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2 years ago

Please help me, i'll give you brainlyist 20 points (maths)

Mathematics

1 answer:

DedPeter [7]2 years ago

4 0

Answer:

Your answer would be 3/5 or 3:5

Step-by-step explanation:

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Which letter has rotational symmetry? Answer Choices: 1. B 2. R 3. J 4. I

Whitepunk [10]

The answer is 4. I but it has to be capital

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3 years ago

13.62<br> Round to the<br> nearest whole<br> number.

julsineya [31]

Answer:

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2 years ago

Decimal equivalent of 143%

daser333 [38]

Answer:

1.43 as a decimal

Step-by-step explanation:

%143 = 1.43

7 0

2 years ago

Sofia has four times as many dollars as her friend jessie has. together the two have $35. how much money do they each have?

garik1379 [7]

4J = s
J + s = J + 4j = 5j = 35

J = 7
s = 28

3 0

3 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

4 0

2 years ago