Given
[tex]P(t)=\frac{1200}{1+9e^{-0.156t}}[/tex]Find
Population after 5 months.
when will population reach 500.
Explanation
We have given
[tex]P(t)=\frac{1200}{1+9e^{-0.156t}}[/tex]so population after 5 months.
that is t =5
[tex]\begin{gathered} P(5)=\frac{1200}{1+9e^{-0.156(5)}} \\ \\ P(5)=\frac{1,200}{1+9e^{-0.78}} \\ \\ P(5)=\frac{1,200}{1+9\times0.458406} \\ \\ P(5)=\frac{1,200}{5.125654} \\ \\ P(5)=234.116466 \end{gathered}[/tex]Population reach 500
[tex]\begin{gathered} 500=\frac{1200}{1+9e^{-0.156t}} \\ 1+9e^{-0.156t}=2.4 \\ 9e^{-0.156t}=1.4 \\ t=\frac{\ln(\frac{1.4}{9})}{-0.156} \\ t=\frac{\ln(0.156)}{-0.156} \\ t=11.90 \end{gathered}[/tex]t = 11.90 months
Final Answer
a) P(5) = 234
b) t = 11.90 months
