PLEASE HELP ME I REALLY NEED HELP PLEASEEEE

Mathematics

1 answer:

kykrilka [37]3 years ago

6 0

Answer:

Step-by-step explanation:

area of a circle = πr²

area = A²

A =√ 12π

A = 2 √ 3π

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What is the greatest common factor of 24, 18, and 36?

slava [35]

Answer:

most likely 6

Step-by-step explanation:

because factors of 36 is 4,2,6 etc factors of 18 is 6 and 24 and it the gcf

8 0

3 years ago

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What's the fraction root of 9 over 49?

irakobra [83]

<span>√9/49

First you figure out the square root of 9 which is 3 since 3 x 3 equals 9. Then you find the square root of 49 which is 7 since 7 x 7 equals 49. Therefore, the answer is 3/7</span>

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2 years ago

The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in

Stolb23 [73]

Answer:

$ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75$

±0.75 ounce

Step-by-step explanation:

Assuming this complete question: The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in that state. To do this, they have selected a random sample of 300 people from the designated population. The following results were recorded: xbar=34.5 ounces, s=7.9 ounces Given this information, if the leaders wish to estimate the mean milk consumption with 90 percent confidence, what is the approximate margin of error in the estimate?

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".

$\bar X=34.5$ represent the sample mean

$\mu$ population mean (variable of interest)

s=7.9 represent the sample standard deviation

n=300 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)

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

$df=n-1=300-1=299$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,199)".And we see that $t_{\alpha/2}=1.653$