Allison measured a line to be 18.6 inches long. If the actual length of the line is 17.9 inches, then what was the percent error of the measurement, *to the nearest tenth of a percent*?


To calculate the percent error of Allison's measurement, we use the formula:

\[
\text{Percent Error} = \left(\frac{\left|\text{Measured Value} - \text{Actual Value}\right|}{\text{Actual Value}}\right) \times 100
\]

Given:

- Measured Value = 18.6 inches
- Actual Value = 17.9 inches

Now, let's calculate it:

\[
\text{Percent Error} = \left(\frac{|18.6 - 17.9|}{17.9}\right) \times 100 = \left(\frac{0.7}{17.9}\right) \times 100 \approx 3.91\%
\]

To the nearest tenth of a percent, the percent error is **3.9%**.

Would you like more details or have any questions?

Here are 5 related questions:
1. What would the percent error be if the measured value was 19.0 inches?
2. How can percent error help in assessing the accuracy of a measurement?
3. How does the size of the actual value affect the percent error?
4. Can a percent error be negative? Why or why not?
5. What is the percent error if the measured value and the actual value are the same?

**Tip:** Percent error gives you an idea of how close your measurement is to the actual value, helping to identify potential errors.

Learn how to calculate percent error with a detailed example. Allison measured a line to be 18.6 inches long, while the actual length is 17.9 inches. Understand the formula for percent error and its application in measurement accuracy assessment. Suitable for Grades 7-9.

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