Passed Exam Today Applied-Algebra

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 " )

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 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 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 " )