the answer is 2B divided by A minus 3
Answer:
P( That it will take over 10 years or more of a year with a rainfall above 50inches) = (0.9938)^10
Step-by-step explanation:
Since the annual rainfall is normally distributed,
Given: that
Mean (µ )= 40
and σ = 4.
Let X be normal random variables of the annual rainfall.
P(that there will be over 10 years or more before a year with a rainfall above 50 inches)
P(>50) = 1-P[X ≤50]
1 - P[X- μ/σ ≤ 50 - 40/4]
=1 - P [Z≤ 5/2]
=1 -Φ(5/2)
=1 - 0.9939
= 0.0062
P( the non occurrence of rainfall above 50 inches)
= 1-0.0062
=0.9938
ASSUMPTION:
P( That it will take over 10 years or more of a year with a rainfall above 50inches) =
Take the fouth square root of both sides
![\sqrt[4]{{x}^{4} } = \sqrt[4]{24}](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B%7Bx%7D%5E%7B4%7D%20%7D%20%20%3D%20%20%5Csqrt%5B4%5D%7B24%7D%20)
so
Answer:
n = (4 i)/sqrt(11) or n = -(4 i)/sqrt(11)
Step-by-step explanation:
Solve for n:
11 n^2 + 16 = 0
Subtract 16 from both sides:
11 n^2 = -16
Divide both sides by 11:
n^2 = -16/11
Take the square root of both sides:
Answer: n = (4 i)/sqrt(11) or n = -(4 i)/sqrt(11)