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Nezavi [6.7K]
2 years ago
13

In a board game, you roll a die to win or lose points, depending on the outcome. The outcomes follow this probability distributi

on.
If a player can only win by accumulating 20 points, which of the following best describes the fairness of the game?

Mathematics
2 answers:
Nuetrik [128]2 years ago
4 0

A, it's not fair because the chance to lose points far outweighs the chance to gain points

MariettaO [177]2 years ago
3 0

Answer with explanation:

→Total faces on the dice and marked numbers =6={1,2,3,4,5,6}

If you are rolling the dice and getting number 1 and 5 , you win 4 and 6 points respectively.

And, If you roll the dice and getting number 2,3,4 and 6 , you loose -5  points .

→→So,suppose the dice is rolled 6 times, and considering each outcome to be equally likely

         then you loose more points than gaining which is equal to

   = 4+(-5)+(-5)+(-5)+6+(-5)

   =10-20

   = -10

And, Average

=\frac{-10}{5}\\\\= -2

So, if number of outcomes are equally likely, you can't accumulate 20 points.

Option B:→ It is not a fair game because weighted average is Negative.

     

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Ivenika [448]
When x=0, y=-8, so that is the y-intercept.

When x increases by 1, y decreases by 2, so the slope is -2.

The equation of the function is
.. y = -2x -8
3 0
3 years ago
If
Tju [1.3M]

Answer:

Step-by-step explanation:

Rationalize the denominator of b. So, multiply the numerator and denominator by \sqrt{x}

b = \frac{(1-2\sqrt{x}) *\sqrt{x}}{\sqrt{x}*\sqrt{x}  }=\frac{1*\sqrt{x} -2\sqrt{x} *\sqrt{x} }{\sqrt{x} *\sqrt{x} }\\\\=\frac{\sqrt{x} -2x}{x}\\

Now, find a +b

a +b = \frac{2x+\sqrt{x} }{x}+\frac{\sqrt{x} -2x}{x}\\\\=\frac{2x+\sqrt{x} +\sqrt{x} -2x}{x}

Combine like terms

= \frac{2x-2x+\sqrt{x} +\sqrt{x} }{x}\\\\=\frac{2\sqrt{x} }{x}

Now find (a + b)²

(a +b)² = (\frac{2\sqrt{x} }{x})^{2}

          = \frac{2^{2}*(\sqrt{x} )^{2}}{x^{2}}\\\\= \frac{4* x}{x^{2}}\\\\= \frac{4}{x}

Hint: \sqrt{x} *\sqrt{x}  =\sqrt{x*x}=x

5 0
2 years ago
A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historic
Rina8888 [55]

Answer:

The probability that none of the LED light bulbs are​ defective is 0.7374.

Step-by-step explanation:

The complete question is:

What is the probability that none of the LED light bulbs are​ defective?

Solution:

Let the random variable <em>X</em> represent the number of defective LED light bulbs.

The probability of a LED light bulb being defective is, P (X) = <em>p</em> = 0.03.

A random sample of <em>n</em> = 10 LED light bulbs is selected.

The event of a specific LED light bulb being defective is independent of the other bulbs.

The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 10 and <em>p</em> = 0.03.

The probability mass function of <em>X</em> is:

P(X=x)={10\choose x}(0.03)^{x}(1-0.03)^{10-x};\ x=0,1,2,3...

Compute the probability that none of the LED light bulbs are​ defective as follows:

P(X=0)={10\choose 0}(0.03)^{0}(1-0.03)^{10-0}

                =1\times 1\times 0.737424\\=0.737424\\\approx 0.7374

Thus, the probability that none of the LED light bulbs are​ defective is 0.7374.

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olga_2 [115]
Cameron would have gotten 182 votes. Sydney would have gotten 547.
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Abby is 4 – feet tall. Craig is 4 feet tall.
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A. Craig is taller
B. The positive is higher than the negative value
6 0
2 years ago
Read 2 more answers
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