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9966 [12]
3 years ago
9

Suppose Alice and Bill make their decisions separately and simultaneously, so that each must decide what to do knowing the avail

able choices and payoffs but not what the other has actually chosen. How many potential equilibria are there? (Hint: To see whether a given combination of strategies is an equilibrium, ask whether either player could get a higher payoff by changing his or her strategy.)
Mathematics
1 answer:
Sophie [7]3 years ago
5 0

Answer:

There are two possible equilibrium.

Step-by-step explanation:

Both of them either buy movie tickets or baseball tickets, because then changing the preference can result in lower payoffs.

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What is equivalent to (26)5
vlabodo [156]

Answer:

10(13)

Step-by-step explanation:

26(5)=130 and 130÷(26÷2)=10 therefore 10(13)=130

4 0
3 years ago
Solve for x: 3x - 5 = 2x + 6​
REY [17]

Answer:

3x - 5 = 2x + 6  

Step one- Subtract 2x to isolate the variable  

X – 5 = 6  

 

Step two- Add 5 to continue to isolate the variable  

X = 11  

 

Your answer is      x = 11  

 

Hopefully this helps! Feel free to mark brainliest!

7 0
3 years ago
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12/6 = 4/2 true or false
alexandr402 [8]

Answer:

This is true

Step-by-step explanation:

You are division by 12/6 by 3 and that =4/2

4 0
3 years ago
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Hi<br> Can u please help<br> i will pick the best one
katrin2010 [14]
AB will be the better choice because of its axle
3 0
2 years ago
Only the a) part please!!!!
choli [55]

(i)\\\\2A = \begin{pmatrix}3x \times 2  & 1 \times 2 & 5 \times 2\\0\times 2 & 2x \times 2 & -3 \times 2\end{pmatrix} = \begin{pmatrix}6x & 2 & 10\\0 & 4x & -6\end{pmatrix}\\\\\\(ii)\\\\2A + B = C\\\\\implies  \begin{pmatrix}6x & 2 & 10\\0 & 4x & -6\end{pmatrix} + \begin{pmatrix}y & 0 & -4\\1 & -3y & 2\end{pmatrix} = \begin{pmatrix}31& 2 & 6\\1 & 17& -4\end{pmatrix}

\implies \begin{pmatrix}6x+y & 2+0 & 10-4\\0+1 & 4x-3y & -6+2\end{pmatrix} =  \begin{pmatrix}31 & 2 & 6\\1 & 17& -4 \end{pmatrix}\\\\\\\implies \begin{pmatrix}6x+y & 2 & 6\\1 & 4x-3y & -4\end{pmatrix} =  \begin{pmatrix}31 & 2 & 6\\1 & 17& -4 \end{pmatrix}\\\\\text{System of equations},\\\\6x +y =31,~~ 4x -3y = 17\\\\\\(iii)\\\\6x +y-31 =0 \\\\4x-3y -17=0\\\\\text{Solving by cross multiplication method}\\\\\\\dfrac x{1(-17) - (-31)(-3)} = \dfrac y{4(-31) - 6(-17)} = \dfrac 1{6(-3) - 4(1)}

\implies \dfrac{x}{-17 -93}  = \dfrac{y}{-124 +102} = \dfrac 1{-18-4}\\\\\\\implies -\dfrac{x}{110} = -\dfrac{y}{22} = -\dfrac 1{22}\\\\ \implies \dfrac{x}{110} = \dfrac{y}{22} = \dfrac 1{22}\\\ \\\text{Hence,}\\\\\\x=\dfrac{110}{22}=5,~~~~ y= \dfrac{22}{22} =1

7 0
2 years ago
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