MATH SOLVE

2 months ago

Q:
# The function f(x)=125(0.9)x models the population of a species of fly in millions after x years.How does the average rate of change between years 11 and 15 compare to the average rate of change between years 1 and 5?The average rate of change between years 11 and 15 is about 13 the rate between years 1 and 5.The average rate of change between years 11 and 15 is about 3 times the rate between years 1 and 5.The average rate of change between years 11 and 15 is about 12 the rate between years 1 and 5 .The average rate of change between years 11 and 15 is about 2 times the rate between years 1 and 5 .

Accepted Solution

A:

Given that the population is modeled by:

f(x)=125(0.9)^x

the average rate of change will be as follows:

Year 11 and 15:

f(11)=125(0.9)^11

=39.226

f(15)=125(0.9)^15

=25.736

rate of change=(25.736-39.226)/(15-11)

=-3.37

year 1 and 5

f(1)=125(0.9)^1

=112.5

f(5)=125(0.9)^5

=73.811

rate of change:

(73.811-112.5)/4

=-9.67

dividing the two results of rate of change we get:

-9.67/-3.37

=2.97-=3

From the above we conclude that the average rate of change between year 1 to 5 is above 3 times that of year 11 to year 15

f(x)=125(0.9)^x

the average rate of change will be as follows:

Year 11 and 15:

f(11)=125(0.9)^11

=39.226

f(15)=125(0.9)^15

=25.736

rate of change=(25.736-39.226)/(15-11)

=-3.37

year 1 and 5

f(1)=125(0.9)^1

=112.5

f(5)=125(0.9)^5

=73.811

rate of change:

(73.811-112.5)/4

=-9.67

dividing the two results of rate of change we get:

-9.67/-3.37

=2.97-=3

From the above we conclude that the average rate of change between year 1 to 5 is above 3 times that of year 11 to year 15