Question: User: Adrian measured a line to be $13.8$ inches long. If the actual length of the line is $13.1$ inches, then what was the percent error of the measurement, to the nearest tenth of a percent?
1. **State the problem:**
We need to find the percent error of a measurement where the measured length is $13.8$ inches and the actual length is $13.1$ inches.
2. **Formula for percent error:**
$$\text{Percent Error} = \left| \frac{\text{Measured Value} - \text{Actual Value}}{\text{Actual Value}} \right| \times 100\%$$
3. **Substitute the values:**
$$\text{Percent Error} = \left| \frac{13.8 - 13.1}{13.1} \right| \times 100\%$$
4. **Calculate the numerator:**
$$13.8 - 13.1 = 0.7$$
5. **Calculate the fraction:**
$$\frac{0.7}{13.1}$$
6. **Simplify the fraction (showing cancellation):**
$$\frac{\cancel{0.7}}{13.1} = \frac{0.7}{13.1}$$ (no common factors to cancel)
7. **Divide:**
$$\frac{0.7}{13.1} \approx 0.0534351145$$
8. **Multiply by 100 to get percent:**
$$0.0534351145 \times 100 = 5.34351145\%$$
9. **Round to the nearest tenth of a percent:**
$$5.3\%$$
**Final answer:** The percent error of the measurement is approximately $5.3\%$.