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laiz [17]
4 years ago
7

P(x)-x^4-3x^2+kx where k is an unknown integer.

Mathematics
1 answer:
timurjin [86]4 years ago
5 0

Answer:4

Step-by-step explanation:

P(2)

(2)

4

−3(2)

2

+k(2)−2

16−3⋅4+2k−2

2+2k

2k

k

​

=10

=10

=10

=10

=8

=4

​

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X + y = -3 & 5x - 2y = -50 ELIMINATION
Karolina [17]
I gotchu luh bro....

7 0
3 years ago
What is the answer plz help
stira [4]
0.334 if you turn it to decimals... still acceptable?
4 0
4 years ago
An auditorium has rows of seats with 8 seats in each row. Kayla knows there are at least 70 seats but fewer than 340 seats in th
Vera_Pavlovna [14]

Answer:

The possible number of rows can be 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, or 42.

Step-by-step explanation:

First of all, if there are 70 seats, there should be 9 rows. (9*8 =72, which has 70 seats). And, 340 seats are there, so 340/8 = 42.5, i.e. 42 rows can be there at maximum. So, the actual number of rows can be between 9 and 42. Therefore the possible number of rows can be 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, or 42.

Hope this helps. :)

8 0
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.

8 0
4 years ago
Place the following scenarios in order by the size of their probability, largest to smallest.
IrinaK [193]

Answer:

3, 1, 2, 4

Step-by-step explanation:

i am not sure but i think this is it

7 0
3 years ago
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