, 2014\n\nThis circuit based project demonstrates the working of 555 timer in astable mode to generate pulses of time period 0. 5 second. This pulse can be further used for anything where we need a pulse such as to blink a LED or to create fashionable blinking lights. Image below shows internal circuitry of NE 555 timer which can be used in astable and monos\n\ntable mode: This circuit of this project makes the use of timer IC NE555 which produces a constant square pulse of a desired frequency. This pulse could be either triggered or could be produced continuously depending upon the mode of 555 we are using. The two mostly used modes of 555 are Monostable and Astable. Here it is used in the astable mode with time period of half second, with high time period of 0. 333 seconds and low time period of 0. 166 seconds. For astable mode total time period is [ln2C(R1+2R2)] with high time period as [ln2(R1+R2)C]and low time period as [ln2(R2C)]. Here R1 is the resistor connected between VCC and pin7 (discharge pin), R2 is between pin7 and pin2 (trigger pin) and C1 is the capacitor connected from pin2 to ground. For 555 to function in astable mode pin2 and pin6 (threshold) pin must be shorted. Reset pin is connected to VCC. The output of 555 is taken at pin3 which is in the form of square wave and is then fed to lighting circuit which glows when output is high and stops when it becomes low; there by producing pattern of blinking lights. By varying the value of R1, R2 and C1 square waves of different time periods can be obtained.\n\nLeave Comment", null, "characters left:\n\n• ## New Circuits\n\n.\n\nPopular Circuits\n\n12KV High Voltage Generator\nBCD seven sections of nixie tubes reveal the decoder circuit\nXENON STROBE\nMultiple Capacitor Charger\n\nTop", null, "" ]	[ null, "http://www.next.gr/templates/nova/img/no-avatar.png", null, "http://www.next.gr/circuits/cron.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.948904,"math_prob":0.96680653,"size":3963,"snap":"2022-27-2022-33","text_gpt3_token_len":861,"char_repetition_ratio":0.10659257,"word_repetition_ratio":0.0,"special_character_ratio":0.22104466,"punctuation_ratio":0.113065325,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9587237,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-19T04:35:05Z\",\"WARC-Record-ID\":\"<urn:uuid:6ee03255-dc89-4c8c-921e-ad58d494c7e7>\",\"Content-Length\":\"83376\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:75b7c799-81cf-4da8-a2a1-e348247206bc>\",\"WARC-Concurrent-To\":\"<urn:uuid:89e20e75-af44-4ad7-8ab3-b546da095331>\",\"WARC-IP-Address\":\"198.211.126.132\",\"WARC-Target-URI\":\"http://www.next.gr/circuits/Circuit-diagram-for-generating-time-delay-with-555-IC-l43623.html\",\"WARC-Payload-Digest\":\"sha1:7RXHHDSB2XP6IZZQ7UBD6XIBW5E4DKTB\",\"WARC-Block-Digest\":\"sha1:ZSI2RNAZ7USXUDQMYSIXK4Y2JUIQXNI3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573623.4_warc_CC-MAIN-20220819035957-20220819065957-00628.warc.gz\"}"}
https://answers.everydaycalculation.com/subtract-fractions/3-5-minus-1-9	[ "# Answers\n\nSolutions by everydaycalculation.com\n\n## Subtract 1/9 from 3/5\n\n3/5 - 1/9 is 22/45.\n\n#### Steps for subtracting fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 5 and 9 is 45\n2. For the 1st fraction, since 5 × 9 = 45,\n3/5 = 3 × 9/5 × 9 = 27/45\n3. Likewise, for the 2nd fraction, since 9 × 5 = 45,\n1/9 = 1 × 5/9 × 5 = 5/45\n4. Subtract the two fractions:\n27/45 - 5/45 = 27 - 5/45 = 22/45\n\n#### Subtract Fractions Calculator\n\n-\n\nUse fraction calculator with our all-in-one calculator app: Download for Android, Download for iOS\n\n© everydaycalculation.com" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.60439813,"math_prob":0.995307,"size":325,"snap":"2019-13-2019-22","text_gpt3_token_len":172,"char_repetition_ratio":0.25856698,"word_repetition_ratio":0.0,"special_character_ratio":0.5538462,"punctuation_ratio":0.06185567,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995129,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-25T15:58:21Z\",\"WARC-Record-ID\":\"<urn:uuid:91a45a51-b982-4e15-8081-6c9a1de385a8>\",\"Content-Length\":\"8452\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:12ba3528-06c7-45db-b4a4-024868250615>\",\"WARC-Concurrent-To\":\"<urn:uuid:e1f10c8b-146f-4dd7-841e-b71698a2b45c>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/subtract-fractions/3-5-minus-1-9\",\"WARC-Payload-Digest\":\"sha1:ARZEQ3HBFNOLZEW253HHPQH2Y7ZZKTEE\",\"WARC-Block-Digest\":\"sha1:S2QFPMUWMOTYHBGJVG6EAGXXF6XT2VE4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232258120.87_warc_CC-MAIN-20190525144906-20190525170906-00228.warc.gz\"}"}
https://papers.nips.cc/paper_files/paper/2020/file/4607f7fff0dce694258e1c637512aa9d-Review.html	[ "NeurIPS 2020\n\n### Review 1\n\nSummary and Contributions: This paper studies active learning with label and comparison queries, extending the works of (Kane et al, 2017, Xu et al, 2017). It gives a new set of fundamental upper and lower bounds on halfspace learning in passive/active, label/label+comparison queries, PAC/RPU(Reliably Probably Useful) models, summarized in Tables 1 and 2 in the paper. - On the lower bound side, it shows that any label-only active learning algorithm must have polynomial query complexity in PAC learning setting, and exponential sample complexity in the RPU learning setting. The main techniques are using existing combinatorial geometry results. - On the upper bound side, it shows that under mild distributional assumptions, label+comparison based active learning algorithms can have logarithmic query complexity in both PAC and RPU learning settings. The main technique is by developing a distribution-dependent variant of inference dimension (Kane et al, 2017) and a novel online-to-batch conversion (Thm 4.11).\n\nStrengths: Overall, I think it is a solid paper that greatly advances the understanding of RPU learning and learning with comparison queries. Specifically: - The development of distribution-dependent inference dimension is an important contribution, which can helps spur more works in distribution-dependent analysis of machine learning (similar to Hanneke's disagreement coefficient / Alexander's capacity function for active and passive learning) - The proof of Theorem 4.10 (distribution-dependent inference dimension for halfspaces) is not trivial, in my understanding. I was initially thinking if some arguments therein can be simplified by using (normalized) VC inequalities, but it does not lead to g(n) = 2^{-Omega(n^1+\\alpha)} for some positive alpha, which is crucial for obtaining the results in this paper.\n\nWeaknesses: - Many of the lower bounds seem to be direct applications of existing results of combinatorial geometry (although the tightness of these results are also discussed, and they nicely extend the early negative results of Kivinen on RPU learning)\n\nCorrectness: They are correct as far as I have checked, although there are a few subtle points that I need the authors' clarification.\n\nClarity: While I can understand the statements in this paper, I think the presentation can be much improved. For example: - As RPU learning implies PAC learning, is there really a need to present Algorithm 1 and Theorem 3.5? Aren't we already happy with Theorem 4.11? - in Algorithm 1, Threshold(S) is only informally defined, and an elaboration is needed. I think basically, the algorithm can successfully approximately recover b if it can find two neighboring + and - examples? - In the proof of Proposition 4.2, no high-level idea of the proof was given, which makes it a bit hard to follow (although I agree that it is correct) - In the second half of Corollary 4.7, is the goal only to identify the labels of all examples S drawn? Also, what is the active learning algorithm used here? I am also confused about the statement that \"the boosting algorithm in KLMZ is reliable even when when given the wrong inference dimension as input\". Are you referring the algorithm in page 16 of that paper (arxiv version)? - In Theorem 4.9, is H here H_{d, \\eta}? - In the proof of Theorem 4.10, what is h \\times [-\\gamma, \\gamma]? Is it {x: h(x) \\in [-\\gamma, \\gamma]}? - At the end of the proof of Theorem 4.10, it is said that \"\\forall h there must exists x, s.t. Q(S_h' - {x}) infers x.\" Would it be still possible that there exist two hypotheses h_1, h_2 that have _very small minimal-ratio_, and they agree with the queries in Q(S_h' - {x}), but disagree on x? - In Algorithm 2, if g is a constant function, can be just replace it with a constant? To align with the terminology of KLMZ as much as possible, I suggest changing the name \"average inference dimension\" to e.g.. \"inference dimension tail bound\". I think people usually use \"dimension\" to denote values that take integers.\n\nRelation to Prior Work: Yes. It also discusses a subsequent paper that benefits from the techniques in this paper.\n\nReproducibility: Yes\n\nAdditional Feedback: I thank the authors for the reply. My opinion has not changed. But I have follow-up questions that I hope the authors can clarify in the final version: 1. In the second half of Corollary 4.7, in the active RPU learning algorithm, do we need to first compute the value of k that minimizes the right hand side of the equation, then apply KLMZ's algorithm with that value of k? It might be interesting to extend KLMZ's algorithm and analysis to achieve adaptivity to the data-dependent inference dimension. 2. In Theorem 4.10, it would be very helpful to point out that what the inference algorithm used here is (it is still not clear to me if the inference algorithm needs to know the minimal-ratio or not - specifically, in the linear program alluded in page 16 of KLMZ, in combination with Claim 4.10 therein, it seems like the knowledge of minimal-ratio is needed). The algorithm right below Theorem 4.11 should be revised accordingly to incorporate this.\n\n### Review 2\n\nSummary and Contributions: This paper studies the power of comparisons in the problem of actively learning (non-homogenous) linear classifiers. There are three main results in the paper: 1) in the PAC learning model, neither active learning nor comparison queries alone provide a significant speed-up over passive learning; 2) in the PUR-learning model, the paper confirms that passive learning with label queries is intractable information-theoretically, and active learning alone provides little improvement; 3) in the PUR-learning model, the comparison oracle provides a significant improvement in both active and passive learning scenarios. In the context of previous work, the techniques of this paper are heavily based on a combination of the inference dimension (Kane, Lovett, Moran, and Zhang ) and the (non-efficient version of) margin-based active learning (Balcan and Long ). The paper also extends the analysis to the s-concave distribution based on a concentration result in Balcan and Zhang .\n\nStrengths: 1. Learning of linear classifiers with both comparison and label oracles were rarely studied in the literature; to the best of my knowledge, existing work includes Kane, Lovett, Moran, and Zhang and Xu, Zhang, Miller, Singh, and Dubrawski , but many fundamental questions remain open in the community. Using existing techniques, this paper answers many of these questions. 2. The techniques are novel for the NeurIPS standard. The theoretical analysis seem solid. Experiments are available as a theoretical paper.\n\nWeaknesses: The paper does not consider computational efficiency and noise tolerance (given that the paper is built upon , which is a computationally-inefficient algorithm), while the techniques of achieving computational efficiency and noise tolerance in the existing work of active learning are available.\n\nCorrectness: The claims and method are correct to me.\n\nClarity: The paper is well-written and easy to understand.\n\nRelation to Prior Work: The paper did a good job on discussing how this work differs from previous contributions.\n\nReproducibility: Yes\n\nAdditional Feedback: ==========after rebuttal========== I have read the rebuttal, and I am happy to recommend acceptance of this paper.\n\n### Review 3\n\nSummary and Contributions: I have reviewed this paper before, the following is an adaptation of my past review. This paper considers the problem of learning non-homogenous linear classifiers using comparison queries over distributions that are weakly concentrated (such as s-concave). The main focus of the paper is on the power of comparison queries for distribution-dependent and Reliable and Probably Useful learning. Comparison queries in addition to the labels reveal which of two instances are closer to the boundary of the classifier. Comparison queries were shown to be useful for improving the query complexity of a learning task. Kane et al.’17 considered distribution-independent setting and showed that under assumptions, such as large margin assumption, comparison queries gain an exponential improvement in query complexity over (label) active learning. The same paper also showed that in some cases, query complexity won’t have an asymptotic improvement over passive or active learning, based on a definition of a notion of “inference dimension”. An example of this is learning halfspaces in a distribution independent setting even in R^3. This paper first shows that learning non-homogenous halfspaces over uniform distribution active or passive-comparison learning individually requires poly(1/eps) queries. Similarly, in the RPU setting passive or active learning individually require (1/eps)^O(d) and comparison passive learning requires (1/eps) and comparison pool based only uses (d polylog(1/eps)).\n\nStrengths: Overall, I like the results of the paper and I think studying comparison and label active learning learning for distribution-dependent settings is very valuable. I also like that the paper studies RPU setting, where the algorithm has know what it’s uncertain about. Here, as mentioned above you need the power of comparison queries to get to d polylog(1/eps) as opposed to 1/eps^d. This model is not as well studied in modern literature, but it’s an important learning model when it comes to robustness learning.\n\nWeaknesses: Don't see a particular weakness.\n\nCorrectness: Seems correct.\n\nClarity: Good\n\nRelation to Prior Work: Yes.\n\nReproducibility: Yes\n\n### Review 4\n\nSummary and Contributions: This paper investigates several scenarios in active learning. Namely, both PAC learning and RPU learning are considered for the model; both pool-based and membership query-based settings are considered for the active learning paradigm; and, both labeling and comparisons are considered for the queries. Several new lower bounds and upper bounds are obtained for the query complexity in those scenarios. Remarkably, it is shown that for comparison queries, a query complexity polynomial in log(1/epsilon) is obtained for both the PAC setting and the RPU setting, under relatively weak assumptions for the distributions. Synthetic experiments are performed for corroborating such results.\n\nStrengths: This is a well-written, well-positioned, and well-motivated paper, with new nontrivial results in active learning. Although the proof for the upper-bound in the PAC setting (Theorem 3.3) is inspired from , the upper bound in the RPU setting (Theorems 3.7 & 3.8) relies on advanced techniques in high-dimensional geometry (notably the concept of average inference dimension). In essence, this is a very good theoretical contribution.\n\nWeaknesses: I found no real weaknesses.\n\nCorrectness: I am not an expert in high-dimensional geometry, and techniques related to the interesting concept of inference dimension. Yet, as far as I could check, the proofs look correct.\n\nClarity: As mentioned above, this paper is very well-written: all notations and definitions are clearly presented, and Section 4 is particularly useful for understanding the main concepts and tools used in the proofs.\n\nRelation to Prior Work: To the best of my knowledge, the paper is well-positioned with respect to related work, and clearly explains the main improvements obtained for different scenarios.\n\nReproducibility: Yes\n\nAdditional Feedback: As a minor comment, I would suggest to mention some perspectives of further research, such as active learning with comparison queries in agnostic settings. But a detailed conclusion is already provided in the extended version of the paper." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93040013,"math_prob":0.6792118,"size":11654,"snap":"2023-40-2023-50","text_gpt3_token_len":2452,"char_repetition_ratio":0.14291845,"word_repetition_ratio":0.0150417825,"special_character_ratio":0.20447914,"punctuation_ratio":0.117036335,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96916175,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-30T23:26:05Z\",\"WARC-Record-ID\":\"<urn:uuid:44fcd532-8763-413f-8da4-f38499713deb>\",\"Content-Length\":\"14373\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:837590ea-e71a-43b9-b05c-d8f74ad55cdd>\",\"WARC-Concurrent-To\":\"<urn:uuid:620b48b8-fbb9-4093-8126-9c6503e5cf90>\",\"WARC-IP-Address\":\"198.202.70.94\",\"WARC-Target-URI\":\"https://papers.nips.cc/paper_files/paper/2020/file/4607f7fff0dce694258e1c637512aa9d-Review.html\",\"WARC-Payload-Digest\":\"sha1:FDPBNEIBN7GZEOLSGKZR3W7XIEPSOWX7\",\"WARC-Block-Digest\":\"sha1:QDPESNZFCQA3CSR4WI3VUWZ657KEOVGU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510730.6_warc_CC-MAIN-20230930213821-20231001003821-00576.warc.gz\"}"}
http://prepare.freshersindia.in/reasoning/questions_answersb215.php?tid=15&typeid=2	[ "Non-Verbal Reasoning - SERIES\n\nYou Are Here :: Home > Non-Verbal > SERIES - General Questions\n\n1.\n\nSelect a figure from amongst the Answer Figures which will continue the same series as established by the five Problem Figures.", null, "A. 1 B. 2 C. 3 D. 4\n\n2.\n\nSelect a figure from amongst the Answer Figures which will continue the same series as established by the five Problem Figures.", null, "A. 1 B. 2 C. 3 D. 4\n\n3.\n\nSelect a figure from amongst the Answer Figures which will continue the same series as established by the five Problem Figures.", null, "A. 1 B. 2 C. 3 D. 4\n\n4.\n\nSelect a figure from amongst the Answer Figures which will continue the same series as established by the five Problem Figures.", null, "A. 1 B. 2 C. 3 D. 4\n\n5.\n\nSelect a figure from amongst the Answer Figures which will continue the same series as established by the five Problem Figures.", null, "A. 1 B. 2 C. 3 D. 4\n\n6.\n\nSelect a figure from amongst the Answer Figures which will continue the same series as established by the five Problem Figures.", null, "A. 1 B. 2 C. 3 D. 4\n\n1\n\n2\n\n3\n\n4\n\n5\n\n6\n\n7\n\n8\n\n9\n\nnext »" ]	[ null, "http://prepare.freshersindia.in/reasoning/images/s1q.png", null, "http://prepare.freshersindia.in/reasoning/images/s2q.png", null, "http://prepare.freshersindia.in/reasoning/images/s3q.png", null, "http://prepare.freshersindia.in/reasoning/images/s4q.png", null, "http://prepare.freshersindia.in/reasoning/images/s5q.png", null, "http://prepare.freshersindia.in/reasoning/images/s6q.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93592566,"math_prob":0.5162896,"size":933,"snap":"2023-40-2023-50","text_gpt3_token_len":207,"char_repetition_ratio":0.14854683,"word_repetition_ratio":0.6826347,"special_character_ratio":0.24008575,"punctuation_ratio":0.082840234,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9834852,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T17:13:45Z\",\"WARC-Record-ID\":\"<urn:uuid:cd9529f3-effa-4f1e-b689-d0afb0d1c8c4>\",\"Content-Length\":\"31168\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:58139973-7d1f-4282-8699-b04f9c6ecf1e>\",\"WARC-Concurrent-To\":\"<urn:uuid:08a05b57-03c8-42b8-bf20-c902c50db99b>\",\"WARC-IP-Address\":\"51.15.58.31\",\"WARC-Target-URI\":\"http://prepare.freshersindia.in/reasoning/questions_answersb215.php?tid=15&typeid=2\",\"WARC-Payload-Digest\":\"sha1:Q4UJ7X32UW7GEWZRPLICDDG5CUYD5J5S\",\"WARC-Block-Digest\":\"sha1:V5TMRQPB42PDFB5QHUOVBIPOPI4EG7BU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100531.77_warc_CC-MAIN-20231204151108-20231204181108-00568.warc.gz\"}"}
https://www.doubtnut.com/question-answer-physics/a-mass-is-suspended-separately-by-two-springs-of-spring-constants-k1-and-k2-in-successive-order-the--15600026	[ "# A mass is suspended separately by two springs of spring constants k1 and k2 in successive order. The time periods of oscillations in the two cases are T1 and T2 respectively. If the same mass be suspended by connecting the two springs in parallel, (as shown in figure) then the time period of oscillations is T. The correct relations is", null, "A\n\nT2=T21+T22\n\nB\n\nT2=T21+T22\n\nC\n\nT1=T11+T12\n\nD\n\nT=T1+T2\n\nUpdated on:21/07/2023\nText Solution\nVerified by Experts" ]	[ null, "https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_31Y_NEET_PHY_C10_E01_028_Q01.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92685485,"math_prob":0.9398787,"size":3275,"snap":"2023-40-2023-50","text_gpt3_token_len":793,"char_repetition_ratio":0.18587588,"word_repetition_ratio":0.32240438,"special_character_ratio":0.24671756,"punctuation_ratio":0.064714946,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9875806,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-25T04:17:07Z\",\"WARC-Record-ID\":\"<urn:uuid:e8d20864-2aad-4e6f-bea2-7bffd2c4b070>\",\"Content-Length\":\"284590\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a3c3e036-73a3-4cd2-8a3b-21652d972452>\",\"WARC-Concurrent-To\":\"<urn:uuid:06d504d0-5ad8-4531-907d-787c776c9c3e>\",\"WARC-IP-Address\":\"99.84.191.108\",\"WARC-Target-URI\":\"https://www.doubtnut.com/question-answer-physics/a-mass-is-suspended-separately-by-two-springs-of-spring-constants-k1-and-k2-in-successive-order-the--15600026\",\"WARC-Payload-Digest\":\"sha1:NQRVI4QG52JRQQQ2WIZMHH6ULPKCLX5E\",\"WARC-Block-Digest\":\"sha1:GMLRT2WOPSR3BUQ2FQDYGO3Z76VHPXWE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506676.95_warc_CC-MAIN-20230925015430-20230925045430-00202.warc.gz\"}"}
https://yonsei.pure.elsevier.com/en/publications/thermal-instability-in-a-rotating-porous-layer-saturated-by-a-non	[ "# Thermal instability in a rotating porous layer saturated by a non-Newtonian nanofluid with thermal conductivity and viscosity variation\n\nDhananjay Yadav, R. Bhargava, G. S. Agrawal, Nirmal Yadav, Jinho Lee, M. C. Kim\n\nResearch output: Contribution to journalArticle\n\n59 Citations (Scopus)\n\n### Abstract\n\nThe stability of a non-Newtonian nanofluid saturated horizontal rotating porous layer subjected to thermal conductivity and viscosity variation is investigated using linear and nonlinear stability analyses. The model used for the non-Newtonian nanofluid includes the effects of Brownian motion and thermophoresis. The Darcy law for the non-Newtonian nanofluid of the Oldroyd type is used to model the momentum equation. The linear theory based on the normal mode method, and the criteria for both stationary and oscillatory modes are derived analytically. A weak nonlinear analysis based on the minimal representation of truncated Fourier series method containing only two terms is used to compute the concentration and thermal Nusselt numbers. The results obtained during the analysis are presented graphically.\n\nOriginal language English 425-440 16 Microfluidics and Nanofluidics 16 1-2 https://doi.org/10.1007/s10404-013-1234-5 Published - 2014 Jan\n\n### Fingerprint\n\nthermal instability\nThermal conductivity\nthermal conductivity\nThermophoresis\nViscosity\nviscosity\nthermophoresis\nBrownian movement\nFourier series\nNonlinear analysis\nNusselt number\nMomentum\nmomentum\nHot Temperature\n\n### All Science Journal Classification (ASJC) codes\n\n• Electronic, Optical and Magnetic Materials\n• Condensed Matter Physics\n• Materials Chemistry\n\n### Cite this\n\nYadav, Dhananjay ; Bhargava, R. ; Agrawal, G. S. ; Yadav, Nirmal ; Lee, Jinho ; Kim, M. C. / Thermal instability in a rotating porous layer saturated by a non-Newtonian nanofluid with thermal conductivity and viscosity variation. In: Microfluidics and Nanofluidics. 2014 ; Vol. 16, No. 1-2. pp. 425-440.\n@article{75e1a28515bd49ebb939bbdc2d5f3247,\ntitle = \"Thermal instability in a rotating porous layer saturated by a non-Newtonian nanofluid with thermal conductivity and viscosity variation\",\nabstract = \"The stability of a non-Newtonian nanofluid saturated horizontal rotating porous layer subjected to thermal conductivity and viscosity variation is investigated using linear and nonlinear stability analyses. The model used for the non-Newtonian nanofluid includes the effects of Brownian motion and thermophoresis. The Darcy law for the non-Newtonian nanofluid of the Oldroyd type is used to model the momentum equation. The linear theory based on the normal mode method, and the criteria for both stationary and oscillatory modes are derived analytically. A weak nonlinear analysis based on the minimal representation of truncated Fourier series method containing only two terms is used to compute the concentration and thermal Nusselt numbers. The results obtained during the analysis are presented graphically.\",\nauthor = \"Dhananjay Yadav and R. Bhargava and Agrawal, {G. S.} and Nirmal Yadav and Jinho Lee and Kim, {M. C.}\",\nyear = \"2014\",\nmonth = \"1\",\ndoi = \"10.1007/s10404-013-1234-5\",\nlanguage = \"English\",\nvolume = \"16\",\npages = \"425--440\",\njournal = \"Microfluidics and Nanofluidics\",\nissn = \"1613-4982\",\npublisher = \"Springer Verlag\",\nnumber = \"1-2\",\n\n}\n\nThermal instability in a rotating porous layer saturated by a non-Newtonian nanofluid with thermal conductivity and viscosity variation. / Yadav, Dhananjay; Bhargava, R.; Agrawal, G. S.; Yadav, Nirmal; Lee, Jinho; Kim, M. C.\n\nIn: Microfluidics and Nanofluidics, Vol. 16, No. 1-2, 01.2014, p. 425-440.\n\nResearch output: Contribution to journalArticle\n\nTY - JOUR\n\nT1 - Thermal instability in a rotating porous layer saturated by a non-Newtonian nanofluid with thermal conductivity and viscosity variation\n\nAU - Bhargava, R.\n\nAU - Agrawal, G. S.\n\nAU - Lee, Jinho\n\nAU - Kim, M. C.\n\nPY - 2014/1\n\nY1 - 2014/1\n\nN2 - The stability of a non-Newtonian nanofluid saturated horizontal rotating porous layer subjected to thermal conductivity and viscosity variation is investigated using linear and nonlinear stability analyses. The model used for the non-Newtonian nanofluid includes the effects of Brownian motion and thermophoresis. The Darcy law for the non-Newtonian nanofluid of the Oldroyd type is used to model the momentum equation. The linear theory based on the normal mode method, and the criteria for both stationary and oscillatory modes are derived analytically. A weak nonlinear analysis based on the minimal representation of truncated Fourier series method containing only two terms is used to compute the concentration and thermal Nusselt numbers. The results obtained during the analysis are presented graphically.\n\nAB - The stability of a non-Newtonian nanofluid saturated horizontal rotating porous layer subjected to thermal conductivity and viscosity variation is investigated using linear and nonlinear stability analyses. The model used for the non-Newtonian nanofluid includes the effects of Brownian motion and thermophoresis. The Darcy law for the non-Newtonian nanofluid of the Oldroyd type is used to model the momentum equation. The linear theory based on the normal mode method, and the criteria for both stationary and oscillatory modes are derived analytically. A weak nonlinear analysis based on the minimal representation of truncated Fourier series method containing only two terms is used to compute the concentration and thermal Nusselt numbers. The results obtained during the analysis are presented graphically.\n\nUR - http://www.scopus.com/inward/record.url?scp=84899456995&partnerID=8YFLogxK\n\nUR - http://www.scopus.com/inward/citedby.url?scp=84899456995&partnerID=8YFLogxK\n\nU2 - 10.1007/s10404-013-1234-5\n\nDO - 10.1007/s10404-013-1234-5\n\nM3 - Article\n\nAN - SCOPUS:84899456995\n\nVL - 16\n\nSP - 425\n\nEP - 440\n\nJO - Microfluidics and Nanofluidics\n\nJF - Microfluidics and Nanofluidics\n\nSN - 1613-4982\n\nIS - 1-2\n\nER -" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8118723,"math_prob":0.5818248,"size":3841,"snap":"2019-51-2020-05","text_gpt3_token_len":954,"char_repetition_ratio":0.13057075,"word_repetition_ratio":0.6678766,"special_character_ratio":0.22390002,"punctuation_ratio":0.11402157,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95179635,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-27T22:09:33Z\",\"WARC-Record-ID\":\"<urn:uuid:8280a2b1-e6d2-4a9b-8701-837e99c6a18d>\",\"Content-Length\":\"41030\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:81ef4b20-5c6d-4549-a106-5f3da0872f34>\",\"WARC-Concurrent-To\":\"<urn:uuid:8e7c5f61-6ae8-48f6-bc69-6ac4c7dc24e4>\",\"WARC-IP-Address\":\"52.220.215.79\",\"WARC-Target-URI\":\"https://yonsei.pure.elsevier.com/en/publications/thermal-instability-in-a-rotating-porous-layer-saturated-by-a-non\",\"WARC-Payload-Digest\":\"sha1:EBO3LSP3QTZY6VKDXCOOTI2CWABAQUJG\",\"WARC-Block-Digest\":\"sha1:PMXH7D5IQOK646MTANEQV4AMF6BIQL3F\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251728207.68_warc_CC-MAIN-20200127205148-20200127235148-00196.warc.gz\"}"}
https://articles.emptycrate.com/2008/03/09/c_templates_using_templates_to_generate_the_fibonacci_sequence.html	[ "# EmptyCrate\n\n#### About Me \| Contact Me \| Training \| C++ Weekly (YouTube) \| Awesome C++ T-Shirts\n\nThe follow code demonstrates a method for generating the Fibonacci Sequence at compile time. Since the sequence is generated at compile time runtime execution of retrieving a number is constant time.\n\nThere is a recursive generation of templated types in this code such that the instantiation of a template: Fib<10> f; generates 11 classes, namely: Fib<0> through Fib<10>.\n\nThe nature of the class generation by the compiler means that previous Fibonacci values are effectively cached, making the compile time go much faster. Therefore, this is pretty much the fastest Fibonacci generator I have seen, even if you include compile time.\n\n``````#include <iostream>\n#include <cassert>\nusing namespace std;\n\ntemplate<int stage>\nstruct Fib\n{\n//Make this value a constant value equal to the (stage-1) + (stage -2)\n//which the compile will generate for us and save in the types of:\n// Fib<stage-1> and Fib<stage-2>. This all works because stage is known at compile\n// time, as all template parameters must be.\nstatic const uint64_t value = Fib<stage-1>::value + Fib<stage-2>::value;\n\nstatic inline uint64_t getValue(int i)\n{\nif (i == stage) // Does the current class hold the given place?\n{\nreturn value; // Return it!\n} else {\nreturn Fib<stage-1>::getValue(i); // Get it from the previous class!\n}\n}\n};\n\ntemplate<> // Template specialization for the 0's case.\nstruct Fib<0>\n{\nstatic const uint64_t value = 1;\n\nstatic inline uint64_t getValue(int i)\n{\nassert(i == 0);\nreturn 1;\n}\n};\n\ntemplate<> // Template specialization for the 1's case\nstruct Fib<1>\n{\nstatic const uint64_t value = 1;\n\nstatic inline uint64_t getValue(int i)\n{\nif (i == 1)\n{\nreturn value;\n} else {\nreturn Fib<0>::getValue(i);\n}\n}\n};\n\nint main(int, char *[])\n{\n//Generate (at compile time) 100 places of the Fib sequence.\n//Then, (at runtime) output the 100 calculated places.\n//Note: a 64 bit int overflows at place 92\nfor (int i = 0; i < 100; ++i)\n{\ncout << \"n:=\" << i << \" => \" << Fib<100>::getValue(i) << endl;\n}\n}\n``````\nMy Videos\n\nRelated" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.71909624,"math_prob":0.9780144,"size":1963,"snap":"2023-40-2023-50","text_gpt3_token_len":500,"char_repetition_ratio":0.13169985,"word_repetition_ratio":0.10670732,"special_character_ratio":0.30565462,"punctuation_ratio":0.1420765,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9860926,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-01T13:09:13Z\",\"WARC-Record-ID\":\"<urn:uuid:219e841b-913c-4d63-8eef-a9910309d37c>\",\"Content-Length\":\"76448\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b547766d-a350-4d72-b609-a151bf01ec60>\",\"WARC-Concurrent-To\":\"<urn:uuid:613ee6ae-f028-43a9-a1ec-1838e427d4ee>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"https://articles.emptycrate.com/2008/03/09/c_templates_using_templates_to_generate_the_fibonacci_sequence.html\",\"WARC-Payload-Digest\":\"sha1:YOWX53FYNKSGPXBSUFIRLFDV2MK3RVJM\",\"WARC-Block-Digest\":\"sha1:GRSUSIWLY3ANBOGKOTNQHK2KXLHHEN5Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100287.49_warc_CC-MAIN-20231201120231-20231201150231-00490.warc.gz\"}"}
https://byjus.com/question-answer/a-glass-sphere-mu-1-5-of-radius-20-cm-has-a-small-air-bubble/	[ "", null, "", null, "Question\n\nA glass sphere $$(\\mu = 1.5)$$ of radius $$20\\ cm$$ has a small air bubble $$4\\ cm$$ below its center. The sphere is viewed from outside and along a vertical line through the bubble. The apparent depth of the bubble below the surface of sphere is (in cm) :\n\nA\n13.33", null, "", null, "B\n26.67", null, "", null, "C\n15", null, "", null, "D\n30", null, "", null, "Solution\n\nThe correct option is B $$26.67$$concept: It is the case of refraction through curved surface formulae used$$\\dfrac{\\mu_2}{v}-\\dfrac{\\mu_1}{u}=\\dfrac{\\mu_2-\\mu_1}{R}$$Given: $$\\mu_2=1$$, $$\\mu_1=1.5$$$$R=-20$$Object distance$$=u=-(20+4)=-24$$Here apparent depth is image distance which is v$$\\dfrac{1}{v}-\\dfrac{1.5}{-24}=\\dfrac{1-1.5}{-20}$$$$\\Rightarrow \\dfrac{1}{v}=\\dfrac{1}{40}-\\dfrac{1}{16}=\\dfrac{-3}{80}$$$$\\Rightarrow v=-\\dfrac{80}{3}=-26.67$$Hence bubble looks at depth of $$26.67$$.", null, "Physics\n\nSuggest Corrections", null, "", null, "0", null, "", null, "Similar questions\nView More", null, "", null, "People also searched for\nView More", null, "" ]	[ null, "data:image/svg+xml;base64,PHN2ZyB3aWR0aD0iNDQiIGhlaWdodD0iNDQiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmVyc2lvbj0iMS4xIi8+", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/svg+xml;base64,PHN2ZyB3aWR0aD0iMjAiIGhlaWdodD0iMjAiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmVyc2lvbj0iMS4xIi8+", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/svg+xml;base64,PHN2ZyB3aWR0aD0iMjAiIGhlaWdodD0iMjAiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmVyc2lvbj0iMS4xIi8+", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/svg+xml;base64,PHN2ZyB3aWR0aD0iMjAiIGhlaWdodD0iMjAiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmVyc2lvbj0iMS4xIi8+", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/svg+xml;base64,PHN2ZyB3aWR0aD0iMjAiIGhlaWdodD0iMjAiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmVyc2lvbj0iMS4xIi8+", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "https://search-static.byjusweb.com/question-images/toppr_ext/questions/1446054_1158732_ans_bcf78b623c6a4973ab507e26cf4aa75d.png", null, "data:image/svg+xml;base64,PHN2ZyB3aWR0aD0iMjQiIGhlaWdodD0iMjQiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmVyc2lvbj0iMS4xIi8+", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/svg+xml;base64,PHN2ZyB3aWR0aD0iNDAiIGhlaWdodD0iNDAiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmVyc2lvbj0iMS4xIi8+", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/svg+xml;base64,PHN2ZyB3aWR0aD0iNDAiIGhlaWdodD0iNDAiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmVyc2lvbj0iMS4xIi8+", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92193145,"math_prob":0.9999331,"size":267,"snap":"2022-05-2022-21","text_gpt3_token_len":68,"char_repetition_ratio":0.1444867,"word_repetition_ratio":0.0,"special_character_ratio":0.29962546,"punctuation_ratio":0.0754717,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99994767,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,1,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-28T02:35:30Z\",\"WARC-Record-ID\":\"<urn:uuid:979a44b6-6b04-4f92-b427-de50695a00c4>\",\"Content-Length\":\"148701\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a1ff3a42-7771-4685-a22c-bd9e6c24d08b>\",\"WARC-Concurrent-To\":\"<urn:uuid:7dbed65a-9a21-4dc9-872f-2bbf0aa4eb3c>\",\"WARC-IP-Address\":\"162.159.129.41\",\"WARC-Target-URI\":\"https://byjus.com/question-answer/a-glass-sphere-mu-1-5-of-radius-20-cm-has-a-small-air-bubble/\",\"WARC-Payload-Digest\":\"sha1:UG6PT74VF3S4562NZLXILTK4D4EIZ7DA\",\"WARC-Block-Digest\":\"sha1:EINDDPYDQJA7XHFLTYYQKM2EJSKQCYBZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320305341.76_warc_CC-MAIN-20220128013529-20220128043529-00487.warc.gz\"}"}
https://discuss.pytorch.org/t/different-optimisers-in-different-layers/72484	[ "# Different Optimisers In Different Layers\n\nHi,\n\nI am currently looking into hybrid neural networks. I would like to apply a different optimizer to quantum circuit compared to the classical layers.\n\nThe quantum circuit has classical input and parameters so it can be optimised by an classical optimiser.\n\nMy network looks like\n\n``````class NeuralNet(nn.Module)\n\ndef __init__(self):\nsuper(NeuralNet, self).__init__()\nself.pre_net = nn.Linear(2828, 4)\nself.q_params = nn.Parameter(0.01 torch.randn(q_depth * 4)) #Quantum circuit parameters\nself.post_net = nn.Linear(4, 10)\n``````\n\n1.) How can I apply the same optimiser to all layers, but have a different step size for the quantum circuit parameters\n\n2.) How can I apply a completely different optimiser to the quantum circuit parameters?\n\nReally appreciate the help", null, "" ]	[ null, "https://discuss.pytorch.org/images/emoji/apple/grinning.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.716384,"math_prob":0.89102477,"size":753,"snap":"2022-27-2022-33","text_gpt3_token_len":175,"char_repetition_ratio":0.14285715,"word_repetition_ratio":0.0,"special_character_ratio":0.2430279,"punctuation_ratio":0.15068494,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97771555,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-27T04:37:56Z\",\"WARC-Record-ID\":\"<urn:uuid:aa2d9d0a-1e41-4803-9dda-44d8df2bdde8>\",\"Content-Length\":\"17618\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8fe3bf74-c2bd-430b-a7fb-cc1510582803>\",\"WARC-Concurrent-To\":\"<urn:uuid:be14e541-8ed0-4e11-bf9a-ede6c1487a5c>\",\"WARC-IP-Address\":\"159.203.145.104\",\"WARC-Target-URI\":\"https://discuss.pytorch.org/t/different-optimisers-in-different-layers/72484\",\"WARC-Payload-Digest\":\"sha1:73DOOHNKRQ4YCLACAHMSBGBBFDWGV2MZ\",\"WARC-Block-Digest\":\"sha1:4NC4U36RSMWT4WAHFNGQV6HNUA6THS6P\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103328647.18_warc_CC-MAIN-20220627043200-20220627073200-00527.warc.gz\"}"}
https://number.academy/16966	[ "# Number 16966\n\nNumber 16,966 spell 🔊, write in words: sixteen thousand, nine hundred and sixty-six . Ordinal number 16966th is said 🔊 and write: sixteen thousand, nine hundred and sixty-sixth. The meaning of number 16966 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 16966. What is 16966 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 16966.\n\n## What is 16,966 in other units\n\nThe decimal (Arabic) number 16966 converted to a Roman number is (X)(V)MCMLXVI. Roman and decimal number conversions.\n\n#### Weight conversion\n\n16966 kilograms (kg) = 37403.2 pounds (lbs)\n16966 pounds (lbs) = 7695.7 kilograms (kg)\n\n#### Length conversion\n\n16966 kilometers (km) equals to 10543 miles (mi).\n16966 miles (mi) equals to 27305 kilometers (km).\n16966 meters (m) equals to 55663 feet (ft).\n16966 feet (ft) equals 5172 meters (m).\n16966 centimeters (cm) equals to 6679.5 inches (in).\n16966 inches (in) equals to 43093.6 centimeters (cm).\n\n#### Temperature conversion\n\n16966° Fahrenheit (°F) equals to 9407.8° Celsius (°C)\n16966° Celsius (°C) equals to 30570.8° Fahrenheit (°F)\n\n#### Time conversion\n\n(hours, minutes, seconds, days, weeks)\n16966 seconds equals to 4 hours, 42 minutes, 46 seconds\n16966 minutes equals to 1 week, 4 days, 18 hours, 46 minutes\n\n### Codes and images of the number 16966\n\nNumber 16966 morse code: .---- -.... ----. -.... -....\nSign language for number 16966:", null, "", null, "", null, "", null, "", null, "Number 16966 in braille:", null, "Images of the number\nImage (1) of the numberImage (2) of the number", null, "", null, "More images, other sizes, codes and colors ...\n\n#### Number 16966 infographic", null, "## Share in social networks", null, "## Mathematics of no. 16966\n\n### Multiplications\n\n#### Multiplication table of 16966\n\n16966 multiplied by two equals 33932 (16966 x 2 = 33932).\n16966 multiplied by three equals 50898 (16966 x 3 = 50898).\n16966 multiplied by four equals 67864 (16966 x 4 = 67864).\n16966 multiplied by five equals 84830 (16966 x 5 = 84830).\n16966 multiplied by six equals 101796 (16966 x 6 = 101796).\n16966 multiplied by seven equals 118762 (16966 x 7 = 118762).\n16966 multiplied by eight equals 135728 (16966 x 8 = 135728).\n16966 multiplied by nine equals 152694 (16966 x 9 = 152694).\nshow multiplications by 6, 7, 8, 9 ...\n\n### Fractions: decimal fraction and common fraction\n\n#### Fraction table of 16966\n\nHalf of 16966 is 8483 (16966 / 2 = 8483).\nOne third of 16966 is 5655,3333 (16966 / 3 = 5655,3333 = 5655 1/3).\nOne quarter of 16966 is 4241,5 (16966 / 4 = 4241,5 = 4241 1/2).\nOne fifth of 16966 is 3393,2 (16966 / 5 = 3393,2 = 3393 1/5).\nOne sixth of 16966 is 2827,6667 (16966 / 6 = 2827,6667 = 2827 2/3).\nOne seventh of 16966 is 2423,7143 (16966 / 7 = 2423,7143 = 2423 5/7).\nOne eighth of 16966 is 2120,75 (16966 / 8 = 2120,75 = 2120 3/4).\nOne ninth of 16966 is 1885,1111 (16966 / 9 = 1885,1111 = 1885 1/9).\nshow fractions by 6, 7, 8, 9 ...\n\n### Calculator\n\n 16966\n\n#### Is Prime?\n\nThe number 16966 is not a prime number. The closest prime numbers are 16963, 16979.\n\n#### Factorization and factors (dividers)\n\nThe prime factors of 16966 are 2 * 17 * 499\nThe factors of 16966 are 1 , 2 , 17 , 34 , 499 , 998 , 8483 , 16966\nTotal factors 8.\nSum of factors 27000 (10034).\n\n#### Powers\n\nThe second power of 169662 is 287.845.156.\nThe third power of 169663 is 4.883.580.916.696.\n\n#### Roots\n\nThe square root √16966 is 130,253599.\nThe cube root of 316966 is 25,695663.\n\n#### Logarithms\n\nThe natural logarithm of No. ln 16966 = loge 16966 = 9,738967.\nThe logarithm to base 10 of No. log10 16966 = 4,229579.\nThe Napierian logarithm of No. log1/e 16966 = -9,738967.\n\n### Trigonometric functions\n\nThe cosine of 16966 is 0,170292.\nThe sine of 16966 is 0,985394.\nThe tangent of 16966 is 5,786504.\n\n### Properties of the number 16966\n\nIs a Friedman number: No\nIs a Fibonacci number: No\nIs a Bell number: No\nIs a palindromic number: No\nIs a pentagonal number: No\nIs a perfect number: No\n\n## Number 16966 in Computer Science\n\nCode typeCode value\n16966 Number of bytes16.6KB\nUnix timeUnix time 16966 is equal to Thursday Jan. 1, 1970, 4:42:46 a.m. GMT\nIPv4, IPv6Number 16966 internet address in dotted format v4 0.0.66.70, v6 ::4246\n16966 Decimal = 100001001000110 Binary\n16966 Decimal = 212021101 Ternary\n16966 Decimal = 41106 Octal\n16966 Decimal = 4246 Hexadecimal (0x4246 hex)\n16966 BASE64MTY5NjY=\n16966 MD5bb2825b78d37fef9ecabb1b91a8a6b88\n16966 SHA1645b71d515f126bf0bf2ce0c0ab76b99b1ff0af4\n16966 SHA224a1ee0d382a6fb4d309139e44afdf84e0bc9158922330cba51692fa89\n16966 SHA256acdf413fc28537565e558fe239ea5b5dd0b5da6447775e4fab75ca8bb3b42c87\n16966 SHA384353425bd6eec6831c20a16c2f28811eb0026c5d0fa3e55c62e132d348a00c6b2663cff2948a73551c8708bd4cdf60bf4\nMore SHA codes related to the number 16966 ...\n\nIf you know something interesting about the 16966 number that you did not find on this page, do not hesitate to write us here.\n\n## Numerology 16966\n\n### Character frequency in number 16966\n\nCharacter (importance) frequency for numerology.\n Character: Frequency: 1 1 6 3 9 1\n\n### Classical numerology\n\nAccording to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 16966, the numbers 1+6+9+6+6 = 2+8 = 1+0 = 1 are added and the meaning of the number 1 is sought.\n\n## Interesting facts about the number 16966\n\n### Asteroids\n\n• (16966) 1998 SM63 is asteroid number 16966. It was discovered by Beijing Schmidt CCD Asteroid Program from Xinglong Station on 9/29/1998.\n\n### Distances between cities\n\n• There is a 10,543 miles (16,966 km) direct distance between Guiyang (China) and Nova Iguaçu (Brazil).\n• There is a 10,543 miles (16,966 km) direct distance between Hanoi (Viet Nam) and Medellín (Colombia).\n• There is a 10,543 miles (16,966 km) direct distance between Meerut (India) and Santiago (Chile).\n\n## Number 16,966 in other languages\n\nHow to say or write the number sixteen thousand, nine hundred and sixty-six in Spanish, German, French and other languages. The character used as the thousands separator.\n Spanish: 🔊 (número 16.966) dieciseis mil novecientos sesenta y seis German: 🔊 (Anzahl 16.966) sechzehntausendneunhundertsechsundsechzig French: 🔊 (nombre 16 966) seize mille neuf cent soixante-six Portuguese: 🔊 (número 16 966) dezesseis mil, novecentos e sessenta e seis Chinese: 🔊 (数 16 966) 一万六千九百六十六 Arabian: 🔊 (عدد 16,966) ستة عشر ألفاً و تسعمائةستة و ستون Czech: 🔊 (číslo 16 966) šestnáct tisíc devětset šedesát šest Korean: 🔊 (번호 16,966) 만 육천구백육십육 Danish: 🔊 (nummer 16 966) sekstentusinde og nihundrede og seksogtreds Dutch: 🔊 (nummer 16 966) zestienduizendnegenhonderdzesenzestig Japanese: 🔊 (数 16,966) 一万六千九百六十六 Indonesian: 🔊 (jumlah 16.966) enam belas ribu sembilan ratus enam puluh enam Italian: 🔊 (numero 16 966) sedicimilanovecentosessantasei Norwegian: 🔊 (nummer 16 966) seksten tusen, ni hundre og seksti-seks Polish: 🔊 (liczba 16 966) szesnaście tysięcy dziewięćset sześćdziesiąt sześć Russian: 🔊 (номер 16 966) шестнадцать тысяч девятьсот шестьдесят шесть Turkish: 🔊 (numara 16,966) onaltıbindokuzyüzaltmışaltı Thai: 🔊 (จำนวน 16 966) หนึ่งหมื่นหกพันเก้าร้อยหกสิบหก Ukrainian: 🔊 (номер 16 966) шiстнадцять тисяч дев'ятсот шiстдесят шiсть Vietnamese: 🔊 (con số 16.966) mười sáu nghìn chín trăm sáu mươi sáu Other languages ...\n\n## News to email\n\nPrivacy Policy.\n\n## Comment\n\nIf you know something interesting about the number 16966 or any natural number (positive integer) please write us here or on facebook." ]	[ null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-1.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-6.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-9.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-6.png", null, "https://numero.wiki/s/senas/lenguaje-de-senas-numero-6.png", null, "https://number.academy/img/braille-16966.svg", null, "https://numero.wiki/img/a-16966.jpg", null, "https://numero.wiki/img/b-16966.jpg", null, "https://number.academy/i/infographics/6/number-16966-infographic.png", null, "https://numero.wiki/s/share-desktop.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.59239507,"math_prob":0.96868277,"size":7159,"snap":"2022-05-2022-21","text_gpt3_token_len":2634,"char_repetition_ratio":0.14772886,"word_repetition_ratio":0.024042742,"special_character_ratio":0.41095126,"punctuation_ratio":0.16179775,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9911332,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,1,null,1,null,1,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-25T22:13:44Z\",\"WARC-Record-ID\":\"<urn:uuid:53f553a0-e4c1-472f-b755-167a72180f67>\",\"Content-Length\":\"40753\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2037cf10-7067-486b-a4b4-f6356e98b31c>\",\"WARC-Concurrent-To\":\"<urn:uuid:b1e60ce1-4fa4-4315-8b4d-9a9a32287c83>\",\"WARC-IP-Address\":\"162.0.227.212\",\"WARC-Target-URI\":\"https://number.academy/16966\",\"WARC-Payload-Digest\":\"sha1:NRSYO7MG636OT7VVDRY26H7ZNEU6FGPT\",\"WARC-Block-Digest\":\"sha1:GSKUTJ4JN7WWB5PTHTHIPMZU7GE2UKIK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662594414.79_warc_CC-MAIN-20220525213545-20220526003545-00252.warc.gz\"}"}
https://www.cuemath.com/ncert-solutions/q-5-exercise-13-1-direct-and-inverse-proportions-class-8-maths/	[ "In the verge of coronavirus pandemic, we are providing FREE access to our entire Online Curriculum to ensure Learning Doesn't STOP!\n\n# Ex.13.1 Q5 Direct and Inverse Proportions Solution - NCERT Maths Class 8\n\nGo back to 'Ex.13.1'\n\n## Question\n\nA photograph of a bacteria is enlarged $$50,000$$ times attains a length of $$5 \\rm{cm}$$ as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged $$20,000$$ times only, what would be its enlarged length?\n\nVideo Solution\nDirect And Inverse Proportions\nEx 13.1 \| Question 5\n\n## Text Solution\n\nWhat is Known?\n\nBacteria enlarged $$50,000$$ times attain a length of $$5 \\,\\rm{cm.}$$\n\nWhat is Unknown?\n\nActual length of the bacteria\n\nif $$20,000$$ times enlarged what will be the length of the bacteria?\n\nReasoning:\n\nTwo numbers $$x$$ and $$y$$ are said in direct proportion if,\n\n\\begin{align}\\frac{x}{y}=k,\\quad x=y\\,k\\end{align}\n\nWhere $$k$$ is a constant.\n\nSteps:\n\n\\begin{align} \\text{Actual length,}\\ l&=\\frac{{{y}_{1}}}{{{x}_{1}}} \\\\ l&=\\frac{5}{50000} \\\\ l&=0.0001\\ \\rm{cm} \\\\ \\end{align}\n\n Number of times enlarged Length attained ${{x_1} = {\\rm{50,000}}}$ ${{y_{\\rm{1}}} = {\\rm{5}}}$ ${{x_2} = {\\rm{20,000}}}$ ${{y_{\\rm{2}}} = {\\rm{?}}}$\n\nThe number of times enlarged is directly proportional to the length attained.\n\nActual length $$= 0.0001 \\,\\rm{cm}$$\n\nEnlarged length will be $$2 \\,\\rm{cm.}$$\n\nLearn from the best math teachers and top your exams\n\n• Live one on one classroom and doubt clearing\n• Practice worksheets in and after class for conceptual clarity\n• Personalized curriculum to keep up with school" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6664737,"math_prob":0.9997849,"size":1157,"snap":"2020-10-2020-16","text_gpt3_token_len":379,"char_repetition_ratio":0.15177797,"word_repetition_ratio":0.0,"special_character_ratio":0.40103716,"punctuation_ratio":0.13478261,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9987008,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-07T20:52:15Z\",\"WARC-Record-ID\":\"<urn:uuid:a9acb66b-a9ae-45fc-9d89-326e8fdddabb>\",\"Content-Length\":\"105828\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:901a846a-76a9-4980-b6bd-9603b004a66d>\",\"WARC-Concurrent-To\":\"<urn:uuid:a504d9c9-0fce-4043-ab87-64691b658696>\",\"WARC-IP-Address\":\"52.74.183.11\",\"WARC-Target-URI\":\"https://www.cuemath.com/ncert-solutions/q-5-exercise-13-1-direct-and-inverse-proportions-class-8-maths/\",\"WARC-Payload-Digest\":\"sha1:HZ4JH2AKNEIRXF3WYZYVC3YT7PWIFGEJ\",\"WARC-Block-Digest\":\"sha1:QB7PVBL7CMZE4NPOLW23KVRNWU3AQ6JH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371805747.72_warc_CC-MAIN-20200407183818-20200407214318-00315.warc.gz\"}"}
https://informatiestandaarden.nictiz.nl/wiki/Sjabloon:Valid/doc	[ "# Sjabloon:Valid/doc\n\nGa naar: navigatie, zoeken\n\n## Purpose\n\nDetermines whether something is valid in a certain context. Currently only implemented for determining if a number is within the precision that Wikipedia expressions can handle.\n\n## Returns\n\ntrue if the argument is valid, false if it is not.\n\n## Examples\n\n`{{valid\|number=A}}` = false (not a number)\n`{{valid\|number=1234}}` = false\n`{{valid\|number=+1234}}` = false\n`{{valid\|number=-1234}}` = false\n`{{valid\|number=(1234)}}` = false (one pair of parenthesis is allowed)\n`{{valid\|number=--1234}}` = false (incorrect sign use)\n`{{valid\|number=1234567890}}` = false\n`{{valid\|number=12345678901234567890}}` = false (too large)\n`{{valid\|number=1.234567890}}` = false\n`{{valid\|number=1.2345678901234567890}}` = false (too many decimals)\n\n## Performance impact\n\nTemplate:Valid returns \"true\" for a valid, single number (2+2 gives \"false\"), and allows scientific notation (such as: -3.45E-07). The precision limit is determined live, for whichever server is formatting the page, typically allowing 14-digit precision (plus trailing zeroes), but it does not detect extreme precision problems dropping minor end-digits:\n\n• cannot reject: -10020030040050000000000000000.70, treated as -1.002003004005E+28\n\nThe template has been written with minimal markup text, and could be used 50,000-13,000 times per page, or less when all 30-digit numbers. Template:Valid is typically used at upper levels of other templates, so it is unlikely to trigger expansion-depth problems. However, it has an expansion-depth of 8 levels, and returns \"false\" if used when nested too deep inside other templates, such as at level 33 when the expansion depth limit is 40." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.63470876,"math_prob":0.97486085,"size":1804,"snap":"2023-14-2023-23","text_gpt3_token_len":458,"char_repetition_ratio":0.16055556,"word_repetition_ratio":0.0,"special_character_ratio":0.30044347,"punctuation_ratio":0.12101911,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9746428,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-20T15:09:57Z\",\"WARC-Record-ID\":\"<urn:uuid:c07f6f0e-519f-473a-bfc1-602a10967d17>\",\"Content-Length\":\"20121\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4dbbd6a9-968d-4fb7-962b-2c6028adabbe>\",\"WARC-Concurrent-To\":\"<urn:uuid:15aaccb2-4277-43fe-86cc-2196ca99f9c0>\",\"WARC-IP-Address\":\"37.61.204.124\",\"WARC-Target-URI\":\"https://informatiestandaarden.nictiz.nl/wiki/Sjabloon:Valid/doc\",\"WARC-Payload-Digest\":\"sha1:QZFTJAFBBKZJ2I6RXKVTOEAIR5B2ZZTH\",\"WARC-Block-Digest\":\"sha1:SZZ7RHJ7ZXE5WBJNF63GVRFPPLFC5AJW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943484.34_warc_CC-MAIN-20230320144934-20230320174934-00479.warc.gz\"}"}
https://includestdio.com/935.html	[ "# Swift的性能：排序数组\n\n``````let n = 1000000\nvar x = [Int](repeating: 0, count: n)\nfor i in 0..<n {\nx[i] = random()\n}\n// start clock here\nlet y = sort(x)\n// stop clock here``````\n\n``xcrun swift -O3 -sdk `xcrun --show-sdk-path --sdk macosx```\n\n``xcrun swift -O0 -sdk `xcrun --show-sdk-path --sdk macosx```\n\n``````let n = 10000000\nprint(nnnnn)\nlet x = [Int](repeating: 10, count: n)\nprint(x[n])``````\n\n``````for i in 0..<n {\nx[i] = x[i] ^ 12345678\n}``````\n\n（这里的异或操作只是为了让我能够更容易地找到汇编代码中的相关循环，我试图选择一个易于识别的操作，但也是“无害”的，因为它不需要任何相关的检查到整数溢出。）\n\n• 随着`-Ofast`我得到我所期望的几乎。相关部分是一个包含5个机器语言指令的循环。\n• 随着`-O3`我得到的东西，出乎我的想象。内循环跨越88行汇编代码。我没有试图理解所有这些，但最可疑的部分是13个“callq _swift_retain”的调用和另外13个“callq _swift_release”的调用。也就是说，在内部循环中调用26个子程序\n\n``````let n = 10000\nvar x = [Int](repeating: 1, count: n)\nfor i in 0..<n {\nfor j in 0..<n {\nx[i] = x[j]\n}\n}``````\n\n• C ++ -O3：0.05秒\n• C ++ -O0：0.4 s\n• Java：0.2秒\n• Python与PyPy：0.5秒\n• Python：12秒\n• 快速-0.05秒\n• Swift -O3：23 s\n• Swift -O0：443 s\n\n（如果您担心编译器可能会完全优化无意义的循环，则可以将其更改为例如`x[i] ^= x[j]`，并添加一个输出的打印语句，`x`但这不会改变任何内容;时间点将非常相似。\n\n• 铛++ -O3：0.06秒\n• swiftc – 快：0.1秒\n• swiftc -O：0.1秒\n• swiftc：4秒\n\n• 铛++ -O3：0.06秒\n• swiftc – 快：0.3秒\n• swiftc -O：0.4 s\n• swiftc：540秒\n\ntl;通过使用默认版本优化级别[-O]的这个基准测试，Swift现在与C一样快。\n\n``````func quicksort_swift(inout a:CInt[], start:Int, end:Int) {\nif (end - start < 2){\nreturn\n}\nvar p = a[start + (end - start)/2]\nvar l = start\nvar r = end - 1\nwhile (l <= r){\nif (a[l] < p){\nl += 1\ncontinue\n}\nif (a[r] > p){\nr -= 1\ncontinue\n}\nvar t = a[l]\na[l] = a[r]\na[r] = t\nl += 1\nr -= 1\n}\nquicksort_swift(&a, start, r + 1)\nquicksort_swift(&a, r + 1, end)\n}``````\n\n``````void quicksort_c(int a, int n) {\nif (n < 2)\nreturn;\nint p = a[n / 2];\nint l = a;\nint r = a + n - 1;\nwhile (l <= r) {\nif (l < p) {\nl++;\ncontinue;\n}\nif (r > p) {\nr--;\ncontinue;\n}\nint t = l;\nl++ = r;\nr-- = t;\n}\nquicksort_c(a, r - a + 1);\nquicksort_c(l, a + n - l);\n}``````\n\n``````var a_swift:CInt[] = [0,5,2,8,1234,-1,2]\nvar a_c:CInt[] = [0,5,2,8,1234,-1,2]\n\nquicksort_swift(&a_swift, 0, a_swift.count)\nquicksort_c(&a_c, CInt(a_c.count))\n\n// [-1, 0, 2, 2, 5, 8, 1234]\n// [-1, 0, 2, 2, 5, 8, 1234]``````\n\n``````var x_swift = CInt[](count: n, repeatedValue: 0)\nvar x_c = CInt[](count: n, repeatedValue: 0)\nfor var i = 0; i < n; ++i {\nx_swift[i] = CInt(random())\nx_c[i] = CInt(random())\n}\n\nlet swift_start:UInt64 = mach_absolute_time();\nquicksort_swift(&x_swift, 0, x_swift.count)\nlet swift_stop:UInt64 = mach_absolute_time();\n\nlet c_start:UInt64 = mach_absolute_time();\nquicksort_c(&x_c, CInt(x_c.count))\nlet c_stop:UInt64 = mach_absolute_time();``````\n\n``````static const uint64_t NANOS_PER_USEC = 1000ULL;\nstatic const uint64_t NANOS_PER_MSEC = 1000ULL NANOS_PER_USEC;\nstatic const uint64_t NANOS_PER_SEC = 1000ULL * NANOS_PER_MSEC;\n\nmach_timebase_info_data_t timebase_info;\n\nuint64_t abs_to_nanos(uint64_t abs) {\nif ( timebase_info.denom == 0 ) {\n(void)mach_timebase_info(&timebase_info);\n}\nreturn abs * timebase_info.numer / timebase_info.denom;\n}\n\ndouble abs_to_seconds(uint64_t abs) {\nreturn abs_to_nanos(abs) / (double)NANOS_PER_SEC;\n}``````\n\n``````[-Onone] no optimizations, the default for debug.\n[-O] perform optimizations, the default for release.\n[-Ofast] perform optimizations and disable runtime overflow checks and runtime type checks.``````\n\nn = 10_000时[-Onone]秒数\n\n``````Swift: 0.895296452\nC: 0.001223848``````\n\n``Swift_builtin: 0.77865783``\n\n``````Swift: 0.045478346\nC: 0.000784666\nSwift_builtin: 0.032513488``````\n\n``````Swift: 0.000706745\nC: 0.000742374\nSwift_builtin: 0.000603576``````\n\n``````Swift: 0.107111846\nC: 0.114957179\nSwift_sort: 0.092688548``````\n\n``````Swift: 142.659763258\nC: 0.162065333\nSwift_sort: 114.095478272``````\n\nLLVM中提供了一个新的优化级别-Ofast，可以进行积极的优化。-Ofast放松了一些保守的限制，主要是为了浮点操作，对于大多数代码来说都是安全的。它可以从编译器中获得显着的高性能。\n\nBETA 3更新：\n\nn = 10_000[-O]\n\n``````Swift: 0.019697268\nC: 0.000718064\nSwift_sort: 0.002094721``````\n\n[-Onone]\n\n``````Swift: 0.678056695\nC: 0.000973914``````\n\n[-O]\n\n``````Swift: 0.001158492\nC: 0.001192406``````\n\n[-Ounchecked]\n\n``````Swift: 0.000827764\nC: 0.001078914``````\n\nTL; DR：是的，只有雨燕语言的实现是缓慢的，现在。如果你需要快速，数字（和其他类型的代码，大概是）代码，只需要与另一个代码。将来，你应该重新评估你的选择。尽管如此，对于大多数应用程序代码来说，它可能已经足够了。\n\n``````", null, "", null, "``````\n\n`-O3`得到我们一堆，`swift_retain`然后`swift_release`打电话，说实话，看起来他们不应该在这个例子。优化器应该已经消除了（大部分）AFAICT，因为它知道大部分有关数组的信息，并且知道它至少有一个强烈的引用。\n\n``````import Cocoa\n\nlet swift_start = NSDate.timeIntervalSinceReferenceDate();\nlet n: Int = 10000\nlet x = Int[](count: n, repeatedValue: 1)\nfor i in 0..n {\nfor j in 0..n {\nlet tmp: Int = x[j]\nx[i] = tmp\n}\n}\nlet y: Int[] = sort(x)\nlet swift_stop = NSDate.timeIntervalSinceReferenceDate();\n\nprintln(\"\\(swift_stop - swift_start)s\")``````\n\nPS：我不是Objective-C的专家，也不是来自Cocoa，Objective-C或Swift运行时的所有设备。我也可能会假设一些我没有写的东西。\n\n`` ``" ]	[ null, "https://includestdio.com/wp-content/uploads/2018/02/ujaJ4aA.png", null, "https://includestdio.com/wp-content/uploads/2018/02/t0s6DsZ.png", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.7657066,"math_prob":0.99669856,"size":6465,"snap":"2020-24-2020-29","text_gpt3_token_len":3994,"char_repetition_ratio":0.09688903,"word_repetition_ratio":0.038404725,"special_character_ratio":0.3512761,"punctuation_ratio":0.17576317,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9914652,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-11T20:53:08Z\",\"WARC-Record-ID\":\"<urn:uuid:6b484eb1-e6c7-4898-ad78-2f79a362c607>\",\"Content-Length\":\"72011\",\"Content-Type\":\"application/http; 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https://learnabout-electronics.org/Digital/dig41.php	[ "# Binary Arithmetic Circuits\n\n• After studying this section, you should be able to:\n• Understand the operation of Binary Adder Circuits.\n• • 4 Bit Parallel Adders.\n• • Twos Complement Overflow.\n• Use Free Software to Simulate Logic Circuit operation.", null, "### Fig 4.1.1 The Half Adder", null, "", null, "### Fig 4.1.1 The Half Adder\n\nBinary arithmetic is carried out by combinational logic circuits, the simplest of which is the half adder, shown in Fig. 4.1.1. This circuit consists, in its most basic form of two gates, an XOR gate that produces a logic 1 output whenever A is 1 and B is 0, or when B is 1 and A is 0. The AND gate produces a logic 1 at the carry output when both A and B are 1. The half adder truth table is shown in Table 4.1.1 and describes the result of binary addition.", null, "### Fig. 4.1.2 The Full Adder Circuit.\n\n1 plus 0 = 12 (110)\n\nand\n\n1 plus 1 = 102 (210)\n\nThe half adder is fine for adding two 1-bit numbers together, but for binary numbers containing several bits, a carry may be produced at some time (as a result of adding 1 and 1) that must be added to the next column. As the half adder has only two inputs it cannot add in a carry bit from a previous column, so it is not practical for anything other than 1-bit additions.", null, "When 2 or more bits are to be added, the circuit used is the Full Adder, shown in Fig 4.1.2, (blue background) together with its simplified block diagram symbol. This circuit simply comprises two half adders, the sum of A and B from the first half adder is used as input A on the second half adder, which now produces a sum of the first half adder sum (S1) plus any ‘carry in’ from the CIN terminal. Any carries produced by the two half adders are then ‘ORed’ together to produce a single COUT. output. The truth table for the circuit is given in Table 4.1.2.", null, "### Fig. 4.1.3 4−Bit Parallel Adder.", null, "Even the full adder is only adding two single bit binary numbers, but full adders may be combined to form parallel adders, which will add two multi−bit numbers. Parallel adders can be built in several forms to add multi−bit binary numbers, each bit of the parallel adder using a single full adder circuit. As parallel adder circuits would look quite complex if drawn showing all the individual gates, it is common to replace the full adder schematic diagram with a simplified block diagram version.\n\nFig 4.1.3 illustrates how a number of full adders can be combined to make a parallel adder, also called a ‘Ripple Carry Adder’ because of the way that any carry appearing at the carry in input (CIN) or produced when adding any of the 4-bit inputs, ‘ripples’ along the adder stages until a final carry out appears at the carry out output (COUT) of the final full adder for bit A3+B3.", null, "### Fig. 4.1.4 8−Bit Twos Complement Adder/Subtractor.", null, "## 8 Bit Twos Complement Adder/Subtractor\n\nTo carry out arithmetic however, it is also necessary to be able to subtract. A further development of the parallel adder is shown in Fig.4.1.4. This is an 8-bit parallel adder/subtractor. This circuit adds in the same way as the adder in Fig. 4.1.3 but subtracts using the twos complement method described in Digital Electronics Module 1.5 (Ones and Twos Complement).\n\nWhen subtraction is required, the control input is set to logic 1, which causes the bit at any particular B input to be complemented by an XOR gate before being fed to input B of the full adder circuit.\n\nTwos complement subtraction in an 8-bit adder/subtractor requires that the 8-bit number at input B is complemented (inverted) and has 1 added to it, before being added to the 8-bit number at input A. The result of this will be an 8-bit number in twos complement format, i.e. with its value represented by the lower 7 bits (bit 0 to bit 6) and the sign represented by the most significant bit (bit 7). The logic 1 on the control input is therefore also fed to the first carry input of the adder to be included in the addition, which for subtraction is therefore:\n\nInput A + Input B + 1\n\n(Here + signifies addition rather than OR)\n\nAlternatively, if addition of A and B is required, then the control input is at logic 0 and number B is fed to the adder without complementing.", null, "### Fig. 4.1.5 XOR Gate Used as a Data Selector.\n\nHow an XOR gate is used here to change the adder into a subtractor by inverting the B inputs can be seen from the truth table for an XOR gate, shown in Table 4.1.3 (in Fig. 4.1.5). Notice that if input A, (used as the CONTROL input) of the XOR gate is at logic 0, then the XOR gate selects input B, but if input A is logic 1, then it selects the inverse of input B (i.e.B).\n\n## Twos Complement Overflow\n\nThe 8-bit adder/subtractor illustrated in Fig. 4.1.4 is designed to add or subtract 8−bit binary numbers using twos complement notation. In this system the most significant bit (bit 7) is not used as part of the number’s value, it is used to indicate the sign of the number (0 = positive and 1 = negative).\n\nNo matter what the word size of a digital system (8-bits 16-bits 32-bits etc.), a given number of bits can only process numbers up to a maximum value that can be held in its designed word length.\n\nDuring arithmetical operations it is possible that adding two numbers (with either positive or negative values) that are both within the system’s limit, can produce a result that is too large for the system’s word length to hold.\n\nFor example, in a twos complement adder such as shown in Fig. 4.1.4, when adding either positive or negative 7-bit values, the result could be larger than 7 bits can accommodate. Therefore the result will need to occupy one extra bit, which means that the calculated value will ‘overflow’ into bit eight, losing a major part (12810) of the value and changing the sign of the result.\n\nTo overcome this problem, it is necessary first to detect that an overflow problem has occurred, and then to solve it either by using additional circuits or, in computing, by implementing a corrective routine in software.\n\nFortunately there is a quite simple method for detecting when an overflow occurs. As shown in Fig. 4.1.5 the overflow detection system consists of a single exclusive or (XOR) gate that takes its inputs from the carry in and carry out connections of the bit 7 (sign bit) adder.\n\nWhen the carry in (CIN) and carry out (COUT) bits of this adder are examined, it can be seen that if an overflow has occurred CIN and COUT will be different, but if no overflow has occurred they will be identical.", null, "### Adding Two Positive (In Range) Numbers\n\nTable 4.1.4 shows the effect of adding two positive values where the sum is within the range that can be held in 7 bits (≤12710). The result of adding two positive numbers has produced a correct positive result with no carry and no overflow.", null, "### Twos Complement Subtraction\n\nTable 4.1.5 shows a twos complement subtraction performed by adding a negative number to a positive number. The result is 3110 (within the range 0 to +12710), the sign bit is 0 indicating positive result, CIN and COUT are both 1, so no overflow is detected and the carry bit will be discarded.", null, "### Adding Twos Complement Negative Numbers\n\nTable 4.1.6 shows the effect of adding two negative values where the sum is less than +12710 therefore a correct negative result of −7310 (in twos complement notation) has been obtained. Both CIN and COUT are logic 1 and no overflow will be signalled. As only 8−bit calculations are being considered, the carry will be discarded.", null, "### Out of Range Result Causes Overflow\n\nWhen the addition of two positive numbers shown in Table. 4.1.7 results in a sum greater than +12710 the sign bit is changed from 0 to 1, incorrectly signifying a negative result. As the ‘carry in’ from bit 6 to bit 7 is 1 and the ‘carry out’ from bit 7 into the Carry bit is 0 an overflow is detected indicating an incorrect answer.\n\nNotice that if the result of 100111012 were to be considered as an unsigned binary value, the addition in Table 4.1.7 would be correct (15710). However as the calculation is using twos complement notation, the answer of −9910 must be considered as wrong.", null, "### Out of Range Addition of Negative Values\n\nTable 4.1.8 shows that adding two negative values can also produce a change in sign and a wrong twos complement result if it is greater than −12810. In this case adding −6310 and −7310 should have produced a negative result of −13610 and not +12010. To check this, the correct answer (although still with the wrong sign) could be obtained if, noting that an overflow had occurred, the answer was complemented and 1 added, giving an unsigned binary result of 100010002 which converts to 128 + 8 = 13610. Overflow errors can be corrected, but this would require either some additional electronics or a software action in response to the overflow signal.\n\nThe adders described in this module are generally called Ripple Carry Adders because of the way that the carry bit is propagated from one stage of the adder to the next, rippling through the chain of full adders until the carry out is produced at the carry out pin of the final stage.\n\nThis process takes some time, which is proportional to the number of bits added. Although this may be a minor problem in small adders, with an increase in the number of bits in the binary words to be added, the time delay before the final carry out is produced becomes unacceptable.\n\nTo overcome this problem, IC manufacturers offer a range of ‘Carry Look Ahead Adders’ in which the addition and carry out are produced simultaneously. The system uses complex combinational logic to assess whether, at each individual adder a carry will be produced, based on the state of the A and B inputs to that stage, and the logic state of the carry in bit to the first stage.", null, "Fig. 4.1.7 shows an arrangement for producing a carry out by splitting the full adder into a partial full adder (grey block), which has two additional outputs, a propagate (P) output that takes a logic 1 output whenever inputs A and B are 1,0 or 0,1 and a generate (G) output that will be logic 1 whenever the A and B inputs are at 1,1. Using this information it is possible to decide on the logic state of the carry out depending on a combination of the CIN state and the A and B states.\n\nIn the carry generator (blue block), the P input is ANDed with the CIN and ORed with the G input to produce a carry out. The carry out is fed to the successive adders in the normal way, but the CIN P and G signals are fed in parallel to the other adder stages, where the state of the carry out for each adder stage can be ascertained from the shared CIN signal and the A and B states for the successive stages, depending on the input states at each stage, rather than waiting for the calculations to complete at all the stages.", null, "", null, "" ]	[ null, "https://learnabout-electronics.org/Digital/images/half-adder.gif", null, "https://learnabout-electronics.org/Digital/images/sim-icon.jpg", null, "https://learnabout-electronics.org/Digital/images/half-adder.gif", null, "https://learnabout-electronics.org/Digital/images/adder-full.gif", null, "https://learnabout-electronics.org/Digital/images/table-4-1-2.gif", null, "https://learnabout-electronics.org/Digital/images/adder-4-bit-parallel.gif", null, "https://learnabout-electronics.org/Digital/images/sim-icon.jpg", null, "https://learnabout-electronics.org/Digital/images/add-sub-8-bit.gif", null, "https://learnabout-electronics.org/Digital/images/sim-icon.jpg", null, "https://learnabout-electronics.org/Digital/images/table-4-1-3.gif", null, "https://learnabout-electronics.org/Digital/images/table-4-1-4.gif", null, "https://learnabout-electronics.org/Digital/images/table-4-1-5.gif", null, "https://learnabout-electronics.org/Digital/images/table-4-1-6.gif", null, "https://learnabout-electronics.org/Digital/images/table-4-1-7.gif", null, "https://learnabout-electronics.org/Digital/images/table-4-1-8.gif", null, "https://learnabout-electronics.org/Digital/images/adder-CLA.gif", null, "https://learnabout-electronics.org/Digital/images/CLA-block.gif", null, "https://learnabout-electronics.org/Digital/images/M14008B.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91859263,"math_prob":0.9621734,"size":11553,"snap":"2021-31-2021-39","text_gpt3_token_len":2813,"char_repetition_ratio":0.1436488,"word_repetition_ratio":0.015230843,"special_character_ratio":0.23881243,"punctuation_ratio":0.09620991,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99527705,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36],"im_url_duplicate_count":[null,6,null,null,null,6,null,3,null,3,null,3,null,null,null,3,null,null,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-24T17:24:14Z\",\"WARC-Record-ID\":\"<urn:uuid:c5edefe4-ed4a-4614-a17b-307b172dcbf3>\",\"Content-Length\":\"28665\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b42649ae-6832-48d7-b239-424512132c0c>\",\"WARC-Concurrent-To\":\"<urn:uuid:0abcaefa-b3fd-4f77-b18e-edb1cbd1d5aa>\",\"WARC-IP-Address\":\"217.160.0.248\",\"WARC-Target-URI\":\"https://learnabout-electronics.org/Digital/dig41.php\",\"WARC-Payload-Digest\":\"sha1:A3VMBXDP3EYV3VQQ6OUKVRRSNS3MOIZJ\",\"WARC-Block-Digest\":\"sha1:UWZJKT7QG5O6JT4OQPQ5O4OROIMOOVT5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046150307.84_warc_CC-MAIN-20210724160723-20210724190723-00204.warc.gz\"}"}
https://www.biostars.org/p/191500/	[ "Question: taking intersection between two networks\n1\nA3.7k wrote:\n\nhi,\n\nI have two networks, i took intersection between them but i cant getting edge list from data.frame\n\nwho know the reason of the error please?\n\nlibrary(igraph)\n\nAT1G01060 AT1G01170 AT1G01260 AT1G01380\n\nAT1G01060 0.0000000 0.6397474 0.0000000 2.6052533\n\nAT1G01170 0.6397474 0.0000000 0.2676777 0.0000000\n\nAT1G01260 0.0000000 0.2676777 0.0000000 0.5220432\n\nAT1G01380 2.6052533 0.0000000 0.5220432 0.0000000\n\nAT1G01490 2.5364855 2.5912206 1.6262291 0.6899495\n\nAT1G01500 0.6131010 0.0000000 0.0000000 1.7374863\n\ndim(clr)\n\n 2857 2857\n\nclr <- as.matrix(clr)\n\nAT1G01060 AT1G01170 AT1G01260 AT1G01380\n\nAT1G01060 0.000000e+00 3.397888e-04 5.573000e-04 8.577027e-05\n\nAT1G01170 2.161158e-03 0.000000e+00 3.510125e-04 3.373863e-06\n\nAT1G01260 4.630123e-07 1.294916e-05 0.000000e+00 9.821657e-05\n\nAT1G01380 2.172965e-05 1.112968e-04 2.950147e-04 0.000000e+00\n\nAT1G01490 1.987599e-03 7.534076e-06 1.634816e-06 3.346604e-05\n\nAT1G01500 8.453009e-05 4.127081e-05 1.531739e-05 7.116557e-05\n\ndim(GENEI3)\n\n 2857 2857\n\ng_sim <- graph.intersection(g_1, g_2, byname = \"auto\", keep.all.vertices = FALSE)\n\nedge_ara <- get.data.frame(g,what = \"edges\")\n\nedge_ara<- edge_ara[order(abs(edge_ara\\$weight),decreasing = T),]\n\nError in abs(edge_ara\\$weight) : non-numeric argument to mathematical function\n\nthank you\n\nR software error • 1.2k views\nwritten 4.0 years ago by A3.7k\n1\n\nHI, I would suggest you to print out the `edge_ara\\$weight` to text file and check whether you find any suspicious non-numeric argument. if there is any, just remove them and run the script.\n\n1\n\nthank you I inspect edge_ara <- get.data.frame(g,what = \"edges\") but file was empty then edge_ara<- edge_ara[order(abs(edge_ara\\$weight),decreasing = T),] gives me error but i don't know why file was empty\n\n2\n\nJust go step by step and check whether the data is there in a variable before you go for subsequent analysis\n\n1\n\nYou need to check values of g_1, g_2 and g_sim and first see what is the overlap !" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.58680487,"math_prob":0.93401843,"size":2300,"snap":"2020-24-2020-29","text_gpt3_token_len":906,"char_repetition_ratio":0.13545296,"word_repetition_ratio":0.028673835,"special_character_ratio":0.4926087,"punctuation_ratio":0.20558882,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9732871,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-27T03:29:57Z\",\"WARC-Record-ID\":\"<urn:uuid:91e56531-a1b1-4bce-abdc-6fe9a0e7319d>\",\"Content-Length\":\"30254\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0605be82-907b-44eb-894d-9cb45667e993>\",\"WARC-Concurrent-To\":\"<urn:uuid:50604421-4e20-4e64-a4a0-92d24a2f10dc>\",\"WARC-IP-Address\":\"69.164.220.180\",\"WARC-Target-URI\":\"https://www.biostars.org/p/191500/\",\"WARC-Payload-Digest\":\"sha1:HUIIS3T6KAZ2TAPW7OKP2PGSX7WZB5SP\",\"WARC-Block-Digest\":\"sha1:KPXFGWMY55WZOQXDZFNN3YPCE3KJ4GVW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347392057.6_warc_CC-MAIN-20200527013445-20200527043445-00083.warc.gz\"}"}
https://metanumbers.com/60925	[ "# 60925 (number)\n\n60,925 (sixty thousand nine hundred twenty-five) is an odd five-digits composite number following 60924 and preceding 60926. In scientific notation, it is written as 6.0925 × 104. The sum of its digits is 22. It has a total of 3 prime factors and 6 positive divisors. There are 48,720 positive integers (up to 60925) that are relatively prime to 60925.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 5\n• Sum of Digits 22\n• Digital Root 4\n\n## Name\n\nShort name 60 thousand 925 sixty thousand nine hundred twenty-five\n\n## Notation\n\nScientific notation 6.0925 × 104 60.925 × 103\n\n## Prime Factorization of 60925\n\nPrime Factorization 52 × 2437\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 12185 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 60,925 is 52 × 2437. Since it has a total of 3 prime factors, 60,925 is a composite number.\n\n## Divisors of 60925\n\n1, 5, 25, 2437, 12185, 60925\n\n6 divisors\n\n Even divisors 0 6 6 0\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 6 Total number of the positive divisors of n σ(n) 75578 Sum of all the positive divisors of n s(n) 14653 Sum of the proper positive divisors of n A(n) 12596.3 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 246.83 Returns the nth root of the product of n divisors H(n) 4.83672 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 60,925 can be divided by 6 positive divisors (out of which 0 are even, and 6 are odd). The sum of these divisors (counting 60,925) is 75,578, the average is 125,96.,333.\n\n## Other Arithmetic Functions (n = 60925)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 48720 Total number of positive integers not greater than n that are coprime to n λ(n) 12180 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 6127 Total number of primes less than or equal to n r2(n) 24 The number of ways n can be represented as the sum of 2 squares\n\nThere are 48,720 positive integers (less than 60,925) that are coprime with 60,925. And there are approximately 6,127 prime numbers less than or equal to 60,925.\n\n## Divisibility of 60925\n\n m n mod m 2 3 4 5 6 7 8 9 1 1 1 0 1 4 5 4\n\nThe number 60,925 is divisible by 5.\n\n• Deficient\n\n• Polite\n\n## Base conversion (60925)\n\nBase System Value\n2 Binary 1110110111111101\n3 Ternary 10002120111\n4 Quaternary 32313331\n5 Quinary 3422200\n6 Senary 1150021\n8 Octal 166775\n10 Decimal 60925\n12 Duodecimal 2b311\n20 Vigesimal 7c65\n36 Base36 1b0d\n\n## Basic calculations (n = 60925)\n\n### Multiplication\n\nn×y\n n×2 121850 182775 243700 304625\n\n### Division\n\nn÷y\n n÷2 30462.5 20308.3 15231.2 12185\n\n### Exponentiation\n\nny\n n2 3711855625 226144803953125 13777872180844140625 839416862617929267578125\n\n### Nth Root\n\ny√n\n 2√n 246.83 39.3488 15.7108 9.05647\n\n## 60925 as geometric shapes\n\n### Circle\n\n Diameter 121850 382803 1.16611e+10\n\n### Sphere\n\n Volume 9.47273e+14 4.66446e+10 382803\n\n### Square\n\nLength = n\n Perimeter 243700 3.71186e+09 86161\n\n### Cube\n\nLength = n\n Surface area 2.22711e+10 2.26145e+14 105525\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 182775 1.60728e+09 52762.6\n\n### Triangular Pyramid\n\nLength = n\n Surface area 6.42912e+09 2.66514e+13 49745.1\n\n## Cryptographic Hash Functions\n\nmd5 ab7516dda6d82b5d330c69b6a9d1a490 3223126ae9f6eec2919b649045a68d0eca671120 306184c7005240b32c89b0abae4e6c8ed0e7b8641ce5ae6d69076c071110622d 63949f348d32156b12419bf1a08253a245b9c7ae99096be17325368030844f0ca07290bf06bc6f171210098267180ed890fd3e4db72fc98ba50b3cc394dfcc3e c17e9a3df9bfac1bae68576af3acb40f264ed6d5" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6175006,"math_prob":0.98531735,"size":4487,"snap":"2021-43-2021-49","text_gpt3_token_len":1586,"char_repetition_ratio":0.11978586,"word_repetition_ratio":0.025757575,"special_character_ratio":0.4582126,"punctuation_ratio":0.07922078,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9953346,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-07T01:07:57Z\",\"WARC-Record-ID\":\"<urn:uuid:ea526820-66b1-4c65-a480-34df624ad496>\",\"Content-Length\":\"38939\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bb134b61-ee45-454e-bc6f-42eb96984ed0>\",\"WARC-Concurrent-To\":\"<urn:uuid:2d6424f9-7706-4ec9-aaf3-fa8b2245ea65>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/60925\",\"WARC-Payload-Digest\":\"sha1:ZPGFXRD2XJ7TTLUMBEXM5IE24PMKOU74\",\"WARC-Block-Digest\":\"sha1:GGNRKZZN53MZLBFNOCXRSZRFVTEDJQDX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363327.64_warc_CC-MAIN-20211206224536-20211207014536-00219.warc.gz\"}"}
https://www.idiap.ch/software/bob/docs/bob/bob.rppg.base/stable/py_api_cvpr14.html	[ "Li’s CVPR14 Python API¶\n\nSignals extraction¶\n\nbuilds a mask on the lower part of the face\n\nThe mask is built using selected keypoints retrieved by a Discriminative Response Map Fitting (DRMF) algorithm. Note that the DRMF is not implemented here, and that the keypoints are loaded from file (and are not provided in the package).\n\nNote also that this function is explicitly made for the keypoints set generated by the Matlab software downloaded from http://ibug.doc.ic.ac.uk/resources/drmf-matlab-code-cvpr-2013/\n\nUpdate: this function also works when using bob.ip.dlib.DlibLandmarkExtraction\n\nParameters\n• image (numpy.ndarray) – The current frame.\n\n• keypoints (numpy.ndarray) – the set of 66 keypoints retrieved by DRMF.\n\n• indent (int) – The percentage of the facewidth [in pixels] by which selected keypoints are shifted inside the face to build the mask. THe facewidth is defined by the distance between the two keypoints located on the right and left edge of the face, at the eyes’ height.\n\n• plot (bool) – If set to True, plots the current face with the selected keypoints and the built mask.\n\nReturns\n\n• mask (numpy.ndarray) – A boolean array of the size of the original image, where the region corresponding to the mask is True.\n\n• mask_points (list of tuple) – The points corresponding to vertices of the mask.\n\nreturns a boolean array where the mask is True.\n\nIt turns mask points into a region of interest and returns the corresponding boolean array, of the same size as the image.\n\nParameters\n• image (numpy.ndarray) – The current frame.\n\n• mask_points (list of tuple) – The points corresponding to vertices of the mask.\n\nReturns\n\nmask – A boolean array of the size of the original image, where the region corresponding to the mask is True.\n\nReturn type\n\nnumpy.ndarray\n\nbob.rppg.cvpr14.extract_utils.get_good_features_to_track(face, npoints, quality=0.01, min_distance=10, plot=False)[source]\n\napplies the openCV function “good features to track”\n\nParameters\n• face (numpy.ndarray) – The cropped face image\n\n• npoints (int) – The maximum number of strong corners you want to detect\n\n• quality (float) – The minimum relative quality of the detected corners. Note that increasing this value decreases the number of detected corners. Defaluts to 0.01.\n\n• min_distance (int) – minimum euclidean distance between detected corners.\n\n• plot (bool) – if we should plot the currently selected features to track.\n\nReturns\n\ncorners – the detected strong corners.\n\nReturn type\n\nnumpy.ndarray\n\nbob.rppg.cvpr14.extract_utils.track_features(previous, current, previous_points, plot=False)[source]\n\nprojects the features from the previous frame in the current frame.\n\nParameters\n• previous (numpy.ndarray) – the previous frame.\n\n• current (numpy.ndarray) – the current frame.\n\n• previous_points (numpy.ndarray) – the set of keypoints to track (in the previous frame).\n\n• plot (bool) – Plots the keypoints projected on the current frame.\n\nReturns\n\ncurrent_points – the set of keypoints in the current frame.\n\nReturn type\n\nnumpy.ndarray\n\nbob.rppg.cvpr14.extract_utils.find_transformation(previous_points, current_points)[source]\n\nfinds the transformation matrix from previous points to current points.\n\nThe transformation matrix is found using estimateRigidTransform (fancier alternatives have been tried, but are not that stable).\n\nParameters\n• previous_points (numpy.ndarray) – Set of ‘starting’ 2d points\n\n• current_points (numpy.ndarray) – Set of ‘destination’ 2d points\n\nReturns\n\ntransformation_matrix – the affine transformation matrix between the two sets of points.\n\nReturn type\n\nnumpy.ndarray\n\nParameters\n• previous_mask_points (numpy.ndarray) – The points forming the mask in the previous frame\n\n• transformation_matrix (numpy.ndarray) – the affine transformation matrix between the two sets of points.\n\nReturns\n\nReturn type\n\nnumpy.ndarray\n\ncomputes the average green color within a given mask.\n\nParameters\n• image (numpy.ndarray) – The image containing the face.\n\n• mask (numpy.ndarray) – A boolean array of the size of the original image, where the region corresponding to the mask is True.\n\n• plot (bool) – Plot the mask as an overlay on the original image.\n\nReturns\n\ncolor – The average RGB colors inside the mask ROI.\n\nReturn type\n\nnumpy.ndarray\n\nbob.rppg.cvpr14.extract_utils.compute_average_colors_wholeface(image, plot=False)[source]\n\ncomputes the average green color within the provided face image\n\nParameters\n• image (numpy.ndarray) – The cropped face image\n\n• plot (bool) – Plot the mask as an overlay on the original image.\n\nReturns\n\ncolor – The average green color inside the face\n\nReturn type\n\nfloat\n\nIllumination rectification¶\n\nbob.rppg.cvpr14.illum_utils.rectify_illumination(face_color, bg_color, step, length)[source]\n\nperforms illumination rectification.\n\nThe correction is made on the face green values using the background green values, so as to remove global illumination variations in the face green color signal.\n\nParameters\n• face_color (numpy.ndarray) – The mean green value of the face across the video sequence.\n\n• bg_color (numpy.ndarray) – The mean green value of the background across the video sequence.\n\n• step (float) – Step size in the filter’s weight adaptation.\n\n• length (int) – Length of the filter.\n\nReturns\n\nrectified color – The mean green values of the face, corrected for illumination variations.\n\nReturn type\n\nnumpy.ndarray\n\nbob.rppg.cvpr14.illum_utils.nlms(signal, desired_signal, n_filter_taps, step, initCoeffs=None, adapt=True)[source]\n\nNormalized least mean square filter.\n\nParameters\n• signal (numpy.ndarray) – The signal to be filtered.\n\n• desired_signal (numpy.ndarray) – The target signal.\n\n• n_filter_taps (int) – The number of filter taps (related to the filter order).\n\n• step (float) – Adaptation step for the filter weights.\n\n• initCoeffs (numpy.ndarray) – Initial values for the weights. Defaults to zero.\n\n• adapt (bool) – If True, adapt the filter weights. If False, only filters.\n\nReturns\n\n• y (numpy.ndarray) – The filtered signal.\n\n• e (numpy.ndarray) – The error signal (difference between filtered and desired)\n\n• w (numpy.ndarray) – The found weights of the filter.\n\nMotion correction¶\n\nbob.rppg.cvpr14.motion_utils.build_segments(signal, length)[source]\n\nbuilds an array containing segments of the signal.\n\nThe signal is divided into segments of provided length (no overlap) and the different segments are stacked.\n\nParameters\n• signal (numpy.ndarray) – The signal to be processed.\n\n• length (int) – The length of the segments.\n\nReturns\n\n• segments (numpy.ndarray) – the segments composing the signal.\n\n• end_index (int) – The length of the signal (there may be a trail smaller than a segment at the end of the signal, that will be discarded).\n\nbob.rppg.cvpr14.motion_utils.prune_segments(segments, threshold)[source]\n\nremove segments.\n\nSegments are removed if their standard deviation is higher than the provided threshold.\n\nParameters\n• segments (numpy.ndarray) – The set of segments.\n\n• threshold (float) – Threshold on the standard deviation.\n\nReturns\n\n• pruned_segments (numpy.ndarray) – The set of “stable” segments.\n\n• gaps (list of length (# of retained segments)) – Boolean list that tells if a gap should be accounted for when building the final signal.\n\n• cut_index (list of tuples) – Contains the start and end index of each removed segment. Used for plotting purposes.\n\nbob.rppg.cvpr14.motion_utils.build_final_signal(segments, gaps)[source]\n\nbuilds the final signal with remaining segments.\n\nParameters\n• segments (numpy.ndarray) – The set of remaining segments.\n\n• gaps (list) – Boolean list that tells if a gap should be accounted for when building the final signal.\n\nReturns\n\nfinal_signal – The final signal.\n\nReturn type\n\nnumpy.ndarray\n\nbob.rppg.cvpr14.motion_utils.build_final_signal_cvpr14(segments, gaps)[source]\n\nbuilds the final signal with remaining segments.\n\nWarning\n\nThis contains a bug !\n\nBuilds the final signal, but reproducing the bug found in the code provided by the authors of [li-cvpr-2014]. The bug is in the ‘collage’ of remaining segments. The gap is not always properly accounted for…\n\nParameters\n• segments (numpy.ndarray) – The set of remaining segments.\n\n• gaps (list) – Boolean list that tells if a gap should be accounted for when building the final signal.\n\nReturns\n\nfinal_signal – The final signal.\n\nReturn type\n\nnumpy.ndarray\n\nFiltering¶\n\nbob.rppg.cvpr14.filter_utils.detrend(signal, Lambda)[source]\n\napplies a detrending filter.\n\nThis code is based on the following article “An advanced detrending method with application to HRV analysis”. Tarvainen et al., IEEE Trans on Biomedical Engineering, 2002.\n\nParameters\n• signal (numpy.ndarray) – The signal where you want to remove the trend.\n\n• Lambda (int) – The smoothing parameter.\n\nReturns\n\nfiltered_signal – The detrended signal.\n\nReturn type\n\nnumpy.ndarray\n\nbob.rppg.cvpr14.filter_utils.average(signal, window_size)[source]\n\nMoving average filter.\n\nParameters\n• signal (numpy.ndarray) – The signal to filter.\n\n• window_size (int) – The size of the window to compute the average.\n\nReturns\n\nfiltered_signal – The averaged signal.\n\nReturn type\n\nnumpy.ndarray" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.718593,"math_prob":0.91655874,"size":9293,"snap":"2022-05-2022-21","text_gpt3_token_len":2182,"char_repetition_ratio":0.1735386,"word_repetition_ratio":0.18423048,"special_character_ratio":0.22490047,"punctuation_ratio":0.15729627,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9885233,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-27T00:11:19Z\",\"WARC-Record-ID\":\"<urn:uuid:7a82282c-d4f2-4936-9897-4cba60f8b2a6>\",\"Content-Length\":\"42472\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ee4cd390-b181-41a7-b43f-797ca5bd3352>\",\"WARC-Concurrent-To\":\"<urn:uuid:791e9605-e250-460e-b584-e7893cf84966>\",\"WARC-IP-Address\":\"192.33.221.201\",\"WARC-Target-URI\":\"https://www.idiap.ch/software/bob/docs/bob/bob.rppg.base/stable/py_api_cvpr14.html\",\"WARC-Payload-Digest\":\"sha1:E4BNCRLYKICS3S2GUQ3ZBHGTCORFOD24\",\"WARC-Block-Digest\":\"sha1:JQLI3T4VS2HQXRZIWQYDWKJTBUUKWZE5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320305006.68_warc_CC-MAIN-20220126222652-20220127012652-00030.warc.gz\"}"}