Answer:
To find the conjugate of a complex number we just change the sign of the imaginary part.
-4 + 5i = -4 -5i
The conclusion is P(pass l morning) = 0.56
P(pass l afternoon) = 0.72
Conclusion: a student taking the test in the afternoon has a greater probability of passing than a student taking the test in the morning.
<h3>What are the probabilities?</h3>
Probability is the odds that a random event would happen. The odds that the event would happen lies between 0 and 1. The closer the probability is to one, the more likely it is for the event to happen.
P(pass l morning) = number of students who took the test in the morning and passed / total number of students that took the exam in the morning
28 / 50 = 0.56
P(pass l afternoon) = number of students who took the test in the afternoon and passed / total number of students that took the exam in the afternoon = 36 / 50 = 0.72
To learn more about probability, please check: brainly.com/question/13234031
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There would be a multiplication sign in between the 8 and the 1
Answer:
Use the view example tool
Step-by-step explanation:
Use the view example tool and just plug in your number's instead of theirs. I did the same and I got it right.
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)
