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kondor19780726 [428]
3 years ago
7

Use the formula R = 6e12.77x, where x is the blood alcohol concentration and R, as a percent, is the risk of having a car accide

nt. What blood alcohol concentration corresponds to a 50% risk of a car accident?
Mathematics
2 answers:
tester [92]3 years ago
6 0
R=6e^{12.77x}
50=6e^{12.77x}
\frac{50}{6}= e^{12.77x}
log \frac{50}{6}=log( e^{12.77x})
0.920818754=12.77x
x=0.07
denis-greek [22]3 years ago
6 0

Answer:

0.1660 ( approx )

Step-by-step explanation:

Given,

The percentage of risk of having a car accident is,

R=6e^{12.77x}----(1)

Where, x is the blood alcohol concentration,

From equation (1),

\frac{R}{6}=e^{12.77x}

Taking natural log both sides,

ln(\frac{R}{6})=ln(e^{12.77x})

ln(\frac{R}{6})=12.77xln(e)

Since, ln(e) = 1,

ln(\frac{R}{6})=12.77x

\implies x = \frac{ln(\frac{R}{6})}{12.77}

R = 50 %,

\implies x=\frac{ln(\frac{50}{6})}{12.77}

=\frac{2.1202635362}{12.77}

=0.16603473267\approx 0.1660

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