Liam uses 3/4 of a cup of butter in his cookie recipe. Gloria uses 2/4 of a cup of butter in her cookie recipe. How much more bu
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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.
Refer to the figure shown below.
Because the question states that the semi circles are at the ends of the rectangle, each semicircle has a radius of 30 m.
The circumference of the oval is
2π(30) + 2*120 = 60π + 240 m = 428.5 m
The area of the oval is
π(30²) + 60*120 = 900π + 7200 m² = 1.0027 x 10⁴ m²
Note:
If the semi circles are placed on the 120-m sides of the rectangle, then similar calculations yield:
Circumference = 120π + 120 m = 497 m
Area = 3600π + 7200 m² = 1.851 x 10⁴ m²
Answer:
If the semi circles are placed on the 60-m sides of the rectangle, then
Circumference = 60π + 240 m, or 428.5 m
Area = 900π + 7200 m², or 1.0027 x 10⁴ m²
if the semi circles are placed on the 120-m sides of the rectangle, then
Circumference = 120π + 120 m, or 497 m
Area = 3600π + 7500 m², or 1.851 x 10⁴ m²
Because the question states that the semi circles are at the ends of the rectangle, each semicircle has a radius of 30 m.
The circumference of the oval is
2π(30) + 2*120 = 60π + 240 m = 428.5 m
The area of the oval is
π(30²) + 60*120 = 900π + 7200 m² = 1.0027 x 10⁴ m²
Note:
If the semi circles are placed on the 120-m sides of the rectangle, then similar calculations yield:
Circumference = 120π + 120 m = 497 m
Area = 3600π + 7200 m² = 1.851 x 10⁴ m²
Answer:
If the semi circles are placed on the 60-m sides of the rectangle, then
Circumference = 60π + 240 m, or 428.5 m
Area = 900π + 7200 m², or 1.0027 x 10⁴ m²
if the semi circles are placed on the 120-m sides of the rectangle, then
Circumference = 120π + 120 m, or 497 m
Area = 3600π + 7500 m², or 1.851 x 10⁴ m²