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Leona [35]
3 years ago
7

3O POINTS!! AND I WILL MARK THE BRAINLEST ANSWER EVER PLZ HELPP MEEE

Mathematics
1 answer:
AysviL [449]3 years ago
4 0

Answer: Use order of operations (PEMDAS). Answer is 45.72.

Step-by-step explanation:

First, solve inside parenthesis:

(5.25*1 1/5 - 4.5*4/5)

Convert 1 1/5 into decimal. It will be easier. 1 1/5 is the same as 1.2

Do multiplication and division first.

5.25*1.2 = 6.3

Now it looks like: (6.3 - 4.5*4/5)

Convert 4/5 to a decimal: 4/5 = 0.8

(6.3 - 4.5*0.8)

Multiply 4.5*0.8 = 3.7

Subtract (6.3 - 3.7) = 2.6

Now the full equation is: 19.6*2 1/5 + (2.6)

2 1/5 as a decimal: 2.2

Multiply 19.6*2.2 = 43.12

Now the equation is: 43.12 + 2.6

Add 43.12 + 2.6 = 45.72

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Answer:

0.4332 = 43.32% probability that the sample mean is between 21 and 22.

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.

According to a report from a business intelligence company, smartphone owners are using an average of 22 apps per month.

This means that \mu = 22

Standard deviation is 4:

This means that \sigma = 4

Sample of 36:

This means that n = 36, s = \frac{4}{sqrt{36}}

What is the probability that the sample mean is between 21 and 22?

This is the p-value of Z when X = 22 subtracted by the p-value of Z when X = 21.

X = 22

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

By the Central Limit Theorem

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

Z = \frac{22 - 22}{\frac{4}{sqrt{36}}}

Z = 0

Z = 0 has a p-value of 0.5.

X = 21

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

Z = \frac{21 - 22}{\frac{4}{sqrt{36}}}

Z = -1.5

Z = -1.5 has a p-value of 0.0668.

0.5 - 0.0668 = 0.4332

0.4332 = 43.32% probability that the sample mean is between 21 and 22.

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Answer:

11,000

Step-by-step explanation:

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