Answer
(a) $12$g
(b) $m(t)=12e^{-0.1733t}$
(c) $7.135$g
(d) $25.3$g
Work Step by Step
(a) Given $h=4$, use the radioactive decay model $m(t)=m_0e^{-rt}$ with $r=ln2/h=ln2/4=0.1733$ and
$n(20)=0.375$, we have $m_0e^{-0.1733\times20}=0.375$ which gives $m_0\approx12$g
(b) With the values given above, we have the function as $m(t)=12e^{-0.1733t}$
(c) Let $t=3$, we get $m(3)=12e^{-0.1733\times3}=7.135$g
(d) Given $m(t)=0.15$, we have $12e^{-0.1733t}=0.15$ which gives $t=-ln(\frac{0.15}{12})/0.1733\approx25.3$g
