Answer:
P(A|D) and P(D|A) from the table above are not equal because P(A|D) = and P(D|A) =
Step-by-step explanation:
Conditional probability is the probability of one event occurring with some relationship to one or more other events
.
P(A|D) is called the "Conditional Probability" of A given D
P(D|A) is called the "Conditional Probability" of D given A
The formula for conditional probability of P(A|D) = P(D∩A)/P(D)
The formula for conditional probability of P(D|A) = P(A∩D)/P(A)
The table
↓ ↓ ↓
: C : D : Total
→ A : 6 : 2 : 8
→ B : 1 : 8 : 9
→Total : 7 : 10 : 17
∵ P(A|D) = P(D∩A)/P(D)
∵ P(D∩A) = 2 ⇒ the common of D and A
- P(D) means total of column D
∵ P(D) = 10
∴ P(A|D) =
∵ P(D|A) = P(A∩D)/P(A)
∵ P(A∩D) = 2 ⇒ the common of A and D
- P(A) means total of row A
∵ P(A) = 8
∴ P(D|A) =
∵ P(A|D) =
∵ P(D|A) =
∵ ≠
∴ P(A|D) and P(D|A) from the table above are not equal
Step-by-step explanation:
The answer is 250. Just divide top by bottom.
203/3.5=58mi/h
243/4.5=54mi/h
(58+54)/2=56mi/h
The answer in 56mi/h
<span>y= 2*3-4x;
differentiate with respect to t we get dy = -4xdt; -- eqn 1
given dt = 4 substitute in eqn 1 we get , dy =-4 x( 4) = -16x; --eqn2
given x=1 then sub then the value of x in eqn 2 we get dy = -16(1)=-16
ie Ans dy = -16</span>
Answer:

Step-by-step explanation:
For this triangle we have to

Now we use the sine theorem to find the length of b:

Then:
