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Effectus [21]
3 years ago
13

W + 2 ≥ 3 solve for w

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
5 0

Answer:

w+2≥3

w≥3-2

w≥1

I hope this helps:

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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

5 0
3 years ago
If f(x)=5x^2 and g(x)=x+1 find (f•g)
LenKa [72]

(f•g)=   f(1)•g(1)

= 5(1)^2 • (1)+1

= 5(1) • 2

= 5 • 2

= 10

Your answer here is 10.

Hope I helped!

8 0
3 years ago
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A grocery store is selling 12 bananas for $2.40.
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B.$2.55 i think but just check
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A meteorologist at a television station reported that a town received 0.38 in. of rain. Which fraction is equivalent to this amo
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Answer:

19/50

Step-by-step explanation:

38 divided by 2 since 100 divided by 2 to get 19/50

please pick brainliest.

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Answer:

6 million cubic feet

Step-by-step explanation:

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4 years ago
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