sofiandreagarcia
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Steven uses a
caliper that has an absolute error of 0.02 millimeters. An object's width is measured
using the caliper, which records a width of 14.5 millimeters. What equation would represent the situation?

A. |x-0.02|=14.5
B. |x-14.5|=0.02
C. |0.02-x|=14.5
D. |14.5-0.02|=x

Respuesta :

fichoh

The relationship between the measured value, error and actual width of the object is |x - 14.5| = 0.02

  • The measurement error = 0.02 millimeters
  • The measured width of caliper = 14.5 millimeters

If the actual width of the caliper is represented as x

  • The measurement error is an absolute value which can be calculated thus :

  • |Actual value - measured value| = absolute error

  • |x - 14.5| = 0.02

Therefore, the equation which represents the scenario described is |x - 14.5| = 0.02

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