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GrogVix [38]
3 years ago
10

Please help ASAP 50 points :(((

Mathematics
2 answers:
Katyanochek1 [597]3 years ago
8 0

Answer:

(f o g) (x) = 36x² + 3

(g o f) (x)  = 6x² + 18

Step-by-step explanation:

f(x) is x² + 3  

g(x) is 6x

(f o g) (x) means for all xs in f replace it with g(x) or 6x

(f o g) (x) = (6x)² + 3  - see 6x is in place of x and 6x is g(x)

(f o g) (x) = 36x² + 3

(g o f) (x)  = 6(x² + 3)  

(g o f) (x)  = 6x² + 18

enot [183]3 years ago
5 0

Answer:

f(g(x)) = 36x^2+3

g(f(x)) = 6x^2+18

Step-by-step explanation:

f(x) =x^2+3

g(x) =6x

f°g(x) = means put g(x) in for x into the function f(x)

f(g(x)) = (g(x))^2 +3

            (6x)^2 +3

           36x^2+3

g°f(x) = means put f(x) in for x into the function g(x)

g(f(x)) = 6(f(x)

           6(x^2+3)

           6x^2+18

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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}

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

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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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