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svetlana [45]
3 years ago
7

Solve for x. Sqrt. x +3 / x+3 = 1

Mathematics
2 answers:
algol [13]3 years ago
6 0

Answer:

(x+3)²=(√x+3)²

x²+6x+9=x+3

x²+5x+6=0

3×2=6

3+2=5

(x+3)×(x+2)=0

x=-2 and x=-3

SashulF [63]3 years ago
3 0

Answer:

x = -3 and  x = -2

Step-by-step explanation:

-3 +3 = 0

-3 + 3 = 1

thus that means 0 + 1 = 1

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Pls help me!!<br> Which two equations are equivalent?<br> y= 1/2x+3<br> y=x+6<br> 2y=x+6
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A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri
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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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