WGU Applied-Algebra WGU Applied Algebra FXO2 PFXP C957 Exam Practice Test

The graph models the number of widgets the team can produce each day.

The horizontal axis represents:

" days "

The vertical axis represents:

" widgets "

The graph gives two labeled points:

(6│3.4)

and

(14│23)

The average rate of change is:

" change in widgets " / " change in days "

Substitute the values:

(23-3.4)/(14-6)

=19.6/8

=2.45

Rounded to the nearest tenth:

2.5

This means the team increased production by an average of approximately:

2.5 " widgets each day "

Therefore, the correct answer is:

▭ ( " B " )

The scatterplot shows data collected over time, and a regression curve is used to model the pattern.

Extrapolation means using a model to make predictions slightly outside the observed data range. In Applied Algebra, extrapolation can be appropriate when:

" the regression model fits the data reasonably well "

and the predicted x-values are not too far outside the observed data.

From the scatterplot, the data points appear to run approximately from:

x=3 " to " x=15

So the observed data range is about:

15-3=12

A reasonable extrapolation range extends a little beyond the data, but not too far. The interval:

x=0 " to " x=18

extends about 3 units beyond each side of the observed data, which is reasonable.

The interval:

x=-3 " to " x=21

extends much farther beyond the data and would be less reliable.

Also, r^2does not have to equal exactly 1 for extrapolation to be useful. A value less than 1 can still represent a strong model.

Therefore, the correct answer is:

▭ ( " A " )

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 " )

This question asks us to compare the heights of two balls at a specific time:

t=0.9

The graph shows:

Ball 1 as the solid blue curve.

Ball 2 as the dashed blue curve.

To determine which ball was lower at t=0.9, we look at the vertical positions of both curves when the time is 0.9seconds.

At approximately t=0.9:

" height of Ball 1 " < " height of Ball 2 "

This means Ball 1 is lower than Ball 2 at that time.

The correct reasoning must say both:

Ball 1 was lower, and

Ball 1’s height was less than Ball 2’s height at t=0.9.

That matches option A.

The function a(x)represents the total value of an investment account.

The horizontal axis represents:

x= " time in years "

The vertical axis represents:

a(x)= " total account value in dollars "

We are asked to find the total value when:

x=4

To answer this from the graph:

Locate x=4on the horizontal axis.

Move vertically upward until reaching the blue graph.

Read the corresponding value on the vertical axis.

From the graph, when x=4, the value is slightly above $1,800, approximately:

a(4)≈1,830.59

So the total value of the investment account after 4 years is approximately:

$1,830.59

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 "

The value

r^2=0.96

indicates a strong fit because it is very close to 1. This means the regression model explains much of the variation in the traffic-frequency data.

However, we also need to check whether the prediction at

x=16.2

is a reasonable extrapolation.

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

x=14

The predicted value is slightly beyond the observed data range, but not extremely far beyond it.

Since x=16.2is still within a reasonable extrapolation range, and the model has a strong r^2value, the prediction is appropriate.

Therefore, the correct answer is:

▭ ( " C " )

This question asks which town’s population is growing at a faster rate.

Since both population graphs are straight lines, we compare their slopes.

The slope of a line represents the rate of change:

" slope " = " change in population " / " change in time "

From the graph:

Town A starts at about 8.0thousand people and increases at a rate of about 1.0thousand people per year.

Town B starts at about 5.5thousand people and increases at a rate of about 1.3thousand people per year.

Even though Town A starts with a larger population, the question asks about the growth rate, not the starting population.

Compare the rates:

1.3 > 1.0

So Town B is growing faster than Town A.

The function is:

f(x)=190× 〖 0.99 〗 ^x

Since xrepresents the number of years since 2003:

2012 ⇒ x=9

2015 ⇒ x=12

The average rate of change from 2012 to 2015 is:

(f(12)-f(9))/(12-9)

First, find f(9):

f(9)=190× 〖 0.99 〗 ^9

f(9)≈173.59

Now find f(12):

f(12)=190× 〖 0.99 〗 ^12

f(12)≈168.37

Now calculate the average rate of change:

(168.37-173.59)/(12-9)

=(-5.22)/3

≈-1.74

Rounded to the nearest whole number:

-2

The negative sign means the number of subscribers is decreasing.

So the correct interpretation is:

" The number of subscribers decreases at an average rate of about 2 subscribers per year. "

This question asks us to identify a correct conclusion from the graph.

The vertical axis represents:

" number of daily homework problems "

The horizontal axis represents:

" school days since the beginning of the term "

From the graph, the lowest point occurs at the far right, around:

" day " 22

The corresponding number of homework problems is approximately:

18

So the minimum number of daily homework problems is 18, and it occurs at day 22.

Therefore, the correct answer is:

▭ ( " D " )

The function M(t)represents the company’s predicted market share after tyears.

The horizontal axis represents:

t= " time in years "

The vertical axis represents:

M(t)= " market share percentage "

We need to find when the company’s market share reaches:

60%

To answer this from the graph:

Locate 60on the vertical axis.

Move horizontally until reaching the blue curve.

Move downward to the horizontal axis to read the corresponding time.

From the graph, the blue curve reaches 60%a little after 6years, approximately at:

t=6.6

So the company should expect to have a market share of 60%after about:

6.6 " years "

The function is:

f(x)=10× 〖 1.09 〗 ^x

This is an exponential growth function because:

1.09 > 1

For an exponential growth function, the rate of increase becomes larger as time passes. That means later intervals usually have faster average increases than earlier intervals.

Since xrepresents years since 2001:

2001 ⇒ x=0

2003 ⇒ x=2

2005 ⇒ x=4

2013 ⇒ x=12

2016 ⇒ x=15

2019 ⇒ x=18

2021 ⇒ x=20

The latest interval is:

2013 " to " 2021

Because the graph is exponential growth and is steepest later in time, this interval has the fastest average increase.

This question asks us to interpret a value from a graph.

The horizontal axis represents:

Days

The vertical axis represents:

Practice duration in minutes

We need to find the practice duration on day 8.

To do this:

Locate 8 on the horizontal axis.

Move vertically until you reach the blue graph.

Read the corresponding value on the vertical axis.

From the graph, when:

x=8

the blue graph is at:

y=115

So the musician practiced for:

115 minutes

We need to compare the two account values after:

t=3

First, find the value of Account 1:

A_1 (3)=6,500× 〖 1.05 〗 ^3

A_1 (3)=6,500×1.157625

A_1 (3)≈7,524.56

Now find the value of Account 2:

A_2 (3)=3,000× 〖 0.96 〗 ^3

A_2 (3)=3,000×0.884736

A_2 (3)≈2,654.21

Now subtract:

7,524.56-2,654.21=4,870.35

So Account 1 is about:

$4,870

greater than Account 2 after 3 years.

Therefore, the correct answer is:

▭ ( " C " )

The function is:

W(x)=740× 〖 0.87 〗 ^x

This is an exponential decay function because the base 0.87is between 0 and 1:

0 < 0.87 < 1

That means the sample’s weight decreases over time.

We need to find the weight when:

x=3

Substitute 3into the function:

W(3)=740× 〖 0.87 〗 ^3

First calculate the exponent:

〖 0.87 〗 ^3=0.87×0.87×0.87

〖 0.87 〗 ^3=0.658503

Now multiply by 740:

W(3)=740×0.658503

W(3)=487.29222

Rounded to two decimal places:

W(3)≈487.29

So the weight of the radioactive sample after 3 years is approximately:

487.29 " oz. "

The graph compares the populations of two towns over time.

The horizontal axis represents:

" Time in years "

The vertical axis represents:

" Population in thousands "

The graph shows:

Town A as the solid blue line.

Town B as the dashed blue line.

To determine which town’s population is growing faster, we compare the slopes of the two lines.

In Applied Algebra, the slope of a line represents the rate of change:

" slope " = " change in population " / " change in time "

From the graph:

" Town A grows at about " 0.9 " thousand people per year "

" Town B grows at about " 1.6 " thousand people per year "

Now compare the growth rates:

1.6 > 0.9

So Town B’s population is growing at a faster rate.

The values 8.0and 6.5describe starting populations, not growth rates. Since the question asks about growing at a faster rate, we must compare the slopes.

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.

This situation can be modeled with a linear relationship because each box has the same weight.

We are given:

4 " boxes " →116 " pounds "

and

9 " boxes " →161 " pounds "

The weight increases because more boxes are added.

First, find the change in total weight:

161-116=45

Then find the change in the number of boxes:

9-4=5

Since 5 additional boxes increased the total weight by 45 pounds, the weight of each box is:

45/5=9

So each box weighs:

9 " pounds "

Check:

If each box weighs 9 pounds, then adding 5 more boxes adds:

5(9)=45

116+45=161

This matches the given information.

We are told that supervisor B has 13 more employees than supervisor A.

Let:

n= " number of employees under supervisor A "

and

F(n)= " number of employees under supervisor B "

The phrase “13 more than” means we add 13:

F(n)=n+13

So, if supervisor A has nemployees, supervisor B has 13 additional employees.

For example, if supervisor A had 20 employees, then supervisor B would have:

F(20)=20+13=33

Therefore, the correct answer is:

▭ ( " C " )

The scatterplot shows a decreasing pattern.

At first, the number of monthly crossings decreases quickly. Then the values begin to level off.

This type of pattern is characteristic of an exponential decay model.

A linear model would show points decreasing at a constant rate, forming an approximately straight line. Here, the decrease is not constant; it is steep at first and then slows down.

A logistic model usually has an S-shaped pattern, which is not shown here.

A polynomial model may curve, but the long-term leveling behavior shown in the scatterplot is best represented by an exponential decay function.

Therefore, the correct answer is:

▭ ( " D " )

This question asks us to interpret a graph showing cost and revenue.

In Applied Algebra, the break-even point is where:

" Revenue " = " Cost "

On a graph, this happens where the revenue line and the cost line intersect.

From the graph:

The solid blue line represents revenue.

The dashed blue line represents cost.

The break-even point occurs where these two lines cross.

Looking carefully at the graph, the two lines intersect at approximately:

x=6

The horizontal axis represents the number of units, meaning the number of down quilts sold.

So the person needs to sell about:

6 " quilts "

to break even.

After selling more than 6 quilts, the revenue line is above the cost line, meaning the person begins making a profit.

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 " )

The function is:

P(t)=24t+185

This is a linear function in the form:

P(t)=mt+b

where:

m= " rate of change "

and

b= " initial value "

In the function:

P(t)=24t+185

the coefficient of tis:

24

So the rate of change is:

24 " fish per month "

Because 24is positive, the fish population is increasing.

The number 185is not the rate of change. It represents the starting fish population at the beginning of the year, when t=0:

P(0)=24(0)+185=185

Therefore, the correct interpretation is:

" The number of fish in the lake is increasing at a constant rate of 24 fish per month. "

So the correct answer is:

▭ ( " B " )

The graph shows:

P(t)= " weekly profit in thousands of dollars "

where:

t= " years since opening "

The graph is a downward-opening curve, so it has a maximum value at its highest point. In Applied Algebra, the highest point of this kind of polynomial graph is called the vertex.

To answer the question, we need the t-value of the vertex because the question asks when the store had its maximum weekly profit.

From the graph, the highest point occurs between:

t=5

and

t=6

More specifically, it is approximately:

t=5.5

So the store had its maximum weekly profit approximately:

5.5 " years after opening "

This situation can be modeled using a linear relationship because each sack has the same weight.

We are given:

After 3sacks are removed, the total weight is 116pounds.

After 6sacks are removed, the total weight is 101pounds.

From 3 sacks removed to 6 sacks removed, the number of removed sacks increases by:

6-3=3

During that time, the total weight decreases from 116pounds to 101pounds:

116-101=15

So removing 3 additional sacks decreases the total weight by 15 pounds.

Now divide to find the weight of one sack:

15/3=5

So each sack weighs:

5 " pounds "

Check:

If 3 more sacks are removed and each sack weighs 5 pounds, the total weight should decrease by:

3(5)=15

116-15=101

This matches the given information.

The graph represents yearly profit as a function of time:

P(t)= " yearly profit "

where:

t= " years since opening "

The graph is a downward-opening curve, so its maximum value occurs at the highest point, also called the vertex.

From the graph, the highest point occurs at approximately:

t=10.5

So the maximum yearly profit happened approximately:

10.5 " years after opening "

The vertical axis measures yearly profit in millions of dollars. From the graph, the maximum value is approximately:

1.41

Since the units are millions of dollars:

1.41×1,000,000=1,410,000

So the maximum yearly profit was approximately:

$1,410,000

Therefore, the correct answer is:

▭ ( " C " )

The value r^2measures how well a regression model fits the data.

Here:

r^2=0.25

This is a low r^2value, meaning the model explains only about:

25%

of the variation in the data.

That indicates a weak fit. When a model has a weak fit, using it for extrapolation is not reliable because predictions outside the data range may be very inaccurate.

Since:

0.25 < 0.3

the best conclusion is that extrapolation is not appropriate because the regression model does not fit the data well enough.

Therefore, the correct answer is:

▭ ( " C " )

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.