Solve.
-A lake is stocked with 578 fish of a new variety. The size of the lake, the availability of food, and the number of other fish restrict growth in the lake to a limiting value of 3612. The population of fish in the lake after time $t$, in months, is given by the function
$$P(t)=\frac{3612}{1+4.98 \mathrm{e}^{-0.32 \mathrm{t}}} .$$
Find the population after 20 months.
A) 3597
B) 3582
C) 3572
D) 3592

Question

Quizplus · Accepted Answer

The Answer of Solve.
-A lake is stocked with 578 fish of a new variety. The size of the lake, the availability of food, and the number of other fish restrict growth in the lake to a limiting value of 3612. The population of fish in the lake after time $t$, in months, is given by the function
$$P(t)=\frac{3612}{1+4.98 \mathrm{e}^{-0.32 \mathrm{t}}} .$$
Find the population after 20 months.
A) 3597
B) 3582
C) 3572
D) 3592

Solve -A Lake Is Stocked with 578 Fish of a New

Solve
-A Lake Is Stocked with 578 Fish of a New