Answer:
P(B|A)=0.25 , P(A|B) =0.5
Step-by-step explanation:
The question provides the following data:
P(A)= 0.8
P(B)= 0.4
P(A∩B) = 0.2
Since the question does not mention which of the conditional probabilities need to be found out, I will show the working to calculate both of them.
To calculate the probability that event B will occur given that A has already occurred (P(B|A) is read as the probability of event B given A) can be calculated as:
P(B|A) = P(A∩B)/P(A)
= (0.2) / (0.8)
P(B|A)=0.25
To calculate the probability that event A will occur given that B has already occurred (P(A|B) is read as the probability of event A given B) can be calculated as:
P(A|B) = P(A∩B)/P(B)
= (0.2)/(0.4)
P(A|B) =0.5
Answer:
No.
Step-by-step explanation:
As x increases 2^x will grow faster than 5 x^2. This is because the x in 2^x is an exponent and its graph grows very steeply compared with 5x^2 , as x increases.
For example when x = 10
5x^2 = 5*10^2 = 500
2^x = 2^10 =1024
When x = 11:
5x^2 = 605
2^x = 2048 and the difference will continue to increase.
Answer:j
Step-by-step explanation:
If it’s closed it would be greater than -2 but since it’s open it’s equal or greater than
Answer: D) 101
Step-by-step explanation:
By linearity, we can break it up into 2 integrals. The integral and derivative of f easily cancel out
I used the table for values of f(x) at 10 and -1. Wouldn't be surprised if this was part of a series of questions about f because I really can't see how you could use the hypothesis that f is twice differentiable on R. Same for the other table values. I'm curious about how you found the answer. Was it a different way?
