Answer:
<h3>For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.</h3>
By De morgan's law

which is Bonferroni’s inequality
<h3>Result 1: P (Ac) = 1 − P(A)</h3>
Proof
If S is universal set then

<h3>Result 2 : For any two events A and B, P (A∪B) = P (A)+P (B)−P (A∩B) and P(A) ≥ P(B)</h3>
Proof:
If S is a universal set then:

Which show A∪B can be expressed as union of two disjoint sets.
If A and (B∩Ac) are two disjoint sets then
B can be expressed as:

If B is intersection of two disjoint sets then

Then (1) becomes

<h3>Result 3: For any two events A and B, P(A) = P(A ∩ B) + P (A ∩ Bc)</h3>
Proof:
If A and B are two disjoint sets then

<h3>Result 4: If B ⊂ A, then A∩B = B. Therefore P (A)−P (B) = P (A ∩ Bc) </h3>
Proof:
If B is subset of A then all elements of B lie in A so A ∩ B =B
where A and A ∩ Bc are disjoint.

From axiom P(E)≥0

Therefore,
P(A)≥P(B)
Answer:
5 km
Step-by-step explanation:
Use the distance formula.
Hopes this helps if right please mark as brainlist:
Answer: -12x^2 + 15x
Answer:
x=19
Step-by-step explanation:
The consecutive integers will be x, x+1 and x+2 since consecutive integers have a pattern.
So, the equation is x+x+1+x+2=60
Solve:
x+x+1+x+2=60
3x+3=60
Subtract 3 on both sides:
3x+3-3=60-3
3x=57
Divide by 3
3x/3=57/3
x=19