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andrew11 [14]
2 years ago
11

I have more than one so first one is

Mathematics
1 answer:
diamong [38]2 years ago
8 0

Answer:

Step-by-step explanation:

1 L = 1,000 mL

52 L = 52,000 mL > 5,200 mL

::::

1 m = 100 cm

14 m = 1,400 cm > 140 cm

You might be interested in
Which subset of real numbers does 0 not belong to?
Margarita [4]

Integers are irrational numbers. A natural number is a rational number. An irrational number is a real number. Zero is a natural number.

I hope this helps!

6 0
2 years ago
4. You want to know if there's an association between college students' spring break destinations and what year they're in. You
Pachacha [2.7K]

Answer:

\chi^2 =27.356

p_v = P(\chi^2_{9} >27.356)=0.00122

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(27.356,9,TRUE)"

Since the p value is lower than the significance level assumed 0.05 we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

Step-by-step explanation:

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".

Assume the following dataset:

                           Amusement Parks     Mexico    Home    Other    Total

Freshman                     23                         21            43          21         108

Sophomore                  34                         23            14          26        97

Junior                           25                         30            23         26        104

Senior                           27                         33             17         19         96

Total                            109                        107            97         92        405

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is independence between the two random variables

H1: There is dependence between the two variables

The level os significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

\chi^2 =\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{total col * total row}{grand total}

And the calculations are given by:

E_{1} =\frac{109*108}{405}=29.07

E_{2} =\frac{107*108}{405}=28.53

E_{3} =\frac{97*108}{405}=25.87

E_{4} =\frac{92*108}{405}=24.53

E_{5} =\frac{109*97}{405}=26.11

E_{6} =\frac{107*97}{405}=25.63

E_{7} =\frac{97*97}{405}=23.23

E_{8} =\frac{92*97}{405}=22.03

E_{9} =\frac{109*104}{405}=27.99

E_{10} =\frac{107*104}{405}=27.48

E_{11} =\frac{97*104}{405}=24.91

E_{12} =\frac{92*104}{405}=23.62

E_{13} =\frac{109*96}{405}=25.84

E_{14} =\frac{107*96}{405}=25.36

E_{15} =\frac{97*96}{405}=22.99

E_{16} =\frac{92*96}{405}=21.81

And the expected values are given by:

                        Amusement Parks     Mexico    Home    Other    Total

Freshman                   29.07                28.53     25.87     24.53      108

Sophomore                26.11                  25.63     23.23     22.03      97

Junior                         27.99                 27.48     24.91      23.62     104

Senior                        25.84                 25.36     22.99      21.81       96

Total                            109                      107         97           92         405

And now we can calculate the statistic:

\chi^2 =27.356

Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(4-1)(4-1)=9

And we can calculate the p value given by:

p_v = P(\chi^2_{9} >27.356)=0.00122

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(27.356,9,TRUE)"

Since the p value is lower than the significance level assumed 0.05 we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

7 0
3 years ago
Michelle baby-sits the neighbors' children. She watches the Rogers' child for $5 per hour
mixas84 [53]

Answer:

What is the objective to answer the question? Is it how much they will get?

4 0
2 years ago
Please help me oh me a lot to me and you get a lot of points I think
Andrei [34K]

Answer:

A. 0.2, 0.25, 0.2

B. the scores are a quarter of the whole

3 0
2 years ago
Read 2 more answers
Ali surveyed 200 students at a school and recorded the eye color and the gender of each student. of the 80 male students who wer
Aloiza [94]
Total number of students surveyed = 200
Number of male students = 80
Number of female students = 200 - 80 = 120

Number of brown eyed male students = 60
Probability of a brown eyed male student = 60 / 80 = 0.75.

Since, <span>eye color and gender are independent, this means that eye color is not affected by the gender. Thus, we expect a similar probability of brown eye for female as we had for male.

Let the number expected of brown eyed females be x, then x / 120 = 0.75.

Thus, x = 120(0.75) = 90.

Therefore, the number female students surveyed expected to be brown eyed is 90.</span>
7 0
3 years ago
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