Free Practice Questions for WGU Applied-Algebra Exam

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 graph is a logistic growth function.

A logistic function often has an S-shape. It begins by increasing slowly, then increases faster and faster, and later levels off as it approaches a maximum carrying capacity.

From year 1 to year 9, the graph is rising and becoming steeper.

That means:

" the wild horse population is increasing "

and the slope is getting larger, so:

" the population is increasing faster and faster "

Therefore, the correct answer is:

▭ ( " A " )

The graph compares the spread of two viruses over time.

The horizontal axis represents:

t= " time in months "

The vertical axis represents:

" percentage of the population affected "

The graph shows:

Virus A as the solid blue curve.

Virus B as the dashed blue curve.

To determine which statement is correct, we compare when each virus reaches the same population percentage.

At 20%:

Virus B reaches 20%earlier, around t≈5months.

Virus A reaches 20%later, around t≈6.5months.

So Virus B affects 20%of the population before Virus A does.

For higher percentages such as 60%, 70%, and 80%, Virus A rises more quickly after about 50%, so Virus A reaches those percentages before Virus B.

Therefore, the correct conclusion is:

" Virus B affects " 20% " of the population before virus A does. "

The graph shows population in thousands, and the horizontal axis represents:

" years since 1995 "

Point Pis located at approximately:

x=12

Since xrepresents years since 1995:

1995+12=2007

So point Pcorresponds to the year:

2007

The instantaneous rate of change at point Pis:

3.74

Because the vertical axis is measured in thousands, this means:

3.74 " thousand animals per year "

Convert thousands to actual animals:

3.74×1000=3740

So in 2007, the population was increasing by approximately:

3,740 " animals per year "

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

The horizontal axis represents:

" Days "

The vertical axis represents:

" Widgets produced each day "

The two labeled points are:

(16│9.1)

and

(18│12.4)

This means:

On day 16, the team could produce about 9.1widgets per day.

On day 18, the team could produce about 12.4widgets per day.

The average rate of change is:

(12.4-9.1)/(18-16)

=3.3/2

=1.65

Rounded to the nearest tenth:

1.7

So from day 16 to day 18, the team’s production increased by an average of about:

1.7 " widgets each day "

This graph represents a logistic growth function, which has an S-shaped curve:

Slow growth at the beginning (near point A)

Rapid growth in the middle (near point B and C)

Slowing growth at the end (near point D)

Key Concept:

Instantaneous rate of change = slope of the tangent line at a single point

Average rate of change = slope between two points

Analyze the graph:

At point A → slope is small (slow increase)

From A to C → increasing but not maximum

At point C → curve is steepest → maximum slope

From C to D → curve flattens → smaller slope

At point D → slope is very small (almost flat)

Conclusion:

The greatest rate of change occurs where the graph is steepest, which is at:

" Point C "

The graph shows a decreasing exponential function.

The vertical axis represents:

" Letters per day "

The horizontal axis represents:

" Time in years "

For an exponential decay graph, the value decreases quickly at first and then begins to level off. The value it approaches in the long run is called the horizontal asymptote.

From the graph, the curve levels off near:

y=2,500

So the number of letters processed per day is trending toward:

2,500

This does not mean the mail center immediately processes exactly 2,500 letters per day. It means that as time continues, the number gets closer and closer to 2,500.

Therefore, the correct answer is:

▭ ( " B " )

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

Here:

t= " hours after 5:00 a.m. "

and

n(t)= " number of animals "

The two given points are:

A=(1,10.6)

and

B=(8,6.59)

Use the average rate of change formula:

(y_2-y_1)/(x_2-x_1 )

Substitute the coordinates:

(6.59-10.6)/(8-1)

=(-4.01)/7

≈-0.573

The negative sign means the number of animals is decreasing.

So the correct interpretation is:

" The number of animals at the station decreases at an average rate of " 0.573 " animals per hour. "

Therefore, the correct answer is:

▭ ( " B " )