Applications

Section 2.C Applications

Objectives

Solve number problems by creating and solving an equation.
Solve problems using scientific notation and exponent properties.
Solve formulas for a given variable.
Solve practical percent problems.
Solve weighted average problems.
Interpret plus or minus notation for intervals.

The previous sections focused on core quantitative and algebric skills. In this section, we use these these skills to solve problems. The key to using algebra to determine a solution in a particular scenario is the ability to translate the essence of the problem into a mathematical relationship. Once the situation is represented as a mathematical equation or inequality, we use algebra to determine a solution for the mathematical expression which we use to find the solution to the problem. In order to translate a problem into symbols, you must understand what the situation is in its essential components. You might ask yourself: What is the relationship between the quantities? Are quantities being added? Subtracted?

Subsection 2.C.1 Linear Equation Applications

These first examples will focus on basic number problems to build your skill at "translating" words into math symbols.

A few important phrases can give us clues for how to set up a problem.

A number (or an unknown, a value, etc) usually becomes our variable
Is (or other forms of is: was, will be, are, etc) often represents equals ( $=$):
- "$x$ is $5$" becomes $x=5$
More than often represents addition:
- "Three more than a number" becomes $x+3$
Less than often represents subtraction:
- "Four less than a number" becomes $x-4$
Times often represents multiplicaton:
- "Six times a number" becomes $6x$
Of often represents multiplicaton in percent problems:
- "Fifty percent of a number" becomes $0.5x$

Example 2.C.1. Number Problem.

If $28$ more than five times a certain number is $233\text{,}$ what is the number?

\begin{align*} \text{Let $x$ equal }``\text{the number"} \amp\amp\amp \text{The variable stands in for the unknown quantity.}\\ 5x\amp\amp\amp\text{Translation of }``\text{five times a certain number"}\\ 5x+28\amp\amp\amp \text{We need }``28\text{ more than $5x$"}\\ 5x+28\amp =233 \amp\amp \text{The quantity }``\text{is" $233$} \end{align*}

Note that the key to solving the problem is to create an equation that describes the situation: $5x+28=233.$ Once the equation is designed, the solution is just a few algebraic steps away.

\begin{align*} 5x+28\amp =233 \amp\amp \text{Now, we solve the equation.}\\ \underline{-28}\amp\underline{\phantom{1}-28} \amp\amp \text{Subtract 28 from both sides}\\ 5x\amp =205\amp\amp \text{The variable $x$ is multiplied by $5$}\\ \overline{\,5\,}\amp\phantom{123}\overline{\,5\,}\amp\amp \text{Divide both sides by $5$}\\ x\amp =41 \amp\amp \text{Our Solution: The number is $41$.}\checkmark \end{align*}

Although the next few examples are fairly artificial, working similar problems will help develop your skills in translating problems into mathematical expressions. Boiling down a situation into a mathematical representation allows us to use algebra tools to determine a solution to the problem.

Example 2.C.2. Translating Words into a Mathematical Expression.

On Saturday, Brooks bought groceries and filled her gas tank. She spent $\$53$ on groceries, which was $\$18$ less than twice what she spent on gasoline. How much did Brooks spend on filling her tank with gas?

Since we don't know how much she spent on gasoline, we will use a variable to represent this unknown amount. Sometimes it is helpful to write out the other quantities described in the problem in terms of the variable.

\begin{align*} \text{Let $d =$ dollars spent on gas} \amp\amp\amp \text{Translate the quantities into mathematical expressions}\\ 2d\amp\amp\amp\text{Translation of }``\text{twice what she spent on gasoline"}\\ \text{$2d-18 =$ cost of groceries} \amp\amp\amp \text{The groceries cost $\$18$ }``\text{less than" twice the gas cost}\\ \text{$\$53 =$ cost of groceries} \amp\amp\amp \text{We already know the cost of the groceries}\\ 2d-18\amp = 53 \amp\amp \text{We have two equal expressions for cost of the groceries} \end{align*}

We have an equation that relates the unknown dollars spent on gas to the situation in the problem. Now, solve the equation.

\begin{align*} 2d-18\amp = 53 \amp\amp \text{Solve the equation for$x$}\\ 2d \amp = 71 \amp\amp \text{Added $18$ to each side}\\ d \amp = 35.5 \amp\amp \text{Divided both sides by 2}\\ \text{She spent $\$35.50$ on gas.} \amp\amp\amp \text{Our Solution}\checkmark \end{align*}

WeBWorK: Entering Currency.

Enter currency with a dollar sign $\$$ in front. If your answer is not in whole dollars, type two decimal places: $35.50 and not 35.5.

If your calculation results in more than two decimal places, round the answer to the nearest penny (hundredths): For $7.386\text{,}$ round to $7.39.

Example 2.C.3. A Problem Involving Scientific Notation.

The number of atoms in $12$ grams of carbon is $6.022\times 10^{23}.$ Determine the mass of one atom of carbon.

\begin{align*} \text{ $6.022\times 10^{23}$ atoms yields $12$ g} \amp\amp\amp \text{Understand what information we are given}\\ \text{Let $x=$ the mass of one atom of carbon} \amp\amp\amp \text{Convert the given information into an equation}\\ (6.022\times 10^{23})x =12 \amp\amp\amp \text{The number of atoms times the mass of each } \\ \text{ } \amp\amp\amp \text{equals the total mass of $12$ grams} \end{align*}

Now we solve the equation.

\begin{align*} (6.022\times 10^{23})x =12 \amp\amp\amp \text{Solve for $x$}\\ (6.022\times 10^{23})x =1.2\times 10^1 \amp\amp\amp \text{Convert $12$ to scientific notation.}\\ x=\dfrac{1.2\times 10^1}{6.022\times 10^{23}} \amp\amp\amp \text{Divide both sides by } 6.022\times 10^{23}\\ \dfrac{1.2}{6.022}=0.19927... \amp\amp\amp \text{Divide the numbers}\\ \dfrac{10^{1}}{10^{23}}=10^{-22} \amp\amp\amp \text{Use quotient rule on the powers}\\ x = 0.19927\times 10^{-22} \amp\amp\amp \text{Write the result in correct scientific notation}\\ x=1.9927 \times 10^{-23} \amp\amp\amp \text{Our Solution:} \end{align*}

The the mass of one atom of carbon is $1.9927 \times 10^{-23}$ gram.$\checkmark$

Subsection 2.C.2 Solving for Variables in Formulas

Solving formulas is much like solving general linear equations. The only difference is we will have several variables in the problem and we will be attempting to solve for one specific variable. For example, we may have a formula such as $A=\pi r^2 + \pi r s$ (formula for surface area of a right circular cone) and we may be interested in solving for the variable $s\text{.}$ This means we want to isolate the $s$ so the equation has $s$ on one side, and everything else on the other. So a solution might look like $s = \dfrac{A-\pi r^2}{\pi r}\text{.}$ This second equation gives the same information as the first which means they are algebraically equivalent. However, one is solved for the area $A\text{,}$ while the other is solved for $s$ (slant height of the cone). In this section we will discuss how we can move from the first equation to the second.

When solving formulas for a variable we need to focus on the one variable we are trying to solve for, all the others are treated just like numbers. This is shown in the following example. Two parallel problems are shown, the first is a one step equation with numbers, the second is a formula that we are solving for $x\text{:}$

Example 2.C.4. Compare Solving for a Variable in an Equation and a Formula.

Solve for $x\text{.}$

\begin{align*} 3x\amp =12\amp\amp\amp wx\amp =z\amp\amp\text{In both problems, $x$ is multiplied by something}\\ \overline{\,3\,}\amp\phantom{123}\overline{\,3\,}\amp\amp\amp\overline{\,w\,}\amp\phantom{12}\overline{\,w\,}\amp\amp\text{To isolate $x$ we divide by $3$ or $w$}\\ x\amp =4\amp\amp\amp x\amp =\dfrac{z}{w}\amp\amp\text{Our Solutions}\checkmark \end{align*}

We use the same process to solve $3x = 12$ for $x$ as we use to solve $wx = z$ for $x\text{.}$ Because we are solving for $x$ we treat all the other variables the same way we would treat numbers. Thus, to get rid of the multiplication we divided by $w\text{.}$ This same idea is seen in the following example.

Example 2.C.5. Solve for a Variable in a Formula.

Solve for $n\text{.}$

\begin{align*} m+n\amp =p\amp\amp\text{Solving for $n$, treat all other variables like numbers}\\ \underline{-m}\amp\underline{\phantom{123}-m}\amp\amp\text{Subtract $m$ from both sides}\\ n\amp =p-m\amp\amp\text{Our Solution}\checkmark \end{align*}

The next example takes more than one step:

Example 2.C.6. Solve for a Variable in a Formula--Two steps.

Solve for $m\text{.}$

\begin{align*} y\amp =mx+b\amp\amp\text{Solving for $m$, focus on addition first}\\ \underline{-b}\amp\phantom{1234567}\underline{-b}\amp\amp\text{Subtract $b$ from both sides}\\ y-b\amp =mx\amp\amp\text{The $m$ is multiplied by $x$, therfore}\\ \overline{\,\,\,\,x\,\,\,\,}\amp\phantom{1234}\overline{\,x\,}\amp\amp\text{Divide both sides by $x$}\\ \dfrac{y-b}{x}\amp =m\amp\amp\text{Our Solution}\checkmark \end{align*}

It is important to note that we know we are done with the problem when the variable we are solving for is isolated or alone on one side of the equation and it does not appear anywhere on the other side of the equation.

The next example is also a two-step equation (it is the problem we started with at the beginning of the section):

Example 2.C.7. Formula for Surface Area of a Right Circular Cone.

Solve for $s\text{.}$

\begin{align*} A\amp =\pi r^2 + \pi r s\amp\amp\text{Solving for $s$, focus on what is added to the term with $s$}\\ \underline{-\pi r^2}\amp\phantom{1}\underline{-\pi r^2}\amp\amp\text{Subtract $\pi r^2$ from both sides}\\ A-\pi r^2\amp =\pi r s\amp\amp\text{The $s$ is multiplied by $\pi r$, so}\\ \overline{\phantom{123}\pi r\phantom{123}}\amp\phantom{123}\overline{\,\pi r\,}\amp\amp\text{Divide both sides by $\pi r$}\\ \dfrac{A-\pi r^2}{\pi r}\amp =s\amp\amp\text{Our Solution}\checkmark \end{align*}

Notice, we cannot reduce the $\pi r$ in the numerator and denominator because of the subtraction in the numerator.

Formulas may involve quite a few variables and sometimes quantities represented in a formula might be related. For example, you have two times to keep track of or three lengths. In these cases, using the same letter for each variable makes it easier to remember what the variable represent. To keep track of these "same named" quantities, we use subscripts on the variables. A subscript is a number or letter which is attached to the variable at the lower left of the variable like $C_t$ or $C_d.$ In this case, $C_t$ might represent the cost of a tablet computer and $C_d$ might be the cost of a desktop computer. Both might occur in a formula.

WeBWorK: Entering subscripted variables.

For $C_t$ type: Ct.

Example 2.C.8. A More Involved Equation.

Solve the equation for $x_1.$

We will need several steps to determine the solution for this problem. Focus on simple steps to bring the $x_1'$s together.

\begin{align*} L + x_1\amp=\dfrac{x_2 - x_1}{k}\amp\amp\text{Solving for $x_1$, focus on getting $x_1$ "loose" on the RHS}\\ k(L + x_1)\amp=k\left(\dfrac{x_2 - x_1}{k}\right)\amp\amp\text{Multiply both sides by $k$}\\ kL + kx_1\amp=x_2 - x_1\amp\amp\text{Distribute the $k$ on the LHS, simplify the RHS}\\ \underline{-kL}\phantom{12345}\amp\phantom{12346}\underline{-kL}\amp\amp\text{Subtract $kL$ from both sides}\\ kx_1 \amp=x_2 - x_1 - kL\amp\amp\text{Simplify}\\ \underline{+x_1}\amp\phantom{123456}\underline{+x_1}\amp\amp\text{Add $x_1$ to both sides}\\ kx_1 + x_1 \amp=x_2 - kL\amp\amp\text{Simplify:}\\ x_1(k + 1) \amp=x_2 - kL\amp\amp\text{Factor out the $x_1$ on the LHS}\\ \overline{\phantom{12}k+1\phantom{12}}\amp\phantom{123}\overline{\phantom{12}k+1\phantom{12}}\amp\amp\text{Divide both sides by $k+1$}\\ x_1\amp =\dfrac{x_2 - kL}{k+1}\amp\amp\text{Our Solution}\checkmark \end{align*}

Example 2.C.9. When the Variable is in the Denominator.

Solve for $z\text{.}$

When working with variables in fractions, you should "clear" the denominators if doing so makes the problem easier to solve.

\begin{align*} \dfrac{2}{z}\amp=\dfrac{5}{v} + \dfrac{x}{z}\amp\amp\text{Solving for $z$}\\ vz\cdot \dfrac{2}{z}\amp=vz\cdot \left(\dfrac{5}{v}+\dfrac{x}{z}\right)\amp\amp\text{Multiply both sides by $vz$ to clear the denominators}\\ 2v\amp= vz\left(\dfrac{5}{v}\right) + vz\left(\dfrac{x}{z}\right)\amp\amp\text{Simplify the LHS, distribute the $vz$ on the RHS}\\ 2v\amp= 5z+xv\amp\amp\text{Simplify the RHS:}\\ \underline{-xv}\amp\phantom{12345}\underline{-xv}\amp\amp\text{Subtract $xv$ from both sides}\\ 2v-xv \amp= 5z\amp\amp\text{Simplify:}\\ \overline{\phantom{123}5\phantom{123}}\amp\phantom{123}\overline{\,5\,}\amp\amp\text{Divide both sides by $5$}\\ \dfrac{2v-xv}{5} \amp= z\amp\amp\text{Rewrite with $z$ on the LHS}\\ z\amp =\dfrac{2v-xv}{5}\amp\amp\text{Our Solution}\checkmark \end{align*}

Subsection 2.C.3 Percent Problems

There are three basic types of percent problems:

Determine a certain percent of a given number.
- For example, find $25\%$ of $640\text{.}$
Determine a percent given two numbers.
- For example, $15$ is what percent of $50\text{?}$
Determine a number from a given percent of another number.
- For example, $37.5\%$ of what number is $57\text{?}$

Let's examine each of the three types of percent problems.

Example 2.C.10. 1. Determine a certain percent of a given number.

What number is $25\%$ of $640\text{?}$

Translate the words into an equation. Let

$x=$ the unknown number

\begin{align*} x\amp =\amp\amp \text{Translate "What number is"}\\ \amp .25\cdot 640\amp\amp \text{Translate: $25\%$ of $640$}\\ x\amp =.25\cdot 640 \amp\amp \text{Equation: $x$ is equal to $25\%$ of $640$}\\ \amp\amp\amp \text{Multiply}\\ x\amp =160 \amp\amp \text{Our Solution}\checkmark \end{align*}

Now we'll address our second item on the list:

Example 2.C.11. 2. Determine a percent given two numbers.

$15$ is what percent of $50\text{?}$

Translate the words into an equation. Let

$x=$ the unknown percent in decimal

\begin{align*} 15\amp = \amp\amp \text{Translation of }``15\text{ is"}\\ x\amp\cdot 50 \amp\amp \text{Translation of }``\text{what percent of $50$"}\\ 15\amp = x\cdot 50\amp\amp\text{Set-up our equation and solve for $x$}\\ \dfrac{15}{50}\amp =\dfrac{x\cdot 50}{50} \amp\amp \text{Divide both sides by $50$}\\ \dfrac{15}{50}\amp =x \amp\amp \text{Simplify: $\dfrac{50}{50}=1$ and $\dfrac{15}{50}=0.3$}\\ 0.3\amp = x \amp\amp \text{Change $0.3$ to a percent.}\\ 30\% \amp\amp\amp \text{Our Solution}\checkmark \end{align*}

Now we'll consider the third item on the list:

Example 2.C.12. 3. Determine a number from a given percent of another number.

$37.5\%$ of what number is $57\text{?}$

Translate the words into an equation. Let

$x=$ the unknown number

Recall of translates to times, and is translates to equals.

\begin{align*} 0.375\cdot x \amp\amp\amp \text{Translate: $37.5\%$ of the number}\\ 0.375\cdot x\amp =57\amp\amp \text{Set up the equation}\\ \dfrac{0.375}{0.375}\cdot x\amp =\dfrac{57}{0.375}\amp\amp \text{Divide both sides by $0.375$ and simplify:}\\ \amp\amp\amp \dfrac{0.375}{0.375}=1 \text{ and } \dfrac{57}{0.375}=152\\ x\amp =152 \amp\amp \text{Our Solution}\checkmark \end{align*}

Subsection 2.C.4 Applications of Percent

Consider the following examples. Recognizing the "type" of percent problem will be the key to translating the problem into an equation.

Example 2.C.13. Mixing a Chemical Solution.

A solution of hydrochloric acid is needed to clean a rusty storage barrel. If $35$ millilitres of a $60$ millilitre solution is hydrochloric acid. What percent of the solution is hydrochloric acid?

First recognize this is the second type of percent problem: 2. Determine a percent given two numbers. Start by translating the words into an equation. Let

$p=$ the unknown percent

The two given numbers are:

$35$ millilitres of hydrochloric acid
$60$ millilitres of solution

\begin{align*} p\cdot 60\amp\amp\amp\text{Translation of }``\text{What percent of the solution"}\\ =35\amp\amp\amp\text{Translation of }``\text{is hydrochloric acid"}\\ p\cdot 60\amp = 35\amp\amp \text{Solve for $p$: Divide both sides by $60$}\\ \overline{60}\amp\phantom{123}\overline{60}\amp\amp \text{Convert the fraction to a decimal}\\ p\amp=\dfrac{35}{60}\approx .583333\amp\amp \text{Convert the decimal to a percentage}\\ p\amp =58.333\%\amp\amp \text{Our Solution}\checkmark \end{align*}

Let us take time to summarize the steps to solve a word problem

When solving a word problem you'll need to

Break the situation down into parts you know how to handle.
Use a variable for the unknown quantity.
Build mathematical expressions from the description in the problem.
Put the expressions together to make an equality which you may solve with algebra operations.

Example 2.C.14. Percent Increase.

When Paul was hired from his internship to a full time position, he was given a $130\%$ salary increase. If his new salary is $\$4,000$ per month, what was his old monthly internship pay?

Let $p=$ the unknown monthly internship pay

\begin{align*} 1.30\amp\amp\amp\text{The decimal equivalent of $130\%$}\\ 1.30\cdot p \amp\amp\amp 130\%\text{ of his old pay (the salary increase)}\\ p + 1.30 p \amp\amp\amp \text{Add the increase to his old pay for his new salary}\\ p + 1.30 p\amp =4000 \amp\amp \text{Set up the equation: both sides represent the new salary}\\ 2.30 p\amp =4000 \amp\amp \text{Combine like terms on the LHS, remember, $p = 1p$}\\ 2.30 p\amp =4000 \amp\amp \text{Divide both sides by $2.30$}\\ p\amp= \dfrac{4000}{2.30} \amp\amp \text{Simplify}\\ p\amp= 1739.1304 \amp\amp \text{Our answer must be in terms of dollars and cents}\\ \text{Paul's internship pay was } \amp \$1,739.13 \amp\amp \text{Our Solution}\checkmark \end{align*}

The basic idea used to solve one problem may be the same in a different setting.

Example 2.C.15. Sales Tax.

Bea orders gardening tools online and $\$532.45$ is charged to her credit card for the order. How much did the garden tools cost if shipping/handling is $8.5\%$ of the purchase? i.e. what was the cost of just the tools?

First recognize this involves the same strategies used in the previous problem. We start by translating the words into an equation. Let

$x=$ the unknown cost of the tools

\begin{align*} 0.085\amp\amp\amp\text{The decimal equivalent of $8.5\%$}\\ 0.085\cdot x \amp\amp\amp 8.5\%\text{ of the tool cost (the shipping/handling charge)}\\ x\amp = 532.45 - 0.085 x\amp\amp \text{Subtract the extra charges from the total to get the tool cost}\\ x + 0.085 x\amp =532.45 \amp\amp \text{Add $0.085x$ to both sides to collect like terms}\\ 1.085 x\amp=532.45 \amp\amp \text{Simplify the RHS by adding $1x$ and $0.085x$}\\ x\amp= \dfrac{532.45}{1.085} \amp\amp\text{Divide both sides by $1.085$}\\ x\amp=490.7373 \amp\amp \text{Round the answer to dollars and cents}\\ \text{The tools alone cost} \amp=\$490.74 \amp\amp \text{Our Solution}\checkmark \end{align*}

Note that the problems in WeBWorK will have different settings, but similar methods for setting up an equation to solve the problem. Also notice that we could have set up the equation for the past example as $x + 0.085 x =532.45$ to start with if since it is equivalent to $x = 532.45 - 0.085 x.$ As long as your equation describes a valid relationship from the problem, you will be able to use it to calculate the solution to the problem.

Subsection 2.C.5 Averages and Weighted Averages

You are all familiar with computing average grades. For example, if your scores on $2$ homework assignments are $89$ and $85\text{,}$ then the average of the two assignments is $\dfrac{89+85}{2}=\dfrac{174}{2}=87\text{.}$ One application involving averages is to determine the minimum value of a third grade based on a desired minimum average. Consider the following example:

Example 2.C.16. Computing Desired Score for a Minimum Average.

Suppose your scores on $2$ homework assignments are $89$ and $85\text{.}$ What minimum score should you achieve on your third assignment if you desire the average of the $3$ assignments to be at least $90\text{?}$

Let

$x$ represent the unknown third score

\begin{align*} \amp\dfrac{89+85+x}{3}\amp\amp\text{Formula for the average of the $3$ scores}\\ \dfrac{89+85+x}{3}\amp \geq 90\amp\amp\text{Compare the average of the $3$ scores to the desired minimum}\\ \dfrac{174+x}{3}\amp \geq 90\amp\amp\text{Simplified addition}\\ 3\cdot\dfrac{174+x}{3}\amp\geq 3\cdot 270\amp\amp\text{Multiply both sides by $3$}\\ 174+x\amp\geq 270\amp\amp\text{Simplify}\\ \underline{-174\phantom{1234}}\amp\underline{\phantom{1}-174}\amp\amp\text{Subtract $174$ from both sides}\\ x\amp \geq 96\amp\amp\text{Our Solution:} \end{align*}

You should earn at least a $96$ on your third assignment if you desire the average of the $3$ assignments to be at least $90\text{.}$$\checkmark$

Let us take a different view of the calculation of the average. We could rearrange the expression in the following manner:

\begin{align*} \dfrac{89+85+96}{3}\amp\amp\amp\text{Rewrite as multiplying by } \frac{1}{3}\\ \frac{1}{3}(89+85+96)\amp\amp\amp\text{Distribute the fraction.}\\ \frac{1}{3}(89)+\frac{1}{3}(85)+\frac{1}{3}(96)\amp\amp\amp\text{Simplify.}\\ 29.6667+28.3333+32\amp = 90\amp\amp\text{ } \end{align*}

This way of calculating the average shows each score has equal value or “weight” in the average. In this case, each score is worth one-third of the calculation.

Frequently, the final grade in a college course is calculuted with a weighted average. Suppose the final course grade in a course was based on attendance, exams and a final, each weighted as follows:

Attendance and Participation has weight $5\%$
Exams have weight $70\%\text{,}$ and the
Final Exam has weight $25\%\text{.}$

Consider the computation of the following example grade:

Example 2.C.17. Computed Weighted Average Grade.

At the end of the semester, a student had the following: Attendance and Participation $=95\text{,}$ Exam Average $=83$ and Final Exam $=72\text{.}$ Compute the overall weighted average for this student.

\begin{align*} 95(.05)+ 83(.70) + 72(.25)\amp\amp\amp\text{Multiply each grade by its corresponding weight}\\ \amp\amp\amp PE\underline{M}DAS:\text{ Multiplication before Addition}\\ 4.75+58.1+18\amp\amp\amp PEMD\underline{A}S:\text{ Add}\\ 75.35\amp\amp\amp\text{Our Solution:} \end{align*}

The student's overall weighted average would be $80.85\%\text{.}$ $\checkmark$

Example 2.C.18. Computing Desired Score for a Minimum Weighted Average.

Suppose you were in a course with four exams. The first three exams are each worth $20\%$ of your grade but that last exam is worth $40\%\text{.}$ Your scores on the first three were $38$ out of $50$ points, $32$ out of $40$ and $44$ out of $55\text{.}$ The last exam is worth $120$ points. What minimum score should you achieve on the last exam if you desire your overall weighted average in the course to be at least $80\text{?}$

The calculation will be simpler if we convert the exam scores to percents.

$x$ represent the unknown percent needed on the fourth exam
$\dfrac{38}{50} = 0.76 = 76\%$ is the average on the first exam
$\dfrac{32}{40} = 0.8 = 80\%$ is the average on the second exam
$\dfrac{44}{55} = 0.8 = 80\%$ is the average on the third exam

\begin{align*} 76(.2) + 80(.2) + 80(.2) + x(.4)\amp\amp\amp\text{The weighted average of the $4$ scores}\\ 76(.2) + 80(.2) + 80(.2) + x(.4) \amp\geq 80\amp\amp\text{Write the inequality}\\ 15.2 + 16 + 16 + .40x \amp \geq 80\amp\amp\text{Solve for $x$: Multiply numbers first}\\ 47.2+.4x\amp\geq 80\amp\amp\text{Add like terms}\\ \underline{-47.2\phantom{123456}}\amp\underline{\phantom{1}-47.2}\amp\amp\text{Subtract $47.2$ from both sides}\\ .4x\amp\geq 32.8\amp\amp\text{Divide both sides by $.4$}\\ x\amp \geq 82\amp\amp\text{} \end{align*}

You will need to earn at least an $82\%$ on the last exam if you want your overall weighted average in the class to be at least $80\%\text{.}$ So, your score on the exam must be at least $0.82(120)=98.4$ points, but if the there is no partial credit on problems, you will need at least $99$ points out of $120\text{.}$

Subsection 2.C.6 Intervals in Context

In this section we introduce a notation that is used in two different ways. The correct interpretation of the notation depends on the context in which it occurs. The notation is called the "plus or minus" sign, $\pm.$

Example 2.C.19. The Plus or Minus sign.

Calculate $5 \pm 2.$

The expression represents two numbers calculated by subtracting and adding $2\text{:}$

\begin{align*} 5-2=3 \amp\amp \text{ and } \amp\amp 5+2=7\\ \amp\amp 3, 7 \amp\amp\amp \text{Our solution }\checkmark \end{align*}

The plus or minus sign is also used to indicate an interval or range of values. For example, you may have seen reports based on opinion polls where the percentage of the population in favor of a particular idea is estimated. This estimate is usually given in the form: $45 \pm 3$ percent. Here, it means that the percentage of people in favor of the idea could range any where from $42\%$ to $48\%\text{.}$ It's the interval $[42, 48]$ where the ends of the interval are calculated by $42=45-3$ and $48=45+3.$ In this case, the $\pm$ symbol is used to represent a range of values, not just two numbers as in the previous example.

Example 2.C.20. Intervals Using the Plus or Minus sign.

A study of trout in a certain mountain lake in Idaho states that the average length of the trout population in the lake is $21.4 \pm 2.7$ cm. Let $L$ represent the average length of a trout in the lake and state an inequality that represents the plus or minus expression given. Give four values within the range of the plus or minus expression.

\begin{align*} 21.4-2.7 \amp\le L \le 21.4+2.7 \amp\amp \text{Subtract and add $2.7$ to get the ends of the interval}\\ 18.7 \amp\le L \le 24.1 \amp\amp \text{An inequality that represents the plus or minus expression}\checkmark\\ \amp 19, 21.4, 22, 24\amp\amp \text{Four values within the interval}\checkmark \end{align*}