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Answer :
at sea level, the pressure is 1013, that's when the altitude is at 0, sea level, let's see
[tex]\bf \textit{Periodic Exponential Decay}\\\\
A=I(1 - r)^{\frac{t}{p}}\qquad
\begin{cases}
A=\textit{accumulated amount}\to &1013\\
I=\textit{initial amount}\\
r=rate\to 11.8\%\to \frac{11.8}{100}\to &0.118\\
t=\textit{meters climbed}\to &0\\
p=period\to &1000
\end{cases}
\\\\\\
1013=I(1-0.118)^{\frac{0}{1000}}\implies 1013=I\cdot 1\implies 1013=I[/tex]
so, the inital amount is 1013, when t = 0,
[tex]\bf \textit{Periodic Exponential Decay}\\\\
A=I(1- r)^{\frac{t}{p}}\qquad
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\to &1013\\
r=rate\to 11.8\%\to \frac{11.8}{100}\to &0.118\\
t=\textit{meters climbed}\to &t\\
p=period\to &1000
\end{cases}
\\\\\\
A=1013(1-0.118)^{\frac{t}{1000}}\implies A=1013(0.882)^{\frac{t}{1000}}[/tex]
now, to check the atmospheric pressure at 4000, simply set t = 4000, to get A.
[tex]\bf \textit{Periodic Exponential Decay}\\\\
A=I(1 - r)^{\frac{t}{p}}\qquad
\begin{cases}
A=\textit{accumulated amount}\to &1013\\
I=\textit{initial amount}\\
r=rate\to 11.8\%\to \frac{11.8}{100}\to &0.118\\
t=\textit{meters climbed}\to &0\\
p=period\to &1000
\end{cases}
\\\\\\
1013=I(1-0.118)^{\frac{0}{1000}}\implies 1013=I\cdot 1\implies 1013=I[/tex]
so, the inital amount is 1013, when t = 0,
[tex]\bf \textit{Periodic Exponential Decay}\\\\
A=I(1- r)^{\frac{t}{p}}\qquad
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\to &1013\\
r=rate\to 11.8\%\to \frac{11.8}{100}\to &0.118\\
t=\textit{meters climbed}\to &t\\
p=period\to &1000
\end{cases}
\\\\\\
A=1013(1-0.118)^{\frac{t}{1000}}\implies A=1013(0.882)^{\frac{t}{1000}}[/tex]
now, to check the atmospheric pressure at 4000, simply set t = 4000, to get A.
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