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Bas_tet [7]
3 years ago
5

Suppose P(E) = 81%; P(E and F) = 9%; and P(E or F) = 65%,

Mathematics
1 answer:
PIT_PIT [208]3 years ago
4 0

Using Venn probabilities, it is found that P(F) = -0.07, which is not a probability, as it is a negative value.

------------------

  • We have two events, E and F.
  • Their probabilities are: P(E) = 0.81, P(E \cap F) = 0.09, P(E \cup F) = 0.65
  • The probabilities are related by the following equation:

P(E \cap F) = P(E) + P(F) - P(E \cup F)

  • Replacing into the equation, we can find P(F).

0.09 = 0.81 + P(F) - 0.65

0.16 + P(F) = 0.09

P(F) = -0.07

P(F) = -0.07, which is not a probability, as it is a negative value.

A similar problem is given at brainly.com/question/21421475

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Given z1 = 2 +StartRoot 3 EndRoot i and z2 = 1 – StartRoot 3 EndRoot i, what is the sum of z1 and z2?
Fofino [41]

Answer:

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Step-by-step explanation:

Since we are given z1 = 2 + √(3)i and z2 = 1 – √(3)i. The sum of z1 + z2 would be:

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8 0
3 years ago
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