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kvasek [131]
3 years ago
14

Function Operations Picture attached

Mathematics
1 answer:
sergeinik [125]3 years ago
6 0

Answer: Second option

Step-by-step explanation:

You need to make the multiplication of the function c(x)=\frac{5}{x-2} and the function d(x)=x+3. Then:

(cd)(x)=(\frac{5}{x-2})(x+3)

You need to apply the Distributive property:

(cd)(x)=\frac{5(x+3)}{x-2}\\\\(cd)(x)=\frac{5x+15}{x-2}

Therefore, the domain will be all the number that make the denominator equal to zero.

Then, make the denominator equal to zero and solve for x:

x-2=0\\x=2

Therefore, the domain is: All real values of x except x=2

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The mean life of a television set is 119119 months with a standard deviation of 1414 months. If a sample of 7474 televisions is
Pepsi [2]

Answer:

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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 normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, 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 random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean

In this problem, we have that:

\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.6275

What is the probability that the sample mean would differ from the true mean by less than 1.11 months?

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

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

By the Central Limit Theorem

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

Z = \frac{120.1 - 119}{1.6275}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

X = 117.9

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

Z = \frac{117.9 - 119}{1.6275}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

3 0
3 years ago
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Answer:Oliver has 5 pieces of string that are each 4 2/12 feet long. Destiny has 4 pieces of string that are each 5 14/16 feet long. Use an estimation strategy to determine who has the most string. Choose the name and number to complete the statement.

Step-by-step explanation:

4 0
2 years ago
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What’s the correct answer for this?
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Answer:

E:

Step-by-step explanation:

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Where

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