Free Practice Questions for WGU Applied-Algebra Exam

We are told that computer B completes 12 more tasks than computer A.

Let:

x= " number of tasks completed by computer A "

and

F(x)= " number of tasks completed by computer B "

The phrase “12 more than” means add 12 to the original amount:

F(x)=x+12

So, if computer A completes xtasks, computer B completes:

x+12

tasks.

For example, if computer A completed 50 tasks, then computer B completed:

F(50)=50+12=62

Therefore, the correct answer is:

▭ ( " B " )

To determine which town’s population is growing faster, compare the slopes of the two lines. In a linear population model, the slope represents the growth rate, while the starting value represents the initial population. The graph shows that Town A and Town B both increase over time, but Town B has the larger slope. Town B grows at about 1.3 thousand people per year, while Town A grows at about 1.1 thousand people per year. Since 1.3 > 1.1, Town B’s population is increasing at a faster rate. The values 6.5 and 5.5 describe starting population amounts, not growth rates. Therefore, the correct answer is A.

The graph shows a logistic regression model for rabbit population data.

The given value is:

r^2=0.94

An r^2value close to 1means the model fits the data well. Since 0.94is close to 1, the regression model is a strong fit.

However, extrapolation should still be done carefully. Extrapolation means using the model to predict values outside the observed data range.

From the scatterplot, the data values appear to run roughly from about:

x=4 " to " x=20

A reasonable extrapolation range should stay close to the observed data. The interval:

x=0 " to " x=24

extends slightly beyond the data on both sides, so it is more reasonable than the much wider interval:

x=-4 " to " x=28

The statements saying extrapolation is not appropriate because r^2 < 1or because r^2 > 0.7are not valid interpretations.

We are told that the number of people expected to audition this year is 4 times the number of people who auditioned last year.

Let:

t=number of people who auditioned last year

and

A(t)=number of people expected to audition this year

Since this year’s number is 4 times last year’s number, the function is:

A(t)=4t

The question says that 330 people auditioned last year, so:

t=330

Now substitute 330 into the function:

A(330)=4(330)

A(330)=1320

So, the correct function notation is:

A(330)=1,320

This means: when 330 people auditioned last year, 1,320 people are expected to audition this year.

Therefore, the correct answer is:

B

The average rate of change measures how much the output changes per one unit of input.

Here:

t= " hours after 8:00 a.m. "

and

n(t)= " number of people at the park "

The two given points are:

A=(2,110)

and

B=(6,182)

This means:

At t=2hours after 8:00 a.m., there were 110people at the park.

At t=6hours after 8:00 a.m., there were 182people at the park.

Now calculate the average rate of change:

" Average rate of change " =(182-110)/(6-2)

=72/4

=18

Because the value is positive, the number of people is increasing.

So the correct interpretation is:

" The number of people at the park increases at an average rate of 18 people per hour. "

The regression function has:

r^2=0.54

This indicates a moderate fit, not a strong fit. The model explains some of the variation in the data, but not as strongly as a model with an r^2value closer to 1.

Now consider the prediction at:

x=20.6

From the scatterplot, the observed data values extend to about:

x=18

The prediction at x=20.6is beyond the observed data, so it is an extrapolation. However, it is not very far beyond the data range.

Because x=20.6is within a reasonable extrapolation range and the model has a moderate fit, the prediction is appropriate.

The function A(t)represents the number of people expected to audition this year.

The input trepresents the number of people who auditioned last year.

The problem says this year’s number is expected to be 3 less than last year’s number. Therefore, the function rule is:

A(t)=t-3

We are told that 280 people auditioned last year, so:

t=280

Substitute 280into the function:

A(280)=280-3

A(280)=277

So, the notation that represents the number of people expected to audition this year is:

A(280)=277

This means if 280 people auditioned last year, then 277 people are expected to audition this year.

Therefore, the correct answer is:

▭ ( " A " )

The function is:

p(m)=5m+5

This is a linear function in the form:

p(m)=mx+b

For a linear function, the average rate of change over any interval is the same as the slope.

The slope is the coefficient of m:

5

So the price increases at a constant rate of:

$5 " per month "

We can also verify using the average rate of change formula:

(p(10)-p(1))/(10-1)

Find p(10):

p(10)=5(10)+5=55

Find p(1):

p(1)=5(1)+5=10

Now calculate:

(55-10)/(10-1)

=45/9

=5

The graph shows the temperature of an object decreasing over time.

The horizontal axis represents:

" Time in minutes "

The vertical axis represents:

" Temperature in degrees Fahrenheit "

The curve decreases and approaches the horizontal axis as time increases.

A horizontal asymptote is a horizontal line that the graph approaches but may not reach. Horizontal lines have equations of the form:

y= " constant "

From the graph, the curve approaches:

y=0

So the horizontal asymptote is:

y=0

The options x=0and x=500are vertical lines, not horizontal asymptotes.

Therefore, the correct answer is:

▭ ( " B " )

From the graph, the horizontal axis represents:

t= " hours after 7:00 a.m. "

The vertical axis represents:

" estimated wait time in minutes "

Point Ais approximately at:

A=(5,88.3)

Point Bis approximately at:

B=(9,65)

The average rate of change from point Ato point Bis calculated using:

(y_2-y_1)/(x_2-x_1 )

Substitute the values:

(65-88.3)/(9-5)

=(-23.3)/4

=-5.825

Rounded to two decimal places:

-5.83

So, the wait time decreases by about:

5.83 " minutes per hour "