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Anton [14]
3 years ago
9

What is the distance?

Mathematics
1 answer:
Anon25 [30]3 years ago
6 0

Answer:

<h3><em>2√41 unit</em>s</h3>

Step-by-step explanation:

<h2>Explanation is in the attachment</h2>

<h3>Hope it is helpful to you ☺️</h3>

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Angle A corresponds to angle____ <br><br> B <br> C<br> E<br> D<br> none of the above
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Answer: none of the above

Explanation:

There is no angle corresponding to angle A
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Dan invests £7600 into his bank account.
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Answer: 8512

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Alice searches for her term paper in her filing cabinet, which has several drawers. She knows thatshe left her term paper in dra
katen-ka-za [31]

You made a mistake with the probability p_{j}, which should be p_{i} in the last expression, so to be clear I will state the expression again.

So we want to solve the following:

Conditioned on this event, show that the probability that her paper is in drawer j, is given by:

(1) \frac{p_{j} }{1-d_{i}p_{i}  } , if j \neq i, and

(2) \frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i}  } , if j = i.

so we can say:

A is the event that you search drawer i and find nothing,

B is the event that you search drawer i and find the paper,

C_{k}  is the event that the paper is in drawer k, k = 1, ..., n.

this gives us:

P(B) = P(B \cap C_{i} ) = P(C_{i})P(B | C_{i} ) = d_{i} p_{i}

P(A) = 1 - P(B) = 1 - d_{i} p_{i}

Solution to Part (1):

if j \neq i, then P(A \cap C_{j} ) = P(C_{j} ),

this means that

P(C_{j} |A) = \frac{P(A \cap C_{j})}{P(A)}  = \frac{P(C_{j} )}{P(A)}  = \frac{p_{j} }{1-d_{i}p_{i}  }

as needed so part one is solved.

Solution to Part(2):

so we have now that if j = i, we get that:

P(C_{j}|A ) = \frac{P(A \cap C_{j})}{P(A)}

remember that:

P(A|C_{j} ) = \frac{P(A \cap C_{j})}{P(C_{j})}

this implies that:

P(A \cap C_{j}) = P(C_{j}) \cdot P(A|C_{j}) = p_{i} (1-d_{i} )

so we just need to combine the above relations to get:

P(C_{j}|A) = \frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i}  }

as needed so part two is solved.

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