Suppose the average automobile insurance claim for the population is $4,560 with standard deviation of $600. What is the percentage of probability that a sample of 100 claim files will yield an average claim of $4,500 or less?

0.15866

Step-by-step explanation:

Given that :

Population mean = μ = 4560

Standard deviation, σ = 600

Sample size, n = 100

Using the relation :

(x - μ) ÷ σ/sqrt(n)

P(x ≤ 4500) :

(4500 - 4560) ÷ 600/sqrt(100)

P(x ≤ 4500) = -60 / (600/10)

P(x ≤ 4500) = - 60 / 60

P(x ≤ 4500) = - 1

P(Z ≤ - 1) :

Using the Z probability calculator ;

P(Z ≤ - 1) = 0.15866

Hence, P(x ≤ 4500) = 0.15866

the answer is (4,3)

since the rule for r0, 90º is (-y,x) and it was already applied to form the image, all you have to do is reverse the answer using this formula.

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15