The decay of $200 \mathrm{mg}$ of an isotope is given by $\mathrm{A}(\mathrm{t})=200 \mathrm{e}^{-0.012 \mathrm{t}}$, where $t$ is time in years since the initial amount of $200 \mathrm{mg}$ was present. Find the amount left (to the nearest $\mathrm{mg}$ ) after 71 years.
A) $84 \mathrm{mg}$
B) $43 \mathrm{mg}$
C) $198 \mathrm{mg}$
D) $85 \mathrm{mg}$

Question

Quizplus · Accepted Answer

To find the amount left after 71 years, substitute $t = 71$ into the decay formula: $A(71) = 200e^{-0.012 \times 71}$. Calculating this gives approximately $85 \mathrm{mg}$.

The Decay Of 200mg200 \mathrm{mg}200mg Of an Isotope Is Given By A(t)=200e−0.012t\mathrm{A}(\mathrm{t})=200 \mathrm{e}^{-0.012 \mathrm{t}}A(t)=200e−0.012t

The Decay Of $200 \mathrm{mg}$ Of an Isotope Is Given By $\mathrm{A}(\mathrm{t})=200 \mathrm{e}^{-0.012 \mathrm{t}}$