Answer
(a) $329.3\ g$
(b) $13.9\ days$
Work Step by Step
Given $A(t)=600e^{-0.05t}$, we have:
(a) for $t=12$, $A(12)=600e^{-0.05(12)}\approx329.3\ g$
(b) for $A=\frac{600}{2}=300$, we have $300=600e^{-0.05t} \longrightarrow e^{-0.05t}=0.5 \longrightarrow -0.05t=ln(0.5)$, thus $t=-\frac{ln(0.5)}{0.05}\approx13.9\ days$
