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yan [13]
1 year ago
10

1. use the following normal-form game to answer the questions below. a. identify the one-shot nash equilibrium (10 pts) b. suppo

se this game is infinitely repeated and the interest rate is 6 percent. can the players achieve payoffs that are better than the one-shot nash equilibrium? what is the payoff? (15 pts)
Mathematics
1 answer:
Bingel [31]1 year ago
7 0

One-Shot Nash equilibrium is (A, C).

Yes the players can achieve payoffs that are better than the one-shot Nash equilibrium.

This is due to the fact that player 2 will choose strategy C if player A chooses option A. Player 2 will immediately choose Plan C if Player A chooses Option B. As a result, C becomes the second player's dominant strategy. The optimal decision for player 1 is to choose strategy A if player 2 chooses option C. Player 1 will select A if Player 2 chooses to choose D, making this Player 1's dominant move.

A Nash equilibrium is necessary for matrix reward games with two players if the row chosen is to maximize the payoff for the row player given the column chosen by the column player, and the column, in turn, is to maximize the payoff for the column player given the row chosen by the row player.

Learn more about Nash equilibrium:

brainly.in/question/4220195

#SPJ4

You might be interested in
Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set
SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange  milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected number of orange  milk chocolate M&M’s in a bag of 55 candies as follows:

E(X)=np

         =55\times 0.20\\=11

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}

                 =0.1968

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

4 0
3 years ago
Cos4theta+cos2theta/ cos4theta-cos2theta= _____
vovangra [49]

\bf \textit{Sum to Product Identities} \\\\ cos(\alpha)+cos(\beta)=2cos\left(\cfrac{\alpha+\beta}{2}\right)cos\left(\cfrac{\alpha-\beta}{2}\right) \\\\\\ cos(\alpha)-cos(\beta)=-2sin\left(\cfrac{\alpha+\beta}{2}\right)sin\left(\cfrac{\alpha-\beta}{2}\right) \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \cfrac{cos(4\theta )+cos(2\theta )}{cos(4\theta )-cos(2\theta )}\implies \cfrac{2cos\left( \frac{4\theta +2\theta }{2} \right)cos\left( \frac{4\theta -2\theta }{2} \right)}{-2sin\left( \frac{4\theta +2\theta }{2} \right)sin\left( \frac{4\theta -2\theta }{2} \right)} \implies \cfrac{cos\left( \frac{6\theta }{2} \right)cos\left( \frac{2\theta }{2} \right)}{-sin\left( \frac{6\theta }{2} \right)sin\left( \frac{2\theta }{2} \right)}

\bf \cfrac{cos(3\theta )cos(\theta )}{-sin(3\theta )sin(\theta )}\implies -\cfrac{cos(3\theta )}{sin(3\theta )}\cdot \cfrac{cos(\theta )}{sin(\theta )}\implies -cot(3\theta )cot(\theta )

8 0
3 years ago
How to find average value of a function over a given interval?
Strike441 [17]
<span><span>f<span>(x)</span>=8x−6</span><span>f<span>(x)</span>=8x-6</span></span> , <span><span>[0,3]</span><span>[0,3]

</span></span>The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.<span><span>(−∞,∞)</span><span>(-∞,∞)</span></span><span><span>{x|x∈R}</span><span>{x|x∈ℝ}</span></span><span><span>f<span>(x)</span></span><span>f<span>(x)</span></span></span> is continuous on <span><span>[0,3]</span><span>[0,3]</span></span>.<span><span>f<span>(x)</span></span><span>f<span>(x)</span></span></span> is continuousThe average value of function <span>ff</span> over the interval <span><span>[a,b]</span><span>[a,b]</span></span> is defined as <span><span>A<span>(x)</span>=<span>1<span>b−a</span></span><span>∫<span>ba</span></span>f<span>(x)</span>dx</span><span>A<span>(x)</span>=<span>1<span>b-a</span></span><span>∫ab</span>f<span>(x)</span>dx</span></span>.<span><span>A<span>(x)</span>=<span>1<span>b−a</span></span><span>∫<span>ba</span></span>f<span>(x)</span>dx</span><span>A<span>(x)</span>=<span>1<span>b-a</span></span><span>∫ab</span>f<span>(x)</span>dx</span></span>Substitute the actual values into the formula for the average value of a function.<span><span>A<span>(x)</span>=<span>1<span>3−0</span></span><span>(<span>∫<span>30</span></span>8x−6dx)</span></span><span>A<span>(x)</span>=<span>1<span>3-0</span></span><span>(<span>∫03</span>8x-6dx)</span></span></span>Since integration is linear, the integral of <span><span>8x−6</span><span>8x-6</span></span> with respect to <span>xx</span> is <span><span><span>∫<span>30</span></span>8xdx+<span>∫<span>30</span></span>−6dx</span><span><span>∫03</span>8xdx+<span>∫03</span>-6dx</span></span>.<span><span>A<span>(x)</span>=<span>1<span>3−0</span></span><span>(<span>∫<span>30</span></span>8xdx+<span>∫<span>30</span></span>−6dx)</span></span><span>A<span>(x)</span>=<span>1<span>3-0</span></span><span>(<span>∫03</span>8xdx+<span>∫03</span>-6dx)</span></span></span>Since <span>88</span> is constant with respect to <span>xx</span>, the integral of <span><span>8x</span><span>8x</span></span> with respect to <span>xx</span> is <span><span>8<span>∫<span>30</span></span>xdx</span><span>8<span>∫03</span>xdx</span></span>.<span><span>A<span>(x)</span>=<span>1<span>3−0</span></span><span>(8<span>∫<span>30</span></span>xdx+<span>∫<span>30</span></span>−6dx)</span></span><span>A<span>(x)</span>=<span>1<span>3-0</span></span><span>(8<span>∫03</span>xdx+<span>∫03</span>-6dx)</span></span></span>By the Power Rule, the integral of <span>xx</span> with respect to <span>xx</span> is <span><span><span>12</span><span>x2</span></span><span><span>12</span><span>x2</span></span></span>.<span>A<span>(x)</span>=<span>1<span>3−0</span></span><span>(8<span>(<span><span>12</span><span>x2</span><span>]<span>30</span></span></span>)</span>+<span>∫<span>30</span></span>−6dx<span>)</span></span></span>
3 0
4 years ago
Complete the table by filling in the elapsed tine.<br> Help pls
Sav [38]

Answer:

RADIUS

==================================

PROBLEM

Mary’s bicycle wheel has a circumference of 226.08 cm². What is its radius?

SOLUTION

We can solve this problem using the circumference formula in which π stands for ( 3.14 ), C stands for circumference itself and r stands for radius.

\bold{Formula \: || \: C = 2πr}Formula∣∣C=2πr

\tt{226.08 = 2(3.14) r}226.08=2(3.14)r

'Now to find the radius,Substitute 226.08 for c which is circumference in the formula.

\begin{gathered} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{C = 2πr} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{226.08 = 2(3.14)\red{r}} \\ \\ \: \: \: \: \: \: \: \: \large \tt{ \frac{226.08}{6.28} = \cancel\frac{6.28 \red{r}}{6.28} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt\green{C = 36}}\end{gathered}

C=2πr

226.08=2(3.14)r

6.28

226.08

=

6.28

6.28r

C=36

To check:

\begin{gathered} \small\begin{array}{|c|}\hline \bold{circumference }\\ \\ \tt{C = 2πr} \\ \tt{C = 2(3.14) (36\:cm) } \\ \tt{C = 2(113.04\:cm) } \\ \underline{\tt \green{C = 226.08\:cm }} \\ \hline \end{array} \end{gathered}

circumference

C=2πr

C=2(3.14)(36cm)

C=2(113.04cm)

C=226.08cm

FINAL ANSWER

If Mary's Bicycle has a circumference of 226.08 cm then the radius is 36.

\boxed{ \tt \red{r = 36}}

r=36

==================================

#CarryOnLearning

7 0
2 years ago
Read 2 more answers
HELPPPP I MARK BRAINLIEST TO CORRECT ANSWERS
azamat

Answer:

I think B option is correct ..I m 95% sure

8 0
3 years ago
Read 2 more answers
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