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3 years ago

Write the equation of the line that has a slope of - 2/3 and a y-intercept of (0,-7)

2 answers:

6 0

Since we have the y-intercept and the slope of the line, let’s use slope-intercept form:

y = mx + b, where m is the slope and b is the y-intercept.

y = -2/3x - 7

earnstyle [38]3 years ago

4 0

You will use the equation y=mx+b.

M is the slope and b is the y-intercept.

So since we already have both, all we have to do is plug it in.

y= - 2/3 -7.

Hope this helps!

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The answer is 6 1/5.

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3 years ago

Which of the following data is used to determine credit scores? A. who stays with you at your current residence B. the location

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The correct option is the length of stay at your current residence.

Step-by-step explanation:

We have been asked that which data is used to determine credit scores.

The correct option is the length of stay at your current residence.

This question is used to determine credit scores. This helps the lender to look up information about your history of payments. When a lender asks this question they are able to determine how long you typically stay in one place, and look up bills at that residence to make sure they are paid on time before giving out a loan. They can also use this information to determine a credit score based on what is being paid on time, late or not at all...

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2 years ago

A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri

nlexa [21]

Answer:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday Tuesday Wednesday Thursday Friday Saturday Total

12 9 11 10 9 9 60

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{60}{6}= 10$ and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

$\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8$

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

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Answer:

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