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lana [24]
3 years ago
11

What is the GFC of 72 and 36

Mathematics
1 answer:
Mashutka [201]3 years ago
6 0
The GCF of 72 and 36 is: 36

List out the factors of each number (72 and 36), and find the common factors between the two numbers.

Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

Looking at the factors above, the common ones are 1, 2, 3, 4, 6, 9, 12, 18, and 36. But... considering we are looking for the "greatest" common factor, our highest number out of the listed ones is 36, making it the greatest common factor of 72 and 36.
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Answer:

a) The 99% confidence interval would be given by (24.409;24.979)  

b) n=464

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Part a

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

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

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:  

df=n-1=47-1=46  

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,46)".And we see that t_{\alpha/2}=2.01  

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

525-2.01\frac{75}{\sqrt{47}}=503.01  

525+2.01\frac{75}{\sqrt{47}}=546.99  

So on this case the 95% confidence interval would be given by (503.01;546.99)

Part b

The margin of error is given by this formula:  

ME=t_{\alpha/2}\frac{s}{\sqrt{n}} (1)  

And on this case we have that ME =7 msec, we are interested in order to find the value of n, if we solve n from equation (1) we got:  

n=(\frac{t_{\alpha/2} s}{ME})^2 (2)  

The critical value for 95% of confidence interval is provided, t_{\alpha/2}=2.01 from part a, replacing into formula (2) we got:  

n=(\frac{2.01(75)}{7})^2 =463.79 \approx 464  

So the answer for this case would be n=464 rounded up to the nearest integer  

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