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ale4655 [162]
3 years ago
7

In a meeting, there are five presenters: Lily, Catherine, Ted, Edward, and Bianca. Bianca must present after Edward but not imme

diately so. There must be one presenter between Bianca and Catherine Catherine must present immediately after Ted. At least two other people must present after Edward and before Ted. When must Lily present?
Mathematics
1 answer:
Marta_Voda [28]3 years ago
8 0

Answer:

Lily would present after Edward

Step-by-step explanation:

Catherine must present after Ted. Therefore we have Ted, Catherine.

There must be one presenter between Bianca and Catherine, therefore we have: Bianca, x, Catherine. x is the name of the presenter between  Bianca and Catherine. Since Ted presents immediately before Catherine, x = Ted. This means we have Bianca, Ted, Catherine.

There must be at least two other people must present after Edward and before Ted. This means we have Edward, y, Bianca, Ted, Catherine. y is the name of the presenter between Edward and Bianca.. Since we have only five presenters, the only presenter remaining is Lily.

Therefore the presenters are arranged as:

Edward, Lily, Bianca, Ted, Catherine.

Lily is the second to present. Lily would present immediately after Edward and before Bianca

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3 years ago
In a math class, 12 people reported their grades on the last test. Here are their responses:
Murrr4er [49]

Answer:

(a) 23, 35, 28, 33, 5, 12, 40, 25, 20, 18, 1, 16

1, 5, 12, 16, 18, 20 23, 25, 28, 33, 35, 40

n = 12 M = (20 + 23)/ 2 = 21.5

Q1 = (12 + 16) / 2 = 14 Q3 = (28 + 33) /2 = 30.5

(b) 22, 33, 25, 28, 5, 12, 35, 23, 20, 18, 1, 40, 16

1, 5, 12, 16, 18, 20 | 22 | 23, 25, 28, 33, 35, 40

n = 13 M = 22 Q1 = 14 Q3 = 30.5

(c) 20, 28, 23, 25, 3, 5, 33, 22, 18, 16, 40, 1, 35, 12

1, 3, 5, 12, 16, 18, 20, 22, 23, 25, 28, 33, 35, 40

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(d) 20, 28, 23, 25, 3, 5, 30, 22, 18, 40, 16, 35, 1, 33, 12

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hope this helps :)

5 0
3 years ago
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:

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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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Answer:

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