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

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 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 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)=24t+185

This is a linear function in the form:

P(t)=mt+b

where:

m= " rate of change "

and

b= " initial value "

In the function:

P(t)=24t+185

the coefficient of tis:

24

So the rate of change is:

24 " fish per month "

Because 24is positive, the fish population is increasing.

The number 185is not the rate of change. It represents the starting fish population at the beginning of the year, when t=0:

P(0)=24(0)+185=185

Therefore, the correct interpretation is:

" The number of fish in the lake is increasing at a constant rate of 24 fish per month. "

So the correct answer is:

▭ ( " B " )

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 " )

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 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.

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 " )

Since xrepresents the number of years since 2003:

2003 ⇒ x=0

2007 ⇒ x=4

The average rate of change from 2003 to 2007 is:

(f(4)-f(0))/(4-0)

Find f(0):

f(0)=280× 〖 1.08 〗 ^0

f(0)=280

Find f(4):

f(4)=280× 〖 1.08 〗 ^4

f(4)≈380.94

Now calculate:

(380.94-280)/4

=100.94/4

≈25.23

Rounded to the nearest whole number:

25

Since the value is positive, the number of subscribers is increasing.

So the correct interpretation is:

" The number of subscribers increases at an average rate of 25 subscribers per year. "

Therefore, the correct answer is:

▭ ( " C " )

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.