Big Ideas Math Integrated I, 2016
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Big Ideas Math Integrated I, 2016 View details
4. Solving Absolute Value Equations
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Exercise 47 Page 33

Practice makes perfect
a Let's take a look at the graph.
Looking at the graph, we can tell that the percent of teens who are in favor of year-round school is 32 %. The error on the graph tells us the actual percent could be 5 % less or 5 % more. In Favor: 32 ± 5 % The difference between the actual percent x and the percent displayed can be expressed algebraically as an absolute value. |x- 32| We use an absolute value because the value of x can differ in either the positive or negative direction. Since the error equals 5 % , we can express this as an equality. |x-32| = 5 To solve for x, we will split our equality into two separate cases.
|x-32|=5

lc x-32 ≥ 0:x-32 = 5 & (I) x-32 < 0:x-32 = - 5 & (II)

lcx-32=5 & (I) x-32=- 5 & (II)

(I), (II): LHS+32=RHS+32

lx=37 x=27
The minimum percent of students in favor is 27 %, and the maximum percent of students in favor is 37 %.
b Our classmate claims that 13 of the student body is in favor of year-round school. To verify that claim, we need to find if 13 falls between the minimum and maximum percent of students in favor we found in Part A. First, let's convert the fraction to percent.
1/3
33.333333 ... %
≈ 33 %
Since 33 % is greater than 27 % but less than 37 %, it falls within the range. This means that it does not conflict with the survey data.