There are approximately 3,000 bass in a lake. The population grows at a rate of 2% per year. Round answers to the nearest tenth.
2 answers:
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Answer:
A. 7,348
Step-by-step explanation:
P = le^kt
intitial population = 500
time = 4 hrs
end population = 3,000
So we have all these variables and we need to solve for what the end population will be if we change the time to 6 hours. First, we need to find the rate of the growth(k) so we can plug it back in. The given formula shows a exponencial growth formula. (A = Pe^rt) A is end amount, P is start amount, e is a constant that you can probably find on your graphing calculator, r is the rate, and t is time.
A = Pe^rt
3,000 = 500e^r4
now we can solve for r
divide both sides by 500
6 = e^r4
now because the variable is in the exponent, we have to use a log
ln(6) = 4r
we can plug the log into a calculator to get
1.79 = 4r
divide both sides by 4
r = .448
now lets plug it back in
A = 500e^(.448)(6 hrs)
A = 7351.12
This is closest to answer A. 7,348