Respuesta :
A(t) = A0e^(kt)
A (t) ?
A0 34000
E constant
K -0.121
T 2015-2000=15 years
A (15)=34,000×e^(−0.121×15)=5,536
A (t) ?
A0 34000
E constant
K -0.121
T 2015-2000=15 years
A (15)=34,000×e^(−0.121×15)=5,536
Answer:
Population of bats in year 2015 is 5536.
Step-by-step explanation:
The given function A(t) = [tex]A_{0}e^{-kt}[/tex] predicts the population of bats after t years.
Here A(t) = population after 't' years
k = growth constant
t = time in years
[tex]A_{0}[/tex] = initial population
In year 2000 population of bats A(t) = 34,000
Since rate of decrease in population k = 0.121 per year
So population in 2015 will be
A(t) [tex]A_{0}e^{(-0.121)(15)}[/tex]
= 34,000 [ [tex]e^{-1.815}[/tex] ]
= 34,000 ( 0.162837)
= 5536 bats
Therefore, population of bats in year 2015 is 5536.