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.

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.