You can
put this solution on YOUR website!A small lake is stocked with a certain species of fish. The fish population is modeled by the function
P =
where, P is the number of fish in thousands and t is measured in years since the lake was stocked.
:
The fish population (correct up to two decimal places) after 3 years is______
(Remark: don't forget to use the correct units for P .)
:
Substitute 3 for t in the given equation:
:
P =
P =
Find e^-2.4 on a good calc:
P =
P =
P = 7.34 thousand
:
The fish population reaches 5000 after years is __________ (correct up to two decimal places).
:
P = 5
5 =
;
Multiply both sides by =
5
= 10
:
Divide both sides by 5:
= 2
:
Subtract 1 from both sides and you have:
= 1
Divide both sides by 4
= .25
-.8t = ln(.25); remember the ln(e) = 1
:
-.8t = -1.3863
t =
t = 1.733 years
:
:
Check solution by using 1.73 for t in the original equation and find P
P =
P =
P =
P =
P = 5 which is 5000 fish
