Applied-Algebra WGU Applied Algebra FXO2 PFXP C957 Questions and Answers

Because the vehicle is traveling away from the town at a fixed rate, the distance can be modeled by a linear function:

d(t)=mt+b

where:

m= " rate of change "

and

b= " initial distance from the town "

We are given two points:

(1│200)

and

(4│395)

These mean:

After 1hour, the vehicle is 200miles away.

After 4hours, the vehicle is 395miles away.

First, find the rate of change:

m=(395-200)/(4-1)

m=195/3

m=65

So the vehicle is moving away from the town at a rate of:

65 " miles per hour "

Now the function has the form:

d(t)=65t+b

Use the point (1│200)to find b:

200=65(1)+b

200=65+b

b=135

Therefore, the function is:

d(t)=65t+135

Check using t=4:

d(4)=65(4)+135

d(4)=260+135

d(4)=395

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 " )

The graph shows the total value of an investment account as time increases.

The horizontal axis represents:

" Time in years "

The vertical axis represents:

" Total value in dollars "

We need to find the value of the account after:

8 " years "

From the graph, when t=8, the curve is slightly above $2,200. The closest answer choice is:

$2,221.12

So the value of the account after 8 years is approximately:

$2,221.12

Therefore, the correct answer is:

▭ ( " D " )

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 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 " )

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 " )

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

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 "

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 function F(n) gives the number of animals in enclosure B.

The input n represents the number of animals in enclosure A.

The problem says enclosure B has 20 more animals than enclosure A. The phrase “20 more than” means we add 20:

F(n)=n+20

For example, if enclosure A had 30 animals, then enclosure B would have:

F(30)=30+20=50

So the function that represents the situation is:

F(n)=n+20

The graph shows a decreasing exponential model.

The vertical axis represents:

" Licensed sellers "

The horizontal axis represents:

" Time in years "

The graph starts near 700 licensed sellers and decreases over time. However, it does not appear to decrease all the way to 0. Instead, it levels off near:

y=100

That horizontal line is the horizontal asymptote.

A horizontal asymptote represents the long-term value the function approaches as time continues.

So the correct interpretation is:

" In the long run, the number of licensed sellers nears 100. "

Therefore, the correct answer is:

▭ ( " C " )

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 " )

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 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 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 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 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. "

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 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

This is also a logistic growth curve.

Analyze the interval from month 2 to month 6:

The graph is increasing during this time.

The curve is becoming steeper, meaning the slope is increasing.

This indicates the rate of change is getting larger over time.

So, the number of affected acres is:

" increasing faster and faster "

This corresponds to the early-to-middle growth phase of a logistic function.

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.

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 " )

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 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 "

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 " )