Step-by-step answer:
Given:
mean, mu = 200 m
standard deviation, sigma = 30 m
sample size, N = 5
Maximum deviation for no damage, D = 100 m
Solution:
Z-score for maximum deviation
= (D-mu)/sigma
= (100-200)/30
= -10/3
From normal distribution tables, the probability of right tail with
Z= - 10/3
is 0.9995709, which represents the probability that the parachute will open at 100m or more.
Thus, by the multiplication rule, the probability that all five parachutes will ALL open at 100m or more is the product of the individual probabilities, i.e.
P(all five safe) = 0.9995709^5 = 0.9978565
So there is an approximately 1-0.9978565 = 0.214% probability that at least one of the five parachutes will open below 100m
The sum is the gcf, if 60 plus 84 is 148 then wouldn't it be 148 that's what they all have in common, you can use a calculator to figure this out like say 148 times 2 equals 296, divide 296 by 60. Does it go in?
Answer: c.50 is the answer
