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BigorU [14]
3 years ago
8

In the casino game of roulette, a gambler can bet on which of 38 numbers the ball will land when the roulette wheel is spun. On

a $2 bet, a gambler gains $70 (so a net profit of $68) if he or she chooses the winning number, but loses the $2 otherwise. The probability of choosing the winning number is 1/38 and the probability of not choosing the winning number is 37/38. Let X denote the net gain from a roulette spin. What is the expected winnings for one spin?
Mathematics
1 answer:
ollegr [7]3 years ago
4 0

Answer:

Expected winnings for one spin = 0.1

Step-by-step explanation:

Expected Value = Sum of [(Probability of X)(Value of X)]

E(X) = Σ [P(X).X]

Two Events : X = +68 , P(X) = 1/38 and X = -2 , P(X) = 37/38

E(X) = [(68) (1/38)] + [(-2)(37/38)]

[(68)(0.03)] + [(-2)(0.97)]

= 2.04 - 1.94

= 0.1

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Which two rational numbers does √14 lie between?
Fiesta28 [93]

First, let's find what √14 equals.

√14 ≈ 3.74

Now, we can solve for all the answer choices.

Option A:

19/6 = 3.1666...

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Option B:

Doesn't need to be solved.

3.17

3.71

Option C:

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√9 = 3

Option D:

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The only option that contains √14 between it, is Option D.

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If we inscribe a circle such that it is touching all six corners of a regular hexagon of side 10 inches, what is the area of the
Brrunno [24]

Answer:

\left(100\pi - 150\sqrt{3}\right) square inches.

Step-by-step explanation:

<h3>Area of the Inscribed Hexagon</h3>

Refer to the first diagram attached. This inscribed regular hexagon can be split into six equilateral triangles. The length of each side of these triangle will be 10 inches (same as the length of each side of the regular hexagon.)

Refer to the second attachment for one of these equilateral triangles.

Let segment \sf CH be a height on side \sf AB. Since this triangle is equilateral, the size of each internal angle will be \sf 60^\circ. The length of segment

\displaystyle 10\, \sin\left(60^\circ\right) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}.

The area (in square inches) of this equilateral triangle will be:

\begin{aligned}&\frac{1}{2} \times \text{Base} \times\text{Height} \\ &= \frac{1}{2} \times 10 \times 5\sqrt{3}= 25\sqrt{3} \end{aligned}.

Note that the inscribed hexagon in this question is made up of six equilateral triangles like this one. Therefore, the area (in square inches) of this hexagon will be:

\displaystyle 6 \times 25\sqrt{3} = 150\sqrt{3}.

<h3>Area of of the circle that is not covered</h3>

Refer to the first diagram. The length of each side of these equilateral triangles is the same as the radius of the circle. Since the length of one such side is 10 inches, the radius of this circle will also be 10 inches.

The area (in square inches) of a circle of radius 10 inches is:

\pi \times (\text{radius})^2 = \pi \times 10^2 = 100\pi.

The area (in square inches) of the circle that the hexagon did not cover would be:

\begin{aligned}&\text{Area of circle} - \text{Area of hexagon} \\ &= 100\pi - 150\sqrt{3}\end{aligned}.

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Step-by-step explanation:

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