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natima [27]
3 years ago
13

Simplify by combining like terms: 3 1/9p + 5 - 1/3p

Mathematics
2 answers:
diamong [38]3 years ago
8 0

Answer:

7\frac{2}{3}

Step-by-step explanation:

3\frac{1}{9}p+5-\frac{1}{3}p

Put it into a different form to make it easier.

3\frac{1}{9}p-\frac{1}{3}p+5

Add the fractions. We need to figure out the LCM to figure out the common denominator. The LCM is 9 in this case.

3\frac{1}{9}-\frac{3}{9} +5

I like to convert into imroper fractions. This is optional though.

\frac{28}{9} -\frac{3}{9} +5

\frac{24}{9} +5

\frac{8}{3}+5

Get a common denominator. The LCM is 3.

\frac{8}{3}+\frac{15}{3}

\frac{23}{3}

Make this fraction into a mixed number.

7\frac{2}{3}

I hope this helps!

Airida [17]3 years ago
6 0

Answer:

5(5p+9)

-----------

9

Step-by-step explanation:

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Step-by-step explanation:

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B, 6

Step-by-step explanation:

In the equation (x - c)² + (y - d)² = r², (c, d) is the center and r is the radius. In this case, c = -2, d = 7 and r = 6 so the center is (-2, 7) and the radius is 6.

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3 years ago
Based on historical data, your manager believes that 37% of the company's orders come from first-time customers. A random sample
fomenos

Answer:

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

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 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}}

37% of the company's orders come from first-time customers.

This means that p = 0.37

A random sample of 225 orders will be used to estimate the proportion of first-time-customers.

This means that n = 225

Mean and standard deviation:

\mu = p = 0.37

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.37*0.63}{225}} = 0.0322

What is the probability that the sample proportion is between 0.26 and 0.38?

This is the pvalue of Z when X = 0.38 subtracted by the pvalue of Z when X = 0.26.

X = 0.38

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

By the Central Limit Theorem

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

Z = \frac{0.38 - 0.37}{0.0322}

Z = 0.31

Z = 0.31 has a pvalue of 0.6217

X = 0.26

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

Z = \frac{0.26 - 0.37}{0.0322}

Z = -3.42

Z = -3.42 has a pvalue of 0.0003

0.6217 - 0.0003 = 0.6214

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

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Answer:

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There's no "possible answers"

Step-by-step explanation:

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