Answer:
the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes is 0.4215
Step-by-step explanation:
Let consider Q to be the opening altitude.
The mean μ = 135 m
The standard deviation = 35 m
The probability that the equipment damage will occur if the parachute opens at an altitude of less than 100 m can be computed as follows:




If we represent R to be the number of parachutes which have equipment damage to the payload out of 5 parachutes dropped.
The probability of success = 0.1587
the number of independent parachute n = 5
the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes can be computed as:
P(R ≥ 1) = 1 - P(R < 1)
P(R ≥ 1) = 1 - P(R = 0)
The probability mass function of the binomial expression is:
P(R ≥ 1) = 
P(R ≥ 1) =
P(R ≥ 1) = 1 - 0.5785
P(R ≥ 1) = 0.4215
Hence, the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes is 0.4215
Part a) is b part b) is b to
2 11/78 you just have to simply multiply 64/143 x2 because since it can't be subtracted. And after you multiply 64/143, you just have to subtract normally which is 128-21 and 286-208. Then since your answer came out to an improper fraction, so you have to simplify by dividing 107 divided by 78. Then you will get your answer which is 2 11/78.
Answer:
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Step-by-step explanation:
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