Select the correct answer.
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.
Park A
Park B
Years since 2005 Swallow Population
0 1,370
1 1,870
2 2,077
3 2,236
4 2,370
5 2,488
6 2,595
7 2,693
8 2,784
9 2,870
10 2,951
A nonlinear graph represents years since 2005 on x-axis and population in hundreds on f of x, with blue line starting at (12, 0) and slowly increasing to (10, 29)
A nonlinear graph represents years since 2005 on x-axis and population in hundreds on f of x, with blue line starting at (12, 0) and slowly increasing to (10, 29)
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.
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.
To find the rates of change in population for both parks between 2009 and 2014, we need to compare the population at those times in both parks.
For Park A:
Population in 2009: 2,870
Population in 2014: 2,951
Change in population = 2,951 - 2,870 = 81
For Park B:
Population in 2009: 2,784
Population in 2014: 2,488
Change in population = 2,488 - 2,784 = -296
The rate of change in population for Park A between 2009 and 2014 is (81/5) = 16.2 swallows per year.
The rate of change in population for Park B between 2009 and 2014 is (-296/5) = -59.2 swallows per year.
Therefore, Park A showed faster growth between 2009 and 2014, at an average rate of 16.2 swallows per year. The correct answer is C.