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slavikrds [6]
4 years ago
7

Allyson loves to eat cheese sticks. A small order comes with 3 cheese sticks, and a large order comes with 6 cheese sticks. Last

month, Allyson ordered cheese sticks a total of 7 times. She received a total of 36 cheese sticks. How many times did she get the large order of cheese sticks?
Mathematics
2 answers:
never [62]4 years ago
8 0

Answer:  Allyson got the large order of cheese sticks 5 times.

Step-by-step explanation:  Given that Allyson loves to eat cheese sticks. A small order comes with 3 cheese sticks, and a large order comes with 6 cheese sticks.

Last month, Allyson ordered cheese sticks a total of 7 times and she received a total of 36 cheese sticks.

We are to find the number of times she got large order of cheese sticks.

Let x and y represent the number of times she got the small order and large order respectively of cheese sticks.

Then, according to the given information, we have

x+y=7~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\3x+6y=36~~~~~~~~~~~~~~~~~~~~~~~~(ii)

Multiplying equation (i) by 3, we have

3x+3y=21~~~~~~~~~~~~~~~~~~~~~~(iii)

Subtracting equation (iii) from equation (ii), we get

(3x+6y)-(3x+3y)=36-21\\\\\Rightarrow 3y=15\\\\\Rightarrow y=5.

Thus, Allyson got the large order of cheese sticks 5 times.

lisabon 2012 [21]4 years ago
4 0

Allyson loves to eat cheese sticks. A small order comes with 3 cheese sticks, and a large order comes with 6 cheese sticks....

Answer : 5

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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

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10

Then add them all together:

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