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Rasek [7]
1 year ago
8

In each turn of a game you toss two coins. if 2 heads come up, you win 2 points and if 1 head comes up you win 1 point. if no he

ads come up, you lose 3 points. what is the expected value of the number of points for each turn?
Mathematics
1 answer:
Blizzard [7]1 year ago
3 0

The expected value of the number of points for each turn is 0.25

<h3>What is expected value?</h3>

Expected value describes the long term average level of a random variable based on its probabilty distribution.

Now, Probability that 2 heads come up is

P(2 is H) = 1/4

Probability that 1 heads come up is

P(1 is H) = 2/4

⇒P(1 is H) = 1/2

Probability that no head will come up

P(no H's) = 1/4

Hence, the expected value for winning of 2points, 1point and  lose of3 points is given as-

Expected value = 2 * P(2 is H) + 1 * P(1 is H)  - 3 * P(no H's)

⇒Expected value = 2 * 1/4 + 1 * 1/2 - 3 * 1/4

= 0.5 + 0.5 - 0.75

⇒Expected value = 0.25

More about Expected value :

brainly.com/question/22715250

#SPJ1

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Answer:  51 %

Step-by-step explanation:

3.5 * 5 = 17.5

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17.5 * 100 = 34x

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x = 1750/34

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The radius of the base of a cylinder is decreasing at a rate of 121212 kilometers per second. The height of the cylinder is fixe
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Answer:

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Step-by-step explanation:

Given that:

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First of all, let us have a look at the formula for volume of a cylinder.

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As per question statement:

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\dfrac{dV}{dt} = \pi h\dfrac{d}{dt} r^2\\\Rightarrow \dfrac{dV}{dt} = \pi h\times 2 r\dfrac{dr}{dt} \\$Putting the values:$\\\Rigghtarrow\dfrac{dV}{dt} = 3.14 \times 2.5\times 2 \times 40\times 12 \\\Rigghtarrow\dfrac{dV}{dt} = 7536\ km^3/sec

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