Question

The population of fish in a certain lake follows the logistic growth function f(t)=25000/1+8.25e^-0.076 where t is the time in years. According to the growth model, what will the populations be after 20 years

Answer 1

Answer:

The population will be approx 8915.

Step-by-step explanation:

The population of fish in a certain lake follows the logistic growth function -

$f(t)=\frac{25000}{1+8.25e^{-0.076t} }$

We have to find the population when t = 20

$f(20)=\frac{25000}{1+8.25e^{-0.076(20)} }$

=> $\frac{25000}{1+8.25e^{-1.52} }$

=> $\frac{25000}{1+8.25(0.21871188)}$

=> $\frac{25000}{1+1.8044}$

=> $\frac{25000}{2.8044}$

= 8914.56 or approx 8915.

The population will be approx 8915.

Answer 2

The population of fish in a certain lake follows the logistic growth function

$f(x)= \frac{25000}{1+8.25e^{-0.076t}}$

where t is the time in years. According to the growth model, what will the populations be after 20 years?

We are given with growth model f(x). In that , t represents the time in years.

To find out the population after 20 years, we plug in 20 for t and find out f(20)

$f(x)= \frac{25000}{1+8.25e^{-0.076*20}}$

=8914.65

The population after 20 years = 8914.65