Alonzo Odom

2022-07-20

The concentration of alcohol in a person's bloodstream is measurable. Suppose that the relative risk R of having an accident while driving a car can be modeled by an equation of the form $R={e}^{kx}$ where x is the percent of concentration of alcohol in the bloodstream andk is a constant. Suppose that a concentration of alcohol in the bloodstream of 0.03 percent results in a relative risk of an accident of 1.4. Find the constant k in the equation.

Kyan Hamilton

Beginner2022-07-21Added 12 answers

To solve the problem, substitute the given values of x and R in the given equation and solve for the value of k.

$1.4={e}^{0.03k}$ Rewrite in natural logarithmic form.

$0.03k=\mathrm{ln}1.4$ Divide both sides by 0.03

$\frac{0.03k}{0.03}=\frac{\mathrm{ln}1.4}{0.03}$

$k=\frac{0.3365}{0.03}$

k=11.22

Thus, the value of the constant k is 11.22

Result:

11.22

$1.4={e}^{0.03k}$ Rewrite in natural logarithmic form.

$0.03k=\mathrm{ln}1.4$ Divide both sides by 0.03

$\frac{0.03k}{0.03}=\frac{\mathrm{ln}1.4}{0.03}$

$k=\frac{0.3365}{0.03}$

k=11.22

Thus, the value of the constant k is 11.22

Result:

11.22

Jadon Melendez

Beginner2022-07-22Added 4 answers

Substitute 0.03 to x and 1.4 to R in the given equation to find the value of k:

$R)={e}^{kx}$

$1.4={e}^{k(0.03)}$

$1.4={e}^{0.03k}$

Use the rule $a={e}^{b}\Rightarrow \mathrm{ln}a=b$ to obtain:

$\mathrm{ln}1.4=0.03k$

$\frac{\mathrm{ln}1.4}{0.03}=\frac{0.03k}{0.03}$

$\frac{\mathrm{ln}1.4}{0.03}=k$

Use a calculator to obtain:

k=11.21574122

$\approx 11.2157$

Thus, $k=\frac{\mathrm{ln}1.4}{0.03}\approx 11.2157$

Result:

$k=\frac{\mathrm{ln}1.4}{0.03}\approx 11.2157$

$R)={e}^{kx}$

$1.4={e}^{k(0.03)}$

$1.4={e}^{0.03k}$

Use the rule $a={e}^{b}\Rightarrow \mathrm{ln}a=b$ to obtain:

$\mathrm{ln}1.4=0.03k$

$\frac{\mathrm{ln}1.4}{0.03}=\frac{0.03k}{0.03}$

$\frac{\mathrm{ln}1.4}{0.03}=k$

Use a calculator to obtain:

k=11.21574122

$\approx 11.2157$

Thus, $k=\frac{\mathrm{ln}1.4}{0.03}\approx 11.2157$

Result:

$k=\frac{\mathrm{ln}1.4}{0.03}\approx 11.2157$

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