bits 1 and 0 are transmitted with equal probability. at the receiver, the pdf of the respective received signals for both bits are as shown below.the optimum threshold to achieve minimum bit error rate (ber) is
4/5
dear student,

each radio station is given a different frequency band and these
radio signals are sent at that particular frequency only
for example fm 93 5 or 94 3 etc are given radio frequencies given
in decimals only

if we see each radio station contains name like fm 93 5 or 94 3
etc

regards, from here [fstrm python receiver](https://github.com/dmachard/python-framestream/blob/main/fstrm/fstrm_codec.py)

# frame streams control frame format - data frame length equals 00 00 00 00

# |||

# | data frame length | 4 bytes |

# ||… remove the server max and min, and add folowing:

min protocol = smb2

max protocol = smb2

client min protocol = smb2

client max protocol = smb2 a 5aa blastocyst is probably one of the best embryos that we can get, but, unfortunately, having a good morphology does not necessarily mean that the embryo is of good quality.  it only means that the chances of having normal genetics and implantation are going to be higher.
when we grade an embryo 5aa, it means that it has almost done the hatching, and both the inner cell mass and trophectoderm are of top quality.  the chances for one of these embryos, though it will depend on the age if we get this quality in egg donation, the pregnancy rate should be around 60-65% with just a single embryo transfer. hi , i try it with this scenario, it work.

[[image]](https://aws1.discourse-cdn.com/business5/uploads/pynq1/original/1x/1ef72ed473166fc2811618e9c1ec0febb7a2ecc8.png "image")

test code:

ov1 = overlay(’/home/xilinx/haddsub/prio.bit’)

pr=ov1.pr\_decoupler\_0

pr.write(0x0, 0) <- default set to zero

// before download partial bitstream set to one

pr.write(0x0, 1)

add\_pb=bitstream(’/home/xilinx/haddsub/pr\_0\_sub.bit’, n… the octal numbering system is very similar in principle to the previous hexadecimal numbering system except that in octal, a binary number is divided up into groups of only 3 bits, with each group or set of bits having a distinct value of between 000 (0) and 111 ( 4+2+1 = 7 ). the octal numeral system, or oct for short, is the base-8 number system, and uses the digits 0 to 7. octal number system is one the type of number representation techniques, in which there value of base is 8. the number system whose base is 8 is known as the octal number system. octal number system is a base-8 number system where the range of numbers is 0,1,2,3,4,5,6,7. octal number system watch more videos at https://www. (a) for digit



expression : 5n + 1

where n = no of digits

for 5 digits

no of segments required = 5(5) + 1 = 25 + 1 = 26

for 10 digits

no of segments required = 5(10) + 1 = 50 + 1 = 51

for 100 digits

no of segments required = 5(100) + 1 = 500 + 1 = 501

(b) for digit



expression : 3n + 1

where n = no of digits

for 5 digits

no of segments required = 3(5) + 1 = 15 + 1 = 16

for 10 digits

no of segments required = 3(10) + 1 = 30 + 1 = 31

for 100 digits

no of segments required = 3(100) + 1 = 300 + 1 = 301

(c) for digit



expression : 5n + 2

where n = no of digits

for 5 digits

no of segments required = 5(5) + 2 = 25 + 2 = 27

for 10 digits

no of segments required = 5(10) + 2 = 50 + 2 = 52

for 100 digits

no of segments required = 5(100) + 2 = 500 + 2 = 502 the worst case should be o(n) if each hash is the same, however it should be noted that by default the stdlib's hashmap and hashset use an hasher that provide resistance against [hashdos](https://cryptanalysis.eu/blog/2011/12/28/effective-dos-attacks-against-web-application-plattforms-hashdos/) attacks, so you should be safe.

[image] tamas:

amortized/average is also interesting, but i assume that's goi… When you say bytes do you really mean the first four bytes or do you mean the first four numbes? I have a suspicion you mean the latter. Total no. of possible outcomes = 30 {1, 2, 3, … 30}  

  

E ⟶ event of getting no. divisible by 3.  

  

No. of favourable outcomes = 10 {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}  

  

Probability, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)  

  

P(E) = 10/30 = 1/3  

  

Bar E ⟶ event of getting no. not divisible by 3.  

  

P(Bar E) = 1 – P(E)  

  

= 1 - 1/3 = 2/3