Answer
(a) $BAC(t) = \frac{1}{2^{t/1.5}}~0.6$
(b) The person can drive home at about 4:20 am
Work Step by Step
(a) $BAC(0) = 0.6$
$BAC(1.5) = 0.3$
$BAC(3) = 0.15$
etc...
$BAC(t) = \frac{1}{2^{t/1.5}}~0.6$
(b) We can see a sketch of the graph below.
On the graph, we can see that the BAC reaches a level of $0.8~mg/mL$ a little before 4.5 hours have passed after midnight.
Therefore, the person can drive home at about 4:20 am
