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:
Option c is right:
(-6 , 2)
Step-by-step explanation:
( -6 , 2)
By <span>(x + 5/x^2 + 9x + 20) you apparently meant the following:
x+5
----------------------
x^2 + 9x + 20
and by
</span><span>(x^2-16/x-4)
x^2-16
you apparently meant ---------------
x - 4
Please use additional parentheses for clarity.
Dividing,
</span> x+5 (x-4)(x+4)
---------------------- * ---------------
x^2 + 9x + 20 x-4
Now, x^2 + 9x + 20 factors into (x+4)(x+5), so what we have now is
(x+5)(x+4)
------------------------- = 1 This is true for all x, so there are no exclusions.
(x+4)(x+5)
solve for "k", to find k or the "constant of variation"
then plug k's value back to
now.... what is "p" when q = 5? well, just set "q" to 5 on the right-hand-side, and simplify, to see what "p" is
