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Andrei [34K]
2 years ago
7

Terrance made a list of his expenses. He is creating a check-off matrix to organize when his bills are due. A check-off matrix i

s a table without dollar amounts. You can use a matrix to indicate what time of the year an expense will occur. Help Terrance complete his check-off matrix by placing an X in the correct cells of the table according to his list of expenses.
mortgage payment: monthly
utilities: monthly
insurance: quarterly
internet: monthly
cell phone: monthly
car loan: semiannually
lawn care: monthly, April–September
food: monthly
day care: every other month beginning with February
gym membership: annually, July
Mathematics
1 answer:
valkas [14]2 years ago
6 0

Complete the chart  by marking each of the expenses in the correct cell depending on whether the expense occurs montlhy, quarterly, annually, etc. as shown in the chart.

<h3>What is a check-off matrix?</h3>

This is a type of chart that is completed by adding X to the cells. In the case of expenses, check-off matrixes help you to visualize how often every expense occurs, and therefore it can be useful to control or monitor expenses.

This type of matrix often includes:

-Columns with the months of the year or the initals of these. June = J.

-Rows that show the expenses a person has.

<h3>How to do a check-off matrix?</h3>
  1. List the expenses
  2. Make an X in each of the cells that the each expense should occur.
  3. Verify the information once the chart is finished.

Below you can find the complete chart:

Learn more about expenses in brainly.com/question/24803457

#SPJ1

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aliya0001 [1]

Answer:

a. We fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

b. The 95% confidence interval for the true difference of means is -2.2468 and 36.2468. There is a probability of 95% that the true difference of means \mu_{1}-\mu_{2} is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

Step-by-step explanation:

Let \mu_{1}-\mu_{2} be the true difference between the average credit score for an adult in Virginia and the average credit score for an adult in North Carolina. We have the large sample sizes n_{1} = 40 and n_{2} = 35, the unbiased point estimate for \mu_{1}-\mu_{2} is \bar{x}_{1} - \bar{x}_{2}, i.e., 699-682 = 17.

The standard error is given by \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}, i.e.,

\sqrt{\frac{(44)^{2}}{40}+\frac{(41)^{2}}{35}} = 9.8198.

a. We want to test H_{0}: \mu_{1}-\mu_{2} = 0 vs H_{1}: \mu_{1}-\mu_{2} \neq 0 (two-tailed alternative). The rejection region is given by RR = {z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles of the standard normal distribution respectively. The test statistic is Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}} and the observed value is z_{0} = \frac{17}{9.8198} = 1.7312. Because 1.7312 does not fall inside RR, we fail to reject the null hypothesis.

b. The endpoints for a 95% confidence interval for \mu_{1}-\mu_{2} is given by 17\pm (z_{0.05/2})9.8198, i.e., 17\pm (z_{0.025})9.8198 where z_{0.025} is the 2.5th quantile of the standard normal distribution, i.e., -1.96, so, we have 17-(1.96)(9.8198) and 17+(1.96)(9.8198), i.e., -2.2468 and 36.2468. There is a probability of 95% that the true difference of means \mu_{1}-\mu_{2} is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

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