Answer:
0.1660 ( approx )
Step-by-step explanation:
Given,
The percentage of risk of having a car accident is,
[tex]R=6e^{12.77x}----(1)[/tex]
Where, x is the blood alcohol concentration,
From equation (1),
[tex]\frac{R}{6}=e^{12.77x}[/tex]
Taking natural log both sides,
[tex]ln(\frac{R}{6})=ln(e^{12.77x})[/tex]
[tex]ln(\frac{R}{6})=12.77xln(e)[/tex]
Since, ln(e) = 1,
[tex] ln(\frac{R}{6})=12.77x[/tex]
[tex]\implies x = \frac{ln(\frac{R}{6})}{12.77}[/tex]
R = 50 %,
[tex]\implies x=\frac{ln(\frac{50}{6})}{12.77}[/tex]
[tex]=\frac{2.1202635362}{12.77}[/tex]
[tex]=0.16603473267\approx 0.1660[/tex]
