In 1995, an aircraft set the around the-world record time for a passenger jet at 31h,27min,49s. Six refueling stops took a total of 8h, 48min,8s.
2006-10-07 13:54:56 · 5 answers · asked by kachilous_2002 1 in Science & Mathematics ➔ Mathematics
The total number of seconds the trip took is
31*3600+27*60+49 = 113269
The total number of seconds spent refueling is
8*3600+48*60+8 = 31688
Therefore, the percentage of time spent in the air is
100*(113269-31688)/113269 = 72.0241%.
2006-10-07 13:58:54 · answer #1 · answered by James L 5 · 0⤊ 0⤋
First turn everything in to seconds, and divide.
31h 27min 49s = 113,269 seconds
8h 48min 8s = 31,688 seconds
Subtract out the ground time
113,269 seconds - 31,688 seconds = 81,581 seconds in the air
divide air seconds by total seconds
81,581 / 113,269 = 0.720241195738
Convert to % by multiplying by 100 = 72.0241195738 %
;-D They had time on the ground to have a cuppa coffee.
2006-10-07 21:07:15 · answer #2 · answered by China Jon 6 · 0⤊ 0⤋
41 secs, 39 mins and 22 hrs.
2006-10-07 20:57:27 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Assuming jas is right about the hours
2006-10-07 20:59:28 · answer #4 · answered by ? 6 · 0⤊ 0⤋
Do the math,and get back with us
2006-10-07 20:56:33 · answer #5 · answered by Anonymous · 0⤊ 0⤋