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Mamont248 [21]
3 years ago
10

A fitness center is interested in finding a 95% confidence interval for the mean number of days per week that Americans who are

members of a fitness club go to their fitness center. Records of 246 members were looked at and their mean number of visits per week was 2.2 and the standard deviation was 2.6. Round answers to 3 decimal places where possible.
a. To compute the confidence interval use a ? distribution.
b. With 95% confidence the population mean number of visits per week is between and visits.
c. If many groups of 246 randomly selected members are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of visits per week and about percent will not contain the true population mean number of visits per week

Mathematics
1 answer:
fgiga [73]3 years ago
4 0

Hey Mate !

Here is your expert answer. If you do like this accurate answer. Please Mark as Brainliest !

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How do you solve this? (#11/#12)
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3 0
3 years ago
Help Q-Q 7/10c =4 1/5
shepuryov [24]

\huge\mathfrak\green{answer}

= q -   \frac{q7}{10} c =  \frac{41}{5}

\red{multiply}

= q -  \frac{cq7}{10} =  \frac{41}{5}

\red{subtract \: q \: from \: both \: sides}

= q -  \frac{cq7}{10}  - q =  \frac{41}{5} - q

\red{now \: simplify}

=  -  \frac{cq7}{10}  =  \frac{41}{5}  - q

\red{multiply \: 10 \: both \: sides}

= 10( -  \frac{cq7}{10} ) = 10. \frac{41}{5}  - 10q

\red{simplify}

=  - cq7 = 82 - 10q

\red{now \: divide}

=  \frac{ - cq7}{ - q7}  -  \frac{82}{ - q7}  -  \frac{10q}{ - q7}

\red{simplify}

= c =   - \frac{82 - 10q}{q7} (q≠0)

Brainliest? :) (I'd really appreciate if you mark me as brainliest)

\huge\mathfrak\green{thank \: you}

5 0
3 years ago
Read 2 more answers
Show work for this scientific notation
Viktor [21]

\frac{5.1*10^{-2}}{4.57*10^{-4}} \\=\frac{5.1*10^{2}}{4.57} \\=\frac{5.1}{4.57}*10^{2}\\=1.1*10^{2}

The answer is rounded to the nearest tenth.

7 0
3 years ago
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