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2 years ago

"There are 15 questions on an exam. In how many ways can the exam be answered with exactly 8 answers correct?"

Mathematics

1 answer:

erastovalidia [21]2 years ago

5 0

Answer:

The exam can be answered with exactly 8 answers correct in 6435 ways.

Step-by-step explanation:

The order is not important.

For example, answering correctly the questions 1,2,3,4,5,6,7,8 is the same outcome as answering 2,1,3,4,5,6,7,8. So we use the combinations formula to solve this problem.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

"There are 15 questions on an exam. In how many ways can the exam be answered with exactly 8 answers correct?"

Combinations of 8 questions from a set of 15. So

$C_{15,8} = \frac{15!}{8!(15-8)!} = 6435$

The exam can be answered with exactly 8 answers correct in 6435 ways.

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Answer:

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Step-by-step explanation:

First, distribute the 2 to the 3, -9, and 5. You will get 6 + 6 - 18 + 10 + (-9). Then, you can start from the beginning and work through combining the terms. The original equation is in bold after every step.

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3 years ago

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Wewaii [24]

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Y^2 - 40.

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Y^2 - 40 = 3Y

If we move all expressions to one side of the equation, what we would have is,

Y^2 - 3Y -40 = 0

(Remember that when a positive value crosses to the other side of an equation it becomes negative and vice versa)

We now have a quadratic equation

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By factorizing we now have

(Y -8) (Y + 5) = 0

Therefore Y - 8 = 0 or

Y + 5 = 0

Hence, Y = 8 or Y = -5

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