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VladimirAG [237]
3 years ago
5

If mc019-1.jpg and n(x) = x – 3, which function has the same domain as mc019-2.jpg?

Mathematics
2 answers:
Free_Kalibri [48]3 years ago
4 0

Answer:

It's C to save you time.

Step-by-step explanation:

Eddi Din [679]3 years ago
3 0
We know that m o n(x) = m(n(x))

This means           

                                     (x-3) + 5       x+2
                   m(n(x)) =  ------------- = --------
                                     (x-3) -1         x-4

The domain of this function is all real numbers - {4}

The function with the same domain is h(x) = 11/x-4
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A farmer has 90 feet of fence with which to make a corral. If he arranges it into a rectangle that is twice as long as it is wid
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Answer:

W = 15 ft. and L = 30 ft.

Step-by-step explanation:

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P = 2(L + W) = 2(2W + W) = 6W

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3 years ago
A recent study found that the average length of caterpillars was 2.8 centimeters with a
pogonyaev

Using the normal distribution, it is found that there is a 0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

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

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by:

\mu = 2.8, \sigma = 0.7.

The probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters is <u>one subtracted by the p-value of Z when X = 4</u>, hence:

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

Z = \frac{4 - 2.8}{0.7}

Z = 1.71

Z = 1.71 has a p-value of 0.9564.

1 - 0.9564 = 0.0436.

0.0436 = 4.36% probability that a randomly selected caterpillar will have a length longer than (greater than) 4.0 centimeters.

More can be learned about the normal distribution at brainly.com/question/24663213

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