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klasskru [66]
2 years ago
15

Find the answer to: 0.8/9.2=x+0.2

Mathematics
2 answers:
Ulleksa [173]2 years ago
3 0

                                                                    <u>0.8 / 9.2 = x + 0.2</u>

Perform the division on the left side:   0.08696 = x + 0.2

Subtract  0.2  from each side:                 <em>-0.113  =  x</em>    (rounded)


Serga [27]2 years ago
3 0
0.8/9.2= x+0<span>.</span>2
x= 0.8/9.2 - 0.2
x= -0.11
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The following data points represent the number of children in each household on Maple Street. \qquad 0, 1, 2, 1, 20,1,2,1,20, co
zaharov [31]

Answer:

The mean number of the children is 1.2

<em></em>

Step-by-step explanation:

Given

Children: 0, 1, 2, 1, 2

Required

Determine the Mean number

The mean of a set is calculated as follows;

Mean = \frac{\sum x}{n}

Where x is the given set and n is the number of sets

In this case, n = 5 children

Hence;

Mean = \frac{0 + 1 + 2 + 1 + 2}{5}

Mean = \frac{6}{5}

Mean = 1.2

<em>Hence, the mean number of the children is 1.2</em>

3 0
3 years ago
The Wall Street Journal Corporate Perceptions Study 2011 surveyed readers and asked how each rated the Quality of Management and
kykrilka [37]

Answer:

The calculated χ² =  <u>17.0281</u>    falls in the critical region χ² ≥  9.49  so we reject the null hypothesis that  the quality of management and the reputation of the company are independent and conclude quality of management and the reputation of the company are dependent

The p-value is 0 .001909. The result is significant at p < 0.05

Part b:

40 > 8.5

35> 7.5

25> 4

Step-by-step explanation:

1) Let the null and alternative hypothesis as

H0: the quality of management and the reputation of the company are independent

against the claim

Ha: the quality of management and the reputation of the company are dependent

2) The significance level alpha is set at 0.05

3) The test statistic under H0 is

χ²= ∑ (o - e)²/ e where O is the observed and e is the expected frequency

which has an approximate chi square distribution with ( 3-1) (3-1)=  4 d.f

4) Computations:

Under H0 ,

Observed       Expected E              χ²= ∑(O-e)²/e

40                      35.00                          0.71

25                      24.50                         0.01

5                         10.50                         2.88  

35                      40.00                         0.62

35                      28.0                          1.75

10                       12.00                           0.33  

25                      25.00                             0.00

10                        17.50                              3.21

<u>15                       7.50                                 7.50  </u>

<u>∑                                                               17.0281</u>

     

     

Column Totals 100 70 30   200  (Grand Total)

5) The critical region is χ² ≥ χ² (0.05)2 = 9.49

6) Conclusion:

The calculated χ² =  <u>17.0281</u>    falls in the critical region χ² ≥  9.49  so we reject the null hypothesis that  the quality of management and the reputation of the company are independent and conclude quality of management and the reputation of the company are dependent.

7) The p-value is 0 .001909. The result is significant at p < 0.05

The p- values tells that the variables are dependent.

Part b:

If we take the excellent row total = 70 and compare it with the excellent column total= 100

If we take the good row total = 70 and compare it with the good column total= 80

If we take the fair row total = 50 and compare it with the fair column total= 30

The two attributes are said to be associated if

Thus we see that ( where (A)(B) are row and columns totals and AB are the cell contents)

AB> (A)(B)/N  

40 > 1700/200

40 > 8.5

35> 1500/200

35> 7.5

25> 800/200

25> 4

and so on.

Hence they are positively associated

8 0
3 years ago
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Answer:

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I hope I helped you^_^

8 0
2 years ago
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Vesnalui [34]

Answer: w=−4

Step-by-step explanation:

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3 0
3 years ago
Does anyone know this?
Anni [7]
7,8, and 9
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5 0
3 years ago
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