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RSB [31]
3 years ago
6

A bag of marbles contains 1 blue, 5 green, and 4 red marbles. If you take out 2 marbles, what is the probability that they are b

oth green? (Enter your answer as a fraction.)
Mathematics
1 answer:
ehidna [41]3 years ago
5 0
If you take out two marbles, and they both end up green, then it would be five marbles out of 10 because there are 5 green marbles but . Then it would be a 5/10 chance of getting two green marbles.






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Divide:(10x²-3x+4)÷(2x-5)
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Answer:

\frac{10x^2-3x+4}{2x-5}=5x+11+\frac{59}{2x-5}

Quotient: 5x+11

Remainder: 59

Step-by-step explanation:

I'm going to do long division.

The bottom goes on the outside and the top goes in the inside.

  Setup:

        ---------------------------------

2x-5 |  10x^2    -3x      +4

  Starting the problem from the setup:

            5x     +11                        (I put 5x on top because 5x(2x)=10x^2)

        ---------------------------------   (We are going to distribute 5x to the divisor)

2x-5 |  10x^2    -3x      +4

        -(10x^2  -25x)                  (We are now going to subtract to see what's left.)

      -----------------------------------

                      22x      +4          (I know 2x goes into 22x, 11 times.)

                                                ( I have put +11 on top as a result.)

                    -(22x     -55)        (I distribute 11 to the divisor.)

                 -----------------------

                                  59          (We are done since the divisor is higher degree.)

The quotient is 5x+11.

The remainder is 59.

The result of the division is equal to:

5x+11+\frac{59}{2x-5}.

We can actually use synthetic division as well since the denominator is linear.

Let's solve 2x-5=0 to find what to put on the outside of the synthetic division setup:

2x-5=0

Add 5 on both:

2x=5

Divide both sides by 2:

x=5/2

Or realize that 2x-5 is the same as 2(x-(5/2)) which you will have to do anyways if you choose this route:

So 5/2 will go on the outside:

5/2  |    10          -3            4

      |                  25         55

         ------------------------------

           10          22        59

So we have:

\frac{10x^2-3x+4}{2x-5}

=\frac{10x^2-3x+4}{2(x-\frac{5}{2})}=\frac{1}{2} \cdot \frac{10x^2-3x+4}{x-\frac{5}{2}}=\frac{1}{2}(10x+22+\frac{59}{x-\frac{5}{2}})

Distribute the 1/2 back:

\frac{10x^2-3x+4}{2x-5}=\frac{10x+22}{2}+\frac{59}{2(x-\frac{5}{2})}

\frac{10x^2-3x+4}{2x-5}=5x+11+\frac{59}{2x-5}

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Answer:

Answer: binomial

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1/2x is one of the terms.

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the 7 is the second term.

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Answer:

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