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Vilka [71]
3 years ago
13

Find the domain for the function f(x) = the quotient of the square root of the quantity x minus 3 and the quantity x minus 5.

Mathematics
2 answers:
Daniel [21]3 years ago
4 0

f(x) =  sqrt [( x - 3) / (x - 5)]

well x cannot be 5  because  that would make  x - 5 = 0.

Also the fraction x - 3 / x - 5 cannot be negative because theres no real square root of a negative.

So  x must be <= 3 or  > 5

In interval notation the domain is ( -∞,3] or (5 , ∞)

dybincka [34]3 years ago
4 0

Answer:

The domain is (-∞, 3] and (5, ∞)

Step-by-step explanation:

The given function is \sqrt{\frac{x -3}{x - 5} }

In order to find the domain, the denominator cannot be zero and quotient must not be a negative number.

Here the denominator is x - 5

Which must be greater than zero.

x - 5 > 0

x > 5

The numerator must be less than or equal to 0.

x - 3 ≤ 0

x ≤ 3

Therefore, the domain is (-∞, 3] and (5, ∞)

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A catering company provides packages for weddings and for showers. The cost per person for small groups
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Using the <em>normal distribution and the central limit theorem</em>, it is found that the probability the mean cost of the weddings is more than the mean cost of the showers is of 0.9665.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

  • It 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, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation s = \frac{\sigma}{\sqrt{n}}.
  • When two variables are subtracted, the mean is the subtraction of the means, while the standard error is the square root of the sum of the variances.

<h3>What is the mean and the standard error of the distribution of differences?</h3>

For each sample, they are given by:

\mu_W = 82.3, s_W = \frac{18.2}{\sqrt{9}} = 6.0667

\mu_S = 65, s_S = \frac{17.73}{\sqrt{6}} = 7.2382

For the distribution of differences, we have that:

\mu = \mu_W - \mu_S = 82.3 - 65 = 17.3

s = \sqrt{s_W^2 + s_S^2} = \sqrt{6.0667^2 + 7.2382^2} = 9.4444

The probability the mean cost of the weddings is more than the mean cost of the showers is P(X > 0), that is, <u>one subtracted by the p-value of Z when X = 0</u>, hence:

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

By the Central Limit Theorem

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

Z = \frac{0 - 17.3}{9.4444}

Z = -1.83

Z = -1.83 has a p-value of 0.0335.

1 - 0.0335 = 0.9665.

More can be learned about the <em>normal distribution and the central limit theorem</em> at brainly.com/question/24663213

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