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zavuch27 [327]
3 years ago
11

A punch recipe that serves 24 people calls for:

Mathematics
2 answers:
Elza [17]3 years ago
8 0

Answer:

6 liters of lemon-lime soda

3 pints of sherbet

9 cups of ice

Step-by-step explanation:

divide the original by 2 the add that to the original

Lapatulllka [165]3 years ago
5 0

Answer:

6 liters of lemon-lime soda

3 pints of sherbet

9 cups of ice

Step-by-step explanation:

Divide the original by 2 add the answer to the original

Hope it's helpful to you

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Write an expression equivalent to `b+b+b+b+b` that is a product of a coefficient and a variable.
mestny [16]

Answer:

b^5

Step-by-step explanation:

5 0
3 years ago
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Please help solve <br><br> -2x - 4 - 7 = - 3x - 15
valentina_108 [34]

Answer:

x = -4

Step-by-step explanation:

-2x-11=-3x-15

x - 11 = -15

x = -4

8 0
3 years ago
What is the geometrick mean of 27 and 8 with the pothagereom theorem
enyata [817]

Answer:

x = 6√6

Step-by-step explanation:

27/x = x/8

x² = 27 × 8

x = √(27 × 8)

x = √(9 × 4 × 6)

x = 3 × 2 × √6

x = 6√6

5 0
2 years ago
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A family buys 6 airline tickets online. The family buys travel insurance that costs $18 per ticket. The
Flauer [41]

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15

Step-by-step explanation:

4 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
3 years ago
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