Answer:
P(A∣D) = 0.667
Step-by-step explanation:
We are given;
P(A) = 3P(B)
P(D|A) = 0.03
P(D|B) = 0.045
Now, we want to find P(A∣D) which is the posterior probability that a computer comes from factory A when given that it is defective.
Using Bayes' Rule and Law of Total Probability, we will get;
P(A∣D) = [P(A) * P(D|A)]/[(P(A) * P(D|A)) + (P(B) * P(D|B))]
Plugging in the relevant values, we have;
P(A∣D) = [3P(B) * 0.03]/[(3P(B) * 0.03) + (P(B) * 0.045)]
P(A∣D) = [P(B)/P(B)] [0.09]/[0.09 + 0.045]
P(B) will cancel out to give;
P(A∣D) = 0.09/0.135
P(A∣D) = 0.667
X is less than -2, so -2 is our largest value of the interval, so it goes on the right. Since there is no lower endpoint (it is ALL values less than -2), we put the negative infinity symbol on the left side. The curved end on -2 indicates an open interval
A + 2c - 5b - c + a - b
We're gonna have to combine like terms.
(a+a)+(-5b-b)+(2c-c)
2a-6b+c This is your answer
Answer:
69 is your answer
Step-by-step explanation:
Have a wonderful day and good luck on you're test!
Parametric form.....
y=3x+ 1/4z
