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FrozenT [24]
3 years ago
12

What is the total number of minutes if there are 3 classes and each last 50 minutes?

Mathematics
2 answers:
Marina86 [1]3 years ago
7 0
To find the total number of minutes, multiply 50 x 3. The answer will be 150 minutes.
Doss [256]3 years ago
4 0
If each class lasts 50 minutes and there are 3 classes, then the answer would be 150. 50x3=150
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1) 0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588    

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b) ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653

And we want 2/3 of the margin of error so then would be: 2/3 ME = 0.00001111

The margin of error is given by this formula:

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So the answer for this case would be n=12 rounded up to the nearest integer

Step-by-step explanation:

Information given

0.082601, 0.082621, 0.082589, 0.082617, 0.082598

We can calculate the sample mean and deviation with the following formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=0.0826052 represent the sample mean

\mu population mean

s=0.000013424 represent the sample standard deviation

n=5 represent the sample size  

Part 1

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The degrees of freedom, given by:

df=n-1=5-1=4

The Confidence level is 0.95 or 95%, and the significance would be \alpha=0.05 and \alpha/2 =0.025, the critical value would be using the t distribution with 4 degrees of freedom: t_{\alpha/2}=2.776

Now we have everything in order to replace into formula (1):

0.0826052-2.776\frac{0.000013424}{\sqrt{5}}=0.082588    

0.0826052+2.776\frac{0.000013424}{\sqrt{5}}=0.0826219    

Part 2

The original margin of error is given by:

ME= 2.776\frac{0.000013424}{\sqrt{5}}=0.0000166653

And we want 2/3 of the margin of error so then would be: 2/3 ME = 0.00001111

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Replacing we got:

n=(\frac{2.776(0.000013424)}{0.00001111})^2 =11.25 \approx 12

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