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Liula [17]
3 years ago
6

Need help I’m getting stuck on this problem

Mathematics
2 answers:
Juliette [100K]3 years ago
6 0

Answer:

y is already isolated in the top equation

x = 4

y = 12

(4,12)

Step-by-step explanation:

y is already isolated in the top equation

plug y = 3x into bottom equation

5x + 2y = 44

5x +2(3x) = 44

5x + 6x = 44

11x = 44

x = 4

plug x = 4 into y = 3x

y = 3(4)

y = 12

mariarad [96]3 years ago
3 0
Plug in y in the bottom equation
5x + 6x = 44
11x = 44
X= 4
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<h3>Answer:  Choice D)  -$22</h3>

You'll lose on average $22 per roll.

====================================================

Explanation:

Normally there is a 1/6 chance to land on any given side of a standard die, but your friend has loaded the die in a way to make it have a 40% chance to land on "1" and an equal chance to land on anything else. Since there's a 40% chance to land on "1", this leaves 100% - 40% = 60% for everything else.  

Let's define two events

  • A = event of landing on "1".
  • B = event of landing on anything else (2 through 6).

So far we know that P(A) = 0.40 and P(B) = 0.60; I'm using the decimal form of each percentage.

The net value of event A, which I'll denote as V(A), is -100 since you pay $100 when event A occurs. So we'll write V(A) = -100. Also, we know that V(B) = 30 and this value is positive because you receive $30 if event B occurs.

To recap things so far, we have the following:

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  • P(B) = 0.60
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  • V(B) = 30

Multiply the corresponding probability and net value items together

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Then add up those products:

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This does not mean that any single die roll you would lose $22; instead it means that if you played the game say 1000 or 10,000 times, then averaging out the wins and losses will get you close to a loss of $22.

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