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Tamiku [17]
2 years ago
7

31. Which equation can be used to describe the relationship between x and y shown in the graph below?

Mathematics
1 answer:
Ipatiy [6.2K]2 years ago
6 0
I’m sorry I don’t see the graph so I would not know.
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What is the exact solution to the system of equations shown on the graph? HELPPPPP !!!​
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The answer is B

Step-by-step explanation:

I made an approximation by looking and using paint 3D to even zoom up the image

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TVs are measured by
kondor19780726 [428]

Answer:

Step-by-step explanation:

Diagonal is 60 inches, height is 24 inches so

60^2 = 24^2 + width^2

6.25 inches

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What is the fraction of 2,00000.900?
dybincka [34]


The fraction of 200000.900 is <span>200,000 9/10 in mixed numbers. </span>

<span>Hope it helps!</span>

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3 years ago
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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

4 0
3 years ago
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