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A Lake Is Stocked with 422 Fish of a New P(t)=26371+5.05e0.39tP ( t ) = \frac { 2637 } { 1 + 5.05 e ^ { - 0.39 t } }

Question 143

Multiple Choice

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) =26371+5.05e0.39tP ( 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

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