The living spaces of all homes in a city have a mean of 2300 square feet and a standard deviation of 500 square feet. Let x be the mean living space for a random sample of 25 homes selected from this city. Find the mean and standard deviation of the sampling distribution of x.

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;

3 3 6, because in every triangle the longer side is always shorter than the sum of the other two

the square root of 16 is 4.

4*4 = 16

answer: the correct option is d.

explanation:

it is given that the triangle j'k'l' shown on the grid below is a dilation of triangle jkl using the origin as the center of dilation.

from the given graph it is noticed that k(4,8) and k'(2,4). let origin is defined by o and scale factor is defined by k, then

    (1)

distance formula,

put these values in equation (1).

therefore scale factor is     and option d is correct.