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%
