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Sladkaya [172]
2 years ago
12

A red die is tossed and a green die is tossed.what is the probability that a red die shows an even number or the green die shows

an even number? Make sure your answer is reduced
Mathematics
1 answer:
AlekseyPX2 years ago
6 0

Answer: 75%

Step-by-step explanation:

Since there are 3 even numbers on each die and 3 odd, each dice has a 50% chance of showing an even number.

An easier way to think about this might be what are the chances that both dice show odd, because that is the only scenario where neither show an even number.

To do this, simply multiply the chances together:

0.5 * 0.5 = .25

Therefore, there is a 25% chance that both die will be odd, meaning there is a 75% chance that one or both of the dice show an even number

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icang [17]

Answer:

no error original system so good by

8 0
3 years ago
Mais family drinks a total of 10 gallons of milk every 6 weeks. How many gallons of milk does the family drink per week? How man
tino4ka555 [31]

Answer:

they drink 1.667 gallons a week. it takes about 6.5 days to drink a whole gallon

Step-by-step explanation:

take 10/6 to get 1.667 then how many days would it take to drink 1 whole gallon

7 0
3 years ago
Part I
tino4ka555 [31]

Answer:

Step-by-step explanation:

I know that the two means are are 11.28 and 12.

I could be wrong but I did the math 5 times.

3 0
3 years ago
Read 2 more answers
0.003 divided by 8.1
dusya [7]

2700  is your answer

Step-by-step explanation:

7 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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