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aleksandr82 [10.1K]
3 years ago
10

Doug lives 7. Block away from his school . Each block is about 265 feet long. What is total round-trip dostance,in miles that Do

ug walks to and from school each day ? A0.35, B0.7, C0.5 , D 0.03
Mathematics
1 answer:
olga_2 [115]3 years ago
5 0

Answer:

B : 0.7

Step-by-step explanation:

Given:

Doug lives 7. Block away from his school.

Each block is about 265 feet long.

Question asked:

What is total round-trip distance,in miles that Doug walks to and from school each day ?

Solution:

Each block is about = 265 feet

7 Block = 265\times7 = 1855 \ feet

As he lives 7 block away from his school that means his round trip distance is,

7 block ( travel during going to school ) + 7 block ( when return from school )

<em>Therefore, his total round trip is 14 block, means 1855 feet + 1855 feet = 3710 feet.</em>

Now, we have to convert it into miles as here asked: (unitary method)

5280 feet = 1 mile

1 feet = \frac{1}{5280}

3710 feet = \frac{1}{5280}\times3710

               = \frac{3710}{5280} = 0.7 \ miles

Thus, total round-trip distance,in miles that Doug walks to and from school each day is 0.7 miles.

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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

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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

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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

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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

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25> 800/200

25> 4

and so on.

Hence they are positively associated

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