WGU Applied-Algebra Exam With Confidence Using Practice Dumps

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 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 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 " )