from the Fermium 257, he's got 100Kgs of Fermium 252, and Fermium 253.
the 252 is 40%, well, 40% of 100 is just 40, so 40 of Fermium 252, with a half-life of 1 day.
and 60% of 100 is just 60, so 60Kgs of Fermium 253 with a half-life of 3 days.
from August 31st to September 15, that'd be 15 days later, so how much is left altogether after 15 days.
[tex]\bf \stackrel{\textit{Fermium 252}}{\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}\dotfill &40\\ t=\textit{elapsed time}\dotfill &15\\ h=\textit{half-life}\dotfill &1 \end{cases} \\\\\\ A=40\left( \frac{1}{2} \right)^{\frac{15}{1}}\implies A=40\left( \frac{1}{2} \right)^{15}\implies A\approx \boxed{0.00122} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]\bf \stackrel{\textit{Fermium 253}}{\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}\dotfill &60\\ t=\textit{elapsed time}\dotfill &15\\ h=\textit{half-life}\dotfill &3 \end{cases}[/tex]
[tex]\bf A=60\left( \frac{1}{2} \right)^{\frac{15}{3}}\implies A=60\left( \frac{1}{2} \right)^5\implies \boxed{A=1.875} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{0.00122+1.875}{1.87622} ~\hfill[/tex]
