Free and Premium WGU Applied-Algebra Dumps Questions Answers

We are told that the number of property sales this year is expected to be 8 times the number of property sales last year.

Let:

x= " number of properties sold last year "

and

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

Since this year’s sales are 8 times last year’s sales, the function rule is:

H(x)=8x

The question says that last year’s property sales were:

x=360

Substitute 360into the function:

H(360)=8(360)

H(360)=2,880

So the correct notation is:

H(360)=2,880

This means: if 360 properties were sold last year, then 2,880 property sales are expected this year.

Therefore, the correct answer is:

▭ ( " C " )

A logistic function often increases quickly in the middle and then begins to level off as it approaches a maximum carrying capacity. From year 7 to year 10, the graph is still increasing, so the bison population is going up. However, the curve is becoming flatter during this interval. A flatter curve means the slope is getting smaller, so the population is increasing at a slower rate. This behavior is described as “increases slower and slower.” It is not decreasing, because the graph continues to move upward. It is also not increasing faster and faster, because the slope is not becoming steeper in this interval. Therefore, the correct interpretation is answer B.

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.

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

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.

This question asks for the cost of each additional shirt, which is the slope of the linear graph. The slope tells how much the total cost increases for each increase of 1 shirt. From the exhibit, the graph shows a starting cost near $39 and a point near (10, 139). The increase in cost from 0 shirts to 10 shirts is 139−39=100 dollars. Dividing by 10 shirts gives a rate of 10 dollars per shirt. The fixed starting cost is not the cost of a shirt; it is the initial fee or setup cost. Since the graph rises by $10 for each additional shirt, the cost per additional shirt is $10.00. Therefore, the answer is B.

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

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 value r

2

=0.89 indicates a strong fit because it is close to 1. A strong fit means the regression function follows the scatterplot data closely enough to support reasonable prediction. The question asks whether a prediction at x=18.6 months is appropriate. This value is outside the observed data range, so the prediction involves extrapolation. For a strong regression model, extrapolation can be acceptable if the new x-value is not too far beyond the maximum observed value. Here, x=18.6 is within 50% of the range beyond the maximum value, so the prediction is considered appropriate. Therefore, the correct answer is C.

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.

The graph shows weekly profit as a function of time.

The horizontal axis represents:

" years since opening "

The vertical axis represents:

" weekly profit in hundreds of dollars "

The graph is an upward-opening curve, so its minimum value occurs at the lowest point, called the vertex.

From the graph, the lowest point occurs at approximately:

t=6

So the minimum weekly profit happened approximately:

6 " years after opening "

The vertical value at this point is approximately:

9.9

Since the profit is measured in hundreds of dollars:

9.9×100=990

So the minimum weekly profit was approximately:

$990

Therefore, the correct answer is:

▭ ( " A " )

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 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 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 regression model has:

r^2=0.99

This means the model is a very strong fit for the data because 0.99is close to 1.

However, a strong r^2value does not automatically make every prediction appropriate. We also have to check whether the x-value is within a reasonable extrapolation range.

The data shown on the graph appear to extend to about:

x=11

The prediction is for:

x=16.4

This is far beyond the observed data range. Even though the model fits the known data very well, predicting too far beyond the data can be unreliable.

The correct statement is that the r^2value indicates a strong fit, but x=16.4is more than 50%of the range beyond the maximum observed value.

Therefore, the correct answer is:

▭ ( " A " )

The function f(n) gives the distance traveled by vehicle B.

The input n represents the distance traveled by vehicle A.

The problem says vehicle B travels 17 more than vehicle A. The phrase “17 more than” means we add 17 to the amount traveled by vehicle A.

So the relationship is:

f(n)=n+17

For example, if vehicle A traveled 50 miles, then vehicle B traveled:

f(50)=50+17=67

So the correct function is:

f(n)=n+17

Therefore, the correct answer is:

B

This is a polynomial function evaluation problem. Since I(m) represents daily income in dollars, the difference between I(7) and I(9) must also be in dollars. First evaluate I(7)=41(49)−490(7)+30=2009−3430+30=−1391. Then evaluate I(9)=41(81)−490(9)+30=3321−4410+30=−1059. The difference is |−1059−(−1391)|=332. Because the question asks for the difference between the two outputs, the correct unit is dollars, not months. So the difference between I(7) and I(9) is approximately $332. Therefore, the correct answer is C.

This question asks us to interpret unusually high values on a graph. The problem states that more customer accounts are opened during snowy weeks than during weeks without snow. Therefore, the likely snowy weeks correspond to the peaks on the graph, where the number of new customer accounts is noticeably higher than surrounding weeks. The graph shows clear spikes at approximately week 7 and week 14. These weeks have higher account openings compared with nearby weeks, making them the best evidence of snowy conditions. Week 1 and week 5 do not match the main peaks shown in the graph. Since the snowy weeks should correspond to the highest customer-account values, the correct answer is D.

For a logistic graph, concavity changes at the inflection point.

Before the inflection point, the graph is usually concave up. This means the function is increasing faster and faster.

After the inflection point, the graph is concave down. This means the function is still increasing, but it is increasing slower and slower.

From the graph, the curve changes concavity around:

x=8

After x=8, the graph begins to level off as it approaches its upper limit.

So one interval where the graph is concave down is:

(8│16)

Therefore, the correct answer is:

▭ ( " A " )

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 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.

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

This question asks for correct function notation. The input x represents the number of properties sold last year, and the output H(x) represents the expected number of property sales this year. Since this year’s number is expected to be 9 more than last year’s number, the function rule is H(x)=x+9. The problem states that 140 properties were sold last year, so the correct input is x=140. Substituting gives H(140)=140+9=149. This means that when last year’s sales were 140, this year’s expected sales are 149. The input must be inside the parentheses, and the output must follow the equals sign. 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 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 coefficient of determination, r

2

, measures how well a regression model fits the data. A value close to 1 indicates a strong fit, while a lower value indicates that the model does not explain the data as reliably. Here, r

2

=0.45, which is below 0.7. That means the model does not have a strong enough fit for reliable extrapolation in this question. Extrapolation means using the model to predict values outside the observed data range, which is riskier than interpolation. Because the model fit is not strong, choosing an extended range of x-values would not be appropriate. The correct reason is that extrapolation is not appropriate because r

2

< 0.7. Therefore, the answer is D.

The function is:

P(t)=462-35t

This is a linear function in the form:

P(t)=mt+b

where:

m= " rate of change "

and

b= " initial value "

In this function:

m=-35

The negative sign means the fish population is decreasing.

The number 35tells us the amount of decrease per month.

So the fish population is decreasing by:

35 " fish per month "

The value 462is not the rate of change. It represents the starting fish population at the beginning of the year, when t=0:

P(0)=462

Therefore, the correct interpretation is:

" The number of fish in the lake is decreasing at a constant rate of 35 fish per month. "

So the correct answer is:

▭ ( " D " )

The average rate of change compares the change in output to the change in input over an interval. From the graph, the two labeled points are approximately (20,58.5) and (26,91.2). The input is days, and the output is widgets produced each day. Use the formula

x

2

​

−x

1

​

y

2

​

−y

1

​

​

Substituting gives

26−20

91.2−58.5

​

=

6

32.7

​

=5.45, which rounds to approximately 5.5. Since this value is an average rate, it means the team’s daily production increased by about 5.5 widgets each day during the interval. It is not the total number of additional widgets over the whole interval. Therefore, the correct answer is A.

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 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 horizontal asymptote describes the long-term behavior of the function. In this problem, the graph shows a decreasing exponential model for the number of licensed sellers. The graph begins near a higher value and then decreases, eventually leveling off near 300. This means the number of licensed sellers approaches 300 over time. It does not mean the number necessarily becomes exactly 300 at 21 months, and it does not mean the sellers increase back to the initial value. The initial value is separate from the horizontal asymptote. Since the graph levels off around 300, the correct interpretation is that in the long run, the number of licensed sellers nears 300. Therefore, the answer is C.

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 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 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.

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 " )

The function is:

f(x)=330× 〖 1.15 〗 ^x

Because xrepresents the number of years since 2005:

2006 ⇒ x=1

2011 ⇒ x=6

The average yearly rate of change from 2006 to 2011 is:

(f(6)-f(1))/(6-1)

First, find f(1):

f(1)=330× 〖 1.15 〗 ^1

f(1)=330×1.15

f(1)=379.50

Now find f(6):

f(6)=330× 〖 1.15 〗 ^6

f(6)≈330×2.31306

f(6)≈763.31

Now calculate the average rate of change:

(763.31-379.50)/(6-1)

=383.81/5

=76.762

Rounded to two decimal places:

76.76

So the average yearly rate of change of the artifact’s value from 2006 to 2011 is approximately:

$76.76 " per year "

Therefore, the correct answer is:

▭ ( " A " )

This is an exponential growth problem because the base 1.13 is greater than 1. The function starts with 2,900 bacteria and grows by a factor of 1.13 each hour, which represents a 13% hourly increase. To find the population after 18 hours, substitute t=18 into the function: f(18)=2900(1.13)

18

Evaluating the exponent and multiplying by 2900 gives approximately 26,170.38. Since bacteria are counted as whole organisms, this rounds to about 26,170 bacteria. The other answer choices do not match the exponential model at t=18. Therefore, the correct answer is C.

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 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 question asks us to read and interpret a value from a graph.

The horizontal axis represents:

" Days since the beginning of the month "

The vertical axis represents:

" Practice duration in minutes "

We need to find the practice duration for:

" day " 9

To do this, locate 9on the horizontal axis, move upward to the blue graph, and then read the corresponding value on the vertical axis.

From the graph, the point at day 9is approximately:

127

So the musician practiced for:

127 " minutes "

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 question is about reading an ordered pair from a graph.

An ordered pair has the form:

(x│y)

where:

x= " time in hours "

and

y= " audience size "

From the graph:

Point Ais at approximately (0│402)

Point Bis at (2│401)

Point Cis at (4│403)

Point Dis at (6│406)

Point Eis at approximately (8│410)

Point Fis at (10│408)

Now compare these graph points with the answer choices.

Choice A says:

(2│401)

This matches point Bon the graph exactly.

The other choices do not match the graph:

(4│404)

is incorrect because at x=4, the graph shows about 403, not 404.

(8│409)

is incorrect because at x=8, the graph is about 410, not 409.

(10│407)

is incorrect because at x=10, the graph shows about 408, not 407.

In this function-notation problem, the input represents last year’s number of people and the output represents this year’s expected number. Because this year’s number is 7 more than last year’s, the rule is A(t)=t+7. The question says 440 people auditioned last year, so the correct input is 440. Substituting gives A(440)=440+7=447. This means if 440 people auditioned last year, then 447 are expected to audition this year. The notation must place 440 inside the parentheses because that is the input value, and 447 must be written as the output value. The other options reverse the input and output or subtract instead of adding. Therefore, option C is correct.

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

The function is:

G(t)=1,600× 〖 1.31 〗 ^t

This is an exponential growth function because the base is greater than 1:

1.31 > 1

We need to find the population after 25 minutes, so substitute:

t=25

into the function:

G(25)=1,600× 〖 1.31 〗 ^25

Evaluate the exponent and multiply:

G(25)≈1,367,421.41

Since the population is a number of bacteria cells, round to the nearest whole number:

G(25)≈1,367,421

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 "