The answer is C.
Vendor A= 456/1700, which is .268 or 27%
Vendor B= 607/1700, which is .357 or 36%
Vendor C= 637/1700, which is .374 or 37%
For a total of 100%
Using the binomial distribution, it is found that there is a 0.125 = 12.5% probability that the team goes to the championship.
<h3>What is the binomial distribution formula?</h3>
The formula is:
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- His team has a p = 0.5 probability of winning each game.
- The team plays 3 games, hence n = 3.
The team moves to the championship if they win all three games, hence the probability is P(X = 3), given as follows:
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 3) = C_{3,3}.(0.5)^{3}.(0.5)^{0} = 0.125](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20C_%7B3%2C3%7D.%280.5%29%5E%7B3%7D.%280.5%29%5E%7B0%7D%20%3D%200.125)
0.125 = 12.5% probability that the team goes to the championship.
More can be learned about the binomial distribution at brainly.com/question/24863377
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Answer:
10
Step-by-step explanation:
h(x)=4x-2 h(x)=3
h(3)=4(3)-2 substitute
h(3)=12-2 multiply
=10 subtract