N+2n+20-4=12+3n
3n+20-4-12=12-12+3n
3n+4=3n
3n/3 + 4= 3n/3
n+4=n
n+4-4=n-4
n=n-4
theres no further simplification, unless i messed up ...
2740/1000= 2.74
So the answer is 2.74 Liters
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
This is most likely a proportional relationship because DOUG regularly mows his neighbors yard
if 60 beats = 1 minute, and a minute is 60 seconds then 1,000,000 beats has to be divided by 60 = 16667 rounded to the nearest hundred thousand and divide that by 1440 (How many seconds in a day) = 11.574 days.
That equates to 11days, 13hours, 46 min, and 40 sec
