WGU Applied-Algebra Dumps

The maximum value of a graph is the highest output value the function reaches. Since P(t) represents yearly profit, the maximum value occurs where the store earns its greatest yearly profit. The graph is a downward-opening curve, so its highest point is the vertex. To answer the question, we need the t-value of the vertex, not the profit amount. From the graph, the highest point occurs between 4 and 5 years after opening, closest to 4.5 years. This means the virtual store reached its maximum yearly profit approximately 4.5 years after it opened. The later values decrease after the vertex, so options involving 7.5 or 14 years are not correct. 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 " )

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

Key behavior of logistic functions:

Initial phase (early time) → growth is slow

Middle phase → growth speeds up (increasing rate)

Later phase → growth slows down as it approaches a maximum

Analyze the interval from day 1 to day 17:

This interval is in the early part of the graph.

The curve is increasing and getting steeper over this range.

That means the rate of increase is growing over time.

So, the number of comments is:

" increasing faster and faster "

The r

2

value of 0.41 indicates a moderate fit, not a strong fit. A moderate regression model may sometimes support limited extrapolation, but only if the prediction stays close to the observed data. Here, the prediction is made at x=19.4, which is beyond the maximum x-value shown in the scatterplot. For a moderate fit, extrapolation should be more conservative than for a strong fit. The value x=19.4 is more than 25% of the data range beyond the maximum observed value, so the prediction is too far outside the data. Because the fit is only moderate and the extrapolation distance is too large, the prediction is not appropriate. Therefore, the answer is B.

To describe “decreasing slower and slower,” the graph must be moving downward while becoming less steep. This means the function values are decreasing, but the rate of decrease is becoming smaller in magnitude. On the graph of x(t), the number of users first increases, then decreases after reaching a local maximum. The interval from t=6.8 to t=7.4 lies on a decreasing part of the curve where the graph is beginning to flatten compared with the steeper earlier decrease. That behavior means the number of users is still going down, but not as quickly as before. The other intervals either occur while the graph is increasing or include behavior that does not consistently match decreasing slower and slower. Therefore, the correct answer is C.

This question asks for the break-even point, which is the point where:

Revenue=Cost

On the graph:

The solid blue line represents revenue.

The dashed blue line represents cost.

The revenue line starts below the cost line, meaning the person is not yet covering costs. The revenue line eventually crosses the cost line at a little more than 4 units.

Since the question asks for the minimum number of livestock feeders that must be sold, we need a whole number of feeders.

The graph shows the intersection occurs between:

4 and 5

So selling 4 feeders is not enough, but selling 5 feeders is enough for revenue to cover the costs.

Therefore, the correct answer is:

C

The graph shows a linear relationship between the amount of red dye, r, and the amount of yellow dye, y.

The rate of change of a line is the slope:

" slope " =( " change in " y)/( " change in " r)

From the graph, the line goes through the origin and rises steeply. A clear point on the line is approximately:

(2│9)

This means when the amount of red dye is 2gallons, the amount of yellow dye is 9gallons.

So the rate of change is:

9/2

That means for every 2gallons of red dye, the factory uses 9gallons of yellow dye.

The relationship can be written as:

y=9/2 r

So the correct interpretation is:

" The amount of yellow dye must be " 9/2 " of the amount of red dye. "

Therefore, the correct answer is:

▭ ( " A " )

The graph compares the spread of two viruses over time.

The horizontal axis represents:

" time in months "

The vertical axis represents:

" percent 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, compare when each virus reaches the same population percentage.

At 20%, Virus A reaches that level earlier than Virus B. The solid curve is already at 20%before the dashed curve reaches 20%.

However, for higher percentages like 70%and 80%, Virus B reaches those levels before Virus A because the dashed curve becomes steeper and rises above the solid curve after about 50%.

So the correct conclusion is:

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

Therefore, the correct answer is:

▭ ( " A " )

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 problem translates a verbal statement into function notation. The variable z represents the number of units in inventory A, and f(z) represents the number of units in inventory B. The phrase “2 more than” tells us to add 2 to the number of units in inventory A. That gives the rule f(z)=z+2. For example, if inventory A has 10 units, then inventory B has f(10)=12 units. Option A subtracts 2, option C multiplies by 2, and option D divides by 2, so those do not match the stated relationship. Because inventory B is always exactly 2 more than inventory A, the correct function is f(z)=z+2. Therefore, the answer is B.

================

The graph represents:

p(t)= " number of active players "

where:

t= " hours after 11:00 a.m. "

The phrase “decreasing faster and faster” means two things are happening:

The graph is going downward, so the number of players is decreasing.

The graph is becoming steeper downward, so the rate of decrease is increasing.

This corresponds to a graph that is decreasing and concave down.

Looking at the graph, the number of active players reaches a maximum around:

t≈9

After that, the graph begins decreasing. Near the far right side of the graph, especially around:

t=11.2 " to " t=11.5

the curve is dropping more and more steeply.

That means the number of players is decreasing faster and faster on that interval.

The graph shows a decreasing exponential function.

In Applied Algebra, a decreasing exponential function may approach a fixed value over time. This fixed value is called the horizontal asymptote.

The horizontal asymptote represents the long-term value that the function gets closer and closer to, but does not necessarily cross or reach exactly.

From the graph, the curve decreases quickly at first, then begins to level off near:

y=1,000

This means that as time continues, the number of letters processed per day approaches:

1,000

So the long-term trend is that the mail center will process about:

1,000 " letters per day "

The graph represents:

p(t)= " number of active players "

where:

t= " hours after 11:00 a.m. "

We need to interpret the concavity between:

t=0.4

and

t=3.6

On this interval, the graph is rising, so the number of players is increasing.

But as the graph rises, it begins to level off near the top. This means the rate of increase is getting smaller.

So the number of players is not increasing faster and faster. Instead, it is increasing, but more slowly as time passes.

That means the graph is concave down on this interval.

The correct interpretation is:

" The number of players is increasing slower and slower. "

From d=0to about d=2, the graph is decreasing, but it is flattening as it approaches a minimum. This means the number of raffle tickets sold is decreasing at a slower rate:

" decreases slower and slower "

After the minimum, from about d=2to d=4.4, the graph increases and becomes steeper. This means the number of raffle tickets sold is increasing at a faster rate:

" increases faster and faster "

So the correct description is:

" The number of raffle tickets decreases slower and slower and then increases faster and faster. "

We are told that this year’s number of property sales is 6 less than last year’s number of property sales.

Let:

x= " number of properties sold last year "

and

H(x)= " number of property sales this year "

Since this year’s sales are 6 less than last year’s sales, the function rule is:

H(x)=x-6

The question says that the number of properties sold last year was:

x=330

Substitute 330into the function:

H(330)=330-6

H(330)=324

So the notation that correctly represents the number of property sales this year is:

H(330)=324

This means: if 330 properties were sold last year, then 324 properties are expected to be sold this year.

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 regression model has r

2

=0.56, which indicates a moderate fit. A moderate fit is not as reliable as a strong fit, but it can still be used for limited prediction when the extrapolation is close to the observed data. The prediction is made at x=10.8 days. This value is slightly beyond the maximum observed x-value, so the prediction is an extrapolation. For a moderate fit, the x-value should remain within about 25% of the range beyond the maximum value. In this case, x=10.8 is within that acceptable range, so the prediction is appropriate. Therefore, the correct answer is B.

The graph shows the water volume decreasing over time.

The vertical axis represents:

" Water volume in gallons "

The horizontal axis represents:

" Time in minutes "

The graph is an exponential decay curve. It decreases quickly at first, then levels off.

From the graph, the curve approaches a horizontal value near:

y=1,000

This means the horizontal asymptote is approximately:

y=1,000

Since the graph approaches 1,000 gallons, it will not decrease to 500 gallons according to this model.

The water volume does not change by a constant rate because exponential functions do not have constant additive change. Also, the graph decreases fastest at the beginning, not later.

The value r

2

=0.99 indicates a very strong regression fit because it is close to 1. A strong fit means the model explains most of the variation in the data, so limited extrapolation can be appropriate. However, extrapolation should stay close to the observed data range. From the scatterplot, the data values appear to run from about x=5 to x=15. A reasonable extrapolation range extends slightly beyond those observed values, not too far outside them. The interval x=2.5 to x=17.5 stays close to the original data and is more appropriate than the wider interval x=0 to x=20. Therefore, the correct answer is B.

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

The horizontal axis represents:

" Time in minutes "

The vertical axis represents:

" Temperature in degrees Fahrenheit "

The curve is decreasing quickly at first and then begins to level off. This is the shape of an exponential decay function.

A horizontal asymptote is a horizontal line that the graph approaches as time increases.

Because a horizontal asymptote is a horizontal line, its equation must have the form:

y= " constant "

From the graph, the temperature approaches about:

25^ ∘ F

So the horizontal asymptote is:

y=25

This means the object’s temperature gets closer and closer to 25^ ∘ Fover time.

The scatterplot shows the number of monthly train crossings decreasing over time. The decrease is not constant from one year to the next, so a linear model is not the best choice. Instead, the data decreases quickly at first and then begins to level off, which is characteristic of exponential decay. Exponential models are appropriate when a quantity changes by a repeated factor or percentage over equal time intervals. A logistic model would usually show an S-shaped pattern with a carrying capacity, and a polynomial model would be more appropriate for turning points or more complex curvature. Since the points show a smooth decreasing curve that flattens over time, an exponential function best models the data. Therefore, the correct answer is B.

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 model has:

r^2=0.42

This indicates a moderate fit, not a strong fit.

The prediction is being made at:

x=19.5

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

x=16

So x=19.5is beyond the observed data range.

Because the model has only a moderate fit and the prediction is too far beyond the observed data, the prediction is not appropriate.

The correct interpretation is:

" No. The " r^2 " value indicates a moderate fit, but " x=19.5 " is more than " 25% " of the range beyond the maximum value. "

This question asks for a description of concavity and rate of change on a specific interval. From d=10 to part of the interval, the elevation is increasing, but the graph becomes less steep as it approaches a local maximum. That means the elevation increases slower and slower. After the peak, the graph begins to decrease, and the downward slope becomes steeper. This means the elevation decreases faster and faster. The correct statement must describe both parts of the interval in order: first increasing at a slowing rate, then decreasing at an increasing rate. The other answer choices either reverse the behavior or describe the wrong direction of change. Therefore, the correct answer is A.

The graph shows the value of the coin increasing over time. Since the curve rises from left to right, the value is not decreasing. The graph also becomes steeper as time increases, which means the rate of increase is getting larger. In Applied Algebra, when a graph is increasing and its slope is increasing, the quantity is described as increasing faster and faster. This is typical of exponential growth, where values grow by a repeated percent or factor rather than by a constant amount. The choices involving decreasing are incorrect because the graph moves upward. The choice “increasing slower and slower” would require the graph to flatten, which it does not. Therefore, the correct answer is B.

Since xrepresents the number of years since 2004:

2013-2004=9

so:

2013 ⇒ x=9

Also:

2018-2004=14

so:

2018 ⇒ x=14

The average yearly rate of change from 2013 to 2018 is:

(f(14)-f(9))/(14-9)

Now evaluate the function:

f(9)=330× 〖 1.04 〗 ^9

f(9)≈469.37

f(14)=330× 〖 1.04 〗 ^14

f(14)≈571.13

Now calculate the average rate of change:

(571.13-469.37)/5

=101.76/5

=20.35

So the painting’s value increased by an average of about:

$20.35 " per year "

Therefore, the correct answer is:

▭ ( " A " )

The graph shows an increasing exponential curve.

The number of subscribers is going up as time passes, so the function is increasing.

Also, the graph becomes steeper as time increases. This means the rate of increase is getting larger.

So the number of subscribers is:

" increasing faster and faster "

This is a key feature of exponential growth.

This situation has a constant rate of change because each sack weighs the same amount. After 5 sacks are removed, the total weight is 135 pounds. After 9 sacks are removed, the total weight is 87 pounds. From 5 to 9 sacks removed is 4 more sacks. Over the same interval, the total weight decreases by 135−87=48 pounds. Since 4 sacks account for a 48-pound decrease, each sack must weigh 48÷4=12 pounds. This uses the linear idea that equal changes in the input produce equal changes in the output. Therefore, the weight of each sack is 12 pounds, which is answer A.

================

The graph represents the value of an investment account over time. The horizontal axis gives time in years, and the vertical axis gives the total account value in dollars. To find the value after 4 years, locate x=4 on the horizontal axis and read the corresponding G(x)-value on the graph. The plotted exponential growth curve is slightly above $1,000 at 4 years, closest to $1,052.87. Because the graph shows an increasing exponential trend, values such as $480.52, $584.62, and $711.28 are too low for the account value at x=4. The correct interpretation is that after 4 years, the investment account is worth approximately $1,052.87. Therefore, the correct answer is D.

This problem asks for an average yearly rate of change over a time interval. Since x represents years since 2005, the year 2013 corresponds to x=8, and the year 2015 corresponds to x=10. Use the average rate of change formula:

10−8

f(10)−f(8)

​

Evaluating the model gives f(8)=240(1.12)

8

≈594.23 and f(10)=240(1.12)

10

≈745.40. The total change is approximately 151.17 dollars over 2 years. Dividing by 2 gives about 75.59 dollars per year. Since the question asks for the average yearly rate, not the total increase, the correct answer is B.

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 function is:

D(t)=48t+23

This is a linear function because it has the form:

D(t)=mt+b

where:

48

is the rate of change, and:

23

is the starting distance from the rest stop.

We need to find:

D(1.7)

That means substitute:

t=1.7

into the function.

D(1.7)=48(1.7)+23

First multiply:

48(1.7)=81.6

Now add 23:

D(1.7)=81.6+23

D(1.7)=104.6

So the vehicle is:

104.6 miles

from the rest stop after 1.7 hours.

The average rate of change tells us how much the output changes, on average, for each 1-unit increase in the input.

Here:

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

and

y= " estimated wait time in minutes "

From the graph, point Ais approximately:

A=(5,88.3)

and point Bis approximately:

B=(9,65)

Now use the average rate of change formula:

" Average rate of change " =(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

The negative sign means the estimated wait time is decreasing from point Ato point B.

So the average rate of change is approximately:

-5.83 " minutes per hour "

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 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 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.

A decreasing exponential function falls quickly at first and then levels off over time. The “slowest average decrease” occurs where the graph is flattest, because the output changes by the smallest amount per year. Earlier intervals, such as 2005 to 2008, occur when the graph is steepest, so those intervals have a faster average decrease. Later intervals occur after the exponential decay has slowed. From the graph, the interval from 2015 to 2025 is farthest to the right and has the least steep downward trend. Therefore, the number of registered voters decreases most slowly during that interval. This is why the correct answer is D.

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 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 "

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.

We are given the function:

d(x)=14+65x

This is a linear function because it has the form:

d(x)=mx+b

where 65is the rate of change and 14is the starting value.

We need to find:

d(1.5)

That means substitute x=1.5into the function:

d(1.5)=14+65(1.5)

Now multiply:

65(1.5)=97.5

Then add:

d(1.5)=14+97.5

d(1.5)=111.5

So the distance from the tower to the object at 1.5seconds is:

111.5 " meters "

The average rate of change measures how much the output changes per one unit of input. Here, the input is time in hours after 6:00 a.m., and the output is the number of vehicles stopped at the toll booth. The two given points are A=(1,49.4) and B=(4,45.8). Use the average rate of change formula: (y

2

​

−y

1

​

)/(x

2

​

−x

1

​

). Substituting the coordinates gives (45.8−49.4)/(4−1). The numerator is −3.6, and the denominator is 3, so the average rate of change is −1.2. The negative sign means the number of vehicles decreased over the interval.