Suppose the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 11 million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 114 million dollars

95.73%

Step-by-step explanation:

Given data:

mean μ= 95

standard deviation, σ = 11

to calculate, the probability that a randomly selected firm will earn less than 114 million dollars;

Use normal distribution formula

Substitute the required values in the above equation;

Therefore, the probability that a randomly selected firm will earn less than 114 million dollars = 95.73%

the answer is a

step-by-step explanation:

because if you put 1 in to the square root of x it equals 300.

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb