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Nutka1998 [239]
2 years ago
9

Which table correctly shows all the sample spaces for tossing a coin to get heads and rolling a number on a six-sided number cub

e? Coin Number cube Outcome Tails 1 Tails, 1 Tails 2 Tails, 2 Tails 3 Tails, 3 Tails 4 Tails, 4 Tails 5 Tails, 5 Tails 6 Tails, 6 Coin Number cube Outcome Heads 1 Heads, 1 Heads 2 Heads, 2 Heads 3 Heads, 3 Heads 4 Heads, 4 Heads 5 Heads, 5 Heads 6 Heads, 6 Coin Number cube Outcome Heads 1 Heads, 2 Heads 2 Heads, 4 Heads 3 Heads, 6 Tails 4 Tails, 2 Tails 5 Tails, 4 Tails 6 Tails, 6 Coin Number cube Outcome Heads 2 Heads, 1 Heads 4 Heads, 3 Heads 6 Heads, 5 Tails 2 Tails, 1 Tails 4 Tails, 3 Tails 6 Tails, 5
Mathematics
2 answers:
Veseljchak [2.6K]2 years ago
8 0

Answer:

I think its all head

Step-by-step explanation:

Coin Number

cube Outcome

Heads  

1

Heads, 1

Heads  

2

Heads, 2

Heads  

3

Heads, 3

Heads  

4

Heads, 4

Heads  

5

Heads, 5

Heads  

6

Heads, 6

Elan Coil [88]2 years ago
4 0

Answer:

I am pretty sure that it is

heads 1

heads 2

heads 3

heads 4

heads 5

heads 6

Step-by-step explanation:

because in the question it says to roll a six sided die and a coin that lands on heads so there are all the numbers on die and heads only

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Complete Question

The complete question is shown on the first uploaded image

Answer:

First Question

Option A is correct

Second  Question

Option C is correct

Third   Question

     D =  A^{-1}  *  B^{-1} *  C^{-1}

Fourth   Question

  So substituting for D in  (ABC) D =  I

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Step-by-step explanation:

From the question we are told that

   A , B and  C are invertible which means that A^{-1} , B^{-1}, C^{-1} exist

Now

 From the question

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Now when we multiply both sides by  B^{-1}  we have  

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Now when we multiply both sides by  C^{-1}  we have  

          C^{-1} * I CD =  A^{-1}  *  B^{-1} *  C^{-1}

              I D =  A^{-1}  *  B^{-1} *  C^{-1}

                 D =  A^{-1}  *  B^{-1} *  C^{-1}

So substituting for D in the above equation

                 (ABC) *  A^{-1}  *  B^{-1} *  C^{-1} =  I

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