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polet [3.4K]
4 years ago
5

A random sample of 12 lunch orders at noodles and company showed a mean bill of $12.99 with a standard deviation of $4.6. find t

he 98 percent confidence interval for the mean bill of all lunch orders. (round your answers to 4 decimal places.)
Business
1 answer:
kow [346]4 years ago
6 0
12.99 + 4.6 = 17.59 / 98
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