You have used the significant level of 0.01 to estimate the average hours of television viewing for the nj students. what is the chance that our interval estimate does not contain the true population mean? a.99% b.10% c.100% d.1% e. none of the above

1%

Step-by-step explanation:

Given : You have used the significant level of 0.01 to estimate the average hours of television viewing for the NJ students.

To Find :What is the chance that our interval estimate does not contain the true population mean?

Solution:

We are given that You have used the significant level of 0.01 to estimate the average hours of television viewing for the NJ students.

So, α=0.01

Significance level indicates a risk of concluding that a difference exists when there is no actual difference.

So, The chance that our interval estimate does not contain the true population mean is 0.01

So, The chance that our interval estimate does not contain the true population mean is

Hence Option D is true

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)