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Elodia [21]
3 years ago
5

What's y^2=0+4 What's the answer and how do I complete this?

Mathematics
1 answer:
asambeis [7]3 years ago
6 0
If you're taking 0+4=y^2
It would be y^2=4
And then you take the square root or four which is 2.
So y=2
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the perimeter of a geometric figure is the sum of its sides. if the perimeter of the following pentagon is 42 meters, find the l
stellarik [79]

Answer: 2.5 divided by 2 equals 1.25 so each side equals 1.25

Step-by-step explanation:

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I need help, also give explantion please
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Answer:

\huge  \: 8 {x}^{3}  \times 3 {x}^{9}  = 24 {x}^{12}

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3 years ago
The box plots below show student grades on the most recent exam compared to overall grades in the class:
ANTONII [103]

Answer:

1)  The class and exam medians are almost the same.

Step-by-step explanation:

1) According to the given data (box-plot) the median of Class is 84 and median of Exam is 85.

Thus, the medians of both class and exam are almost same.

Hence, this is the correct option.

2) Since the median of Class is 84 and median of Exam is 85, Thus, we can't say median of exam is much high than median of class.

Hence, this option is not correct.

3) Since Q3 of Class is 89 and Q3 of exam is 94, which is not equal.

Hence, this option is also not correct.

4) Since the median of Class is 84 and median of Exam is 85, Thus, median of exam is not lower than median of Class.

Hence, this option is also not correct.

3 0
3 years ago
What does 1/8 + 1/6 equal?​
GuDViN [60]

Answer:

7/24

Step-by-step explanation:

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3 years ago
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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
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