Respuesta :
Solution
Given:
[tex]\begin{gathered} \text{The model } \\ p(t)=65,000(1-e^{-0.0009t}) \end{gathered}[/tex]Where t = number of hours
We are required to find the number of hours until 20% and 60% of the population have heard the rumor.
(A) Find the number of hours until 20% of the population have heard the rumor:
20% of the population =
[tex]\begin{gathered} \frac{20}{100}\text{ x total population} \\ =\frac{20}{100}\text{ x 65,000} \\ =13,000 \end{gathered}[/tex][tex]\begin{gathered} \text{Substitute p(t) = 13000 into the model} \\ p(t)=65,000(1-e^{-0.0009t}) \\ 13,000=65,000(1-e^{-0.0009t}) \\ \frac{13000}{65000}=1-e^{-0.0009t} \\ 0.2=1-e^{-0.0009t} \\ 0.2-1=-e^{-0.0009t} \\ -0.8=-e^{-0.0009t} \\ e^{-0.0009t}=0.8 \\ \text{take ln of both sides} \\ \ln e^{-0.0009t}=\ln 0.8 \end{gathered}[/tex][tex]\begin{gathered} -0.0009t\ln e=-0.22314 \\ -0.0009t(1)=-0.22314 \\ t=\frac{-0.22314}{-0.0009} \\ t=247.933 \\ t=248\text{ hours (nearest hour)} \end{gathered}[/tex]The number of hours until 20% of the population have heard the rumor is 248 hours
(b) Find the number of hours until 60% of the population have heard the rumor
60% of the population =
[tex]\begin{gathered} \frac{60}{100}\text{ x 65000} \\ =39000 \end{gathered}[/tex][tex]\begin{gathered} \text{Substitute p(t) =39000 into the model} \\ p(t)=65,000(1-e^{-0.0009t}) \\ 39000=65,000(1-e^{-0.0009t}) \\ \text{Divide both sides by 65000} \\ \frac{39000}{65000}=\frac{65,000}{65000}(1-e^{-0.0009t}) \\ \\ 0.6=1-e^{-0.0009t} \\ 0.6-1=-e^{-0.0009t} \\ -0.4=-e^{-0.0009t} \\ 0.4=e^{-0.0009t} \\ e^{-0.0009t}=0.4 \\ \text{take ln of both sides} \\ \ln e^{-0.0009t}=\ln 0.4 \\ -0.0009t\ln e=\ln 0.4 \\ -0.0009t(1)\text{ = -0.9162} \\ -0.0009t\text{= -0.9162} \\ t=\frac{-0.9162}{-0.0009} \\ t=1018.1 \\ t=1018hours\text{ (nearest hour)} \end{gathered}[/tex]The number of hours until 60% of the population have heard the rumor is 1018 hours
