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Gekata [30.6K]
3 years ago
11

Mary and Lamar establish a tutoring service at a local mall. They tutor college students in math and English. They charge $40 pe

r hour for tutoring. In the month of January they charged for 200 tutoring hours. The expenses of their business are given in the table below. Expense Amount/Month Rent of Space $4000 Electricity $325 Advertising $375 Using the formula Profit = $40(number of hours) – (expenses) , calculate their profit for the month of January.
Mathematics
1 answer:
Hatshy [7]3 years ago
7 0

The profit is $3300 for the month of January.

Step-by-step explanation:

Per hour charge = $40

Total number of tutoring hours = 200

Rent of space = $4000

Electricity = $325

Advertising = $375

Total expenses = 4000+325+375 = $4700

Profit = $40(number of hours) - (expenses)

Profit=\$40(200)-4700\\Profit=\$8000-4700\\Profit=\$3300

The profit is $3300 for the month of January.

Keywords: profit, addition

Learn more about addition at:

  • brainly.com/question/4279146
  • brainly.com/question/4354581

#LearnwithBrainly

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95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

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n_2 = sample of females = 1000

p_1 = population proportion of males having blood disorder

p_2 = population proportion of females having blood disorder

<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>

<u>So, 95% confidence interval for the difference between the population proportions, </u><u>(</u>p_1-p_2<u>)</u><u> is ;</u>

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P( -1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < {(\hat p_1-\hat p_2)-(p_1-p_2)} < 1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

P( (\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < (p_1-p_2) < (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

<u>95% confidence interval for</u> (p_1-p_2) =

[(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }, (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }]

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