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3 years ago

A dinner plate fits tightly into a box whose square base has a perimeter of 28 inches.What is the circumference of the plate?

Mathematics

1 answer:

ZanzabumX [31]3 years ago

5 0

The diameter of the plate would be equal to the length of a side of the box.

Find the length of the box by dividing the perimeter by 4:

28 / 4 = 7 inches.

The diameter is 7 inches.

Circumference is found by multiplying the diameter by pi:

The exact circumference is 7PI

Using 3.14 for pi, the decimal answer is:

3.14 x 7 = 21.98 inches ( round answer as needed.)

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Mathematics, Help, not sure about my answer on this one

olga nikolaevna [1]

Answer:

F. 2 1/9

Step-by-step explanation:

Step 1: Substitute q

$r = 3(\frac{1}{3^n} )+2$

When you plug in n = 1, 2, 3:

n(1) = 3

n(2) = 2.11111

n(3) = 2.3333

So our answer is F.

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3 years ago

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algol13

Answer:

Yes.

Step-by-step explanation:

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3 years ago

How is circumference and area the same?

V125BC [204]

The area and the circumference of a circle. Make sure that you remember that acircumference is a length and so is measured , , etc. Remember as well that area is measured in square units, , etc. All of these calculations will involve the use of Pi ( ) which you can take as being equal to .

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3 years ago

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A union of restaurant and foodservice workers would like to estimate the mean hourly wage, μ, of foodservice workers in the U.S.

pav-90 [236]

Answer:

$n=(\frac{1.960(2.25)}{0.35})^2 =158.76 \approx 159$

So the answer for this case would be n=159 rounded up to the nearest integer

Step-by-step explanation:

Assuming this complete question: A union of restaurant and foodservice workers would like to estimate the mean hourly wage, , of foodservice workers in the U.S. The union will choose a random sample of wages and then estimate using the mean of the sample. What is the minimum sample size needed in order for the union to be 95% confident that its estimate is within $0.35 of ? Suppose that the standard deviation of wages of foodservice workers in the U.S. is about $2.15 .

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)

$\sigma=2.25$ represent the population standard deviation

n represent the sample size

Solution to the problem

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

And on this case we have that ME =0.35 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (b) we got:

$n=(\frac{1.960(2.25)}{0.35})^2 =158.76 \approx 159$

So the answer for this case would be n=159 rounded up to the nearest integer

3 0

3 years ago