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2 years ago

Convert the angle 1.5 radians to degrees, rounding to the nearest 10th.

Mathematics

2 answers:

Keith_Richards [23]2 years ago

8 0

Answer:

85.9 (to the nearest hundredth)

Step-by-step explanation:

1 radian = 180°/ $\pi$

Therefore, 1.5 radians = 1.5 x (180°/ $\pi$ ) = 270/ $\pi$ = 85.9 (to the nearest hundredth)

Vaselesa [24]2 years ago

7 0

Answer:

85.9

Step-by-step explanation:

1.5rad × 180/π = 85.9°

Thank You!

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The percentage of body fat of a random sample of 36 men aged 20 to 29 found a sample mean of 14.42. Find a 95% confidence interv

Rina8888 [55]

Answer:

$14.42-1.96\frac{6.95}{\sqrt{36}}=12.150$

$14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690$

So on this case the 95% confidence interval would be given by (12.150;16.690)

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".

$\bar X=14.42$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=6.95$ represent the population standard deviation

n=36 represent the sample size

Solution to the problem

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

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

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: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

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

$14.42-1.96\frac{6.95}{\sqrt{36}}=12.150$

$14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690$

So on this case the 95% confidence interval would be given by (12.150;16.690)

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3 years ago

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Galina-37 [17]

Answer:

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I got this by doing:

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fgiga [73]

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