A researcher wishes to estimate the proportion of adults who have high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.040.04 with 9999% confidence if (a) she uses a previous estimate of 0.540.54? (b) she does not use any prior estimates?

(a) The sample size required is 1034.

(b) The sample size required is 1040.

Step-by-step explanation:

The confidence interval for population proportion is:

The margin of error is:

Given:

The critical value of z for 99% confidence level is:

*Use a standard normal table.

(a)

Compute the sample size required if the previous estimate is as follows:

Thus, the sample size required is 1034.

(b)

Compute the sample size required if there was no previous estimate as follows:

Assume that the estimated proportion be 0.50.

Thus, the sample size required is 1040.

3.how is solving radical equations different than solving linear equations?

step-by-step explanation:

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)