What is the answer to this question
1 answer:
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Answer:
Step-by-step explanation:
Whenever we are doing unit conversions, we have to remember to make the first ratio and then each successive ratio should be put <em>in such a way</em> that the top and bottom cancels out with each unit and whatever we want, remains.
A simple example would be if we were to just convert from cm to mm. Suppose we want to convert 5 centimeters to millimeters. We can write the ratios is 2 ways:
First:
Second:
The first one is correct since we want mm, mm should be on top and cm should be on bottom form cm and cm to get canceled.
Now, for our problem, we want mm on top and day on bottom. So, cm and year should cancel. Looking at the first choice, the conversion factors are correct & skimming through, we see that
cm and cm cancels (top and bottom), also year and year cancels (top and bottom), it will leave us with mm on top and days on bottom, <em>which is what we want</em>.
First answer choice is right.
1) We have that there are in total 6 outcomes If we name the chips by 1a, 1b, 3 ,5 the combinations are: 1a,3 \ 1b, 3 \1a, 5\ 1b, 5\ 3,5\1a,1b. Of those outcomes, only one give Miguel a profit, 1-1. THen he gets 2 dollars and in the other cases he lose 1 dollar. Thus, there is a 1/6 probability that he gets 2$ and a 5/6 probability that he loses 1$.
2) We can calculate the expected value of the game with the following: E=
. In general, the formula is 
where E is the expected value, p the probability of each event and V the value of each event. This gives a result of E=2/6-5/6=3/6=0.5$ Hence, Miguel loses half a dollar ever y time he plays.
3) We can adjust the value v of the winning event and since we want to have a fair game, the expecation at the end must be 0 (he must neither win or lose on average). Thus, we need to solve the equation for v:
0=
. Multiplying by 6 both parts, we get v-5=0 or that v=5$. Hence, we must give 5$ if 1-1 happens, not 2.
4) So, we have that the probability that you get a red or purple or yellow sector is 2/7. We have that the probability for the blue sector is only 1/7 since there are 7 vectors and only one is blue. Similarly, the 2nd row of the table needs to be filled with the product of probability and expectations. Hence, for the red sector we have 2/7*(-1)=-2/7, for the yellow sector we have 2/7*1=2/7, for the purple sector it is 2/7*0=0, for the blue sector 1/7*3=3/7. The average payoff is given by adding all these, hence it is 3/7.
5) We can approach the problem just like above and set up an equation the value of one sector as an unknown. But here, we can be smarter and notice that the average outcome is equal to the average outcome of the blue sector. Hence, we can get a fair game if we make the value of the blue sector 0. If this is the case, the sum of the other sectors is 0 too (-2/7+0+2/7) and the expected value is also zero.
6) We want to maximize the points that he is getting. If he shoots three points, he will get 3 points with a probability of 0.30. Hence the average payoff is 0.30*3=0.90. If he passes to his teammate, he will shoot for 2 points, with a success rate of 0.48. Hence, the average payoff is 0.48*2=0.96. We see that he should pass to his player since 0.06 more points are scored on average.
7) Let us use the expections formula we mentioned in 1. Substituting the possibilities and the values for all 4 events (each event is the different profit of the business at the end of the year).
E=0.2*(-10000)+0.4*0+0.3*5000+0.1*8000=-2000+0+1500+800=300$
This is the average payoff of the firm per year.
8) The firm goes even when the total profits equal the investment. Suppose we have that the firm has x years in business. Then x*300=1200 must be satisfied, since the investment is 1200$ and the payoff per year is 300$. We get that x=4. Hence, Claire will get her investment back in 4 years.
2) We can calculate the expected value of the game with the following: E=
3) We can adjust the value v of the winning event and since we want to have a fair game, the expecation at the end must be 0 (he must neither win or lose on average). Thus, we need to solve the equation for v:
0=
4) So, we have that the probability that you get a red or purple or yellow sector is 2/7. We have that the probability for the blue sector is only 1/7 since there are 7 vectors and only one is blue. Similarly, the 2nd row of the table needs to be filled with the product of probability and expectations. Hence, for the red sector we have 2/7*(-1)=-2/7, for the yellow sector we have 2/7*1=2/7, for the purple sector it is 2/7*0=0, for the blue sector 1/7*3=3/7. The average payoff is given by adding all these, hence it is 3/7.
5) We can approach the problem just like above and set up an equation the value of one sector as an unknown. But here, we can be smarter and notice that the average outcome is equal to the average outcome of the blue sector. Hence, we can get a fair game if we make the value of the blue sector 0. If this is the case, the sum of the other sectors is 0 too (-2/7+0+2/7) and the expected value is also zero.
6) We want to maximize the points that he is getting. If he shoots three points, he will get 3 points with a probability of 0.30. Hence the average payoff is 0.30*3=0.90. If he passes to his teammate, he will shoot for 2 points, with a success rate of 0.48. Hence, the average payoff is 0.48*2=0.96. We see that he should pass to his player since 0.06 more points are scored on average.
7) Let us use the expections formula we mentioned in 1. Substituting the possibilities and the values for all 4 events (each event is the different profit of the business at the end of the year).
E=0.2*(-10000)+0.4*0+0.3*5000+0.1*8000=-2000+0+1500+800=300$
This is the average payoff of the firm per year.
8) The firm goes even when the total profits equal the investment. Suppose we have that the firm has x years in business. Then x*300=1200 must be satisfied, since the investment is 1200$ and the payoff per year is 300$. We get that x=4. Hence, Claire will get her investment back in 4 years.