8a)
[tex]\bf \textit{Amount of Population Growth}\\\\
A=Ie^{rt}\qquad
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\\
\end{cases}\qquad
\begin{cases}
year\ 0\\
-----\\
t=0\\
A=42679
\end{cases}
\\\\\\
42679=Ie^{k0}\implies 42679=I\cdot 1\implies 42679=I\qquad thus
\\\\\\
\boxed{A=42679e^{rt}}[/tex]
now, let's fast-forward 8 years later, t = 8, the population has dropped to 33247, so A = 33247, what's "r", or in your paper, it uses "k", anyhow, is just the rate.
[tex]\bf A=42679e^{rt}\qquad
\begin{cases}
t=8\\
A=33247
\end{cases}\implies 33247=42679e^{r8}
\\\\\\
\cfrac{33247}{42679}=e^{8r}\impliedby \textit{let's now take \underline{ln} to both sides}
\\\\\\
ln\left( \frac{33247}{42679} \right)=ln(e^{8r})\implies ln\left( \frac{33247}{42679} \right)=8r\implies \cfrac{ln\left( \frac{33247}{42679} \right)}{8}=r
\\\\\\
-0.0312178\approx r\qquad now\ 100\cdot r\approx -3.1\%[/tex]
is a negative rate, because, the population is decreasing.
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8)
how many days does it take for 60% of the population to know about the crime? well, what's 60% of 1,150,000? well (60/100) * 1,150,000, or 69000.
[tex]\bf P(t)=1150000e^{-0.03t}\implies 690000=1150000e^{-0.03t}
\\\\\\
\cfrac{690000}{1150000}=e^{-0.03t}\implies \cfrac{3}{5}=e^{-0.03t}
\\\\\\
\textit{again, we take \underline{ln} to both sides}
\\\\\\
ln\left( \frac{3}{5} \right)=ln(e^{-0.03t})\implies ln\left( \frac{3}{5} \right)=-0.03t
\\\\\\
\cfrac{ln\left( \frac{3}{5} \right)}{-0.03}=t\implies 17.0275\approx t[/tex]
so.. roughly about 17 days.
