2 years ago

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Mathematics

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Maurinko [17]2 years ago

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I dont see a question.

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During the first stages of an epidemic, the number of sick people increases exponentially with time. Suppose that at = 0 days th

nydimaria [60]

T = days passed

r = rate of growth

by 0 day, or t = 0, there are 2 folks sick,

$\bf \qquad \textit{Amount for Exponential Growth}
\\\\
A=P(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &2\\
P=\textit{initial amount}\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\to &0\\
\end{cases}
\\\\\\
2=P(1+r)^0\implies 2=P\cdot 1\implies 2=P\qquad \boxed{A=2(1+r)^t}$

by the third day, t = 3, there are 40 folks sick,

$\bf \qquad \textit{Amount for Exponential Growth}
\\\\
A=P(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &40\\
P=\textit{initial amount}\to &2\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\to &3\\
\end{cases}
\\\\\\
40=2(1+r)^3\implies 20=(1+r)^3\implies \sqrt[3]{20}=1+r
\\\\\\
\sqrt[3]{20}-1=r\implies 1.7\approx r\qquad \boxed{A=2(2.7)^t}$

how many folks are there sick by t = 6? $\bf \stackrel{that~many}{A=2(2.7)^6}$