Answer
(a).$m(t)=10\times2^{-t/30}$
(b).$m(t)=10e^{-0.02311t}$
(c).. $m(80)=10\times2^{-2.66}=1.5749$
(d).$t=69.64$
Work Step by Step
$m(t)=m_0e^{-rt}$. Whereas,$m(t)$ is the mass of radioactive substance after time $t$, $m_0$ is the Initial mass of radioactive substance, $r=\frac{\ln 2}{h}$ is the rate of decay while $h$ is the half-life and $t$ is time.
(a).$m_0=10$
$m(t)=m_02^{-t/h}=10\times2^{-t/30}$
(b). $r=\frac{\ln2}{h}=0.02311$
$m(t)=m_0e^{-rt}=10e^{-0.02311t}$
(c). $m(80)=10\times2^{-80/30}=10\times2^{-2.66}=1.5749$
(d). $2=10\times e^{-0.02311t}$,
$0.2=e^{-0.02311t}$,
$\ln(0.2)=-0.02311t$,
$t=69.64$
