Question

Several students measured a 25-mm-long nail
and wrote the measurements shown in the table
below. Whose measurement had the greatest
percent error? Round to the nearest percent.
Student Measurement
(mm)
Percent
Error
Layne 26 %
Tenicia 23 %
Juan 25 %

Answer 1

To find the percent error for each student's measurement, we use the formula for percent error:

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

In this case, the true value of the length of the nail is 25 mm.

Now, let's calculate the percent error for each student.

### Layne's Measurement
- Measured Value = 26 mm
- True Value = 25 mm

\[
\text{Percent Error} = \left( \frac{|26 - 25|}{25} \right) \times 100\% = \left( \frac{1}{25} \right) \times 100\% = 4\%
\]

### Tenicia's Measurement
- Measured Value = 23 mm
- True Value = 25 mm

\[
\text{Percent Error} = \left( \frac{|23 - 25|}{25} \right) \times 100\% = \left( \frac{2}{25} \right) \times 100\% = 8\%
\]

### Juan's Measurement
- Measured Value = 25 mm
- True Value = 25 mm

\[
\text{Percent Error} = \left( \frac{|25 - 25|}{25} \right) \times 100\% = \left( \frac{0}{25} \right) \times 100\% = 0\%
\]

Now we can summarize the percent errors:

- Layne: 4%
- Tenicia: 8%
- Juan: 0%

### Conclusion
Tenicia had the greatest percent error with a value of 8%.