Karen measures the width of a garden plot and records that it is 44.25 meters. It's actual width is 45.5. What is the percent error in the measurements, to the nearest tenth of a percent?

In this question, Karen measures the garden width result in 44.25 m while it's actual width is 45.5m. Then the amount of error that the Karen did would be: 45.5m-44.25m = 1.25m

The percent error would be:

percent error = amount of error / actual measurement

percent error = 1.25m / 45.5m * 100%=2.7%