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svetoff [14.1K]
3 years ago
12

Hellppppppppppppppppppppppppppppppppppppppppppppppppppppppppp

Mathematics
2 answers:
vodka [1.7K]3 years ago
7 0
The answer should be 4/1
const2013 [10]3 years ago
5 0

Answer: y = 4 and x = 1

Step-by-step explanation:

when y is 2, x is 0.5

when y is 4, x is 1

                .(+ 2)   .(+ 0.5)

                .(+ 2)   .(+ 0.5)

                .(+ 2)   .(+ 0.5)

                .(+ 2)   .(+ 0.5)

the rule is that y is 4 times larger than x

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A rat is trapped in a maze. Initially he has to choose one of two directions. If he goes to the right, then he will wander aroun
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Answer:

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Step-by-step explanation:

The rat has two directions to leave the maze.

The probability of selecting any of the two directions is, \frac{1}{2}.

If the rat selects the right direction, the rat will return to the starting point after 3 minutes.

If the rat selects the left direction then the rat will leave the maze with probability \frac{1}{3} after 2 minutes. And with probability \frac{2}{3} the rat will return to the starting point after 5 minutes of wandering.

Let <em>X</em> = number of minutes the rat will be trapped in the maze.

Compute the expected value of <em>X</em> as follows:

E(X)=[(3+E(X)\times\frac{1}{2} ]+[2\times\frac{1}{6} ]+[(5+E(X)\times\frac{2}{6} ]\\E(X)=\frac{3}{2} +\frac{E(X)}{2}+\frac{1}{3}+\frac{5}{3} +\frac{E(X)}{3} \\E(X)-\frac{E(X)}{2}-\frac{E(X)}{3}=\frac{3}{2} +\frac{1}{3}+\frac{5}{3} \\\frac{6E(X)-3E(X)-2E(X)}{6}=\frac{9+2+10}{6}\\\frac{E(X)}{6}=\frac{21}{6}\\E(X)=21

Thus, the expected number of minutes the rat will be trapped in the maze is 21 minutes.

3 0
4 years ago
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A

Step-by-step explanation:

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