WGU Applied-Algebra Exam With Confidence Using Practice Dumps

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 function is:

P(t)=462-35t

This is a linear function in the form:

P(t)=mt+b

where:

m= " rate of change "

and

b= " initial value "

In this function:

m=-35

The negative sign means the fish population is decreasing.

The number 35tells us the amount of decrease per month.

So the fish population is decreasing by:

35 " fish per month "

The value 462is not the rate of change. It represents the starting fish population at the beginning of the year, when t=0:

P(0)=462

Therefore, the correct interpretation is:

" The number of fish in the lake is decreasing at a constant rate of 35 fish per month. "

So the correct answer is:

▭ ( " D " )

The graph shows a decreasing exponential function.

In Applied Algebra, a decreasing exponential function may approach a fixed value over time. This fixed value is called the horizontal asymptote.

The horizontal asymptote represents the long-term value that the function gets closer and closer to, but does not necessarily cross or reach exactly.

From the graph, the curve decreases quickly at first, then begins to level off near:

y=1,000

This means that as time continues, the number of letters processed per day approaches:

1,000

So the long-term trend is that the mail center will process about:

1,000 " letters per day "