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Sagot :
Question A.
The initial population ocurrs at t=0. Then, by substituting this value into the given model ,we get
[tex]N(0)=\frac{2040}{1+39e^0}[/tex]which gives
[tex]\begin{gathered} N(0)=\frac{2040}{1+30} \\ N(0)=\frac{2040}{40} \\ N(0)=51 \end{gathered}[/tex]then, the answer is 51 owls.
Question B.
The limits when t approaches to + infinity is
[tex]\begin{gathered} N(0)=\frac{2040}{1+39e^{-\infty}} \\ N(0)=\frac{2040}{1+0} \\ N(0)=\frac{2040}{1}=2040 \end{gathered}[/tex]then, the answer is 2040 owls.
Question 15.
In this case, we need to find t when N(t) is 950, that is,
[tex]950=\frac{2040}{1+39e^{-0.5t}}[/tex]By moving the denominator to the left hand side, we have
[tex](1+39e^{-0.5t})950=2040[/tex]then, by moving 950 to the right hand side, we obtain
[tex]\begin{gathered} (1+39e^{-0.5t})=\frac{2040}{950} \\ 39e^{-0.5t}=\frac{2040}{950}-1 \end{gathered}[/tex]which is
[tex]39e^{-0.5t}=1.147368[/tex]so, we get
[tex]\begin{gathered} e^{-0.5t}=\frac{1.147368}{39} \\ e^{-0.5t}=0.029419 \end{gathered}[/tex]By applying natural logarithms to both sides, we have
[tex]\begin{gathered} -0.5t=\ln (0.029419) \\ t=\frac{-\ln(0.029419)}{0.5} \end{gathered}[/tex]then, the answer is
[tex]t=7.05[/tex]By rounding o the neares interger, the answer is 7 years
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