Update all PDFs

# Choosing appropriate units

Alignments to Content Standards: 8.EE.A.4

1. A computer has 128 gigabytes of memory. One gigabyte is $1 \times 10^9$ bytes. A floppy disk, used for storage by computers in the 1970's, holds about 80 kilobytes. There are 1000 bytes in a kilobyte. How many kilobytes of memory does a modern computer have? How many gigabytes of memory does a floppy disk have? Express your answers both as decimals and using scientific notation.
2. George told his teacher that he spent over 21,000 seconds working on his homework. Express this amount using scientific notation. What would be a more appropriate unit of time for George to use? Explain and convert to your new units.
3. A certain swimming pool contains about $3 \times 10^7$ teaspoons of water. Choose a more appropriate unit for reporting the volume of water in this swimming pool and convert from teaspoons to your chosen units.
4. A helium atom has a diameter of about 62 picometers. There are one trillion picometers in a meter. The diameter of the sun is about 1,400,000 km. Express the diameter of a helium atom and of the sun in meters using scientific notation.  About many times larger is the diameter of the sun than the diameter of a helium atom?

## IM Commentary

The purpose of this task is to use scientific notation in the context of choosing units to report quantities. The teacher may wish to have students reflect on how the choice of units in each part either helps or hinders our ability to grasp the quantity under consideration. There is always a wide choice of units available and we should try to choose units which are most appropriate for communicating the size of an object without further calculations or conversions. Scientific notation is particularly useful when the quantities being described are very small or very large (though a good choice of units can often remove the need for these small or large quantities).

In part (a), students have an intuition, based on experience, for how much 1 gigabyte of storage is and so expressing the computer memory as 128 gigabytes is helpful. When we use scientific notation and write this in terms of bytes or kilobytes, the number is very large and does not carry much useful information except for an expert. For the situations presented in (b) and (c), neither seconds nor teaspoons is an appropriate unit of measure to give a good intuition for the quantities in question. There are many reasonable choices. For (b), hours is probably the best choice but even minutes or days would be better than seconds. For part (c), students will likely need a conversion factor. In metric units, there are about 4.9 ml in a teaspoon. In metric units, the best choice for giving an intuition for the size of the swimming pool is probably cubic meters. In standard units, gallons are too small: cubic feet or yards would be appropriate but the conversion will be difficult.

In extreme situations like the helium atom example in part (d), any choice of units will have both advantages and disadvantages. Picometers are a natural choice for measuring an atom in the sense that the number of picometers needed is a reasonable size number. On the other hand, we have little physical experience that allows us to grasp what a picometer is. If we choose the more comfortable unit of meters, then we can see that it takes many, many helium atom diameters to make a meter: unfortunately, it is difficult to imagine billions of atoms and so this also has a downside. When dealing with very large or very small quantities, this dilemma always exists and it is only by gaining familiarity with the given context that the units and quantities become more familiar and comfortable.

## Solution

1. If one gigabyte is $1 \times 10^9$ bytes then 128 gigabytes is $128 \times 10^9$ bytes. Rewriting this number in appropriate scientific notation we find

\begin{align} 128 \times 10^9 &= 1.28 \times 100 \times 10^9 \\ &= 1.28 \times 10^{11} \end{align} for the computer storage. Since there are 1000 bytes in a kilobyte we can find the number of kilobytes in the computer memory by dividing the number of bytes by 1000 or $1 \times 10^3$. So the computer has $1.28 \times 10^{8}$ kilobytes of memory. This is 128,000,000 or 128 million kilobytes.

The floppy disk has 80 kilobytes of storage and there are 1000 bytes in a kilobyte so this is 80,000 or $8 \times 10^4$ bytes of storage. There are $1 \times 10^9$ bytes in a gigabyte so the floppy disk storage is $$\frac{8 \times 10^4 \text{ bytes}}{1 \times 10^9 \text{ bytes per gigabyte}}.$$ This is $8 \times 10^{4-9} = 8 \times 10^{-5}$ gigabytes. There are 0.00008 gigabytes of storage on a floppy disk.

2. We have \begin{align}21,000 &= 2.1 \times 10,000 \\ &= 2.1 \times 10^4 \end{align}so it took George $2.1 \times 10^4$ seconds to do his homework. This is not very helpful for giving an intuition of how long George worked on the homework. A more appropriate unit would probably be hours. To check, there are 60 minutes per hour and 60 seconds per minute so there are 60 $\times$ 60 = 3600 seconds per hour. This means that George worked on the homework for 21,000 $\div$ 3600 hours. This is a little less than 6 hours.
3. For a swimming pool, cubic meters or yards would be an appropriate unit. In metric units, there are 100 centimeters in a meter and so $100^3 = 1 \times 10^6$ cubic centimeters in a cubic meter. One teaspoon of water takes up about 4.9 cubic centimeters of space so $3 \times 10^7$ teaspoons would take up about $3 \times 4.9 \times 10^7$ cubic centimeters (or about $1.5 \times 10^8$ cubic centimeters). So, converting from cubic centimeters to cubic meters, our water takes up about $1.5 \times 10^2$ (or 150) cubic meters of space. This is reasonable as a swimming pool 2 meters deep, 5 meters wide, and 15 meters long would have this volume of water.
4. One trillion is 1,000,000,000,000 or, in scientific notation, $1 \times 10^{12}$. So one trillionth is $1 \times 10^{-12}$. One picometer is $1 \times 10^{-12}$ meters and the diameter of a helium atom is about $62 \times 10^{-12}$ or $6.2 \times 10^{-11}$ meters. The diameter of the sun is about 1,400,000 km and there are 1000 meters in a kilometer. So the diameter of the sun is about 1,400,000,000 meters or $1.4 \times 10^{9}$ meters. To compare these two diameters, we compute the quotient $1.4 \times 10^9 \div 6.2 \times 10^{-11} \approx 0.23 \times 10^{20}$ or, writing this in proper scientific notation, $2.3 \times 10^{19}$. The diameter of the sun is more than  $1 \times 10^{19}$ as large as a helium atom, a fantastically large number!
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9513887763023376, "perplexity": 619.5023440486019}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084890928.82/warc/CC-MAIN-20180121234728-20180122014728-00633.warc.gz"}