Free Practice Questions for WGU Applied-Algebra Exam

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. "

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 graph shows customer visits by day.

To determine which days likely featured new releases, we should look for the days with unusually high customer counts.

From the graph:

Day 8 has about 40customers.

Day 11 also has about 40customers.

These are the highest values shown on the graph.

Since the problem says more customers show up on days when new releases are featured, the days with the highest customer counts are the most likely new-release days.

Therefore, the correct answer is:

▭ ( " D " )

The graph is a logistic growth curve.

A logistic growth curve has two main concavity regions:

" concave up before the inflection point "

and

" concave down after the inflection point "

From the graph, the function increases faster and faster until about:

x=6

After x=6, the graph continues increasing, but it begins to flatten out. That means the number of members is increasing slower and slower.

This is concave down behavior.

Therefore, one interval where the graph is concave down is:

(6│11)

So the correct answer is:

▭ ( " D " )

The function c(t)gives the number of active shopping carts on a website.

Because c(t)represents active carts, the difference between c(6)and c(8)will be measured in:

" active carts "

not hours.

First evaluate c(6):

c(6)=-0.01(6)^4+0.25(6)^3-2.33(6)^2+9.02(6)+8

c(6)=19.28

Now evaluate c(8):

c(8)=-0.01(8)^4+0.25(8)^3-2.33(8)^2+9.02(8)+8

c(8)=18.08

Now find the difference:

19.28-18.08=1.20

This is approximately:

1 " active cart "

Therefore, the correct answer is:

▭ ( " B " )

The graph shows a decreasing exponential function.

The horizontal axis represents:

" Time in months "

The vertical axis represents:

" New user accounts per day "

The graph starts near 7,000new user accounts per day and decreases over time. However, it does not decrease toward 0. Instead, the graph levels off near a horizontal value around:

y=2,500

This horizontal value is the horizontal asymptote.

A horizontal asymptote shows the long-term value the function approaches. Since the asymptote is above 0, the model predicts that the number of daily new user accounts approaches a positive value, not zero.

So the justified statement is:

" The number of daily new user accounts will not decrease to 0. "

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 " )

The graph represents the number of user accounts over time.

The horizontal axis represents:

" Time in months "

The vertical axis represents:

" Number of user accounts "

We need to find when the number of user accounts reached:

4,500

On the vertical axis, 4,500is halfway between:

3,000

and

6,000

So we look for the point where the blue curve reaches that height.

From the graph, the curve reaches approximately 4,500user accounts at:

x≈2.0

That means the website had about 4,500user accounts after approximately:

2.0 " months "

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 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 " )