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Wittaler [7]
3 years ago
10

So the answer is (y-1)^3

Mathematics
1 answer:
ivanzaharov [21]3 years ago
7 0
Yes yes yes yes yes yes yes
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Solve the following equation -6p+7=3(2p-3)-4(-10+4p)<br> a.p=6<br> b.p=5<br> c.p=7<br> d.p=12
Ilia_Sergeevich [38]
Solve for p by simplifying both sides of the equation, then isolating the variable.
p=6
7 0
3 years ago
Freeeeeeeeeeeeeeeeee again​
Paraphin [41]

Answer:

cool

Step-by-step explanation:

3 0
3 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
2 years ago
704÷46 what the answer tell me im in the
olasank [31]
15?im not sure...but I think thats right
3 0
3 years ago
Read 2 more answers
What is the slope of the line?
rodikova [14]

Answer:

Step-by-step explanation:

If you look careful,the line is passing throw the middle of the square which that means 0,5=>1/2

3 0
3 years ago
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