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stiv31 [10]
3 years ago
7

There is a machine with 8 toys in it that each cost between 25 cents and 2 dollars, with each toy being 25 cents more expensive

than the next most expensive one. Each time Sam presses the big red button on the machine, the machine randomly selects one of the remaining toys and gives Sam the option to buy it. If Sam has enough money, he will buy the toy, the red button will light up again, and he can repeat the process. If Sam has 8 quarters and a ten dollar bill and the machine only accepts quarters, what is the probability that Sam has to get change for the 10 dollar bill before he can buy his favorite toy- the one that costs $1.75? Express your answer as a common fraction.
This problem is quite long and seems really tedious, but apparently there's a "simple" way to solve it using complementary probability.
Mathematics
1 answer:
aksik [14]3 years ago
8 0

It is actually quite simple: they are asking for the probability:

there are different types of items and each types there is always at least a 25 cent difference in between (if you go up the or down the scale there will be a 25 cent difference to each type of item. They say that Sam wants to get one of the 8 items. The price at this point doesn't matter. If he wants 1 of the 8 types all you have to do is convert it into a fraction: 1/8. That is the answer

Hope I helped...


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Answer:

  k = -1

Step-by-step explanation:

Put the given value of x in the equation, and solve the resulting equation for k.

  2(5 -3) +k(1 +2·5) = k - 5 - 1

  2(2) +k(11) = k -6 . . . . simplify a bit

  10k = -10 . . . . . . . . . . add -4-k to both sides

  k = -1 . . . . . . . . . . . . . divide by 10

The value of k is -1.

_____

<em>Check</em>

Use k = -1 in the original equation and solve for x.

  2(x -3) -(1 +2x) = -1 -x -1

  2x -6 -1 -2x = -x -2 . . . . eliminate parentheses

  x = 7 -2 = 5 . . . . . . add x+7; answer checks OK

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