Answer:
57
Step-by-step explanation:

Your answer is 57.
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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)
The x-intercept of f(x) = (x + 6)(x - 3)
f(x) = 0 → (x + 6)(x - 3) = 0 → x + 6 =0 or x - 3 =0
x = -6 or x = 3
Therefore the x-intercepts are: (-6; 0) and (3; 0)
Your answer is (-6; 0)
R could equal any number, but it has to be higher than 4, because 3+4=7.