The number of hours worked per year per person in a state is normally distributed with a standard deviation of 39. A sample of 15 people is selected at random, and the number of hours worked per year per person is given below. Calculate the 98% confidence interval for the mean hours worked per year in this state. Round your answers to the nearest integer and use ascending order.

The 98% confidence interval for the population mean number of hours worked per year per person is (2146, 2193).

Step-by-step explanation:

The question is incomplete.

The number of hours registered in the sample are:

205120612162216721692171

218021832186219521962198

220522102211

The sample mean can be calculated as:

We have to calculate a 98% confidence interval for the mean.

The population standard deviation is know and is σ=39.

The sample mean is M=2169.67.

The sample size is N=15.

As σ is known, the standard error of the mean (σM) is calculated as:

The z-value for a 98% confidence interval is z=2.326.

The margin of error (MOE) can be calculated as:

Then, the lower and upper bounds of the confidence interval are:

The 98% confidence interval for the population mean is (2146, 2193).

hello from mrbilldoesmath!

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

discussion:

in general, the factorization of a^3 + b^3 is

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

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

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

you,

mrb

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