WGU Applied-Algebra - WGU Applied Algebra FXO2 PFXP C957

The function V(t)represents the number of travelers in the travel depot.

The input trepresents:

" hours after 7:00 a.m. "

The output V(t)represents:

" number of travelers "

So V(2)means the number of travelers in the depot:

2 " hours after 7:00 a.m. "

From the graph, when:

t=2

the value of V(t)is approximately:

79.2

Since V(t)measures travelers, not hours, V(2)represents:

79.2 " travelers "

Therefore, the correct answer is:

â–­ ( " D " )

The function w(t)gives the number of workers at the job site thours after 6:00 a.m.

We need to find the difference between:

w(4)

and

w(8)

Because w(t)represents workers, the difference will be measured in workers, not hours.

First, evaluate w(4):

w(4)=0.03(4)^4-0.81(4)^3+7.55(4)^2-26.62(4)+44

w(4)=0.03(256)-0.81(64)+7.55(16)-106.48+44

w(4)=7.68-51.84+120.8-106.48+44

w(4)=14.16

Now evaluate w(8):

w(8)=0.03(8)^4-0.81(8)^3+7.55(8)^2-26.62(8)+44

w(8)=0.03(4096)-0.81(512)+7.55(64)-212.96+44

w(8)=122.88-414.72+483.2-212.96+44

w(8)=22.4

Now find the difference:

22.4-14.16=8.24

This is approximately:

8 " workers "

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 function P(t)=31t+438 is linear and follows the form P(t)=mt+b. In this form, m is the rate of change and b is the initial value. Here, the coefficient of t is 31, so the fish population changes by 31 fish per month. Because 31 is positive, the population is increasing rather than decreasing. The number 438 is the initial fish population at the start of the year, not the rate of change. Therefore, the correct interpretation is that the number of fish in the lake is increasing at a constant rate of 31 fish per month. That matches option A.

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The graph shows a curved, downward-opening function. This type of graph is commonly associated with a quadratic polynomial function.

The maximum value of a downward-opening parabola occurs at its highest point, called the vertex.

From the graph, the highest point occurs at approximately:

t=9.5

This means the museum reaches its maximum daily profit approximately:

9.5 " years after opening "

The vertical axis represents daily profit in hundreds of dollars. From the graph, the maximum P(t)-value is approximately:

16.5

Since the profit is measured in hundreds of dollars:

16.5×100=1650

So the maximum daily profit is approximately:

$1,650

Therefore, the correct interpretation is:

" Approximately 9.5 years after opening, a maximum daily profit of approximately " $1,650 " was earned. "

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 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.

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.

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. "

The function a(x)represents the total value of an investment account.

The horizontal axis represents:

x= " time in years "

The vertical axis represents:

a(x)= " total account value in dollars "

We are asked to find the total value when:

x=4

To answer this from the graph:

Locate x=4on the horizontal axis.

Move vertically upward until reaching the blue graph.

Read the corresponding value on the vertical axis.

From the graph, when x=4, the value is slightly above $1,800, approximately:

a(4)≈1,830.59

So the total value of the investment account after 4 years is approximately:

$1,830.59