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kondor19780726 [428]
2 years ago
15

Find the value of each variable

Mathematics
1 answer:
4vir4ik [10]2 years ago
5 0

Answer:

x = 10\sqrt{2} , y = 10\sqrt{6}

Step-by-step explanation:

using the sine and cosine ratios in the right triangle and the exact values

cos60° = \frac{1}{2} , sin60° = \frac{\sqrt{3} }{2} , then

cos60° = \frac{adjacent}{hypotenuse} = \frac{x}{20\sqrt{2} } = \frac{1}{2} ( cross- multiply )

2x = 20\sqrt{2} ( divide both sides by 2 )

x = 10\sqrt{2}

and

sin60° = \frac{opposite}{hypotenuse} = \frac{y}{20\sqrt{2} } = \frac{\sqrt{3} }{2} ( cross- multiply )

2y = 20\sqrt{2} × \sqrt{3} = 20\sqrt{6} ( divide both sides by 2 )

y = 10\sqrt{6}

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A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural freq
oksano4ka [1.4K]

Answer:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

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 represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n=5 represent the sample size  

Solution to the problem

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

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

And for this case we know that the 95% confidence interval is given by:

\bar X=\frac{233.002 +229.266}{2}= 231.134

And the margin of error is given by:

ME = \frac{233.002 -229.266}{2}= 1.868

And the margin of error is given by:

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

The degrees of freedom are given by:

df = n-1 = 5-1=4

And the critical value for 95% of confidence is t_{\alpha/2}= 2.776

So then we can find the deviation like this:

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

s = \frac{1.868* \sqrt{5}}{2.776}= 1.506

And for the 99% confidence the critical value is: t_{\alpha/2}= 4.604

And the margin of error would be:

ME = 4.604 *\frac{1.506}{\sqrt{5}}= 3.098

And the interval is given by:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

6 0
3 years ago
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