Chapter 10.3, Problem 21CCR

a.

To determine

To find:What is the percent error in using $0.1$ in a calculation instead of the correct value of $\frac{1}{3}$ ?

Answer to Problem 21CCR

The percent error in using $0.1$ in a calculation instead of correct value of $\frac{1}{3}$ is $70\%$ .

Explanation of Solution

Given: Baseball data presentations may show the decimals $0.1$ and $0.2$ to represent $\frac{1}{3}$ and $\frac{2}{3}$ , respectively. For example, $7\frac{2}{3}$ innings would be displayed as $7.2$ .

Calculation:

Finding percent error,

  $\begin{array}{l}\text{percent}\text{error=}\frac{\text{amount}\text{of}\text{error}}{\text{actual}\text{error}}\times \text{100}\\ \text{Amount of error}=\left|\frac{1}{10}-\frac{1}{3}\right|=\frac{7}{30}\\ \text{percent}\text{error}=\frac{\frac{7}{30}}{\frac{1}{3}}\\ \text{percent}\text{error}=\frac{7}{10}\\ \text{percent}\text{error}=70\%\end{array}$

Thus we can find the percent errorerror in using $0.1$ in a calculation instead of correct value of $\frac{1}{3}$ is $70\%$ .

b.

To determine

To find: What is the percent error in using $6.2$ in a calculation instead of the correct value of $6\frac{2}{3}$ ?

Answer to Problem 21CCR

The percent error in using $6.2$ in a calculation instead of correct value of $6\frac{2}{3}$ is $7\%$ .

Explanation of Solution

Given: Baseball data presentations may show the decimals $0.1$ and $0.2$ to represent $\frac{1}{3}$ and $\frac{2}{3}$ , respectively. For example, $7\frac{2}{3}$ innings would be displayed as $7.2$ .

Calculation:

Finding percent error,

  $\begin{array}{l}\text{percent}\text{error=}\frac{\text{amount}\text{of}\text{error}}{\text{actual}\text{error}}\times \text{100}\\ \text{Amount of error}=\left|6\frac{2}{10}-6\frac{2}{3}\right|=\frac{7}{15}\\ \text{percent}\text{error}=\frac{\frac{7}{15}}{6\frac{2}{3}}\\ \text{percent}\text{error}=\frac{7}{100}\\ \text{percent}\text{error}=7\%\end{array}$

Thus we can find the percent errorerror in using $6.2$ in a calculation instead of correct value of $6\frac{2}{3}$ is $7\%$ .

c.

To determine

To compare:Percent errors in part A and part B.

Answer to Problem 21CCR

The percent error in approximating $\frac{1}{3}$ as $0.1$ is greater than approximating $6\frac{2}{3}$ as $6.2$ .

Explanation of Solution

Given: Baseball data presentations may show the decimals $0.1$ and $0.2$ to represent $\frac{1}{3}$ and $\frac{2}{3}$ , respectively. For example, $7\frac{2}{3}$ innings would be displayed as $7.2$ .

Comparison:

Percent error in approximating $\frac{1}{3}$ as $0.1$ is $70\%$ from part A.

Percent error in approximating $6\frac{2}{3}$ as $6.2$ is $7\%$ from part B.

Therefore, the percent errorin approximating $\frac{1}{3}$ as $0.1$ is greater than approximating $6\frac{2}{3}$ as $6.2$ .