Question

jaquan measures the length of his school’s football field, which is expected to be exactly 100 yards long. he records 99 yards as the field’s actual length. what is the percent of error in the football field’s length, to the nearest hundredth?
%

Explanation:

Step1: Find the absolute error

The expected length (accepted value) is 100 yards, and the measured length (experimental value) is 99 yards. The absolute error is \( |100 - 99| = 1 \) yard.

Step2: Calculate percent error

The formula for percent error is \( \text{Percent Error} = \frac{|\text{Accepted Value} - \text{Experimental Value}|}{\text{Accepted Value}} \times 100\% \). Substituting the values, we get \( \frac{1}{100} \times 100\% = 1.00\% \)? Wait, no, wait: Wait, the accepted value is 100, experimental is 99. So \( \frac{|100 - 99|}{100} \times 100\%=\frac{1}{100}\times 100 = 1.00\% \)? Wait, but let's check again. Wait, the formula is (|measured - actual| / actual) 100? Wait, no, percent error is (|accepted - measured| / accepted) 100. So accepted is 100, measured is 99. So |100 - 99| =1. Then 1/100 = 0.01, times 100 is 1.00%? Wait, but that seems off? Wait, no, wait, maybe I mixed up. Wait, the problem is: expected (accepted) is 100, he measured 99. So the error is 100 - 99 =1 (since he measured less than expected). So percent error is (error / accepted) 100. So (1/100)100 =1.00%? Wait, but let's confirm the formula. Percent error formula: \( \text{Percent Error} = \frac{|\text{True Value} - \text{Measured Value}|}{\text{True Value}} \times 100\% \). True value is 100, measured is 99. So \( \frac{|100 - 99|}{100} \times 100\% = \frac{1}{100} \times 100\% = 1.00\% \). Wait, but that seems correct? Wait, no, wait, maybe I made a mistake. Wait, no, the absolute error is 1, divided by true value 100, times 100 is 1 percent. So the percent error is 1.00%? Wait, but let's check again. Wait, 100 - 99 is 1, so 1/100 is 0.01, times 100 is 1. So yes, 1.00%? Wait, but maybe the question is different? Wait, no, the problem says "what is the percent of error in the football field’s length, to the nearest hundredth?". So the calculation is (|100 - 99| / 100)100 = 1.00%? Wait, but that seems too straightforward. Wait, maybe I misread the problem. Let me check again. The football field is expected to be exactly 100 yards long. He records 99 yards as the actual length. So the error is 100 - 99 =1. Percent error is (1/100)100 =1.00%. So that's the calculation.

Answer:

1.00%