<span>0.3046% chance of damaging at least one of five payloads. First, determine how many standard deviations it takes for the parachute to open at 100 meters. So 200 m - 100 m = 100 m 100 m / 31 m = 3.2258 Looking up on a standard normal table, the probability of having that deviation or higher is 0.00061, or 0.061% Since you're looking for the probability of damaging at least one of five, the easiest way of calculating that is to subtract from 1 the probability of having all five parachutes successfully deploy. So 1 - (1 - 0.00061)^5 = 1 - 0.99939^5 = 1 - 0.996953719 = 0.003046281 = 0.3046%</span>

8 0

3 years ago

Al goes 8 mph; Ben goes 22 mph in the opposite direction. How many hours has Ben traveled when they are 180 miles apart? 15 6 12

Triss [41]

Your answer is 6 I believe

3 0

3 years ago