[tex]\bf \textit{Amount for Exponential Decay using Half-Life}\\\\
A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{initial amount}\to &371\\
t=\textit{elapsed time}\to &45\\
h=\textit{half-life}\to &15
\end{cases}
\\\\\\
A=371\left( \frac{1}{2} \right)^{\frac{45}{15}}\implies A=371\left( \frac{1}{2} \right)^3[/tex]
and surely you know how much that is.
