Answer:sum: -1/6
Step-by-step explanation:
Answer:
Hence proved △ABE∼△CBF.
Step-by-step explanation:
Given,
ABCD is a parallelogram.
BF ⊥ CD and
BE ⊥ AD
To Prove : △ABE∼△CBF
We have drawn the diagram for your reference.
Proof:
Since ABCD is a parallelogram,
So according to the property of parallelogram opposite angles are equal in measure.
⇒1
And given that BF ⊥ CD and BE ⊥ AD.
So we can say that;
⇒2
Now In △ABE and △CBF
∠A = ∠C (from 1)
∠E = ∠F (from 2)
So by A.A. similarity postulate;
△ABE∼△CBF
a. By definition of conditional probability,
P(C | D) = P(C and D) / P(D) ==> P(C and D) = 0.3
b. C and D are mutually exclusive if P(C and D) = 0, but this is clearly not the case, so no.
c. C and D are independent if P(C and D) = P(C) P(D). But P(C) P(D) = 0.2 ≠ 0.3, so no.
d. Using the inclusion/exclusion principle, we have
P(C or D) = P(C) + P(D) - P(C and D) ==> P(C or D) = 0.6
e. Using the definition of conditional probability again, we have
P(D | C) = P(C and D) / P(C) ==> P(D | C) = 0.75
8.6*7=60.2, 16*1.4=22.4, 12*14=168, 7.3*20=146