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3 years ago

Which of the following relations is a function?

Mathematics

2 answers:

kolbaska11 [484]3 years ago

7 0

The answer to your question is b

Klio2033 [76]3 years ago

5 0

$A.\\(\boxed{3}, 1),\ (-3, 4),\ (-5, 1)\, (\boxed{3}, -5)-NO\\\\B.\\(-5, 1),\ (-3, -5),\ (3, 5)\, (6, 1)-YES\\\\C.\\(\boxed{-5}, 4),\ (-3, 6),\ (\boxed{-5}, 3),\ (6, 2)-NO\\\\D.\\(-5, 1),\ (\boxed{-3}, 4),\ (3, -5),\ (\boxed{-3}, 6)-NO$

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Answer:

$182.167-2.03\frac{114.05}{\sqrt{36}}=143.580$

$182.167+2.03\frac{114.05}{\sqrt{36}}=220.754$

So on this case the 95% confidence interval would be given by (143.580;220.754)

Step-by-step explanation:

Assuming the following dataset:

77, 349,417,349, 167 , 225, 265, 360,205

145,335,40,139, 177,108, 163, 202, 22

123,439, 125,135, 86,43, 217,49, 156

119,178, 151, 61, 350, 312, 91, 89,89

We can calculate the sample mean with the followinf formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}= 182.167$

And the sample deviation with:

$s = \sqrt{\frac{\sum_{i=1}^n (X_i-\bar X)^2}{n-1}}=114.05$

The sample size on this case is n =36.

Previous concepts

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".

$\bar X=182.167$ represent the sample mean

$\mu$ population mean (variable of interest)

s=114.05 represent the sample standard deviation

n=36 represent the sample size

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

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

The point estimate of the population mean is $\hat \mu = \bar X =182.167$

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

$df=n-1=36-1=35$

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(0.025,35)".And we see that $t_{\alpha/2}=2.03$

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

$182.167-2.03\frac{114.05}{\sqrt{36}}=143.580$

$182.167+2.03\frac{114.05}{\sqrt{36}}=220.754$

So on this case the 95% confidence interval would be given by (143.580;220.754)

5 0

3 years ago