# Measurement Study Guide for the Math Basics

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### Measuring Temperature

*Temperature* is measured in degrees. There are three main ways to measure: Celsius, Fahrenheit, and Kelvin. The first two are used around the world by the general population while Kelvin is used primarily in science.

*Celsius* is perhaps the easiest scale to understand, which is probably why most of the world uses it. It’s based on the freezing and boiling points of water at sea level. Water freezes at \(0^\circ \;C\) and boils at \(100^\circ \;C\). Room temperature is about \(23^\circ \;C\).

The United States, and three or four other countries use *Fahrenheit*. The origin of the scale is still up for debate. It’s helpful to know that water freezes at \(32^\circ\; F\) and boils at \(212^\circ \; F\). Room temperature is around \(73^\circ \;F\).

If you’re a frequent traveller, you might be pretty scared when the pilot tells you the weather in Spain is \(29^\circ\) and you didn’t pack any winter clothing. Of course, Spain uses the Celsius scale, but what exactly does \(29^\circ \;C\) mean in Fahrenheit?

To switch from Celsius to Fahrenheit, use the formula:

\[F=\frac{9}{5}C + 32\]or equivalently

\[F=1.8C+32\]So, what does \(29^\circ \;C\) mean in Fahrenheit:

\[F =1.8 \cdot 29 + 32\] \[F=52.2 + 32\] \[F=84.4^\circ \; F\]If you want to switch from Fahrenheit to Celsius, use the formula:

\[C=(F-32)\cdot \frac{5}{9}\]or equivalently

\[C=\dfrac{F-32}{1.8}\]A European flying into the USA might want to know what \(50^\circ \;F\) is in Celsius:

\[C=(50-32) \cdot \frac{5}{9}\] \[C=18 \cdot \frac{5}{9}\] \[C=10^\circ \; C\]Finally, the *Kelvin* scale, used primarily in science, is based on *absolute zero*: the temperature which is so cold that all thermal (molecular) motion ceases. It has the same scale as Celsius, but all temperatures are \(273^\circ\) higher. Absolute zero (\(-273^\circ \;C\)) is \(0^\circ \;K\). Water freezes (\(0^\circ \;C\)) at \(273^\circ \;K\) and boils (\(100^\circ \;C\)) at \(373^\circ \;K\).

This table shows conversions of some common temperatures:

\[\begin{array}{|c|c|c|c|} \hline \text{} & \textbf{Fahrenheit} & \textbf{Celsius} & \textbf{Kelvin} \\ \hline \text{Absolute zero} & -460^\circ\;F & -273^\circ\;C & 0^\circ \;K \\ \hline \text{Freezing Point of Water} & 32^\circ \;F & 0^\circ \;C & 273^\circ \;K \\ \hline \text{Room Temperature} & 73^\circ \;F & 23^\circ \;C & 296^\circ \;K \\ \hline \text{Boiling Point of Water} & 212^\circ \;F & 100^\circ \;C & 373^\circ \;K \\ \hline \end{array}\]### Counting Money

Many jobs will require the ability to count money in a professional manner: server, cashier, banker, etc. First, know what denominations are available in the U.S.A.

\[\begin{array}{|c|c|} \hline \textbf{Amount} & \textbf{Name} \\ \hline \$100 & \text{100 dollar bill} \\ \hline \$50 & \text{50 dollar bill} \\ \hline \$20 & \text{20 dollar bill} \\ \hline \$10 & \text{10 dollar bill} \\ \hline \$5 & \text{5 dollar bill} \\ \hline \$1 & \text{dollar bill} \\ \hline \$0.25 \text{ or } 25\text{¢} & \text{quarter} \\ \hline \$0.10 \text{ or } 10\text{¢} & \text{dime} \\ \hline \$0.05 \text{ or } 5\text{¢} & \text{nickel} \\ \hline \$0.01 \text{ or } 1\text{¢} & \text{penny} \\ \hline \end{array}\]There are two main ways to count money back to a customer. The first way is to simply count the amount that should be given as change displayed at the register. For instance, if a purchase is $7.43 and the customer pays with at $20 bill, the register will display $12.57. So, if you were the cashier, you’d pick a $10 bill, two $1 bills, two quarters (50 ¢), one nickel (5 ¢), and two pennies (2 ¢). You should hand back the change first, followed by the bills. You can count it however you wish using this method.

The second way doesn’t require a register or subtraction. Just count from $7.43 up to $20. Start with two pennies to get to $7.45, then pick up a nickel to get to $7.50 and two quarters to get to $8. Now, get two dollar bills to get up to $10 and finally one ten dollar bill to get up to $20. Count it back in the same way you grabbed the money from the register.

### Dimensional Analysis

Dimensional Analysis is the method of tracking units through a series of multiplication steps to ensure proper unit conversion. Use this method if you need to convert units. At the heart of this method is the multiplicative identity: multiplying anything by 1 doesn’t change its value. Also, any value divided by itself is 1. Here is an example:

\[\dfrac{2}{2} = 1\]It works the same with units:

\[\dfrac{2 \text{ cups}}{2 \text{ cups}} = 1\]Now, we know (from the section about liquid measurement) that \(2 \text{ cups}= 1 \text{ pint}\), so substitute \(1 \text{ pint}\) in for either part of the fraction:

\[\dfrac{1 \text{ pint}}{2 \text{ cups}} = 1\]or

\[\dfrac{2 \text{ cups}}{1 \text{ pint}} = 1\]Therefore, any of the conversions from this study guide can be written as ratios that equal \(1\). Here are some examples:

\[\dfrac{1 \text{ foot}}{12 \text{ inches}} = 1\] \[\dfrac{1,000 \text{ meters}}{1 \text{ kilometer}} =1\] \[\dfrac{60 \text{ minutes}}{1 \text{ hour}}=1\]Now let’s move on to how to use these ratios. Maybe you’re given a measurement of 8 feet and you need to know how many inches it is. Start by writing this equation:

\[\dfrac{8 \text{ ft}}{1} \cdot \text{?} = x \text{ in}\]Now, look through your list of known conversions and you’ll find the one for feet and inches. You need to replace the “?” with the right ratio, either \(\dfrac{1 \text{ ft}}{12 \text{ in}}\) or \(\dfrac{12 \text{ in}}{1 \text{ ft}}\). The correct one is the latter, and we’ll see why shortly.

\[\dfrac{8 \text{ ft}}{1} \cdot \dfrac{12 \text{ in}}{1 \text{ ft}} = x \text{ in}\]If you notice, on the left side of the equation, the “ft” unit appears at the top and the bottom of the fraction. Therefore, it cancels out:

\[\require{cancel}\] \[\dfrac{8 \cancel{\text{ ft}}}{1} \cdot \dfrac{12 \text{ in}}{1 \cancel{\text{ ft}}} = x \text{ in}\]Now, multiply straight across:

\[\dfrac{8 \cdot 12 \text{ in}}{1 \cdot 1} = \dfrac{96 \text{ in}}{1} = 96 \text{ in}\]Let’s try another example. Convert 5,400 grams to kilograms. Start by writing:

\[\dfrac{5400 \text{ g}}{1} \cdot \text{?} = x \text{ kg}\]Now, which ratio to use? You’ll see a direct conversion exists between grams and kilograms.

Either \(\dfrac{1,000 \text{ g}}{1 \text{ kg}}\) or \(\dfrac{1 \text{ kg}}{1,000 \text{ g}}\).

You need to choose the one that cancels the “g” on top, so choose the second one (with “g” on the bottom).

\[\dfrac{5,400 \text{ g}}{1} \cdot \dfrac{1 \text{ kg}}{1,000 \text{ g}} = x \text{ kg}\]Now, cancel, multiply, and simplify.

\[\dfrac{5,400 \cancel{\text{ g}}}{1} \cdot \dfrac{1 \text{ kg}}{1,000 \cancel{\text{ g}}} = x \text{ kg}\] \[\dfrac{5,400 \cdot 1 \text{ kg}}{1 \cdot 1,000} =x \text{ kg}\] \[\dfrac{5,400 \text{ kg}}{1000} =5.4 \text{ kg}\]Occasionally, you may not know the direct conversion. For instance, convert 400 fluid ounces to liters. We need to find a stepping stone to bridge the gap. In this case, that would be milliliters, because we know how many milliliters make up a fluid ounce and how many milliliters make up a liter. So, convert to milliliters first:

\[\dfrac{400 \text{ fl oz}}{1} \cdot \text{?} = x \text{ ml}\] \[\dfrac{400 \cancel{\text{ fl oz}}}{1} \cdot \dfrac{30 \text{ ml}}{1 \cancel{\text{ fl oz}}} = x \text{ ml}\] \[\dfrac{ 400 \cdot 30 \text{ ml}}{1}=\dfrac{12,000 \text{ ml}}{1} = 12,000 \text{ ml}\]Now, convert to liters:

\[\dfrac{12,000 \text{ ml}}{1} \cdot \text{?} = x \text{ L}\] \[\dfrac{12,000 \cancel{\text{ ml}}}{1} \cdot \dfrac{ 1 \text{ L}}{1,000 \cancel{\text{ ml}}} = x \text{ L}\] \[\dfrac{ 12,000 \cdot 1 \text{ L}}{1 \cdot 1000} = \dfrac{12,000 \text{ L}}{1,000} = 12 \text{ L}\]So,

\[400 \text{ fl oz} = 12,000 \text{ ml} = 12 \text{ L}\]This process can be sped up by just inserting both ratios into the conversion from fluid ounces to liters.

\[\dfrac{400 \text{ fl oz}}{1} \cdot \text{?} = x \text{ L}\] \[\dfrac{400 \cancel{\text{ fl oz}}}{1} \cdot \dfrac{30 \cancel{\text{ ml}}}{1 \cancel{\text{ fl oz}}} \cdot \dfrac{ 1 \text{ L}}{1,000 \cancel{\text{ ml}}} = \dfrac{x \text{ L}}{1}\] \[\dfrac{400 \cdot 30 \cdot 1 \text{ L}}{1 \cdot 1 \cdot 1000} = \dfrac{12,000 \text{ L}}{1,000} = 12 \text{ L}\]All **Study Guides for the Math Basics** are now available as downloadable PDFs