Respuesta :

[tex]\textit{Amount for Exponential Decay using Half-Life} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{current amount}\\ P=\textit{initial amount}\dotfill &6368\\ t=\textit{elapsed time}\dotfill &542\\ h=\textit{half-life}\dotfill &271 \end{cases} \\\\\\ A=6368\left( \frac{1}{2} \right)^{\frac{542}{271}}\implies A=6368\left( \frac{1}{2} \right)^2\implies A=1592[/tex]

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