Chapter 5.4, Problem 133AYU

(a)

To determine

The constant $k$ in the equation of the relative risk $R=e^{kx}$ , when the concentration of alcohol in the bloodstream of $0.03$ percent results in a relative risk $R$ of an accident of $1.4$

Answer to Problem 133AYU

Solution:

The constant $k$ in the equation of the relative risk $R=e^{kx}$ is $11.2157$

Explanation of Solution

Given information:

The relative risk $R$ of having an accident while driving a car can be modelled by an equation of the form $R=e^{kx}$ , where $x$ is the percent concentration of alcohol in the bloodstream and $k$ is the constant

Here, the concentration of alcohol in the bloodstream of $0.03$ percent results in a relative risk $R$ of an accident of $1.4$ .

The constant $k$ is found by substituting the values $x=0.03$ and $R=1.4$ in the equation $R=e^{kx}$

  $1.4=e^{0.03k}$

Applying the logarithm on both sides

  $\ln (1.4)=\ln (e^{0.03k})$

Apply the law of logarithm given as $\log (a^m)=m(\log \text{}a)$

  $0.3364722366=0.03k\ln (e)$ ,

  $\ln e=1$

  $0.3364722366=0.03k$

Divide both sides by $0.03$

  $\frac{0.3364722366}{0.03}=\frac{0.03k}{0.03}$

  $\Rightarrow k=11.2157$

The constant $k$ in the equation of the relative risk, $R=e^{kx}$ is $11.2157$

(b)

To determine

The relative risk of having an accident if the concentration of alcohol in bloodstream is $0.17$ percent.

Answer to Problem 133AYU

Solution:

The relative risk of having an accident is $6.7315$ , when the concentration of alcohol in bloodstream is $0.17$ percent

Explanation of Solution

Given information:

The relative risk $R$ of having an accident while driving a car can be modelled by an equation of the form $R=e^{kx}$ , where $x$ is the percent concentration of alcohol in the bloodstream.

The concentration of alcohol in bloodstream is $0.17$ percent.

From part(a) the relative risk is given as $R=e^{11.2157x}$

The concentration is given as $0.17$ percent

Plug $x=0.17$ in $R=e^{11.2157x}$ ,

  $R=e^{11.2157(0.17)}$

  $R=e^{1.906669}$ ,

  $R=6.73063$

The relative risk of having an accident is $6.73063$ , when the concentration of alcohol in bloodstream is $0.17$ percent

(c)

To determine

The concentration of alcohol in bloodstream when relative risk is $100$.

Answer to Problem 133AYU

Solution:

Concentration of alcohol in bloodstream is $0.41$ percent when relative risk is $100$.

Explanation of Solution

Given information:

The relative risk $R$ of having an accident while driving a car can be modelled by an equation of the form $R=e^{kx}$ , where $x$ is the percent concentration of alcohol in the bloodstream.

From part(a) as $k=11.2157$ , the relative risk is modelled by an equation as $R=e^{11.2157x}$

As the relative risk is $R=100$

Therefore, $100=e^{11.2157x}$

Applying the logarithm on both sides

  $\ln (100)=\ln (e^{11.2157x})$

Apply the law of logarithm which is given as $\log (a^m)=m(\log \text{}a)$

  $4.60517=11.2157x\ln (e)$

As $\ln e=1$

  $4.60517=11.2157x$

Divide both sides by $11.2157$

  $\frac{4.60517}{11.2157}=\frac{11.2157x}{11.2157}$

  $\Rightarrow x=0.4106$

Concentration of alcohol in the bloodstream is $0.4106$ percent, when relative risk is $100$

(d)

To determine

At what concentration of alcohol in the bloodstream, the driver should be arrested and charged with a DUI.

Answer to Problem 133AYU

Solution:

For $0.1434$or greater percent, concentration of alcohol in the bloodstream, the driver should be arrested and charged with a DUI.

Explanation of Solution

Given information:

The relative risk $R$ of having an accident while driving a car can be modelled by an equation of the form $R=e^{kx}$ , where $x$ is the percent concentration of alcohol in the bloodstream.

The law asserts that anyone with a relative risk of having an accident of 5 or more should not have driving privileges.

From part (a) $k=11.2157$ ; therefore, the relative risk is modelled by an equation as $R=e^{11.2157x}$

The law asserts that anyone with a relative risk of having an accident of 5 or more should not have driving privileges.

Hence, $R=5$

  $\Rightarrow 5=e^{11.2157x}$

Applying the logarithm on both sides

  $\ln (5)=\ln (e^{11.2157x})$

Apply the law of logarithm which is given as $\log (a^m)=m(\log \text{}a)$

  $1.6094=11.2157x\ln (e)$

  $\ln e=1$

  $1.6094=11.2157x$

Divide both sides by $11.2157$

  $\frac{1.6094}{11.2157}=\frac{11.2157x}{11.2157}$

  $\Rightarrow x=0.1434$

For $0.1434$ or greater concentration of alcohol in bloodstream the driver should be arrested and charged with a DUI.

(e)

To determine

Which situation you would support if you are the lawmaker.

Answer to Problem 133AYU

Solution:

If I would be a lawmaker I would support that the limit of alcohol concentration should be $0.14$percent.

Explanation of Solution

Given information:

The relative risk $R$ of having an accident while driving a car can be modelled by an equation of the form $R=e^{kx}$ , where $x$ is the percent concentration of alcohol in the bloodstream.

The law asserts that anyone with a relative risk of having an accident of 5 or more should not have driving privileges.

For the relative risk of having an accident, the alcohol percentage in the bloodstream should not be greater than $0.14$ percent

Thus, if I would be a lawmaker, I would suggest that for $0.14$ percent or greater than $0.14$ percent concentration of alcohol in the bloodstream, the driver should be arrested and charged with a DUI.