Using the function below, how do you solve all parts below of the problem?
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The DEEP is trying to make a Walleye Pike fishery in Tyler Lake. Stocking began in 2006 and the population is expected to be given by the function: p(t)=2500/1+e^-.06t (just so u know, "-.06t" is the exponent to e) a.)How many walleyes were stocked originally? b.)What is the current population? c.)What will the population be in 2025? d.)Will the population ever reach 2000? When? e.)How large will the population get?
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Answer:
a. 2500/2 = 1250 <------- b. 2500/(1+e^(-.06*6)) = 1473 <------- c. 2500/(1+e^(-.06*19)) = 1894 <------- d. 1+e^(-.06t) = 2500/2000 = 1.25, e^(-.06t) = .25 t = ln .25 / -.06 ≈ 23 years, ie 2029 <------ e. 2500 is the limiting population <-------
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Other answers
a. 2500/2 = 1250 <------- b. 2500/(1+e^(-.06*6)) = 1473 <------- c. 2500/(1+e^(-.06*19)) = 1894 <------- d. 1+e^(-.06t) = 2500/2000 = 1.25, e^(-.06t) = .25 t = ln .25 / -.06 ≈ 23 years, ie 2029 <------ e. 2500 is the limiting population <-------
M3
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