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Lera25 [3.4K]
3 years ago
7

A person playing a game of chance has a 0.25 probability of winning. If the person plays the game 20 times and wins half of that

number of times, what is the difference between the theoretical probability and the experimental probability of that person winning ?
Mathematics
1 answer:
antoniya [11.8K]3 years ago
3 0
The theoretical probability is what is expected to happen which is 0.25 or 25%. The experimental probability is the actual or observed during the play. Since the person won half of 20 games, the experimental probability is 10/20 or 50%.

The difference between the theoretical and experimental probability is then

50%-25%=25% or 0.25

We note that the experimental probability is higher than the theoretical. However, as more games are played, the fraction of games won becomes closer and closer to the theoretical probability.
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The vertex of the function f(x) exists (1, 5), the vertex of the function g(x) exists (-2, -3), and the vertex of the function f(x) exists maximum and the vertex of the function g(x) exists minimum.

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If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.
nydimaria [60]

Answer:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

Step-by-step explanation:

Given

\cos(\theta) = -\frac{2}{3}

\theta \to Quadrant III

Required

Determine \tan(\theta) \cdot \cot(\theta) + \csc(\theta)

We have:

\cos(\theta) = -\frac{2}{3}

We know that:

\sin^2(\theta) + \cos^2(\theta) = 1

This gives:

\sin^2(\theta) + (-\frac{2}{3})^2 = 1

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Collect like terms

\sin^2(\theta)  = 1 - \frac{4}{9}

Take LCM and solve

\sin^2(\theta)  = \frac{9 -4}{9}

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Take the square roots of both sides

\sin(\theta)  = \±\frac{\sqrt 5}{3}

Sin is negative in quadrant III. So:

\sin(\theta)  = -\frac{\sqrt 5}{3}

Calculate \csc(\theta)

\csc(\theta) = \frac{1}{\sin(\theta)}

We have: \sin(\theta)  = -\frac{\sqrt 5}{3}

So:

\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}

\csc(\theta) = \frac{-3}{\sqrt 5}

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So, we have:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta)

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Substitute: \csc(\theta) = \frac{-3\sqrt 5}{5}

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Take LCM

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

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