After experiencing a period of rapid increase beginning in 2005, the populations of swallows in two different national parks level off. The approximate populations are modeled here, where x is time, in years, since 2005. Find the rates of change in population for both parks between 2009 and 2014, and determine which park showed faster population growth during those years. HINT: It's not C. GRAPHS BELOW

A. Park B showed faster growth between 2009 and 2014, at an average rate of 100 swallows per year.
B. Park A showed faster growth between 2009 and 2014, at an average rate of 100 swallows per year.
C. Park A showed faster growth between 2009 and 2014, at an average rate of 120 swallows per year.
D. Park B showed faster growth between 2009 and 2014, at an average rate of 120 swallows per year.

Answer: D. Park B showed faster growth between 2009 and 2014, at an average rate of 120 swallows per year.

Step-by-step explanation:

I did the math

The Park B showed faster growth between 2009 and 2014, at an average rate of 100 swallows per year i.e. option A.

The average rate is defined as a measure of how much the function changed per unit, on average, over that interval.

We have,

x is time, in years, since 2005,

and,

f(c) is population in hundreds.

Now,

For Park A,

at x = 4 ,

Swallows Population = 2370

and,

at x = 9,

Swallows Population = 2870

So,

Average Rate [tex]=\frac{2870 - 2370}{10} = 50[/tex]

Now,

For Park B,

at x = 4,

Swallows Population = 2200

and,

at x = 9,

Swallows Population = 2800

So,

Average Rate [tex]=\frac{2800- 2200}{10} = 60[/tex]

So,

Difference in average rate = 60 - 50 = 10

From the average rate of of both park A and Park B,

We can see that,

Park B has faster growth rate of 100 per year.

Hence, we can say that the Park B showed faster growth between 2009 and 2014, at an average rate of 100 swallows per year i.e. option A.

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