What is b. If the number as a fraction is 56/90.
1 answer:
The answer is: " 28 " .
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" Twenty-eight (28) students said they would vote for Brianna."
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Explanation:
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We are given, in the problem, that there are "45 students total" .
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The fraction:
can be reduced as follows:
![\frac{56}{90} = (56÷2) / (90÷2) = [tex] \frac{28}{45}](https://tex.z-dn.net/?f=%20%5Cfrac%7B56%7D%7B90%7D%20%3D%20%2856%C3%B72%29%20%2F%20%2890%C3%B72%29%20%3D%20%5Btex%5D%20%5Cfrac%7B28%7D%7B45%7D%20)
.
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Thus, "28 students out of 45 students voted for Brianna" .
___________________________________________________
The answer is: "28 students" .
___________________________________________________
Hope this helps!
___________________________________________________
___________________________________________________
" Twenty-eight (28) students said they would vote for Brianna."
___________________________________________________
Explanation:
___________________________________________________
We are given, in the problem, that there are "45 students total" .
___________________________________________________
The fraction:
can be reduced as follows:
___________________________________________________
Thus, "28 students out of 45 students voted for Brianna" .
___________________________________________________
The answer is: "28 students" .
___________________________________________________
Hope this helps!
___________________________________________________
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Answer:
The step 4 and 5 describe the errors Stephen made.
Step-by-step explanation:
Given : Stephen evaluated . His work is shown below.
To find : Which two statements describe the errors Stephen made?
Solution :
Step 1 -
Step 2 -
Step 3 -
Step 4 -
Error is the step as 4 sign is negative.
So correct step is
Step 5 -
Error is the step as there is 4 in place of 3.
So correct step is
The step 4 and 5 describe the errors Stephen made.
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.