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David measures a line to be 8.33 in long. If the actual measurement is 8 in, find David's relative error to the nearest thousandth.
Answer:

David measures a line to be 88.3333 in long. If the actual measurement is 88 in, find David's relative error to the nearest thousandth.\newlineAnswer:

Full solution

Q. David measures a line to be 88.3333 in long. If the actual measurement is 88 in, find David's relative error to the nearest thousandth.\newlineAnswer:
  1. Understand relative error: Understand the concept of relative error. Relative error is the absolute error divided by the actual measurement, often expressed as a percentage or a decimal. The absolute error is the difference between the measured value and the actual value.
  2. Calculate absolute error: Calculate the absolute error.\newlineThe absolute error is the difference between David's measurement and the actual measurement.\newlineAbsolute error = Measured ValueActual Value|\text{Measured Value} - \text{Actual Value}|\newlineAbsolute error = 8.33 in8 in|8.33 \text{ in} - 8 \text{ in}|\newlineAbsolute error = 0.33 in|0.33 \text{ in}|
  3. Calculate relative error: Calculate the relative error.\newlineThe relative error is the absolute error divided by the actual measurement.\newlineRelative error = Absolute errorActual measurement\frac{\text{Absolute error}}{\text{Actual measurement}}\newlineRelative error = 0.33in8in\frac{0.33 \, \text{in}}{8 \, \text{in}}\newlineRelative error = 0.041250.04125
  4. Convert to nearest thousandth: Convert the relative error to the nearest thousandth.\newlineTo express the relative error to the nearest thousandth, we round it to three decimal places.\newlineRelative error (to the nearest thousandth) = 0.0410.041

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