The number of bass in a lake is given by

Question

P(t) = 2800/1+6e^-0.05t

where t is the number of months that have passed since the lake was stocked with bass.
(a) How many bass were in the lake immediately after it was stocked?
? bass

(b) How many bass were in the lake 1 year after the lake was stocked? (Round your answer to the nearest whole number.)
? bass

(c) What will happen to the bass population as t increases without bound?
1.The bass population will converge to its original size.

2.The bass population will increase without bound.

3.The bass population will increase, approaching 2800.

4.The bass population will stay the same.

5.The bass population will decrease, approaching 0.

Answer 1

Do you mean:

P(t) = 2800/(1+6e^-0.05t)

????

Answer 2

If so try:

http://www.wolframalpha.com/input/?i=plot++2800%2F%281%2B6%28e^-0.05t%29+%29

when t = 0, e^0 = 1
so 2800/7

when t-->oo
2800/( 1 +6/big) = 2800

Answer 3

http://www.wolframalpha.com/input/?i=plot++2800%2F%281%2B6%28e^-%280.05t%29%29+%29

Answer 4

note for t = 400, there are about 2800

http://www.wolframalpha.com/input/?i=plot++2800%2F%281%2B6%28e^-%280.05t%29%29+%29++for+t+%3D+400