Since xrepresents the number of years since 2003:
2003 # x=0
2007 # x=4
The average rate of change from 2003 to 2007 is:
(f(4)-f(0))/(4-0)
Find f(0):
f(0)=280× # 1.08 # ^0
f(0)=280
Find f(4):
f(4)=280× # 1.08 # ^4
f(4)#380.94
Now calculate:
(380.94-280)/4
=100.94/4
#25.23
Rounded to the nearest whole number:
25
Since the value is positive, the number of subscribers is increasing.
So the correct interpretation is:
" The number of subscribers increases at an average rate of 25 subscribers per year. " Therefore, the correct answer is:
# ( " C " )

The function is:
P(t)=24t+185
This is a linear function in the form:
P(t)=mt+b
where:
m= " rate of change "
and
b= " initial value "
In the function:
P(t)=24t+185
the coefficient of tis:
24
So the rate of change is:
24 " fish per month "
Because 24is positive, the fish population is increasing.
The number 185is not the rate of change. It represents the starting fish population at the beginning of the year, when t=0:
P(0)=24(0)+185=185
Therefore, the correct interpretation is:
" The number of fish in the lake is increasing at a constant rate of 24 fish per month. " So the correct answer is:
# ( " B " )

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 " )

This is also a logistic growth curve.
Analyze the interval from month 2 to month 6:
The graph is increasing during this time.
The curve is becoming steeper, meaning the slope is increasing.
This indicates the rate of change is getting larger over time.
So, the number of affected acres is:
" increasing faster and faster "
This corresponds to the early-to-middle growth phase of a logistic function.