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Ipatiy [6.2K]
3 years ago
9

What is the solution of x=2+x-2squared

Mathematics
1 answer:
Hitman42 [59]3 years ago
7 0
X = 2 + (x-2)sqrt 
<=> x - 2 = (x - 2)sqrt
<=> x-2 = 0 OR x - 2 = 1
=> x = 2 OR x = 3
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Cheese in my pockets velvita
likoan [24]

Answer:

1. A

2. C

Step-by-step explanation:

1.The mean is the best measure when the data is not skewed and when it has no outliers. If it is skewed or has outliers, we need to use something like the median or mode. In this case, the most clustered and normal data, with no outliers, is choice A.

Why not B: It has an outlier at 3.

Why not C: It's all over the place, near 70 and then near 25ish.

Why not D: There's a huge outlier at 301 that will bring up the mean, which isn't accurate to the actual center.

2.The mode is most helpful when there is repeating data. In this case, the one that works is choice C, because it has a clear mode of 36.

Why not A or D: They don't have clear modes, because they have two values that have more than one occurence.

Why not B: It has no mode.

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3 years ago
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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Answer:

Thx

Step-by-step explanation:

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