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David measured the distance from his house to a soccer field as 3 km. The actual distance from his house to the soccer field is 2.84 km. What is the approximate percent error in David’s measurement? Round to the nearest tenth of a percent, if necessary.

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taskmasters

Distance as measured by David from his home to soccer field = 3 km
The actual distance from the house of David to the soccer field = 2.84 km
Then
Amount of error made by David = (3 - 2.84) km
                                                   = 0.16 km
Then
Percentage of error made by David = (0.16/3) * 100
                                                          = 16/3
                                                           = 5.33 percent
So 5.33% error was made by David. I hope the procedure is not very complicated for you to understand and attempt similar problems on your own.

igorburak93

3-2,84 = 0.16
0.16/2.84 * 100% = 5.633 %

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