For example, you might be measuring the length of a football field. You know that the actual, or accepted length of a professional American football field is 360 feet (including both end zones). So, you would use 360 as the actual value:Δx=x0−360{\displaystyle \Delta x=x_{0}-360}.
For example, if you measure the football field and find that it is 357 feet long, you would use 357 as the measured value:Δx=357−360{\displaystyle \Delta x=357-360}.
For example, since Δx=357−360=−3{\displaystyle \Delta x=357-360=-3}, the absolute error of your measurement is 3 feet.
For example, if you know that the relative error is . 025, your formula will look like this: . 025=x0−xx{\displaystyle . 025={\frac {x_{0}-x}{x}}}.
For example, if you know that the actual value is 360 ft, your formula will look like this: . 025=x0−360360{\displaystyle . 025={\frac {x_{0}-360}{360}}}.
For example:. 025=x0−360360{\displaystyle . 025={\frac {x_{0}-360}{360}}}. 025×360=x0−360360×360{\displaystyle . 025\times 360={\frac {x_{0}-360}{360}}\times 360}9=x0−360{\displaystyle 9=x_{0}-360}
For example:9=x0−360{\displaystyle 9=x_{0}-360}9+360=x0−360+360{\displaystyle 9+360=x_{0}-360+360}369=x0{\displaystyle 369=x_{0}}
For example, if the measured value is 369 ft, and the actual value is 360 feet, you would subtract 369−360=9{\displaystyle 369-360=9}. So, the absolute error is 9 feet.
For example, if the measured length of a building is stated as 357 feet, you know that the building was measured to the nearest foot. So, the measuring unit is 1 foot.
For example, if the unit of measure is a foot, the maximum possible error is . 5 ft. So you might see that the measurement of a building is 357±. 5ft{\displaystyle 357\pm . 5ft}. This means that the actual value of the building’s length could be . 5 ft less or . 5 ft more than the measured value. If it was any less/more, the measured value would have been 356 or 358 feet.
For example, if you find the measurement of a building to be 357±. 5ft{\displaystyle 357\pm . 5ft}, the absolute error is . 5 ft.
