Y = T+3 (i)
T = H+2 (ii)
Y = 2*H (iiii)
T = H+2 therefore H = T-2
Substitute H=T-2 into (iii)
Y=2*H
Y=2*(T-2)
Y=2T-4
Now substitute that Y into (i)
(i) says Y = T+3
2T-4=T+3
2T-T-4=3
T-4=3
T=3+4
T=7
Then from (i) Y=T+3 = 7+3 =10
Y=10
And from (iii) Y=2*H
If Y = 10, then H = 5
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
Answer:
65ft
Step-by-step explanation:
Answer:
b is equal to negative a and a is also equal to negative b
Step-by-step explanation:
i think the answer is -50 but im
not sure
