Respuesta :

[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.

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