The accuracy of a certain stopwatch varies by 5%, meaning that the time it shows may be off as much as 5% from the actual time. If the stopwatch shows a time of 3:26 (3 minutes, 26 seconds) for an athlete’s 800m run, then x is the actual time and 0.05x is the potential amount of error. Use this information to write your equation and determine the range of possible times that the run may actually have taken.
Accepted Solution
A:
Answer:3 minutes, 16.19 s ≤ Actual time ≤ 3 minutes, 36.842 sStep-by-step explanation:The accuracy of the stopwatch varies as much of 5%This means that the time it shows can beTime*0.95 ≤ Time ≤ Time*1.05if x is the actual timex- 0.05*x ≤ x ≤ x+ 0.05*xx *0.95 ≤ x ≤ x*1.05If the stopwatch shows 3 minutes, 26 sec = 206 secondsThere are two possible ranges for the actual runx *0.95 = 206 sx = 216.842 sorx *1.05 = 206 sx = 196.190196.190 s ≤ Actual time ≤ 216.842 s3 minutes, 16.19 s ≤ Actual time ≤ 3 minutes, 36.842 s