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ruslelena [56]
3 years ago
9

The amount of paint that David needs to cover a cube is directly proportional to the surface area. If David can completely cover

a cube of side length 2 feet with exactly 16 quarts of paint, how big a cube (in terms of edge length in feet) can David cover with 169 quarts of paint?
Mathematics
1 answer:
Inessa05 [86]3 years ago
7 0

Answer:

The cube must be 21.125 feet long.

Step-by-step explanation:

David can completely cover a cube of side length equal to 2 feet, using 16 quarts of paint.

With this we make a conversion factor:

\frac{2}{16} feet/quarts = 0.125 feet/quarts

This means that with each quart of paint you can paint a 0.125-foot long cube

We need to know how big is the length of a cube that can be covered with 169 quarts of paint

Then we use the conversion factor found earlier

\169quarts * \frac{2pies}{16quarts} = 21.125 feet.

The cube must be 21.125 feet long.

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Answer:

Explained below.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 \mu_{\hat p}= p

The standard deviation of this sampling distribution of sample proportion is:

 \sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}

(a)

The sample selected is of size <em>n</em> = 450 > 30.

Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{450}}=0.0204

So, the sampling distribution of sample proportion is \hat p\sim N(0.75,0.0204^{2}).

(b)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

P(p-0.04

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Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.95.

(c)

The sample selected is of size <em>n</em> = 200 > 30.

Then according to the central limit theorem the sampling distribution of sample proportion is normally distributed.

The mean and standard deviation are:

\mu_{\hat p}=p=0.75\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.75(1-0.75)}{200}}=0.0306

So, the sampling distribution of sample proportion is \hat p\sim N(0.75,0.0306^{2}).

(d)

Compute the probability that the sample proportion will be within 0.04 of the population proportion as follows:

P(p-0.04

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Thus, the probability that the sample proportion will be within 0.04 of the population proportion is 0.81.

(e)

The probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 450 is 0.95.

And the probability that the sample proportion will be within 0.04 of the population proportion if the sample size is 200 is 0.81.

So, there is a gain in precision on increasing the sample size.

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