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USPshnik [31]
3 years ago
12

A roulette wheel contains the integers 1 through 36, 0, and 00. suppose that you spin the wheel 6 times and that each time you b

et on a single number. what is the probability (rounded to the nearest hundredth) that you win on at least one bet?
Mathematics
1 answer:
kompoz [17]3 years ago
5 0
The probabability of winning on at least 1 bet is equal to 1 less the probaility of not winning on either of the 6 bets.

The probability of not wining on any bet is independent of winning or not winning on any of the bets, so the combined probability is calculated as the product of each individual probability.

Each indivitual probability of not winning the is:

(number of not winning outcomes) / (number of possible outcomes) = 37 / 38.

Then, the combined probability of not winning the six times is: (1/38)*(37/38)*(37/38)*(37/38)*(37/38)*(37/38) =(37/38)^6

Therefore, the probability of winning at least one bet is:

= 1 - (37/38)^6 ≈ 1 - 0.973684 ≈ 0.03.

Answer: 0.03.
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Which of the following is being constructed in the image.
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  3. a line perpendicular to a given line through a point not on the line

Step-by-step explanation:

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3 years ago
Brad and Angelina can mow their yard together with two lawn mowers in 30 min. When brad works alone it takes him 45 min. How lon
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3 years ago
What degree of rotation is represented on this matrix
Korvikt [17]

Answer:

Option B is correct

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Step-by-step explanation:

A rotation matrix is a matrix that is used to perform a rotation in Euclidean space.

To find the degree of rotation using a standard rotation matrix i.e,

R = \begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}

Given the matrix: \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}

Now, equate the given matrix with standard matrix we have;

\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} =  \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}

On comparing we get;

\cos \theta = 0       and -\sin \theta =1  

As,we know:

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\cos \theta = \cos(90^{\circ}) = \cos( -90^{\circ})

we get;

\theta = -90^{\circ}

and

\sin \theta =- \sin (90^{\circ}) = \sin ( -90^{\circ})

we get;

\theta = -90^{\circ}

Therefore, the degree of rotation is, -90^{\circ}

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3 years ago
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