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3 years ago

I need help to solve this math problem 12 yd = __ ft

Mathematics

2 answers:

wlad13 [49]3 years ago

5 0

Answer:

36 feet

Step-by-step explanation:

There are 3 feet in a yard, so all you have to do is multiply 12 by 3. 12x3=36

Have a great day!!! I hope this helps C:

andriy [413]3 years ago

4 0

Answer:

36 feet

Step-by-step explanation:

12inches =1 yard

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4 years ago

Find the dimensions of a rectangle with area 512 m2 whose perimeter is as small as possible. (If both values are the same number

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Answer:

Step-by-step explanation:

Length the length and breadth of the rectangle be x and y.

Area of the rectangle A = Length * breadth

Perimeter P = 2(Length + Breadth)

A = xy and P = 2(x+y)

If the area of the rectangle is 512m², then 512 = xy

x = 512/y

Substituting x = 512/y into the formula for calculating the perimeter;

P = 2(512/y + y)

P = 1024/y + 2y

To get the value of y, we will set dP/dy to zero and solve.

dP/dy = -1024y⁻² + 2

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y⁻² = 1/512

1/y² = 1/512

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From xy = 512

x(√512) = 512

x = 512/√512

On rationalizing, x = 512/√512 * √512 /√512

x = 512√512 /512

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a) By the central limit theorem we know that if the random variable X="quantitative scores for young women", and we know that this distribution is normal.

$X \sim N(\mu, \sigma=60)$

And we are interested on the distribution for the sample mean $\bar X$ , we know that distribution for the sample mean is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And with that we have the assumptions required to apply the normal distribution to create the confidence interval since the distribution for $\bar X$ is normal.

b) $270.283\leq \mu \leq 279.717$

c) We are confident (99%) that the true mean for the quantitative scores for young women is between (270.283;3279.717)

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".

$\bar X=275$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=60$ represent the population standard deviation

n=1077 represent the sample size

Part a

By the central limit theorem we know that if the random variable X="quantitative scores for young women", and we know that this distribution is normal.

$X \sim N(\mu, \sigma=60)$

And we are interested on the distribution for the sample mean $\bar X$ , we know that distribution for the sample mean is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And with that we have the assumptions required to apply the normal distribution to create the confidence interval since the distribution for $\bar X$ is normal.

Part b

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

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

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: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

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

$275-2.58\frac{60}{\sqrt{1077}}=270.283$

$275+2.58\frac{60}{\sqrt{1077}}=279.717$

So on this case the 99% confidence interval would be given by (270.283;3279.717)

$270.283\leq \mu \leq 279.717$

Part c

We are confident (99%) that the true mean for the quantitative scores for young women is between (270.283;3279.717)

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3 years ago