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% 

Answer: 900°

good luck on your studies! : )

david edward.