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Izabella goes on a tour of a chocolate factory. They aim to produce chocolate bars weighing $93$ grams. A machine at the factory weighs chocolates chosen at random. The bars must not deviate from the predetermined weight by more than $5$ grams. Otherwise, the machine has to send them back.

Answer

a $88\le w\le 98$

b Graph:

c $|w-93|\le 5$

Solution

a For a chocolate bar to be sent back, the difference between the weight of the chocolate bar and the predetermined weight is more than $5$ grams. Therefore, an acceptable weight $w$ for a chocolate bar, not to be sent back, must be greater than or equal to $93-5=88$ grams.

$$\begin{array}{c}88\le w\end{array}$$

At the same time, it must weigh less than or equal to $93+5=98$ grams.

$$\begin{array}{r}w\le 98\end{array}$$

Consequently, the combination of these individual inequalities will result in a compound inequality describing the range of acceptable weights.

$$\begin{array}{c}88\le w\text{ and }w\le 98\\ \Updownarrow \\ 88\le w\le 98\end{array}$$

b The inequality $88\le w$ is the set of values which are greater than or equal to $88.$ Since the inequality sign is non-strict, $88$ itself is included.

The other inequality represents all the points to the left of $98.$ Again, the inequality sign is non-strict, so $98$ is included.

Because the inequalities are combined using word and, the solution set of the resulting compound inequality is equal to the union of the solution sets of the individual inequalities. Therefore, the graph of the acceptable weight range is the combination of the graphs.

c Any chocolate bar that varies from the predetermined weight by more than $5$ grams is sent back. Therefore, the absolute value of the difference between the weight of a bar $w$ and the predetermined weight should be less than or equal to $5.$ This can be expressed as follows.

$$\begin{array}{c}|w-93|\le 5\end{array}$$

Since this absolute value inequality can be rewritten as the compound inequality obtained in the previous steps, both have the same set of solutions.

Izabella, looking around the chocolate factory, discovers a room full of chocolate fondue fountains on sale for special events! The prices are high and they vary significantly. She decides to make a list of the prices to see which is a fair price.

Solution

a To find the average price $p$ of the chocolate fondues, the sum of the prices will be divide by the number of prices, $8.$

$p=\frac{\text{Sum of Values}}{\text{Number of Values}}$

SubstituteValues

Substitute values

$p=\frac{450+476+560+327+358+325+600+480}{8}$

Evaluate right-hand side

AddTerms

Add terms

$p=\frac{3576}{8}$

CalcQuot

Calculate quotient

$p=447$

Since Izabella thinks a fair amount is within $\$100$ of the average price of $\$447,$ the difference between the amount $x$ Izabella thinks is fair and the average price should be less than $\$100.$ Therefore, the absolute value of the difference between $x$ and the average price is less than $100.$

$$\begin{array}{c}\underline{\text{Absolute Value Inequality}}\\ |x-447|<100\end{array}$$

By solving this absolute value inequality, the chocolate fondues satisfying this condition can be determined. To do so, first rewrite it as a compound inequality and then solve for $x.$

$$\begin{array}{c}|x-447|<100\\ \Downarrow \\ \text{-}100<x-447<100\\ \Downarrow \\ 347<x<547\end{array}$$

The chocolate fondues whose prices are within this range are fair prices.

As seen, $4$ prices meet Izabella's condition.

$$\begin{array}{c}\$450,\$476,\$358,\text{ and }\$480\end{array}$$

b In the previous part, the average price of the fondue's was found to be $\$447.$ Now, the range is narrowed to $\$20.$ The absolute value of the difference between $x$ and $447$ should be less than or equal to $20.$

$$\begin{array}{c}|x-447|\le 20\end{array}$$

By solving the inequality, the fondue prices satisfying this condition can be determined.

$$\begin{array}{c}|x-447|\le 20\\ \Downarrow \\ \text{-}20\le x-447\le 20\\ \Downarrow \\ 427\le x\le 467\end{array}$$

The fondue prices are within this range are those which Izabella would consider to be priced fairly.

There is only one price that meets Izabella's condition.

$$\begin{array}{c}\$450\end{array}$$

At this special chocolate factory the average salary for a Chocolatier is a whopping $\$45700.$

As a company policy, a new Chocolatier's actual salary can only differ from the company average by less than $\$1250.$

Answer

a Absolute Value Inequality: $|s-45700|<1250$

Range: $44450<s<49950$

Graph:

b Absolute Value Inequality: $|s-45700|\ge 1250$

Solution

a This special chocolate factory pays a new Chocolatier's salary within $\$1250$ of the company's average salary of $\$45700.$ This means that the absolute value of the difference between the salary $s$ for a new employee and the average salary of any Chocolatier is less than $1250.$

$$\begin{array}{c}\underline{\text{Absolute Value Inequality}}\\ |s-45700|<1250\end{array}$$

By solving this absolute value inequality, the range for the possible salaries can be found. To do so, the inequality will be rewritten.

$$\begin{array}{c}|s-45700|<1250\\ \Downarrow \\ \text{-}1250<s-45700<1250\\ \Downarrow \\ 44450<s<46950\end{array}$$

The solution set of the inequality is the set of values between $44450$ and $46950.$ Since the inequality sign is strict, those numbers are not included.

b Consider the graph of the absolute value inequality found in Part A.

The values that are more than $1250$ units away from the average salary in the number line, need to be represented by an absolute value inequality.

These values can be written as the following absolute value inequality.

$$\begin{array}{c}|s-45700|\ge 1250\end{array}$$

In this lesson, the relationship between absolute value inequalities and compound inequalities has been explained using real-world examples. Considering those examples, the challenge presented at the beginning can now be solved seamlessly.

It is known that, at atmospheric pressure, water freezes at $32^{\circ }\text{F}$ and vaporizes at $212^{\circ }\text{F}.$

Answer

a $t<32$ or $t>212$

b Graph:

c $|t-122|>90$

Solution

a Water is not liquid when it is in frozen form nor when it is vapor form. Noting the given information, it is known that, at atmospheric pressure, water is not liquid for temperatures less than $32^{\circ }\text{F}$ or greater than $212^{\circ }\text{F}$

$$\begin{array}{rl}\text{Solid: }& t<32\\ \text{Gas: }& t>212\end{array}$$

The values $32^{\circ }F$ and $212^{\circ }F$ are not included because at these temperatures water starts to change its physical state and the liquid form of water can be present in both states of the transition. Under these conditions, the following compound inequality shows the temperatures in which water is not a liquid.

$$\begin{array}{c}t<32\text{ or }t>212\end{array}$$

b The compound inequality is the combination of two inequalities. The inequality $t<32$ shows the set of values less than $32.$ Since the inequality sign is strict, the value $32$ is not included.

The other inequality represents all of the points to the right of $212.$ Again, the inequality sign is strict, so $212$ is not included.

Since the inequalities were combined with the word or, the solution set of the resulting compound inequality is the union of the solutions sets of the two individual inequalities. Therefore, the graph of the range is the combination of the above graphs.

c To write the absolute value equation, the midpoint between $32$ and $212$ on the number line should be found. To do so, half the difference between these numbers will be calculated.

$$\begin{array}{c}\frac{212-32}{2}=\frac{180}{2}\Rightarrow \frac{180}{2}=90\end{array}$$

The midpoint is $90$ units away from the endpoints. Therefore, the midpoint is $32+90=122.$

The points that are further away in units than the calculated distance of $90$ represent the range, which can be written as the following absolute inequality.

$$\begin{array}{c}|t-122|>90\end{array}$$