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Citrus2011 [14]
2 years ago
15

12%of72 is what number

Mathematics
1 answer:
bazaltina [42]2 years ago
3 0

Given:

A number 72.

To Find:

12% of 72.

Solution:

The given problem can be solved using the concepts of percentages and their applications.

1. The given number is 72.

2. x% of a number n is given by the formula,

=> x% of n = ,

3. x% of a number can also be defined as x parts of 100 multiplied by the given number.

4. 12% of 72 can be calculated using the above number,

=> 12% of 72 =

=> 0.12 x 72,

=> 864/100,

=> 8.64.

Therefore, 12% of 72 is 8.64

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Answer:

3/16

Step-by-step explation

1/2 x 3/8

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2 years ago
3+(-7)<br> What is the answer
stepladder [879]

Answer:

-4

Step-by-step explanation:

3 + (-7) = 3 + -7 = 3-7 = -4

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Solve using substitution.<br> 8x + 5y = 4<br> x = 3<br> What is x and y?
Sloan [31]

Answer: y=-4

Step-by-step explanation:

8(3)+5y=4

24+5y=4

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divide by 5

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6 0
2 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
Are the triangles Similar? (yes or no). If yes, by what
kakasveta [241]

Answer:

The triangles are not similar

Step-by-step explanation:

we know that

If two triangles are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent

<em>Triangle X.Y.Z</em>

\angle X=61^o

\angle Z=90^o

\angle Y=90^o-61^o=29^o ---> angle X and angle Y are complementary angles

<em>Triangle H.G.J</em>

\angle H=90^o-34^o=56^o ---> angle H and angle J are complementary angles

\angle G=90^o

\angle J=34^o

so

X.Y.Z is a61^o-90^o-39^o triangle

H.G.J is a56^o-90^o-34^o triangle

The measure of its corresponding angles are not congruent

therefore

The triangles are not similar

8 0
3 years ago
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