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mixer [17]
3 years ago
11

Find the distance between points S(6,−3)

Mathematics
1 answer:
Gelneren [198K]3 years ago
4 0

Answer:

.....................

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can ppl pls look at my three recent questions on my page and help me?? ill mark brainliest and there is pictures!
andrew-mc [135]
We can’t look at recent questions
5 0
2 years ago
Read 2 more answers
Wat is 320,000,000 written in scientific notation
vazorg [7]

Answer:

3.2 x 108

Step-by-step explanation:

it just is lol. also the 8 is to the power of ten

4 0
3 years ago
Between which two consecutive whole number does the square root of 38 lie
Dahasolnce [82]
<span>So rather than think about how to find the whole numbers, let's try to find the two perfect squares nearest to 38. There is one greater than it and one smaller than it. The square root will lie between the squared value of each. (e.g. 64 is the perfect square, 8 is the squared value i refer to)
</span><span>
6 ,7. 6squared is 36 and 7 squared is 42

therefore 38 lies between these two numbers</span>
5 0
2 years ago
Read 2 more answers
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
What is the value of the expression (-8)(6) - (-4)(-5) + (-3)(-6)?
Vlad1618 [11]

-77 is the correct answer. Hope this helped. Comment if you would like a more detailed answer.

5 0
3 years ago
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