Answer:
0.3891 = 38.91% probability that only one is a second
Step-by-step explanation:
For each globet, there are only two possible outcoes. Either they have cosmetic flaws, or they do not. The probability of a goblet having a cosmetic flaw is independent of other globets. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![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)
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![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)
And p is the probability of X happening.
17% of its goblets have cosmetic flaws and must be classified as "seconds."
This means that ![p = 0.17](https://tex.z-dn.net/?f=p%20%3D%200.17)
Among seven randomly selected goblets, how likely is it that only one is a second
This is P(X = 1) when n = 7. So
![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 = 1) = C_{7,1}.(0.17)^{1}.(0.83)^{6} = 0.3891](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20C_%7B7%2C1%7D.%280.17%29%5E%7B1%7D.%280.83%29%5E%7B6%7D%20%3D%200.3891)
0.3891 = 38.91% probability that only one is a second
B) p=9t
he gets paid $9/hr, and his pay depends how many hours he works.
It is (2/3d)^3
Make sure the exponent is on the outside. This is because you are raising 2/3d. Not just D
Answer:
the answer is 8.5
Step-by-step explanation:
divide 85 by 10 to get 8.5 for all the dimension
51.36 is the answer.
The equation is total=original(1+% in decimal form)