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3 years ago

Determine the scale factor for triangle ABC to Triangle A'B'C'.

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

3 0

D. 1/2 due to the fact that 10 would be over 20 so 10/20 and simplified is 1/2.

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2 years ago

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2 years ago

If Upper X overbar equals 62, Upper S equals 8, and n equals 36, and assuming that the population is normally distributed, c

marishachu [46]

Answer:

The 99% confidence interval would be given by (58.373;65.627)

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

Step-by-step explanation:

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".

$\bar X=62$ represent the sample mean

$\mu$ population mean (variable of interest)

s=8 represent the sample standard deviation

n=36 represent the sample size

Part a: Confidence interval

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:

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

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,35)".And we see that $t_{\alpha/2}=2.72$

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

$62-2.72\frac{8}{\sqrt{36}}=58.373$

$62+2.72\frac{8}{\sqrt{36}}=65.627$

So on this case the 99% confidence interval would be given by (58.373;65.627)

We are 99% confident that the true mean for the variable of interest is between 58.373 and 65.627.

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3 years ago