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
X=number of miles
$329=$175+$0.70x
329-175=.70x
154=.70x
154/.70=x
220miles =x
Answer:
B
Step-by-step explanation:
all you need is to solve the first term
2x² - (-3x²) = 5x² ,
b is the only answer with 5x² in it
Coordinates: (3, -3)
this point R would be the same distance that point P is from point Q, fulfilling the statement of it being an isosceles triangle
