WGU Updated Applied-Algebra Exam Questions and Answers by ewan

The graph shows population in thousands, and the horizontal axis represents:

" years since 1995 "

Point Pis located at approximately:

x=12

Since xrepresents years since 1995:

1995+12=2007

So point Pcorresponds to the year:

2007

The instantaneous rate of change at point Pis:

3.74

Because the vertical axis is measured in thousands, this means:

3.74 " thousand animals per year "

Convert thousands to actual animals:

3.74×1000=3740

So in 2007, the population was increasing by approximately:

3,740 " animals per year "

We are given the function:

d(x)=14+65x

This is a linear function because it has the form:

d(x)=mx+b

where 65is the rate of change and 14is the starting value.

We need to find:

d(1.5)

That means substitute x=1.5into the function:

d(1.5)=14+65(1.5)

Now multiply:

65(1.5)=97.5

Then add:

d(1.5)=14+97.5

d(1.5)=111.5

So the distance from the tower to the object at 1.5seconds is:

111.5 " meters "

From week 1 to week 7, the graph is increasing and becoming steeper.

In Applied Algebra, when a graph is increasing and its slope is getting larger, the quantity is said to be:

" increasing faster and faster "

This interval is on the early increasing part of the logistic curve, before the graph begins to level off.

Therefore, the correct answer is:

▭ ( " B " )

The value r^2measures how well a regression model fits the data.

Here:

r^2=0.25

This is a low r^2value, meaning the model explains only about:

25%

of the variation in the data.

That indicates a weak fit. When a model has a weak fit, using it for extrapolation is not reliable because predictions outside the data range may be very inaccurate.

Since:

0.25 < 0.3

the best conclusion is that extrapolation is not appropriate because the regression model does not fit the data well enough.

Therefore, the correct answer is:

▭ ( " C " )