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poizon [28]
2 years ago
13

1. Try It #1 A company sells doughnuts. They incur a fixed cost of $25,000 for rent, insurance, and other expenses. It costs $0.

25 to produce each doughnut.
a. Write a linear model to represent the cost C of the company as a function of x, the number of doughnuts produced.
Mathematics
1 answer:
Pie2 years ago
3 0

Answer:

In the question, it is given that the fixed cost is $25000 and the expense to produce each doughnut is $0.25.

It is required to represent the cost C of the company as a function of x, the number of doughnuts produced.

To do so, multiply the expense of each doughnut by the number of doughnuts then add the fixed expenses of the doughnut selling company to find the total cost of the company.

Step-by-step explanation:

The expense of each doughnut is $0.25.

The expenses for x doughnuts are $0.25x.

The fixed cost is $25000.

The total cost of the company is the sum of fixed cost and the expenses for the doughnut, C = 0.25x + 25000.

The total cost of the company is, C = 0.25x + 25000.

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Answer:

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Step-by-step explanation:

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A bank with a branch located in a commercial district of a city has the business objective of developing an improved process for
tatiyna

Answer:

(a) The test statistic value is -4.123.

(b) The critical values of <em>t</em> are ± 2.052.

Step-by-step explanation:

In this case we need to determine whether there is evidence of a difference in the mean waiting time between the two branches.

The hypothesis can be defined as follows:

<em>H₀</em>: There is no difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ = 0.

<em>Hₐ</em>: There is a difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ ≠ 0.

The data collected for 15 randomly selected customers, from bank 1 is:

S = {4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50, 6.10, 0.38, 5.12, 6.46, 6.19, 3.79}

Compute the sample mean and sample standard deviation for Bank 1 as follows:

\bar x_{1}=\frac{1}{n_{1}}\sum X_{1}=\frac{1}{15}[4.21+5.55+...+3.79]=4.29

s_{1}=\sqrt{\frac{1}{n_{1}-1}\sum (X_{1}-\bar x_{1})^{2}}\\=\sqrt{\frac{1}{15-1}[(4.21-4.29)^{2}+(5.55-4.29)^{2}+...+(3.79-4.29)^{2}]}\\=1.64

The data collected for 15 randomly selected customers, from bank 2 is:

S = {9.66 , 5.90 , 8.02 , 5.79 , 8.73 , 3.82 , 8.01 , 8.35 , 10.49 , 6.68 , 5.64 , 4.08 , 6.17 , 9.91 , 5.47}

Compute the sample mean and sample standard deviation for Bank 2 as follows:

\bar x_{2}=\frac{1}{n_{2}}\sum X_{2}=\frac{1}{15}[9.66+5.90+...+5.47]=7.11

s_{2}=\sqrt{\frac{1}{n_{2}-1}\sum (X_{2}-\bar x_{2})^{2}}\\=\sqrt{\frac{1}{15-1}[(9.66-7.11)^{2}+(5.90-7.11)^{2}+...+(5.47-7.11)^{2}]}\\=2.08

(a)

It is provided that the population variances are not equal. And since the value of population variances are not provided we will use a <em>t</em>-test for two means.

Compute the test statistic value as follows:

t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}

  =\frac{4.29-7.11}{\sqrt{\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}}}

  =-4.123

Thus, the test statistic value is -4.123.

(b)

The degrees of freedom of the test is:

m=\frac{[\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}]^{2}}{\frac{(\frac{s_{1}^{2}}{n_{1}})^{2}}{n_{1}-1}+\frac{(\frac{s_{2}^{2}}{n_{2}})^{2}}{n_{2}-1}}

   =\frac{[\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}]^{2}}{\frac{(\frac{1.64^{2}}{15})^{2}}{15-1}+\frac{(\frac{2.08^{2}}{15})^{2}}{15-1}}

   =26.55\\\approx 27

Compute the critical value for <em>α</em> = 0.05 as follows:

t_{\alpha/2, m}=t_{0.025, 27}=\pm2.052

*Use a <em>t</em>-table for the values.

Thus, the critical values of <em>t</em> are ± 2.052.

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