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Naya [18.7K]
3 years ago
14

Find the domain of the function. f(x)= x+2/1-X

Mathematics
1 answer:
Ymorist [56]3 years ago
3 0

Answer:

all real numbers expect  1.

Step-by-step explanation: This is a rational function. when we  usually the find the domain of a rational function we would look at the bottom because you don't want the denomonatior to be zero because dividing by zero in math makes the range undefined. so we use the equation and set it up to zero to find the domain. 1-x=0 -x=-1, divide by -1 and we get  1.  So x can be any numbers expect 1. Interval notation: (-∞,1)∪(1,+∞)

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Which statement does not use probability correctly to describe an event?
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A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

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 are 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, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

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Which expression is equivalent to 18x + 45
xz_007 [3.2K]
Maybe 3(6x+15)

not sure what you're looking for
6 0
3 years ago
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