Answer
sample answer:
The percentage should be lower.
E.G. For $ 10\%$ risk,
the blood alcohol concentration is about 0.02.
Work Step by Step
(sample answer)
17 is about one-sixth of 100.
So the expentancy is that a driver with a 0.08 blood alcohol concentration will be in an accident once every six times he sits at the wheel.
The odds feel dangerous, so I would prefer this to be lower, say once in TEN,
which is $ 10\%$ risk.
In example 9, p.487,
a model is given for the risk based on blood alcohol concentration:
$R=6e^{12.77x} $
which , for x=0.08 yields R=17 (percent)
If we want $R=10,$ solve for x:
$10=6e^{12.77x}\quad/\div 6$
$\displaystyle \frac{10}{6}=e^{12.77x}\qquad$... apply ln( ) to both sides
$\displaystyle \ln\frac{10}{6}=12.77x\qquad/\div 12.77$
$ x=\displaystyle \frac{\ln\frac{10}{6}}{12.77}\approx$0.0173726507139$\approx 0.02$