A simple random sample of 125 SAT scores has a mean of 1522. Assume that all SAT scores have a standard deviation of 333. Construct a 99% confidence interval of the mean SAT score.
Z D/2 = 2.58

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step-by-step explanation:

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answer: 5 17/33 % (to the tenth)

step-by-step explanation:

he guessed: 34.5 pounds

actual weight is 32.7

there for he is off:

34.5 -32.7 = 1.8

32.7 is the 100%

1.8 represents %? of 32.7

therefore:

32.7   - 100%

1.8     -     x%

x= 1.8*100%/32.7% = 18*100%/327%

x=1800%/327%   = (5 165/327)%

to the tenth:

5 17/33 %

true

step-by-step explanation:

it is more then likely a review to you prepare for an upcomign test or quiz