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Hunter-Best [27]
3 years ago
5

Suppose Jacques and Kyoko are playing a game in which both must simultaneously choose the action Left or Right. The payoff matri

x that follows shows the payoff each person will earn as a function of both of their choices. For example, the lower-right cell shows that if Jacques chooses Right and Kyoko chooses Right, Jacques will receive a payoff of 5 and Kyoko will receive a payoff of 3Kyoko Left Right 3,72,6 4,5 Left Jacques Right 3,8 The only dominant strategy in this game is for to choose The outcome reflecting the unique Nash equilibrium in this game is as follows: Jacques chooses_and Kyoko chooses

Business
1 answer:
ddd [48]3 years ago
7 0

Answer: 1. Jacques picks Right

2. Jacques picks Right and Kyoko picks Right.

Explanation:

Hello.

I wasn't quite clear on your question so I added an attachment with the full question.

1. The only dominant strategy in this game is for ____Jacques____ to choose ___Right____.

The Dominant strategy for a player is that strategy that will result in the highest payoff independent of the actions of the other player.

If Jacques plays Right, they will have more or equal payouts but never less than Left regardless of what Kyoko does. Therefore choosing Right is Jacques's Dominant strategy.

2. The outcome reflecting the unique Nash equilibrium in this game is as follows: Jacques chooses___Right_______and Kyoko chooses ____Right_____.

Jacques will go with their dominant strategy of picking Right. This will make Kyoto pick the alternative of Right that results in the higher payoff. They make a payoff of 8 if they pick Right as well so that is what they will do.

If you need any clarification do comment. Cheers.

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3 years ago
Resolve (7x+19)/(x+1)(x+5) into partial fraction​
Volgvan

Answer:

\dfrac{7x+19}{(x+1)(x+5)}=\dfrac{3}{x+1}+\dfrac{4}{x+5}

Explanation:

The given expression is  

\dfrac{7x+19}{(x+1)(x+5)}

We need to resolve this into partial fraction.

The form of the partial fraction decomposition is

\dfrac{7x+19}{(x+1)(x+5)}=\dfrac{A}{x+1}+\dfrac{B}{x+5}        ...(1)

\dfrac{7x+19}{(x+1)(x+5)}=\dfrac{A(x+5)+B(x+1)}{(x+1)(x+5)}

7x+19=Ax+5A+Bx+B

7x+19=(A+B)x+(5A+B)

On comparing both sides, we get

A+B=7      ...(2)

5A+B=19    ...(3)

Subtract (2) from (3), we get

4A=12

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Put A=3 in (1).

3+B=7

B=4

Put A=3 and B=4 in (1).

\dfrac{7x+19}{(x+1)(x+5)}=\dfrac{3}{x+1}+\dfrac{4}{x+5}

Therefore, \dfrac{7x+19}{(x+1)(x+5)}=\dfrac{3}{x+1}+\dfrac{4}{x+5} .

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