Vanessa can make 5 portions
The denominator of a fraction is 1 more than 3 times the numerator. If the denominator is doubled and the numerator is increased by 2, the value of the resulting fraction is 1/4. Find the original fraction.
Answer:
0.0025 = 0.25% probability that both are defective
Step-by-step explanation:
For each item, there are only two possible outcomes. Either they are defective, or they are not. Items are independent of each other. 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.
5 percent of these are defective.
This means that ![p = 0.05](https://tex.z-dn.net/?f=p%20%3D%200.05)
If two items are randomly selected as they come off the production line, what is the probability that both are defective
This is P(X = 2) when n = 2. 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 = 2) = C_{2,2}.(0.05)^{2}.(0.95)^{0} = 0.0025](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20C_%7B2%2C2%7D.%280.05%29%5E%7B2%7D.%280.95%29%5E%7B0%7D%20%3D%200.0025)
0.0025 = 0.25% probability that both are defective
Answer:
11
Step-by-step explanation:
3+(−3)^2−(9+7)^0
=3+9−(9+7)^0
=12−(9+7)^0
=12−16^0
=12−1
=11
any number to the power of 0 is ALWAYS 1
The answer is A
Hopes this helps