Math Review

Least Common Multiple
Greatest Common Factor
Converting Units
Ratios
Proportions
Rates
Scale Factors
Properties of Circles
Volumes
Scientific Notation
Solving Equations
Fractions, Decimals, and Percents
Percent Change
Positive and Negative Numbers in Addition and Subtraction
Exponents
Properties of Exponents
Squares and Square Roots
Converting Temperature Units
Probability
Averaging Data

Math Review

Least Common Multiple

A multiple of a number is a number that can be completely divided by the number, such that the remainder is zero. For example, 2, 4, 6, 8, and 10 are all divisible by 2. So, they are all multiples of 2.

The least common multiple (LCM) is the smallest multiple that is common to two or more numbers. Consider the numbers 4 and 6. To find the LCM of these two numbers, follow these steps:

1. Write down at least the first 10 multiples of both numbers:
   Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40
   Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60
   
2. Now find the multiples that are common to 4 and 6. If you don’t find any common multiples at this point, you may need to continue writing some more multiples until you find one that is common.
   
3. We see that 12, 24, and 36 are common to 4 and 6.
   
4. Of these, 12 is the smallest number.
   
5. Therefore, 12 is the least common multiple (LCM) of the numbers 4 and 6.

Greatest Common Factor

Factors are numbers that are multiplied to find a product. For example, the numbers 3 and 4 are factors of 12 (that is, 3 × 4 = 12). Here’s a complete list of the factors of 12 in the natural number system:

Factors of 12: 1, 2, 3, 4, 6, 12

A prime factor is a factor that can be divided evenly only by 1 and by itself. The prime factors in the list above are 2 and 3 (note that 1 is not considered a prime factor). We can write the prime factorization of 12 using only its prime factors. Each prime factor may appear more than once in the factorization:

Prime factorization of 12: 2 × 2 × 3 = 12

The greatest common factor (GCF) of two numbers is the product of all the prime factors common to both numbers. We can use prime factorization to arrive at the GCF for a pair of numbers.

Example: Find the greatest common factor of 12 and 18.

1. Begin by writing the prime factorization of both numbers:
   Prime factorization of 12: 2 × 2 × 3
   Prime factorization of 18: 2 × 3 × 3
   
2. Identify all the factors common to both numbers. Here, the numbers 2 and 3 appear in the prime factorization for 12 and 18:
   Prime factorization of 12: 2 × 2 × 3
   Prime factorization of 18: 2 × 3 × 3
   
3. The greatest common factor of 12 and 18 is the product of 2 and 3. The answer is 6:
   2 × 3 = 6
   
4. So, the greatest common factor (GCF) of 12 and 18 is 6.

Converting Units

The table shows prefixes from the International System of Units (SI).

Prefix and Symbol	Value
kilo (k)	1,000
hecto (h)	100
deka (da)	10
deci (d)	0.1
centi (c)	0.01
milli (m)	0.001

Example 1

How many grams are in 3 kilograms?

1. Locate the prefix in the table. One kilogram is the same as 1,000 grams.
   
2. Determine the ratio to use.
   $\frac{1,000\:\textrm{g}}{1\:\textrm{kg}}$ or $\frac{1\:\textrm{kg}}{1,000\:\textrm{g}}$
   
3.  Multiply the ratio by the given value and solve.
   $3\:\textrm{kg} \times \frac{1,000\:\textrm{g}}{1\:\textrm{kg}} = 3,000 \:\textrm{g}$

Example 2

How many millimeters are in 8 centimeters?

1. Each value in the table decreases by a factor of 10, from top to bottom. Locate the prefixes milli and centi in the table. There are 10 millimeters in 1 centimeter.
   $\frac{0.01}{0.001} = 10$
   
2. Determine the ratio to use.
   $\frac{10\: \textrm{mm}}{1\: \textrm{cm}}$ or $\frac{1\: \textrm{cm}}{10\: \textrm{mm}}$
   
3. Multiply the ratio by the given value and solve.
   $8 \:\textrm{cm} \times \frac{10\: \textrm{mm}}{1\: \textrm{cm}} = 80 \:\textrm{mm}$

Ratios

A ratio is a comparison between two numbers. Ratios can be written three different ways. Like fractions, ratios can be written in simplest form. The order of the numbers is important. Here are equivalent ways of writing the ratio "two to seven."

$\begin{matrix} \frac{2}{7} & 2\: \textrm{to}\: 7 & 2:7 \end{matrix}$

Example

Samantha adds 9 grams of salt to 45 grams of water. What is the ratio of the mass of salt to the mass of water? Write the answer as a fraction in simplest form.

1. Write the ratio as a fraction, noting the order of the numbers.
   $\frac{\textrm{mass of salt}}{\textrm{mass of water}} = \frac{9}{45}$
2. Simplify.
   $\frac{9}{45} = \frac{1}{5}$
3. The ratio of the mass of salt to the mass of water is 1 to 5. 

Proportions

A proportion is a statement that shows two ratios as equal. One example of a proportion is $\frac{30}{66} = \frac{5}{11}$.

Cross-multiplying is one way to solve for an unknown quantity in a proportion.

Example

Dalton is studying the map of a popular state park. One inch on the map is the same as 40 yards in the real world. The distance between the ranger station and the public beach on the map is 4.6 inches. How far apart are the ranger station and the beach in reality?

1. Set up the proportion. Assign a variable to the unknown value. In this case, assume the unknown value is d.
   $\frac{1\:\textrm{in}}{40\:\textrm{yd}} = \frac{4.6 \:\textrm{in}}{d}$
2. Cross-multiply.
   $d \times 1 = 40 \times 4.6$
3. Simplify.
   $d = 184$
   
4. The distance between the ranger station and the public beach is 184 yards.

Rates

A rate is a special ratio that compares quantities that have different units.

For example, we might say that the speed of a car is 60 miles per hour. The speed gives the rate at which the car moves. The rate can be expressed as $\frac{60\:\textrm{miles}}{1\: \textrm{hour}}$, where the units are measures of distance and time.

We might pay a grocer $13.50 for 3 pounds of apples. The rate can be expressed as $\frac{\$13.50}{3\:\textrm{pounds}}$. Dividing the numerator and denominator by 3, this rate simplifies to $\frac{\$4.50}{1\:\textrm{pound}}$. Whenever the denominator of a rate equals 1, it is called a unit rate. The unit rate is $4.50 per pound, in which the units are measures of currency and weight.

Example 1

A train travels at a rate of 65 miles per hour. How many miles will the train travel in 3 hours?

1. Recognize that speed is a rate. It’s a ratio of distance to time.
   $\textrm{speed} = \frac{\textrm{distance}}{\textrm{time}}$
2. Rearrange the formula to solve for distance.
   $\textrm{distance} = \textrm{speed}\:\times\:\textrm{time}$
3. Substitute the given speed and time into the formula.
   $\textrm{distance} = \frac{65\:\textrm{miles}}{1\:\textrm{ hour}}\times\:3\:\textrm{ hours} = \textrm{195 miles}$
4. The train will travel 195 miles in 3 hours.

Example 2

Bella exercises by walking briskly on a treadmill. On average, she walks 1 mile in 15 minutes. What is Bella’s average speed measured in miles per hour? One hour is the same as 60 minutes.

1. Recognize that speed is a rate. It’s a ratio of distance to time.
   $\textrm{speed} = \frac{\textrm{distance}}{\textrm{time}}$
2. Substitute the given distance and time into the formula.
   $\textrm{distance} = \frac{1\:\textrm{mile}}{15\:\textrm{minutes}}$
3. Convert the rate miles per minute to the rate miles per hour.
   $\frac{1\:\textrm{mile}}{15\:\textrm{minutes}}\:\times\:\frac{60\:\textrm{minutes}}{1\:\textrm{hour}} = 4\:\frac{\textrm{miles}}{\textrm{hour}}$
4. Bella walks at an average treadmill speed of 4 miles per hour.

Example 3

Newton works in a quarry selling gravel. He makes $225 for every 5 tons of gravel he sells. Baxter is one of Newton’s new clients. He has $6,500 to spend on gravel. How many tons of gravel can Baxter buy? Assume that Newton sells only by the ton.

1. Write a proportion that involves two equivalent rates. Call them rate A and rate B. Each rate is a measure of weight to currency.
   $\frac{\textrm{weight A}}{\textrm{currency A}} = \frac{\textrm{weight B}}{\textrm{currency B}}$
2. Substitute the given weights and currency into the proportion.
   $\frac{5\:\textrm{tons}}{\$255} = \frac{\textrm{weight B}}{\$6500}$
3. Cross multiply to solve the proportion. In this case, we’re solving for weight B.
   $\textrm{weight B} = \frac{\$6500\times\:5\:\textrm{tons}}{\$225} \approx 28.89\:\textrm{tons}$
4. Because Newton sells only by the ton, Baxter can buy up to 28 tons of gravel. 

Scale Factors

Scientists often need to create two- or three-dimensional models of real-world objects on working surfaces, such as computer screens or paper. Some objects, such as the planets in our solar system, are too large to fit onto a computer screen. Other objects, such as atoms and molecules, would be too small to see if they were drawn at their actual size. To replicate these objects, scientists use a scale factor to reduce or enlarge each real-world object.

For an enlargement, the scale factor is greater than one. For a reduction, the scale factor is less than one. This table shows several examples of different scale factors:

Example 1: Reduction

An environmental scientist wants to draw a solar cell panel on her computer screen. The panel is 64 inches long and 40 inches wide. If she uses a scale factor of 0.125, what will the length and width of the panel be on her computer screen?

1. Multiply each of the real-world dimensions separately by the scale factor.
   64 in. × 0.125 = 8 in.
   40 in. × 0.125 = 5 in.
2. Check to be sure that the scaled object is smaller than the real-world object.
3. The dimensions of the panel on the computer screen will be 8 inches by 5 inches.

Example 2: Enlargement

A biologist is sketching a ladybug on a piece of paper. The diameter of the ladybug is 3 millimeters. If the biologist uses a scale factor of 25, what will the diameter of the ladybug be on the piece of paper? Express your answer in centimeters. One centimeter is the same as 10 millimeters.

1. Multiply the dimension by the scale factor.
   3 mm × 25 = 75 mm
2. Check to be sure the scaled object is larger than the real-world object.
3. Express the answer in centimeters, as the question states.
   $75\:\textrm{mm}\:\times\:\frac{1\:\textrm{cm}}{10\:\textrm{mm}} = 7.5 \:\textrm{cm}$
   
4. The diameter of the ladybug on the piece of paper will be 7.5 centimeters.

Example 3: Finding a Scale Factor

An archaeologist has excavated a ditch that passes through the Arabian Desert. She’s collected artifacts from the ditch and is marking the locations on her computer screen. In the real world, the ditch is 15 meters long. On her computer screen, the ditch is only 12 centimeters long. What scale factor did the archaeologist use to draw the ditch? One centimeter is the same as 100 meters.

1. Recognize whether the scale factor represents an enlargement or a reduction. In this case, the scale factor models a reduction, so it will be less than one.
2. Convert the dimensions of the real-world object and the scaled object so that they are the same.
   $15\:\textrm{m}\:\times\:\frac{100\:\textrm{cm}}{1\:\textrm{m}} = 1500 \:\textrm{cm}$
   
3. To achieve a scale factor less than one, divide the dimension of the scaled object by the dimension of the real-world object.
   12 cm ÷ 1500 cm = 0.008
4. The scale factor is 0.008, or $\frac{1}{125}$. This means that one unit of length on the computer screen represents 125 units of length in the real world.

Properties of Circles

A circle is a set of points located the same distance from a central point. The radius (r) of a circle is the distance from the center to any point on the circle’s edge. The diameter (d) of a circle is two times the circle’s radius (d = 2r). The diameter passes through the center.
The circumference is the distance around a circle. The circumference C of a circle is 2πr or πd.
The area A of a circle is πr² where π is approximately 3.14.

Example 1: Radius and Diameter

The radius of a circular pond is 16 meters. What is the diameter of the pond?

1. Multiply the radius by 2 to find the diameter.
   d = 2r = 2 × 16 = 32
2. The diameter of the pond is 32 meters.

Example 2: Circumference

Liam is collecting samples of algae from a circular pond. The pond has a radius of 16 meters. If Liam walks one time around the edge of the pond, how far will he have walked? Round the answer to the nearest meter. Use 3.14 for the value of π.

1. The problem is about circumference. Because the radius is given, use the formula C = 2πr.
2. Substitute r = 16 in the formula and solve.

C	=	2π(16)
	=	32π
	=	32(3.14)
	=	100.48

3. The circumference of the pond rounded to the nearest meter is 100 meters.

Example 3: Area of a Circle

Liam is graphing the rate of evaporation from the surface of a circular pond. He needs to know the area of the surface. If the pond has a radius of 16 meters, what is the area of the pond’s surface? Round the area to the nearest whole number.

1. Because the pond is circular, use the formula A = πr².
2. Substitute r = 16 in the formula and solve.

A	=	π(16)2
	=	π(16 × 16)
	=	π(256)
	=	803.84

3. The area of the pond rounded to the nearest whole number is 804 square meters (m²).

Volumes

Given radius r and height h, the volume, V, of a cylinder can be found using this formula:

V = π × r² × h, where π is about 3.14.

If diameter d is given, the radius is $\frac{1}{2}d$.

Given length l, width w, and height h, the volume, V, of a rectangular prism can be found using this formula:

V = l × w × h.

Example 1: Volume of a Cylinder

Find the volume of a cylinder with a height of 5 centimeters and a diameter of 4 centimeters.

1. Find the radius, r:
   $r=\frac{1}{2}\:d=\frac{1}{2} \times 4=2$ cm
2. Substitute the values for r and h into the formula and solve:
   V = π × r² × h
      = 3.14 × 2² × 5
      = 3.14 × 4 × 5
      = 62.8 cm³

The volume of the cylinder is 62.8 cubic centimeters.

Example 2: Volume of a Rectangular Prism

Find the volume of a rectangular prism that is 20 centimeters long, 4 centimeters wide, and 5 centimeters high.

Substitute the values for l, w, and h into the formula and solve:

V = l × w × h
   = 20 × 4 × 5
   = 400 cm³

The volume of the rectangular prism is 400 cubic centimeters.

Scientific Notation

Scientific notation is a way to write numbers using powers of 10. This notation is used to represent very large values, such as the distance from Earth to Mars (1.4 × 10⁸ kilometers). It is also used to represent very small values, such as the weight of one type of bacteria (9.5 × 10^-13 grams).

This table shows some powers of 10 and what they mean.

Exponent Form	Expanded Form	Value	Place Value
10-4	0.1 × 0.1 × 0.1 × 0.1	0.0001	ten thousandth
10-3	0.1 × 0.1 × 0.1	0.001	thousandth
10-2	0.1 × 0.1	0.01	hundredth
10-1	0.1	0.1	tenth
100	1	1	one
101	10	10	ten
102	10 × 10	100	hundred
103	10 × 10 × 10	1,000	thousand
104	10 × 10 × 10 × 10	10,000	ten thousand

Scientific notation is written as the product of a number between 1 and 10 and a power of ten. In the number 1.4 × 10⁸, the number between 1 and 10 is 1.4 and the power of ten is 10⁸. In the number 9.5 × 10^-13, the number between 1 and 10 is 9.5 and the power of ten is 10^-13.

To write 1.4 × 10⁸ in standard form, recognize that it’s a very large number. We need to move the decimal point to the right eight places, the same number that appears in the exponent:

In standard form, 1.4 × 10⁸ is 140,000,000.

To write 9.5 × 10^-13 in standard form, recognize that it’s a very small number. We need to move the decimal point to the left thirteen places, the same number that appears in the exponent:

In standard form, 9.5 × 10^-13 is 0.00000000000095.

Solving Equations

Solving equations might require one step or many steps. To solve for an unknown variable, get the variable on one side of the equation by itself. Perform opposite operations on both sides of the equation, one step at a time, until the variable is alone.

Example 1: One-Step Equation

Solve for x.

x + 21 = 6

1. Perform an opposite operation on both sides of the equation. Here, the only operation is addition: adding 21. The opposite of adding 21 is subtracting 21. To solve, subtract 21 from both sides.
   x + 21 – 21 = 6 – 21
2. Simplify.
   x + 0 = 6 – 21
3. Simplify again.
   x = -15
4. The solution to x + 21 = 6 is x = -15.

Example 2: Two-Step Equation
Solve for n.
34 = 3n – 5

1. Perform an opposite operation on both sides of the equation. Here, the first operation to look at is subtraction: subtracting 5. The opposite of subtracting 5 is adding 5. Add 5 to both sides to begin.
   34 + 5 = 3n – 5 + 5
2. Simplify.
   34 + 5 = 3n + 0
3. Simplify again.
   39 = 3n
4. Next, perform an opposite operation on both sides of the equation. Here, the operation is multiplication: multiplying 3. The opposite of multiplying 3 is dividing 3. Divide both sides by 3.
   39 ÷ 3 = 3n ÷ 3
5. Simplify.
   39 ÷ 3 = n
6. Simplify one last time.
   13 = n
7. The solution
   34 = 3n – 5 is n = 13.

Fractions, Decimals, and Percents

Fractions, decimals, and percents are all expressions that involve the ratio of two numbers.

A fraction shows a part to whole (or part to part) relationship. A fraction is expressed as a ratio of two numbers. A fraction can represent a number smaller than a whole, such as $\frac{1}{3}$ (one third). It can also represent a number larger than a whole, such as $\frac{9}{5}$ (nine fifths).

Decimal numbers are a way to represent fractions using a base-10 system. This table represents some common fractions and their base-10 equivalent.

Fraction	Decimal	Word Form
$\frac{1}{10}$	0.1	one tenth
$\frac{1}{100}$	0.01	one hundredth
$\frac{1}{1,000}$	0.001	one thousandth
$\frac{1}{10,000}$	0.0001	one ten thousandth

Example 1: Converting Fractions to Decimals

To convert a fraction to a decimal, divide the numerator by the denominator. To find the decimal form of $\frac{5}{8}$, divide 5 by 8 using long division or a calculator.

The result is 0.625, or six hundred twenty-five thousandths.

Example 2: Converting Decimals to Fractions

To convert a decimal to a fraction, express the decimal in word form. Then write the related fraction and simplify. For example, consider the decimal number 0.8. Recognize that the digit 8 is in the tenths place. So, the number is eight tenths. Expressed as a fraction, the number is $\frac{8}{10}$. Written in simplest form, $\frac{8}{10} = \frac{4}{5}$. The fraction equivalent of 0.8 is $\frac{4}{5}$.

The word percent means “per one hundred.” It represents a ratio of some number to 100. Percents are written using the percent sign, %. For instance, the number 55% is read as "55 percent" or "55 per one hundred." "Fifty-five percent" can be written as $\frac{55}{100}, 0.55,$ or, $\frac{11}{20}$.

Example 3: Converting Percents to Fractions and Decimals

Convert 60% to a fraction and to a decimal.

1. Recognize that percent means "per one hundred." So, 60% is 60 per 100 or $\frac{60}{100}$. In simplest form, $\frac{60}{100}$ is written as $\frac{3}{5}$.
2. Because percent means "per one hundred," simply move the decimal point two places to the left to express the number as a decimal. In other words, divide 60 by 100 to arrive at 0.60.
   

Percent Change

Percent change can refer to either a percent increase or percent decrease in a certain quantity. There are several real-world concepts, such as finance and population growth, that use percent increase and decrease.

We define percent change as the ratio of the difference between a current value and the original value to the original value. This value is then multiplied by 100 so that it can be expressed as a percent.

$\textrm{% change} = \frac{\textrm{final value}\:-\:\textrm{inital value}}{\textrm{inital value}} \cdot 100$

Percent Increase

If the final value of a quantity is greater than its initial value, the quantity has experienced a percent increase. Percent increase is always positive.

Example:

Pam doesn’t feel well and is tracking her fever. When she checked it an hour ago, her temperature was 101°C. But after checking it again, the thermometer now reads 102°C.

Calculate the percent increase in Pam’s temperature.

1. Use the formula for percent change, and plug in the respective values.
   $\%\:\textrm{change} = \frac{\textrm{final value}\:-\:\textrm{inital value}}{\textrm{inital value}} \cdot 100$
   
   $\%\:\textrm{change} = \frac{\textrm{102}\:-\:\textrm{101}}{\textrm{101}} \cdot 100$
2. Simplify
   $\%\:\textrm{change} = \frac{1}{\textrm{101}} \cdot 100$
3.  Divide
   $\%\:\textrm{change} = 0.0099 \cdot 100$
4. Multiply
   $\%\:\textrm{change} = 0.99\%$
5. Because the result is positive, we have verified that Pam's temperature increased by 0.99%.

Example:

A company discovered that its net worth increased from $1,000,000 to $2,500,000 within a year. Find the percent increase in the company’s net worth.

1. Use the formula for percent change, and plug in the respective values.
   $\%\:\textrm{change} = \frac{\textrm{final value}\:-\:\textrm{inital value}}{\textrm{inital value}} \cdot 100$
   
   $\%\:\textrm{change} = \frac{\textrm{2,500,000}\:-\:\textrm{1,000,000}}{\textrm{1,000,000}} \cdot 100$
2. Simplify
   $\%\:\textrm{change} = \frac{1,500,000}{1,000,000} \cdot 100$
   
3. Divide
   $\%\:\textrm{change} = 1.5\:\cdot\:100$
4. Multiply
   $\%\:\textrm{change} = 150\%$
5. Because the result is positive, we have verified that the company's net worth increased by 150%.

Percent Decrease

If the final value of a quantity is less than its initial value, the quantity has experienced a percent decrease. Percent decrease is always negative.

$\%\:\textrm{change} = \frac{\textrm{final value}\:-\:\textrm{inital value}}{\textrm{inital value}} \cdot 100$

Example:

A species of bacteria had a population size of 1 million yesterday. However, today the population size had decreased to half its initial value. Find the percent decrease in the bacteria population.

1. Use the formula for percent change and plug in the respective values.
   $\%\:\textrm{change} = \frac{\textrm{final value}\:-\:\textrm{inital value}}{\textrm{inital value}} \cdot 100$
   
   $\%\:\textrm{change} = \frac{\textrm{500,000}\:-\:\textrm{1,000,000}}{\textrm{1,000,000}} \cdot 100$
2. Simplify
   $\%\:\textrm{change} = \frac{\textrm{500,000}}{\textrm{1,000,000}} \cdot 100$
3. Divide
   $\%\:\textrm{change} = \textrm{-}0.5\:\cdot\:100$
4. Multiply
   $\%\:\textrm{change} = \textrm{-}50\%$
5. Because the result is negative, we have verified that the bacteria population decreased by 50%.

Positive and Negative Numbers in Addition and Subtraction

Here are some rules for adding positive and negative numbers:

• When adding a positive number and a negative number, use the sign of the larger number and subtract the smaller number from the larger number:
  3 + (-5) = -(5 − 3) = -2
• When adding two negative numbers, use the negative sign and add the two numbers:
  -5 + (-3) = -(5 + 3) = -8
• When subtracting a positive number from a negative number, use the negative sign and add the two numbers:
  -3 − (5) = -(3 + 5) = -8
• When subtracting a negative number from a positive number, use the positive sign and add the two numbers:
  3 − (-5) = +(3 + 5) = 8
• When subtracting a negative number from a negative number, the second number becomes positive. Then follow the rule for adding a positive number and a negative number:
  -3 − (-5) = -3 + 5 = 5 − 3 = 2

Example

Jason recorded the maximum and minimum temperatures for an Arctic region for one month. The maximum temperature was -7°F, and the minimum temperature was -21°F. By how many degrees does the maximum temperature exceed the minimum temperature?

1. Write an expression or an equation to represent the problem:
   maximum temperature − minimum temperature = ?
2. Plug in the numbers from the problem and simplify:
   -7 − (-21) = ?
                    = -7 + 21
                    = 14

The maximum temperature exceeds the minimum temperature by 14°F.

Exponents

In science and math, a number is sometimes multiplied by itself. For example, 4 × 4 = 16. Exponents are another way to show this relationship: 4² = 4 × 4 = 16. In this case, the base of the exponent is 4 and the power of the exponent is 2. The power represents the number of times the base appears as a factor.

Example 1
Evaluate the expression 3⁴.

1. Recognize that the base is 3 and the power is 4.
2. The base appears four times in the multiplication expression.
   3 × 3 × 3 × 3
3. Multiply to find the solution.
   3 × 3 × 3 × 3 = 81

Example 2
Evaluate the expression 5³.

1. Recognize that the base is 5 and the power is 3.
2. The base appears three times in the multiplication expression.
   5 × 5 × 5
3. Multiply to find the solution.
   5 × 5 × 5 = 125

Properties of Exponents

Properties of exponents can be used to solve problems involving exponents. The properties are particularly useful when working with numbers written in scientific notation. For the properties listed below, assume that x and y are real numbers and m and n are integers.

Product of Powers
When two exponents with the same base are multiplied, the powers of the exponents are added.
x^m • xⁿ = x^(m+n)

Example 1

33 • 34	=	3(3+4)
9 • 81	=	37
2,187	=	2,187 ü

Example 2

This example involving scientific notation uses the associative property along with the product of powers.

$(5.4\:\times\:10^3)(3.2\:\times\:10^6)$	$=$	$(5.4\:\times\:3.2)\:\times\:10^{(3\:+\:6)}$
	$=$	$(17.28)\:\times\:10^9$
	$=$	$1.728\:\times\:10^{10}$

Quotient of Powers

When an exponent is divided by an exponent with the same base, the powers are subtracted.

$\frac{x^m}{x^n} = x^{(m\:-\:n)}$, $x \neq 0$

Example 1

$\frac{2^5}{2^3}$	$=$	$2^{(5\:-\:3)}$
	$=$	$2^2$
	$=$	$4$

Example 2

This example involving scientific notation uses the associative property along with the quotient of powers.

$\frac{1.6\:\times\:10^4}{8.1\:\times\:10^{\textrm{-}3}}$	$=$	$\left ( \frac{1.6}{8.1} \right )\:\times\:10^{(4\:+\:3)}$
	$\approx$	$0.2\:\times\:10^7$
	$\approx$	$2\:\times\:10^6$

Here are some other properties of exponents you might find useful:

Power of a Power
$(x^m)^n = (x)^{mn}$

Power of a Product
$(xy)^m = x^m \cdot y^m$

Negative Exponent
$x^{\textrm{-}m} = \frac{1}{x^m}$

Zero Exponent
$x^0 = 1$, $x \neq 0$

Power of a Quotient
$\frac{x^m}{y^m} = \left ( \frac{x}{y} \right )^m$, $y \neq 0$

Squares and Square Roots

The square of a number is the number times itself. Squares can be written in exponential form. For example, squaring the number 4 gives 4² = 4 × 4 = 16. In word form, four squared equals sixteen.

Taking the square root of a number is the opposite operation to squaring the number. Think of the square root as the number used to make the square. In the example above, the square root of 16 equals 4. The solution is written this way: $\sqrt{16} = 4$. Most calculators have a square root function to find the square root of a number.

Example 1
Evaluate the expression $\sqrt{81}$.

1. Think of the square root as the number to square to make 81.
2. We know that 9 × 9 = 81.
3. So, the square root of 81 is 9.

Example 2
Evaluate the expression $\sqrt{144}$.

1. Think of the square root as the number to square to make 144.
2. We know that 12 × 12 = 144.
3. So, the square root of 144 is 12.

Converting Temperature Units

Temperature is commonly expressed in units of Fahrenheit and Celsius. But temperature can also be expressed in units of Kelvin and Rankine. Here are formulas and examples for converting from one temperature to another.

Example 1: Fahrenheit to Celsius

$\textrm{T}_{(^{\circ}\textrm{C})} = [\textrm{T}_{(^{\circ}\textrm{F})} - 32] \times \frac{5}{9}$

Convert 59°F to degrees Celsius using the formula.

1. Substitute $59$ for $\textrm{T}_{(^{\circ}\textrm{F})}$ and solve.
   

$\textrm{T}_{(^{\circ}\textrm{C})}$	$=$	$[59 - 32] \times \frac{5}{9}$
	$=$	$27 \times \frac{5}{9}$
	$=$	$\frac{135}{9}$
	$=$	$15$

2. A temperature of 59°F is the same as 15°C. 

Example 2: Celsius to Fahrenheit

T_(°F) = [1.8 × T_(°C)] + 32

Convert 28°C to degrees Fahrenheit using the formula.

1. Substitute 28 for T_(°C) and solve.

T(°F)	=	[1.8 × 28] + 32
	=	50.4 + 32
	=	82.4

2. A temperature of 28°C is the same as 82.4°F.

Example 3: Celsius to Kelvin

T_(K) = T_(°C) + 273.15

Convert 10°C to kelvins using the formula.

1. Substitute 10 for T_(°C) and solve.

T(°K)	=	10 + 273
	=	283

2. A temperature of 10°C is the same as 283 K.

Example 4: Fahrenheit to Rankine

T_(°R) = T_(°F) + 460

Convert 71°F to degrees Rankine using the formula.

1. Substitute 71 for T_(°F) and solve.

T(°R)	=	71 + 460
	=	531

2. A temperature of 71°F is the same as 531°R.

Probability

Probability is the likelihood that an event will take place. Probability can be expressed as a fraction, a decimal, or a percent. The general formula for probability is the number of successful outcomes divided by the total number of outcomes:

$\textrm{probability} (p) = \frac{\textrm{number of successful outcomes}}{\textrm{total number of outcomes}}$

The formula gives a numerical value for probability. This table assigns certain words to the likelihood that an event will occur.

Probability as a Decimal	Probability as a Fraction	Probability as a Percent	Description
$0.0$	$0$	$0\%$	impossible
$0.0 < p < 0.25$	$0 < p < \frac{1}{4}$	$0\% < p < 25\%$	very unlikely
$0.25 \leq p < 0.5$	$\frac{1}{4} \leq p < \frac{1}{2}$	$25\% \leq p < 50\%$	unlikely
$0.5$	$\frac{1}{2}$	$50\%$	neither likely nor unlikely (even)
$0.5 < p \leq 0.75$	$\frac{1}{2} < p \leq \frac{3}{4}$	$50\% < p \leq 75\%$	likely
$0.75 < p < 1.0$	$\frac{3}{4} < p < 1$	$75\% < p < 100\%$	very likely
$1.0$	$1$	$100\%$	certain

Example 1

Joaquin places 12 marbles inside a bag: 3 red, 5 blue, and 4 yellow. He reaches in the bag and pulls out a marble at random. What is the probability that Joaquin removed a yellow marble? How would you describe the probability of pulling a yellow marble?

1. Use the formula, $\textrm{probability}(p) = \frac{\textrm{number of successful outcomes}}{\textrm{total number of outcomes}}$.
2. A successful outcome is removing a yellow marble. The number of successful outcomes is 4.
3. The total number of outcomes is the number of marbles in all. The total number of outcomes is 12.
4. Substitute and solve.
   $\textrm{probability} (p) = \frac{4}{12}$
5. Simplify.
   $p = \frac{1}{3}$
6. The probability that Joaquin chooses a yellow marble is $\frac{1}{3}$ or about 33%. According to the table, the chance of pulling a yellow marble at random is unlikely.

Example 2

Joaquin places 20 marbles inside a bag: 2 red, 8 blue, 4 yellow, and 6 green. He reaches in the bag and pulls out a marble at random. What is the probability that Joaquin removed a blue marble or a green marble? How would you describe the probability of pulling either a blue or green?

1. Use the formula, $\textrm{probability}(p) = \frac{\textrm{number of successful outcomes}}{\textrm{total number of outcomes}}$.
2. A successful outcome is removing a blue or a green marble. The number of successful outcomes is 8 + 6 = 14. 
3. The total number of outcomes is the number of marbles in all. The total number of outcomes is 20.
4. Substitute and solve.
   $\textrm{probability} (p) = \frac{14}{20}$
5. Simplify.
   $p = \frac{7}{10}$
6. The probability that Joaquin chooses a blue or a green marble is $\frac{7}{10}$ or 70%. According to the table, the chance of pulling either a blue or a green is likely.

Averaging Data

An average for a data set is the ratio of the sum of the values of all observations to the total number of observations. It’s also called the arithmetic mean. Based on the average, certain conclusions can be made about the data. This is the general formula for averaging data:

$\textrm{average} = \frac{\textrm{sum of values of all observations}}{\textrm{total number of observations}}$

Example 1

Jason measured the heights of basketball players in his class. The table shows the data he collected (in centimeters).

166	167	176	168	162
170	164	162	160	165

What is the average height of basketball players in Jason’s class?

1. Find the sum of the values of all observations:
   166 + 167 + 176 + 168 + 162 + 170 + 164 + 162 + 160 + 165 = 1,660 cm
2. Count the number of observations.
   In this case, the number of observations is 10.
3. Use the formula: $\textrm{average} = \frac{\textrm{sum of values of all observations}}{\textrm{total number of observations}}$.
4. Substitute the corresponding values in the formula and simplify:

$\textrm{average}$	$=$	$\frac{1,\textrm{660}\:\textrm{cm}}{10}$
	$=$	$166$ cm

5. The average height of basketball players in Jason’s class is 166 cm.

Example 2

Ron works at a deli. He wants to find the average amount spent per table in an hour.
The table shows the bill amounts (in dollars) Ron recorded that hour.

17.50	18.00	17.00	16.50
15.50	19.50	16.00	14.00

 What is the average amount spent per table during that hour?

1. Find the sum of the values of all observations:
   17.5 + 18 + 17 + 16.5 + 15.5 + 19.5 + 16 + 14 = $134
   
2. Count the number of observations.
   In this case, the number of observations is 8.
3. Use the formula: $\textrm{average} = \frac{\textrm{sum of values of all observations}}{\textrm{total number of observations}}$.
4. Substitute the corresponding values in the formula and simplify.
   

   $\textrm{average}$	$=$	$\frac{{\$}134}{8}$
   	$=$	${\$}16.75$

The average amount spent per table that hour was $16.75.