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3 years ago

11

PLEASE HELP! ( Drag and Drop USA Test Prep )

1 answer:

Alex17521 [72]3 years ago

4 0

-4x+7>= 27
-4x >= 20
x <= -5

answer is B
-------------------

3x-10<-25
3x < -15
x<-5
answer is A
----------------------
5(x-2)>-15
5x - 10 >-15
5x >-5
x>-1
answer is D
-----------------
x/1 - 9 < -10
x-9<-10
x<-1

answer is C

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4.068×10 to the 2 power<br><br><br>please help me

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4 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:

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

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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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4 years ago

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