A sample of 71 eggs yields a sample mean weight of 2.00 ounces. Assuming that the population standard deviation = 0.590 ounces, find the standard error of the mean in estimating a confidence interval estimate for the true population mean weight at the 95% level of confidence. Round your answer to three decimal places

0.070 is the standard error of the mean weight in estimating a confidence interval estimate        

Step-by-step explanation:

We are given the following in the question:

Sample mean, =  2.00 ounces

Sample size, n = 71

Alpha, α = 0.05

Population standard deviation, σ = 0.590 ounces

Formula for standard error =

Putting values, we get,

Thus, 0.070 is the standard error of the mean weight in estimating a confidence interval estimate

25.2

step-by-step explanation:

2+4x6-8/10= 25.2

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