A publisher wants to estimate the mean length of time (in minutes) all adults spend reading newspapers. To determine this estimate, the publisher takes a random sample of 15 people and obtains the results below. From past studies, the publisher assumes the standard deviation is 2.2 minutes and that the population of times is normally distributed.

Results: 9 11 8 9 7 11 12 6 10 10 10 6 12 8 12

Construct the 90% and 99% confidence intervals for the population mean.

A. The 90% confidence level is ___ , ___ . (Round to 1 decimal place as needed)
B. The 99% confidence level is ___ , ___ (Round to 1 decimal place as needed)
C. Which interval is wider?

step-by-step explanation:

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d. 8

step-by-step explanation:

the given radical expression is:

we can rewrite this as:

we express as square numbers to get;

the exponents will now cancel out to give.

the correct answer is d

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!

answer: 12   they all have at least 28 days

step-by-step explanation: haha