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Licemer1 [7]
3 years ago
5

Fractions uncommon denominators

Mathematics
2 answers:
Maurinko [17]3 years ago
4 0
so so so so sos so true
Nookie1986 [14]3 years ago
3 0

Answer:

ye says he sho wi lo min ointe

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A sample of 56 research cotton samples resulted in a sample average percentage elongation of 8.63 and a sample standard deviatio
ss7ja [257]

To solve for the confidence interval for the true average percentage elongation, we use the z statistic. The formula for confidence interval is given as:

Confidence interval = x ± z σ / sqrt (n)

where,

x = the sample mean = 8.63

σ = sample standard deviation = 0.79

n = number of samples = 56

 

From the standard distribution tables, the value of z at 95% confidence interval is:

z = 1.96

 

Therefore substituting the known values into the equation:

Confidence interval = 8.63 ± (1.96) (0.79) / sqrt (56)

Confidence interval = 8.63 ± 0.207

Confidence interval = 8.42, 8.84

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The decimal equivalent 3/5 is a repeating decimal true or false?
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Step-by-step explanation:

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