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ValentinkaMS [17]
3 years ago
8

(3p^3q) 3/4 in radical form

Mathematics
1 answer:
enot [183]3 years ago
3 0
Convert the expression to radical form using the formula <span><span><span>a<span>xn</span></span>=<span>n<span>√<span>ax</span></span></span></span><span><span>a<span>xn</span></span>=<span><span>ax</span>n</span></span></span>..If <span>nn</span> is a positive integer that is greater than <span>xx</span> and <span>aa</span> is a real number or a factor, then <span><span><span>a<span>xn</span></span>=<span>n<span>√<span>ax</span></span></span></span><span><span>a<span>xn</span></span>=<span><span>ax</span>n</span></span></span>.<span><span><span>a<span>xn</span></span>=<span>n<span>√<span>ax</span></span></span></span><span><span>a<span>xn</span></span>=<span><span>ax</span>n</span></span></span>Use the rule to convert <span><span>x<span>34</span></span><span>x<span>34</span></span></span> to a radical, where <span><span>a=</span><span>a=</span></span>, <span><span>x=</span><span>x=</span></span>, and <span><span>n=</span><span>n=</span></span>.<span>4<span>√<span>x<span>3</span></span></span></span>
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b) 50% probability that the sample proportion of smart phone users is greater than 0.33.

c) 33.39% probability that the sample proportion is between 0.19 and 0.31

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question, we have that:

p = 0.33, n = 100

a) What would the standard deviation of the sampling distribution of the proportion of the smart phone users​ be?

s = \sqrt{\frac{0.33*0.67}{100}} = 0.047

b) What is the probability that the sample proportion of smart phone users is greater than 0.33?

This is 1 subtracted by the pvalue of Z when X = 0.33. So

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{0.33 - 0.33}{0.047}

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50% probability that the sample proportion of smart phone users is greater than 0.33.

c) What is the probability that the sample proportion is between 0.19 and 0.31​?

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Z = -2.97

Z = -2.97 has a pvalue of 0.0015

0.3354 - 0.0015 = 0.3339

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