The number of bass in a lake is given by
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?
400 bass
(b) How many bass were in the lake 1 year after the lake was stocked? (Round your answer to the nearest whole number.) __________
I've got the answer to A just not B.
and (c) is 2800
but (b) you actually have to calculate
first you do not mean what you typed
you mean
P(t) = 2800/ ( 1 + 6e^−0.05t )
now it t = 12 months
p = 2800 / ( 1 + 6 /e^.6)
= 2800 / 4.29
= 652
USE PARENTHESES for denominators with multiple terms
To find the number of bass in the lake 1 year after it was stocked, we substitute t = 12 into the equation and evaluate it:
P(t) = 2800 / (1 + 6e^(-0.05t))
P(12) = 2800 / (1 + 6e^(-0.05*12))
= 2800 / (1 + 6e^(-0.6))
≈ 2800 / (1 + 6*0.5488)
≈ 2800 / (1 + 3.2928)
≈ 2800 / 4.2928
Now, we can calculate the value:
P(12) ≈ 651.4177
Rounding this to the nearest whole number, we get:
P(12) ≈ 651
Therefore, there were approximately 651 bass in the lake 1 year after it was stocked.
To find the number of bass in the lake 1 year after it was stocked, we need to substitute the value of t as 12 months in the given equation:
P(t) = 2800 / (1 + 6e^(-0.05t))
P(12) = 2800 / (1 + 6e^(-0.05 * 12))
Now, we can calculate this value using the order of operations (PEMDAS/BODMAS) to evaluate the expression within the brackets first:
P(12) = 2800 / (1 + 6e^(-0.6))
Next, calculate the value of e^(-0.6) using the exponential function:
e^(-0.6) ≈ 0.54881
Now, substitute this value back into the equation:
P(12) = 2800 / (1 + 6 * 0.54881)
Next, multiply 6 by 0.54881:
P(12) = 2800 / (1 + 3.29286)
Add 1 to 3.29286:
P(12) = 2800 / 4.29286
Finally, divide 2800 by 4.29286:
P(12) ≈ 652.23
Therefore, the number of bass in the lake 1 year after it was stocked (rounded to the nearest whole number) is approximately 652.