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 \pm 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

$x - 32 \geq 0 : x - 32 = 5 x - 32 < 0 : x - 32 = - 5 (I) (II)$

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

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

x = 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

\frac{1}{3}

of the student body is in favor of year-round school. To verify that claim, we need to find if

\frac{1}{3}

falls between the minimum and maximum percent of students in favor we found in Part A. First, let's convert the fraction to percent.

\frac{1}{3}

Convert to percent

33.333333 \dots %

Round to nearest integer

\approx 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.

Communicate Your Answer p.27

Monitoring Progress p.28-31

Exercises p.32-34

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