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Anna [14]
2 years ago
15

Given P(A) = 0.51, P(B) = 0.79, and P(A or B) = 0.66, are events A and B mutually exclusive?​

Mathematics
1 answer:
pentagon [3]2 years ago
4 0

Answer: No, the events are not mutually exclusive

Work Shown:

P(A and B) = P(A) + P(B) - P(A or B)

P(A and B) = 0.51 + 0.79 - 0.66

P(A and B) = 0.64

Since the result is not zero, this means the events are not mutually exclusive.

Mutually exclusive events are ones that cannot happen at the same time. Example: getting a "2" and a "3" on the same roll of a number cube.

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pochemuha

Answer:

10x -  \frac{1}{x}  = 3 \\ 10x = 3 +  \frac{1}{x}  \\ 10x =  \frac{3x + 1}{x}  \\ 10x \times x = 3x + 1 \\ 10 {x}^{2}  = 3x + 1 \\ 10 {x}^{2}  - 3x - 1 = 0 \\ 10 {x}^{2}  - 5x + 2x - 1 = 0 \\ 5x(2x - 1) + 1(2x - 1) = 0 \\ (5x + 1)(2x - 1) = 0 \\  \\ 5x + 1 = 0 \\ 5x =  - 1 \\ x =  \frac{ - 1}{5}  \\  \\ 2x - 1 = 0 \\ 2x = 1 \\ x =  \frac{1}{2}

<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>you</em><em>.</em>

<em>Have</em><em> </em><em>a</em><em> </em><em>nice</em><em> </em><em>day</em><em>!</em>

4 0
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Factorise fully 18e^2f^3-12e^2f
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Answer:

18e^2f^3-12e^2f = 6e^2f(3f^2-2)

Step-by-step explanation:

Given

18e^2f^3-12e^2f

Required

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18e^2f^3-12e^2f

Express 18 as 6 * 3 and 12 as 6 * 2

18e^2f^3-12e^2f = 6*3e^2f^3-6*2e^2f

Factor out 6

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Factor out e^2

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3 years ago
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Answer:

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