A random sample of 42 textbooks has a mean price of $114.50 and a standard deviation of $12.30.
Find a 98% confidence interval for the mean price of all textbooks.

Step-by-step explanation:

We want to determine a 98% confidence interval for the mean price for textbooks.

Number of sample, n = 42

Mean, u = $114.5

Standard deviation, s = $12.3

For a confidence level of 98%, the corresponding z value is 2.33.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

114.5 ± 2.33 × 12.3/√42

= 114.5 ± 2.33 × 1.898

= 114.5 ± 4.42

The lower end of the confidence interval is 114.5 - 4.42 = 110.08

The upper end of the confidence interval is 114.5 + 4.42 = 118.92

Therefore, with 98% confidence interval, the mean price of all textbooks is between $110.08 and $118.92

the length of o'p' is .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

the scale factor is 1.5.

multiply both sides by op.

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is .

yes

step-by-step explanation:

they have to be opposite and reciprocal of eachother.