Pass the WGU Courses and Certificates Applied-Algebra Questions and answers with ExamsMirror

From t=6.2to about t=8.5, the graph is increasing, but it is becoming less steep as it approaches a local maximum.

That means the number of customers is:

" increasing slower and slower "

After the local maximum, from about t=8.5to t=10.1, the graph begins to decrease and becomes steeper downward.

That means the number of customers is:

" decreasing faster and faster "

So the full description is:

" The number of customers increases slower and slower and then decreases faster and faster. "

Therefore, the correct answer is:

▭ ( " D " )

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 function M(t)represents the company’s market share after tyears.

The horizontal axis represents:

t= " time in years "

The vertical axis represents:

M(t)= " market share percentage "

We need to find when the market share reaches:

40%

From the graph, locate 40on the vertical axis. Then move horizontally to the blue curve and read the corresponding value on the horizontal axis.

The curve reaches 40%at approximately:

t=7.4

So the company should expect to have a market share of 40%after about:

7.4 " years "

Therefore, the correct answer is:

▭ ( " C " )

The function is:

p(m)=5m+5

This is a linear function in the form:

p(m)=mx+b

For a linear function, the average rate of change over any interval is the same as the slope.

The slope is the coefficient of m:

5

So the price increases at a constant rate of:

$5 " per month "

We can also verify using the average rate of change formula:

(p(10)-p(1))/(10-1)

Find p(10):

p(10)=5(10)+5=55

Find p(1):

p(1)=5(1)+5=10

Now calculate:

(55-10)/(10-1)

=45/9

=5

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 P(t), where:

t= " years since opening "

and

P(t)= " daily profit in hundreds of dollars "

The graph is a downward-opening curve, so its maximum value occurs at the highest point of the graph, also called the vertex.

From the graph, the highest point occurs at approximately:

t=9

So the maximum daily profit happened approximately:

9 " years after opening "

The vertical value at this maximum point is approximately:

P(t)=29

But the graph measures profit in hundreds of dollars, so:

29×100=2900

Therefore, the maximum daily profit was approximately:

$2,900

So the correct interpretation is:

" Approximately 9 years after opening, a maximum daily profit of approximately " $2,900 " was earned. "

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. "

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.

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.

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

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.