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Dennis_Churaev [7]
3 years ago
10

In the game of​ roulette, when a player gives the casino $21 for a bet on the number 33​, the player has a 37/38 probability of

losing $21 and a 1/38 probability of making a net gain of $735.​(The prize is ​$756​, but the​ player's $21 bet is not​ returned, so the net gain is $735​.) If a player bets $21 that the outcome is an odd​ number, the probability of losing ​$21 is 20/38 and the probability of making a net gain of $21 is 18/38. ​(If a player bets $21 on an odd number and​ win, the player is given ​$42 that includes the​ bet, so the net gain is $21​.)
If a player bets $21 on the number 33​, what is the​ player's expected​ value?

The expected value is _______ dollars. (round to the nearest cent as needed)

________________________________________________________________________________
Mathematics
1 answer:
ohaa [14]3 years ago
4 0

Answer:

Step-by-step explanation:

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If x metres of string are cut from a piece y metres long how much is left?
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Step-by-step explanation:

If we have a piece of string that is y metres long, and we cut x metres of strings from it, then the new length of the string will be equal to the difference between the initial length and the amount that we cut off.

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What is the answer to this?
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5

Step-by-step explanation:

8 0
3 years ago
F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa
DerKrebs [107]

a. Parameterize C by

\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath

with 0\le t\le1.

b/c. The line integral of \vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath over C is

\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt

=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt

=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt

=\displaystyle10\int_0^1\mathrm dt=\boxed{10}

d. Notice that we can write the line integral as

\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)

By Green's theorem, the line integral is equivalent to

\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy

where D is the triangle bounded by C, and this integral is simply twice the area of D. D is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

4 0
2 years ago
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