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2 years ago

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What is the following product? Assume x greater-than-or-equal-to 0 and y greater-than-or-equal-to 0 StartRoot 5 x Superscript 8

Baseline y squared EndRoot times StartRoot 10 x cubed EndRoot times StartRoot 12 y EndRoot

1 answer:

earnstyle [38]2 years ago

7 0

Answer:

The value of the expression is $10\sqrt{6}x^{3}y$ .

Step-by-step explanation:

The expression provided is:

$\sqrt{5x^{3}y}\times \sqrt{10x^{3}}\times \sqrt{12y}$

It is provided that <em>x</em> ≥ 0 and <em>y</em> ≥ 0.

Rules of exponent:

$a^{m}\times a^{n}=a^{m+n}$

Compute the value of the expression as follows:

$\sqrt{5x^{3}y}\times \sqrt{10x^{3}}\times \sqrt{12y}=\sqrt{(5x^{3}y)\times (10x^{3})\times (12y)}$

$=\sqrt{(5\times 10\times 12)\times (x^{3}\times x^{3})\times (y\times y)}\\\\=\sqrt{600\times x^{6}\times y^{2}}\\\\=\sqrt{600}\times \sqrt{x^{6}}\times \sqrt{y^{2}}\\\\=10\sqrt{6}x^{3}y$

Thus, the value of the expression is $10\sqrt{6}x^{3}y$ .

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A hospital finds that 22% of its accounts are at least 1 month in arrears. A random sample of 425 accounts was taken. What is th

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Answer:

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

Step-by-step explanation:

For each account, there are only two possible outcomes. Either they are at least 1 month in arrears, or they are not. The probability of an account being at least 1 month in arrears is independent from other accounts. So the binomial probability distribution is used to solve this question.

However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$p = 0.22, n = 425$

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$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{425*0.22*0.78} = 8.54$

What is the probability that fewer than 82 accounts in the sample were at least 1 month in arrears

This probability is the pvalue of Z when X = 82. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{82 - 93.5}{8.54}$

$Z = -1.35$

$Z = -1.35$ has a pvalue of 0.0885.

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

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2 years ago

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with $0\le u\le3$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

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Then by Stokes' theorem we have

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$\displaystyle\int_0^{2\pi}\sin v\cos^3v\,\mathrm dv=0$

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AlladinOne [14]

Answer:

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Step-by-step explanation:

You are told that ...

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and you are told that ...

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so the ordered pair is ...

(x, y) = (3, 10) . . . . Point A

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Answer: x+y =12

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