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Talja [164]
4 years ago
10

A student worked out the following problem to find

Mathematics
1 answer:
polet [3.4K]4 years ago
7 0
We have that

y=x²<span>+3x-5
</span>y=4x+1

using a graph tool
see the attached figure

the solution are the points
(-2,-7)
(3,13)

<span>the solution obtained by the student
</span>(-2,0)
(3,0)<span>
is incorrect because the values ​​of the coordinate y are incorrect
</span>the values of x are correct
x1=-2
x2=3

<span>with those values ​​of x obtained, the student had to substitute it in any of the two equations to obtain the value of y
</span>so
for x=3
y=4x+1------> 4*(3)+1-----> y=13

for x=-2
y=4x+1------> 4*(-2)+1-----> y=-7

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A bag contains $3$ white chips and $3$ red chips. you repeatedly draw a chip at random from the bag. if it's white, you set it a
Natasha2012 [34]

Answer:

8.5

Step-by-step explanation:

We can divide this experiment in three parts.

  • Before removing any white chip
  • After ramoving one white chip, and before removing two
  • After removing two white chips

For the first experiment, for each extraction we will always have 3 white chips and 3 red chips, becuase if we extract a red chip, then we put it back in the bag, and if it is a white chip, then the experiment ends there. The probability of taking out a white chip is 1/2.

For the second experiment, we will have always 2 white chips, and 3 red chips. So the probability of success is 2/5 = 0.4

For the third experiment, we will have always 1 white chip and 3 red chips, so the probability of success if 1/4 = 0.25.

We want to know how many extractions we need for each experiment until we pick a white chip. Note that each experiment is independent of each other, and each one has geometric distribution, the first one with probability of success 0.5, the second one with probability of success 0.4, and the third one with p = 0.25.

The total experiment, X, is the sum of this random variables, so the expected value of the total experimet is the sum of the expected value from each of its parts, lets call them, X₁, X₂ and X₃. Thus,

E(X) = E(X_1) + E(X_2) + E(X_3) = \frac{1}{0.5}+\frac{1}{0.4} +\frac{1}{0.25} = 2 + 2.5 + 4 = 8.5

The expected number of draws is 8.5.

I hope i could help you!  

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