a. [tex]4\leq x\leq 12[/tex] and [tex]x\geq 19[/tex]
b. x ≅ 15 and x ≅ 7 because the plot of [tex]p(x)[/tex] intersects or meets the horizontal line [tex]p= 100[/tex] at both intervals
c. 12. 5
The population increased at an average rate of 12,500 fruit bats per year between 1994 and 2002
d. -3. 5
The population decreased at an average rate of 3,500 fruit bats per year between 1990 and 2010
How to determine the statements
a. This is easily determined from examining the plot. is clearly increasing for and (more or less).
[tex]4\leq x\leq 12[/tex] and [tex]x\geq 19[/tex]
b. The population was 10000 wherever the plot of [tex]p(x)[/tex] intersects or meets the horizontal line [tex]p= 100[/tex] . This happens twice at x ≅ 15 and x ≅ 7 .
c. The years 1994 and 2002 correspond to x = 4 and x = 12 , respectively.
From the plot, we have p(4) ≅ 50 and p(12)≅ 150
The average rate of change in the population is:
[tex]\frac{p(12) - p(4)}{12 -4}[/tex]
= [tex]\frac{150 - 50}{12 - 4}[/tex]
= 12. 5
Thus explains that the population increased at an average rate of 12,500 fruit bats per year between 1994 and 2002.
d. We have p(0) ≅ 180 and p(20)≅ 110, so the average rate of change over the duration of the study is;
= [tex]\frac{p(20)- p(0)}{20 - 0}[/tex]
= [tex]\frac{110 - 180}{20}[/tex]
= -3. 5
Hence, the population decreased at an average rate of 3,500 fruit bats per year between 1990 and 2010
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