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Alex777 [14]
3 years ago
10

What is the probability that a client chooses Greece, or Italy, or both countries, on a tour?

Mathematics
1 answer:
Jlenok [28]3 years ago
5 0

Answer:

Both

Step-by-step explanation:

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The points (19,-2) and (0,0) fall on a particular line. What’s is it’s equation in slope intercept form?
Mice21 [21]

Answer:

y=-2/19x

Step-by-step explanation:

slope=(y1-y2)/(x1-x2)=-2/19

y-intercept=0

y=-2/19x+0

3 0
2 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
2x+8y=5<br> 24x-4y=-15<br> the solution to the system is ( , ) ( , )
kicyunya [14]

[-½, ¾]; use either "Substitution" or "Elimination" to do this, and you'll see that this answer is correct.

3 0
3 years ago
This probability distribution shows
Charra [1.4K]

Answer:

The probability that a student earns a grade of A is 1/7.

Let E be an event and S be the sample space. The probability of E, denoted by P(E) could be computed as:

P(E) = n(E) / n(S)

As the total number of students = n(S) = 35

Students getting the grade A = n(E) = 5

So, the probability that a student earns a grade of A:

                     P(E) = n(E) / n(S)

                             = 5/35

                             = 1/7

Hence, the probability that a student earns a grade of A is 1/7.

Keywords: probability, sample space, event

7 0
2 years ago
Who do historians believe created the first American flag
bonufazy [111]

Answer:

Besty Ross

Step-by-step explanation:

Brainllest

8 0
2 years ago
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