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Eric measures a line to be 2.92 in long. If the actual measurement is 3 in, find Eric's relative error to the nearest hundredth. Answer:

Eric measures a line to be 22.9292 in long. If the actual measurement is 33 in, find Eric's relative error to the nearest hundredth.\newlineAnswer:

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Full solution

Q. Eric measures a line to be 22.9292 in long. If the actual measurement is 33 in, find Eric's relative error to the nearest hundredth.\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. The absolute error is the difference between the measured value and the actual value.
  2. Calculate absolute error: Calculate the absolute error.\newlineAbsolute error = ∣Measured Value−Actual Value∣|\text{Measured Value} - \text{Actual Value}|\newlineAbsolute error = ∣2.92 in−3 in∣|2.92 \text{ in} - 3 \text{ in}|\newlineAbsolute error = ∣−0.08 in∣|-0.08 \text{ in}| (Taking the absolute value)\newlineAbsolute error = 0.08 in0.08 \text{ in}
  3. Calculate relative error: Calculate the relative error.\newlineRelative error = Absolute errorActual Value\frac{\text{Absolute error}}{\text{Actual Value}}\newlineRelative error = 0.08 in3 in\frac{0.08 \, \text{in}}{3 \, \text{in}}\newlineRelative error = 0.026666…0.026666\ldots
  4. Round relative error: Round the relative error to the nearest hundredth.\newlineRelative error (to the nearest hundredth) = 0.030.03

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