The distribution of life spans for gas ovens (G) has a mean of 15.0 years and a standard deviation of 4.2 years. The distribution of life spans for electric ovens (E) a mean of 13.4 years and a standard deviation of 3.7 years. Both distributions are moderately skewed to the right. Suppose we select a simple random sample of 35 gas ovens and a second simple random sample of 40 electric ovens. What is the probability that the sample mean lifespan for the electric ovens is greater than the sample mean lifespan of the gas ovens?

answer: approximately 206 bones

i hope this u out a branlist and a would be greatly appreciated :

17

step-by-step explanation:

51/3=17

thus 17 is your answer.

x^2 + 81

we can factor this using difference of square binomials.

rewrite the equation:

x^2 - (-81)

factor:

(a^2 - b^2) = (√a^2 + √b^2)(√a^2 - √b^2)

(x^2 - (-81) = (x + √-81)(x - √-81)

we can remove the - from under the radical by multiplying it by i.

(x^2 - (-81) = (x + i√81)(x - i√81)

(x^2 - (-81) = (x + 9i)(x - 9i)

we can check our answer by using foil:

x^2 - 9ix + 9ix - 81i^2

x^2 - 81^2

i^2 = -1

x^2 + 81