Given:
[tex]A=\frac{276}{1+11e^{-0.35t}}[/tex]Where A is the number of deer expected in the herd after t years.
We will find the following:
(a) How many deer will be present after 3 years?
So, substitute t = 3 into the given equation:
[tex]A=\frac{276}{1+11e^{-.35*3}}\approx56.9152[/tex]Rounding to the nearest whole number
So, the answer will be A = 57
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(b) How many years will it take for the herd to grow to 50 deer?
substitute A = 50 then solve for t
[tex]\begin{gathered} 50=\frac{276}{1+11e^{-.35t}} \\ 1+11e^{-.35t}=\frac{276}{50} \\ \\ 11e^{-.35t}=\frac{276}{50}-1=4.52 \\ e^{-.35t}=\frac{4.52}{11} \\ -0.35t=ln(\frac{4.52}{11}) \\ \\ t=\frac{ln(\frac{4.52}{11}_)}{-0.35}=2.54 \end{gathered}[/tex]Round your answer to the nearest whole number.
So, the answer will be t = 3
