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Neko [114]
3 years ago
15

Dave has 10 poker chips, 5 of which are red and the other 5 of which are white. Dave likes to stack his chips and flip them over

as he plays. How many different 10-chip stacks can Dave make if two stacks are not considered distinct if one can be flipped to appear identical to the other?
Mathematics
2 answers:
sergey [27]3 years ago
7 0

Answer:

\frac{10!}{5!*5!}  -  \frac{5!}{3!*2!}  

Step-by-step explanation:

Given:

10 poker chips:

  • 5 red
  • 5 white

Let the stack has positions (1,2,3,4,5,6,7,8,9,10)

So we can find all possible outcomes of the stack is: \frac{10!}{5!*5!} =

and the possibility of a stack to be identical while flipping is: \frac{5!}{3!*2!}

So the different 10-chip stacks can Dave make if two stacks are not considered distinct:

\frac{10!}{5!*5!}  - \frac{5!}{3!*2!}  

max2010maxim [7]3 years ago
7 0

Answer:

10!/(5!× 5!) - 5!/(3! × 2!)

Step-by-step explanation:

From the question,Dave has 10 poker chips, 5 of which are red and the other 5 of which are white. Dave likes to stack his chips and flip them over as he plays. How many different 10-chip stacks can Dave make if two stacks are not considered distinct if one can be flipped to appear identical to the other?

-----Here we have 10 chips

-----5 are red and the rest is white

We are now asked to find out how many different 10-chip stacks can Dave make if two stacks are not considered distinct if one can be flipped to appear identical to the other.

Let the stack has positions (1,2,3,4,5,6,7,8,9,10)

All possibility of stacks is

10!/(5! × 5!)

and the possibility of a stack to be identical while flipping is

5!/(3! × 2!)

No of different stacks is

10!/(5! × 5!) - 5!/(3! × 2!)

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Complete Question

The complete question is shown on the first uploaded image

Answer:

the null hypothesis is  H_o  :  \mu =  122

the alternative hypothesis is H_a  :  \mu \ne  122

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Step-by-step explanation:

From the question we are told that

   The population mean is  \mu  = 122

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Generally the null hypothesis is  H_o  :  \mu =  122

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Generally the test statistics is mathematically evaluated as

         t =  \frac { \= x - \mu }{\frac{ \sigma }{ \sqrt{n} } }

substituting values

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         t =  - 1.761

The p-value is mathematically represented as

      p =  P(Z  <  t )

From the z- table  

     p =  P(Z  <  t ) = 0.039119

So  

     p  \ge 0.01

 

         

     

           

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