Estimate the value of √97.5÷1.96

The downtime per day for a computing facility has mean 4 hours and standard deviation .8 hour. Suppose that we want to compute p

xz_007 [3.2K]

Answer:

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

3 0

3 years ago

T= 1,-1 u= 2, -4

Monica [59]

Answer: 3.16 units (Option C)

Step-by-step explanation:

The key to this problem is to use the distance formula, which is:

Distance = $\sqrt{(x₁-x₂)²+(y₁-y₂)²}$

The first point, T = (x₁,y₁), and the second point, U = (x₂,y₂).

Plugging the two points into the equation, we get:

Distance = $\sqrt{(1-2)^{2}+(-1-(-4))^{2} }$

The values within the parenthesis are subtracted:

Distance = $\sqrt{(-1)^{2}+3^{2} }$

The values are then squared:

Distance = $\sqrt{1+9}$

Finally, they are added together:

Distance = $\sqrt{10}$

$\sqrt{10}$ can be approximated as 3.16, so the distance between the two points is 3.16 units.

7 0

3 years ago

Laura can weed the garden in 1 hour and 20 minutes and her husband can weed it in 1 hour and 30 minutes. How long will they take

Harrizon [31]

<h2>Hello!</h2>

The answer is:

It will take 42.35 minutes to weed the garden together.

To solve the problem, we need to use the given information about the rate for both Laura and her husband. We know that she can weed the garden in 1 hour and 20 minutes (80 minutes) and her husband can weed it in 1 hour and 30 minutes (90 minutes), so we need to combine both's work and calculate how much time it will take to weed the garden together.

So, calculating we have:

Laura's rate:

$\frac{1garden}{80minutes}$