The downtime per day for a computing facility has mean 4 hours and standard deviation .8 hour. Suppose that we want to compute probabilities about the average daily downtime for a period of 30 days. i. What assumptions must be true to use the result of Theorem 7.4 to obtain a valid approximation for probabilities about the average daily downtime?
ii. Under the assumptions described in part (i), what is the approximate probability that the average daily downtime for a period of 30 days is between 1 and 5 hours?

1

Step-by-step explanation:

Given :

Mean, μ = 4

Standard deviation, s = 0.8

Sample size, n = 30

The distribution is independent.

Z = (x - μ) / s /sqrt(n)

Probability that downtime period is between 1 and 5

P(1≤ x ≤ 5) :

[(x - μ) / (s /sqrt(n))] ≤ Z ≤ [(x - μ) / (s /sqrt(n))]

[(1 - 4) / (0.8 /sqrt(30))] ≤ Z ≤ [(5 - 4) / (0.8 /sqrt(30)]

[-3 / 0.1460593] ≤ Z ≤] 1 / 0.1460593]

P(-20.539602 ≤ Z ≤ 6.8465342)

P(Z ≤ 6.8465342) - P(Z ≤ - 20.5396)

P(Z ≤ 6.8465342) = 1 (Z probability calculator)

P(Z ≤ - 20.5396) = 0 (Z probability calculator)

1 - 0 = 1

hello from mrbilldoesmath!

answer:

x - 3

discussion:

x^2 - 9 = (x-3) (x+3)   so

(x^2-9)/ (x + 3) = x-3

you,

mrb

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240