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mafiozo [28]
3 years ago
6

Suppose a package delivery company purchased 14 trucks at the same time. Five trucks were purchased from manufacturer A, four fr

om manufacturer B, and five from manufacturer C. The cost of maintaining each truck was recorded. The company used ANOVA to test if the mean maintenance costs of the trucks from each manufacturer were equal. To apply the F test, how many degrees of freedom must be in the denominator?
Mathematics
1 answer:
nikklg [1K]3 years ago
5 0

Answer:

14

Step-by-step explanation:

Given that a package delivery company purchased 14 trucks at the same time. Five trucks were purchased from manufacturer A, four from manufacturer B, and five from manufacturer C. The cost of maintaining each truck was recorded.

No of companies  = 3

No of items in total = 14

Total df = 17-1=16

degrees of freedom between groups = 3-1 =2

Hence degrees of freedom for denominator

=16-2 =14

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Lady_Fox [76]

Answer:

a) 1,518,000

b) 2,284,880

c) 60,720

Step-by-step explanation:

a) a. if the first letter must be Upper C comma Upper X comma Upper T comma or Upper M and no letter may be​ repeated?

We draw 5 boxes, and based on that we will see the total possible cases. There are 26 alphabets

The first box should have C or X or T or M .No letter may be repeated.

Any                    Any             Any                 Any                    Any

5 alphabets    of the            of the              of the                 of the

C,X , T , M      remaining     remaining       remaining          remaining

                   25 alphabets   24 alphabets   23 alphabets   22 alphabets

Therefore; total possible call letters = 5 × 25 × 24 × 23 × 22 = 1,518,000

b)

The first box should have C or X or T or M Repeats as allowed

Any                    Any             Any                 Any                    Any

5 alphabets    of the            of the              of the                 of the

C,X , T , M      remaining     remaining       remaining          remaining

                   26 alphabets   26 alphabets   26 alphabets   26 alphabets

Therefore Total possible call letters = 5 × 26 × 26 × 26 × 26 = 2,284,880

c)   The first box should have   C,X , T , M  and end with S

So the last place if fixed, and we now have 25 alphabets. The first box can go in 5 ways. The next box then will have only 24 letters to choose from, as the first box has taken a letter and the last box already has S in it. Repetition not allowed

Any                    Any             Any                 Any                       S

5 alphabets    of the            of the              of the                  is fixed

C,X , T , M      remaining     remaining       remaining           here

                   24 alphabets   23 alphabets   22 alphabets  

Therefore Total possible call letters = 5 × 24 × 23 × 22  × 1 =  60,720

3 0
3 years ago
I don’t know how to find X or Y very urgent, need to figure this out how to solve plz help
Lena [83]

Answer:

x = 9

y = 9√3 = 15.6

Step-by-step explanation:

The triangle given is a right triangle, therefore:

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Adjacent = x

Thus:

Cos (60) = x/18

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x = 9

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y² = 18² - x² (pythagorean theorem)

y² = 18² - 9² (substitution)

y² = 243

y = √243

y = √(81*3)

y = 9√3 = 15.6

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