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tensa zangetsu [6.8K]
1 year ago
8

Check all of the ordered pairs that satisfy the equation below.y=2/5x

Mathematics
1 answer:
zvonat [6]1 year ago
3 0

We are going to do this step by step:

Let's start with option A

A.

X = 25 and Y = 5/2

y=\frac{2}{5}\cdot x=\frac{2}{5}\cdot25=\frac{2\cdot25}{5}=\frac{50}{5}=\frac{5\cdot10}{5\cdot1}=10

In this case, when X = 25 , then Y = 10 ,which is different to 5/2

B.

X = 14 , Y = 35

y=\frac{2}{5}\cdot14=\frac{28}{5}

In case, when X = 14, then Y = 28/5, which is different to 35

C.

X = 40 , Y = 24

y=\frac{2}{5}\cdot40=\frac{80}{5}=16

Similarly, in this case, when X = 40, then Y = 16, which is different to 24

D.

X = 10 , Y=4

y=\frac{2}{5}\cdot10=\frac{20}{5}=4

Now, in this case, we can see that when X = 10, then Y = 4 which is the same as the given value of Y

E.

X = 50 , Y = 20

y=\frac{2}{5}\cdot50=\frac{100}{5}=20

In this case, the values of Y are also the same.

F.

X = 30 , Y = 12

y=\frac{2}{5}\cdot30=\frac{60}{5}=12

Again, in this case, the values of Y are the same, so the pair satisfies the equation.

In conclusion: options D, E and F satisfy the equation

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There are 4 green marbles and 2 red marbles in the jar. You just randomly draw one by one without replacement and stop when you
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Then the probability distribution is:

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EV = 1.33...

Step-by-step explanation:

We have a total of 6 marbles in the jar.

The probability of getting a red marble in the first try  (X = 0) is equal to the quotient between the number of red marbles and the total number of marbles, this is:

P(0) = 2/6 = 1/3

The probability of drawing one green marble (X = 1)

is:

First, you draw a green marble with a probability of 4/6

Then you draw the red one, but now there are 5 marbles in the jar (2 red ones and 3 green ones), then the probability is 2/5

The joint probability is:

P(1) = (4/6)*(2/5) = (2/3)*(2/5) = 4/15

The probability of drawing two green marbles (X  = 2)

Again, first we draw a green marble with a probability of 4/6

Now we draw again a green marble, now there are 3 green marbles and 5 total marbles in the jar, so this time the probability is 3/5

Now we draw the red marble (there are 2 red marbles and 4 total marbles in the jar), with a probability of 2/4

The joint probability is:

P(2) = (4/6)*(3/5)*(2/4) = (2/6)*(3/5) = 1/5

The probability of drawing 3 green marbles (X = 3)

At this point you may already understand the pattern:

First, we draw a green marble with a probability 4/6

second, we draw a green marble with a probability 3/5

third, we draw a green marble with a probability 2/4

finally, we draw a red marble with a probability 2/3

The joint probability is:

P(3) = (4/6)*(3/5)*(2/4)*(2/3) = (2/6)*(3/5)*(2/3) = (1/5)*(2/3) = (2/15)

Finally, the probability of drawing four green marbles (X = 4) is given by:

First, we draw a green marble with a probability 4/6

second, we draw a green marble with a probability 3/5

third, we draw a green marble with a probability 2/4

fourth, we draw a green marble with a probability 1/3

Finally, we draw a red marble with a probability 2/2 = 1

The joint probability is:

P(4) = (4/6)*(3/5)*(2/4)*(1/3)*1 = (1/5)*(1/3) = 1/15

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P(0) = 1/3

P(1) = 4/15

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P(3)  = 2/15

P(4) = 1/15

The expected value will be:

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EV = 1*(4/15) + 2*( 1/5) + 3*( 2/15) + 4*(1/15 ) = 1.33

So we can expect to draw 1.33 green marbles in this experiment.

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