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MaRussiya [10]
3 years ago
8

In the game of​ roulette, a player can place a ​$5 bet on the number 3 and have a startfraction 1 over 38 endfraction probabilit

y of winning. if the metal ball lands on 3​, the player gets to keep the ​$5 paid to play the game and the player is awarded an additional ​$175. ​otherwise, the player is awarded nothing and the casino takes the​ player's ​$5. what is the expected value of the game to the​ player? if you played the game 1000​ times, how much would you expect to​ lose? note that the expected value is the​ amount, on​ average, one would expect to gain or lose each game. the expected value is ​$ nothing. ​(round to the nearest cent as​ needed.) the player would expect to lose about ​$ nothing. ​(round to the nearest cent as​ needed.)
Mathematics
1 answer:
Misha Larkins [42]3 years ago
3 0
The expected value per game is -0.26.  Over 1000 games, you can expect to lose $263.16.

To find the expected value, we multiply the probability of winning by the amount of winnings, the probability of losing by the amount of loss, and adding those together.

We have a 1/38 chance of winning; 1/38(175) = $4.61.  We also have a 37/38 chance of losing; 37/38(5) = $4.87.

$4.61-$4.87 = -$0.26 (rounded)

To five decimal places, our answer is -0.26136; multiplied by 1000 games, this is $261.36 lost.
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hichkok12 [17]

Answer:

\mathbf{\dfrac{\pi}{6}[5 \sqrt{5}-1]}

Step-by-step explanation:

Given that:

The surface area (S.A) z = x^2 +y^2

Hence the S.A is of form z = f(x,y)

Then the S.A can be represented with the equation

A(S) = \iint _D \sqrt{1+ (\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2} \ dA

where :

D = cylinder x^2 +y^2 =1

In polar co-ordinates:

D = {(r, θ): 0≤ r ≤ 1, 0 ≤ θ ≤ 2π)

Similarly, \dfrac{\partial z}{\partial x} = 2x and \dfrac{\partial z}{\partial y} = 2y

Therefore;

S.A = \iint_D \sqrt{1+4x^2+4y^2} \ dA

= \iint_D \sqrt{1+4(x^2+y^2)} \ dA

= \int^{2 \pi}_{0} \int^{1}_{0}  \sqrt{1+4r^2} \ r \ dr \d \theta

= [\theta]^{2 \pi}_{0} \dfrac{1}{8}\times \dfrac{2}{3}\begin {bmatrix} (1+4r^2)^{\dfrac{3}{2}}\end {bmatrix}^1_0

= 2 \pi \times \dfrac{1}{12}[5^{\dfrac{3}{2}} - 1]

\mathbf{=\dfrac{\pi}{6}[5 \sqrt{5}-1]}

6 0
3 years ago
Use compatible numbers to estimate the quotient <br> 546÷6
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Answer:

Well alright already 546 divided by 6 is

is 91 just by evaluate Hope this helps :)

Step-by-step explanation:


6 0
3 years ago
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Answer: You have a wonderful day as well!

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That should be it...

Just multiply 2 by t

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