A lake is stocked with 422 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 2637. The population of fish in the lake after time t, in
Months, is given by the function P(t) $P ( t ) = \frac { 2637 } { 1 + 5.05 e ^ { - 0.39 t } }$ . Find the population after 1 months.
A) 597 fish
B) 612 fish
C) 587 fish
D) 607 fish

Question

Quizplus · Accepted Answer

To find the population after 1 month, substitute t = 1 into the function: $P(1) = \frac{2637}{1 + 5.05 e^{-0.39 \times 1}}$. Calculating this gives approximately 597 fish.

A Lake Is Stocked with 422 Fish of a New P(t)=26371+5.05e−0.39tP ( t ) = \frac { 2637 } { 1 + 5.05 e ^ { - 0.39 t } }P(t)=1+5.05e−0.39t2637​

A Lake Is Stocked with 422 Fish of a New $P ( t ) = \frac { 2637 } { 1 + 5.05 e ^ { - 0.39 t } }$