The Half-Life of Phosphorus-32 Is 14 $F=A e^{-0.0485 t}$ , Where $F$

Short Answer

The half-life of phosphorus-32 is 14.29 days and decays according to the formula $F=A e^{-0.0485 t}$ , where $F$ is the final amount, $A$ is the initial amount and $t$ is the time in days. How much phosphorus- 32 is left after 60 days, if the initial amount was $25.0 \mathrm{mg}$ ?