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4 years ago

Explain how to simplify one of these operations with radicals? (45pt)

Mathematics

1 answer:

Vesna [10]4 years ago

8 0

Answer:

-36x z^2 ( 10 y^2 z^2) ^ 1/3

Step-by-step explanation:

16. -9 ( 640 x^3 y^2 z^8) ^ 1/3

^1/3 is the cubed root

First I will separate

(ab)^y = a^y * b^y

-9 ( 640 ) ^ 1/3 x^3 ^ 1/3 y^2 ^ 1/3 z^8 ^ 1/3

Then we know a^b^c = a^(b*c)

-9 (64) ^ 1/3 (10)^ 1/3 x^(3 * 1/3) y^(2 *1/3) z^(8* 1/3)

Simplify

-9 (64) ^ 1/3 (10)^ 1/3 x^(1) y^(2 /3) z^(8/3)

-9* 4 (10)^ 1/3 x y^(2 /3) z^(8/3)

When the exponent is greater than 1, we can take out the whole number

z^ 8/3 = x^2 * x^2/3 for example

-9 *4 (10)^ 1/3 x y^(2 /3) z^2 z^(2/3)

Move everything to the left without fractional exponents

-36x z^2 ( 10 y^2 z^2) ^ 1/3

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Ede4ka [16]

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:

$P(p-0.04$

$=P(-1.96$

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:

$P(p-0.04$

$=P(-1.31$

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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Answer:

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Step-by-step explanation:

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