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2 years ago

Divide. Express your answer in simplest form. 2 1/2 ÷ 2 1/6 13/15 5 5/12 26/30 1 2/13

Mathematics

2 answers:

Tom [10]2 years ago

8 0

Answer:

$2 \frac{1}{2} /2 \frac{1}{6} = \frac{5}{2} / \frac{13}{6}$

$= \frac{5}{2} \times \frac{6}{13 } = \frac{5 \times 6}{2 \times 13}$

$= \frac{5 \times 3}{1 \times 13}$

$= \frac{15}{13}$

$= 1 \frac{2}{13}$

Step-by-step explanation:

Thank you!!!!!!!!!!!

Marianna [84]2 years ago

7 0

Answer:

$\frac{15}{13} = 1 \frac{2}{13}$

Step-by-step explanation:

$2 \frac{1}{2} \div 2 \frac{1}{6} \\ \\ \\ \frac{5}{2} \div \frac{13}{6} \\ \\ \\ \frac{5}{2} \times \frac{6}{13} \\ \\ \\ \frac{5}{1} \times \frac{3}{13} \\ \\ \\ = \frac{15}{13} \\ \\ \\ = 1\frac{2}{13}$

I hope I helped you^_^

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A statistician calculates that 8% of Americans own a Rolls Royce. If the statistician is right, what is the probability that the

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Answer:

0.007 = 0.7% probability that the proportion of Rolls Royce owners in a sample of 595 Americans would differ from the population proportion by more than 3%

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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 $s = \sqrt{\frac{p(1-p)}{n}}$

A statistician calculates that 8% of Americans own a Rolls Royce.

This means that $p = 0.08$

Sample of 595:

This means that $n = 595$

Mean and standard deviation:

$\mu = p = 0.08$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.08*0.92}{595}} = 0.0111$

What is the probability that the proportion of Rolls Royce owners in a sample of 595 Americans would differ from the population proportion by more than 3%?

Proportion above 8% + 3% = 11% or below 8% - 3% = 5%. Since the normal distribution is symmetric, these probabilities are equal, and so we find one of them and multiply by 2.

Probability the proportion is less than 5%:

P-value of Z when X = 0.05. So

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

By the Central Limit Theorem

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

$Z = \frac{0.05 - 0.08}{0.0111}$

$Z = -2.7$

$Z = -2.7$ has a p-value of 0.0035

2*0.0035 = 0.0070

0.007 = 0.7% probability that the proportion of Rolls Royce owners in a sample of 595 Americans would differ from the population proportion by more than 3%

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