in 2009 there was an endangered population of 270 cranes in a western state. Due to wildlife efforts, the population is increasing at a rate of 5% per year.

AWhat exponential function would be a good model for this population of cranes?
BIf this trend continues, how many cranes will there be in this population in 2020?

Verified Answer :

assuming compount interest format

\(A=P(r+1)^t\) for compounded per 1 year
A=future amount
P=present amount
r=rate in decimal
t=time in years


given
P=270
r=5%=0.05

the equaton is
\(A=270(0.05+1)^t\) or
\(A=270(1.05)^t\)
for any year, 2009, is year 0, so if you wanted to input the year then
\(A=270(1.05)^{t-2009}\) would be for t=what year it was

A. \(f(x)=270(1.05)^t\)


b. 2009 to 2020
2020-2009=11 years
t=11
\(f(11)=270(1.05)^{11}\)
f(11)=461.792
about 462 cranes


A. \(f(x)=270(1.05)^t\)
B. about 462