WGU Updated Applied-Algebra Exam Questions and Answers by jimmy

The graph shows the temperature of an object changing over time.

The horizontal axis represents:

" Time in minutes "

The vertical axis represents:

" Temperature in degrees Fahrenheit "

The curve is decreasing quickly at first and then begins to level off. This is the shape of an exponential decay function.

A horizontal asymptote is a horizontal line that the graph approaches as time increases.

Because a horizontal asymptote is a horizontal line, its equation must have the form:

y= " constant "

From the graph, the temperature approaches about:

25^ ∘ F

So the horizontal asymptote is:

y=25

This means the object’s temperature gets closer and closer to 25^ ∘ Fover time.

The scatterplot shows data collected over time, and a regression curve is used to model the pattern.

Extrapolation means using a model to make predictions slightly outside the observed data range. In Applied Algebra, extrapolation can be appropriate when:

" the regression model fits the data reasonably well "

and the predicted x-values are not too far outside the observed data.

From the scatterplot, the data points appear to run approximately from:

x=3 " to " x=15

So the observed data range is about:

15-3=12

A reasonable extrapolation range extends a little beyond the data, but not too far. The interval:

x=0 " to " x=18

extends about 3 units beyond each side of the observed data, which is reasonable.

The interval:

x=-3 " to " x=21

extends much farther beyond the data and would be less reliable.

Also, r^2does not have to equal exactly 1 for extrapolation to be useful. A value less than 1 can still represent a strong model.

Therefore, the correct answer is:

▭ ( " A " )

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 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 " )