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Maru [420]
3 years ago
11

Choose all that apply :)

Mathematics
1 answer:
Veronika [31]3 years ago
7 0

Answer:

1  Multiply the first equation by -1 then add the equations together

3. Multiply the second equation by -1 then add the equations together

4. Multiply the first equation by -2 then add the equations together

5. Multiply the first equation by 2 the second by -1 then add the equations together

Step-by-step explanation:

x+y = 7

2x+y = 5


1  Multiply the first equation by -1 then add the equations together

-x+-y = -7

2x+y = 5

---------------

x    =-2   then solve for y


2.  Multiply the second equation by -1 first equation by -1 then add the equations together

-x+-y = -7

-2x-y = -5

---------------

-3x -2y = -12

Still have both variables


3. Multiply the second equation by -1 then add the equations together

x+y = 7

-2x-y =- 5

----------------------

-x = 2


4. Multiply the first equation by -2 then add the equations together

-2x+-2y = -14

2x+y = 5

--------------

    -y = -9


5. Multiply the first equation by 2 the second by -1 then add the equations together

2x+2y = 14

-2x-y = -5

--------------

y = 9

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A catering company provides packages for weddings and for showers. The cost per person for small groups
Tomtit [17]

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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