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ozzi
3 years ago
7

Jeanne babysits for $6 per hour. She also works as a reading tutor for $10 per hour. She is only allowed to work 20 hours per we

ek. This week, her goal is to make at least $75.
A. Use a system of inequalities to model the scenario above. Let x represent babysitting hours and y represent tutoring hours.

B. Use the model created in part A to create a graph representing Jeanne’s probable income earned and possible number of hours worked this week.

C. Analyze the set of coordinate values that represent solutions for the model created in part A. Choose one of the coordinates within the solution and algebraically prove that the coordinate represents a true solution for the model.


i didnt see a model so you dont have to worry about B
Mathematics
1 answer:
joja [24]3 years ago
8 0
Use a system of inequalitiesto modelthe scenario above so the correct answer is letter A
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Using the z-distribution, it is found that since the p-value of the test is less than 0.05, we reject the null hypothesis and find that there is enough evidence to conclude that the true proportion of all life insurance claims made to this company that are settled within 30 days is less than 85%.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if there is not enough evidence that the proportion is below 85%, that is:

H_0: p \geq 0.85

At the alternative hypothesis, it is tested if there is enough evidence that the proportion is below 85%, that is:

H_0: p < 0.85

<h3>What is the test statistic?</h3>

The test statistic is given by:

z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}

In which:

  • \overline{p} is the sample proportion.
  • p is the proportion tested at the null hypothesis.
  • n is the sample size.

For this problem, the parameters are:

p = 0.85, n = 250, \overline{p} = \frac{203}{250} = 0.812

Hence the test statistic is:

z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}

z = \frac{0.812 - 0.85}{\sqrt{\frac{0.85(0.15)}{250}}}

z = -1.68

<h3>What is the p-value and the conclusion</h3>

Using a z-distribution calculator, with a left-tailed test, as we are testing if the proportion is less than a value, and z = -1.68, the p-value is of 0.0465.

Since the p-value of the test is less than 0.05, we reject the null hypothesis and find that there is enough evidence to conclude that the true proportion of all life insurance claims made to this company that are settled within 30 days is less than 85%.

More can be learned about the z-distribution at brainly.com/question/16313918

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