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aliina [53]
3 years ago
14

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 er

ror in the measurements, to the nearest tenth of a percent?
Mathematics
1 answer:
larisa [96]3 years ago
3 0
In this question, <span>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:
</span>percent error = amount of error/ actual measurement
percent error = 1.25m/ 45.5m *100%=2.7% 
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