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Amanda [17]
2 years ago
5

g European roulette. The game of European roulette involves spinning a wheel with 37 slots: 18 red, 18 black, and 1 green. A bal

l is spun onto the wheel and will eventually land in a slot, where each slot has an equal chance of capturing the ball. Gamblers can place bets on red or black. If the ball lands on their color, they double their money. If it lands on another color, they lose their money. (a) Suppose you play roulette and bet $3 on a single round. What is the expected value and standard deviation of your total winnings
Mathematics
1 answer:
butalik [34]2 years ago
6 0

Answer:

The expected value and standard deviation of your total winnings are -$0.081 and $3 respectively.

Step-by-step explanation:

We are given that the game of European roulette involves spinning a wheel with 37 slots: 18 red, 18 black, and 1 green.

Gamblers can place bets on red or black. If the ball lands on their color, they double their money. If it lands on another color, they lose their money.

Let the probability of the ball landing on red slot = \frac{18}{37}

The probability of the ball landing on black slot = \frac{18}{37}

The probability of the ball landing on green slot = \frac{1}{37}

Now, it is stated that Gambler can place bets only on the red or black slot, so;

The probability of winning the bet will be = \frac{18}{37}

and the probability of losing the bet will be = \frac{18}{37}+\frac{1}{37}

                                                                        = \frac{19}{37}

If the gambler wins he gets $3 and if he loses he will get -$3.

So, the expected value of gambler's total winnings is given by;

E(X) = \sum X \times P(X)

        = \$3 \times \frac{18}{37} + (-\$3 \times \frac{19}{37})

        = \$3 \times (-\frac{1}{37})  = -$0.081

Now, the standard deviation of gambler's total winnings is given by;

S.D.(X) = \sqrt{(\sum X^{2}  \times P(X))-(\sum X \times P(X))^{2} }

So, E(X^{2})=\sum X^{2}  \times P(X)

                 = \$3^{2}  \times \frac{18}{37} + (-\$3^{2}  \times \frac{19}{37})

                 = \$9 \times (\frac{18}{37}+\frac{19}{37})  = $9

Now, S.D.(X) = \sqrt{\$9-(-\$0.081)^{2} }

                     = \sqrt{8.993}  = $2.99 ≈ $3

Hence, the expected value and standard deviation of your total winnings are -$0.081 and $3 respectively.

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Solve x − 5y = 6 for x.
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Answer:

All real numbers

Step-by-step explanation:

Solve a system of equations by elimination

With the equation given, you can solve for y to get: y = \frac{1}{5}x - \frac{6}{5}

Then, substitute this value in on the equation.

x - 5(\frac{1}{5}x - \frac{6}{5}) = 6

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At a football game, 69 out of 92 students were wearing the color red to support the home team. What percent of the students were
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LK is tangent to circle J at point K. Circle J is shown. Line segment J K is a radius. Line segment K L is a tangent that inters
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Answer:

(B)r=\dfrac{85}{12}

Step-by-step explanation:

Given a circle centre J

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From the diagram attached

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Theorem: The angle between a tangent and a radius is 90 degrees.

By the theorem above, Triangle JLK forms a right triangle with LJ as the hypotenuse.

Using Pythagoras Theorem:

(6+r)^2=r^2+11^2\\(6+r)(6+r)=r^2+121\\36+6r+6r+r^2=r^2+121\\12r=121-36\\12r=85\\r=\dfrac{85}{12}

The length of the radius, r=\dfrac{85}{12}

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Answer:

132 m².

Step-by-step explanation:

What's the surface area of each lateral face of this pyramid?

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The base of this pyramid is a square. As a result, all four lateral sides are congruent. The area of each of these triangle is thus

\displaystyle \frac{1}{2}\times \text{Base}\times \text{Height} = \rm \frac{1}{2} \times 6 \times 8 = 24\; m^{2}.

The base of this pyramid is a square. The length of a side of this square is 6 meters. The area of the base will be

\text{Side}^{2} = \rm 6^{2} = 36\; m^{2}.

Put the five faces together to get the total surface area of this square pyramid:

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Let D be the smaller cap cut from a solid ball of radius 8 units by a plane 4 units from the center of the sphere. Express the v
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Answer:

Step-by-step explanation:

The equation of the sphere, centered a the origin is given by x^2+y^2+z^2 = 64. Then, when z=4, we get

x^2+y^2= 64-16 = 48.

This equation corresponds to a circle of radius 4\sqrt[]{3} in the x-y plane

c) We will use the previous analysis to define the limits in cartesian and polar coordinates. At first, we now that x varies from -4\sqrt[]{3} up to 4\sqrt[]{3}. This is by taking y =0 and seeing the furthest points of x that lay on the circle. Then, we know that y varies from -\sqrt[]{48-x^2} and \sqrt[]{48-x^2}, this is again because y must lie in the interior of the circle we found. Finally, we know that z goes from 4 up to the sphere, that is , z goes from 4 up to \sqrt[]{64-x^2-y^2}

Then, the triple integral that gives us the volume of D in cartesian coordinates is

\int_{-4\sqrt[]{3}}^{4\sqrt[]{3}}\int_{-\sqrt[]{48-x^2}}^{\sqrt[]{48-x^2}} \int_{4}^{\sqrt[]{64-x^2-y^2}} dz dy dx.

b) Recall that the cylindrical  coordinates are given by x=r\cos \theta, y = r\sin \theta,z = z, where r corresponds to the distance of the projection onto the x-y plane to the origin. REcall that x^2+y^2 = r^2. WE will find the new limits for each of the new coordinates. NOte that, we got a previous restriction of a circle, so, since \theta[\tex] is the angle between the projection to the x-y plane and the x axis, in order for us to cover the whole circle, we need that [tex]\theta goes from 0 to 2\pi. Also, note that r goes from the origin up to the border of the circle, where r has a value of 4\sqrt[]{3}. Finally, note that Z goes from the plane z=4 up to the sphere itself, where the restriction is \sqrt[]{64-r^2}. So, the following is the integral that gives the wanted volume

\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta. Recall that the r factor appears because it is the jacobian associated to the change of variable from cartesian coordinates to polar coordinates. This guarantees us that the integral has the same value. (The explanation on how to compute the jacobian is beyond the scope of this answer).

a) For the spherical coordinates, recall that z = \rho \cos \phi, y = \rho \sin \phi \sin \theta,  x = \rho \sin \phi \cos \theta. where \phi is the angle of the vector with the z axis, which varies from 0 up to pi. Note that when z=4, that angle is constant over the boundary of the circle we found previously. On that circle. Let us calculate the angle by taking a point on the circle and using the formula of the angle between two vectors. If z=4 and x=0, then y=4\sqrt[]{3} if we take the positive square root of 48. So, let us calculate the angle between the vectora=(0,4\sqrt[]{3},4) and the vector b =(0,0,1) which corresponds to the unit vector over the z axis. Let us use the following formula

\cos \phi = \frac{a\cdot b}{||a||||b||} = \frac{(0,4\sqrt[]{3},4)\cdot (0,0,1)}{8}= \frac{1}{2}

Therefore, over the circle, \phi = \frac{\pi}{3}. Note that rho varies from the plane z=4, up to the sphere, where rho is 8. Since z = \rho \cos \phi, then over the plane we have that \rho = \frac{4}{\cos \phi} Then, the following is the desired integral

\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{\frac{4}{\cos \phi}}^{8}\rho^2 \sin \phi d\rho d\phi d\theta where the new factor is the jacobian for the spherical coordinates.

d ) Let us use the integral in cylindrical coordinates

\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta=\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} r (\sqrt[]{64-r^2}-4) dr d\theta=\int_{0}^{2\pi} d \theta \cdot \int_{0}^{4\sqrt[]{3}}r (\sqrt[]{64-r^2}-4)dr= 2\pi \cdot (-2\left.r^{2}\right|_0^{4\sqrt[]{3}})\int_{0}^{4\sqrt[]{3}}r \sqrt[]{64-r^2} dr

Note that we can split the integral since the inner part does not depend on theta on any way. If we use the substitution u = 64-r^2 then \frac{-du}{2} = r dr, then

=-2\pi \cdot \left.(\frac{1}{3}(64-r^2)^{\frac{3}{2}}+2r^{2})\right|_0^{4\sqrt[]{3}}=\frac{320\pi}{3}

3 0
2 years ago
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