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kondor19780726 [428]
3 years ago
14

Need help on my math!!!!!!!

Mathematics
1 answer:
Natalka [10]3 years ago
3 0
(x^2+5x-36)/(x^2-16)
=(x^2+9x-4x-36)/(x^2-4^2)
=x(x+9)-4(x+9)/(x+4)(x-4)
=(x-4)(x+9)/(x+4)(x-4)
=x+9/x+4
Hope this helps.
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There is a machine with toys in it that each cost between cents and dollars, with each toy being cents more expensive than the n
Tju [1.3M]

The required probability that sam has to get change for the 10-dollar bill before he can buy his favorite toy is 6/7.

<h3>What is probability?</h3>

The fraction of favorable outcomes to the total number of outcomes of an event is said to be its probability of occurrence.

P(N) = n(N)/n(S)

N- an event; S- total outcomes;

<h3>Calculation:</h3>

It is given that,

n(s) = 8

So, the probability of buying the favorite toy first is 1/8

and the probability of buying the favorite toy a second is (1/8)(1/7) = 1/56

For these probabilities, Sam won't need to change the 10-dollar bill.

Then,

The probability that Sam has to get change for the 10-dollar bill before he can buy his favorite toy is

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= 1 - 1/7

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Learn more about probability here:

brainly.com/question/24756209

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Disclaimer: The given question on the portal is incomplete. Here is the complete question.

Question: There is a machine with 8 toys in it that each costs 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.

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3 years ago
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ioda

Answer:

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