A number is randomly selected from the set {1, 2, 3, 4, 5, 6, 7, 8}. If the number chosen is greater than 3, what is the probability it is odd?

Hello!

The probability that the number chosen is greater then 3 and odd is 2/8

There are 5 numbers greater then 3, and of those 5, five and seven are the only odd numbers. So that is 2/8

Step-by-step explanation:

There are 5 numbers greater then 3, and of those 5, five and seven are the only odd numbers. So that is 2/8

If you are only talking about the numbers that are actually greater then 3 as your population, it is 2/5 :)

(I would appreciate Brainliest, please and thanks)

step-by-step explanation:

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(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0