Question

Akira is redecorating her bedroom. She misread the markings on her ruler and thought a picture frame she bought was 5.2 centimeters wide. The frame's actual width was 4.8 centimeters. What is the percent error for her measurement?
If necessary, round your answer to the nearest tenth of a percent.
____%

Answer 1

To calculate the percent error for Akira's measurement, we can use the formula for percent error:

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

In this case:
- The measured value (the width Akira thought) = 5.2 cm
- The actual value (the true width of the frame) = 4.8 cm

Now we will substitute these values into the formula:

1. Calculate the absolute error:
\[
| \text{Measured Value} - \text{Actual Value} | = | 5.2 - 4.8 | = 0.4 \text{ cm}
\]

2. Now, substitute into the percent error formula:
\[
\text{Percent Error} = \left( \frac{0.4}{4.8} \right) \times 100
\]

3. Calculate \( \frac{0.4}{4.8} \):
\[
\frac{0.4}{4.8} = \frac{4}{48} = \frac{1}{12} \approx 0.0833
\]

4. Now calculate the percent error:
\[
\text{Percent Error} \approx 0.0833 \times 100 \approx 8.33\%
\]

Rounding to the nearest tenth of a percent, we get:
\[
\text{Percent Error} \approx 8.3\%
\]

Thus, the percent error for Akira's measurement is **8.3%**.