CPhill

OK......no prob......

Total possible committees   =  C(9,4)  = 126

Those possible with 3 technical people  =  C(5,3)*C(4,1)  = 40

Those possible with 4 technical people  =  C(5,4)  = 5

P[ at least three technical persons ]  =  [ 40 + 5] / 126 = 

45 / 126  =  15/42  = 5 / 14 ≈ .357  ≈  35.7%

The possible permuted sets of choosing 2 b***s from 7  = P(7,2)  = 42

But the only ones we are interested in are  {2,6} or {6,2}

So.....

2 / 42  =  1 / 21   ≈ 4.76%

To see Melody's answer....

The possible number of  sets consist of the possible permutes of choosing any 3 of the 5 books  =  P(5,3)  = 60

We want to choose the sets    {3, a, b}    where a,b  are the possible permutes  of choosing two of the other 4 books  = P(4,2)  = 12

So....12 / 60   = 1 / 5  .

.25555.....=

[25 - 2 ] / 90  =  23 / 90

51 / 300 = .17  =  17 / 100  =  17%

What are the coordinates of M ....????.....you only listed one......

Thanks, Alan......!!!!

y = -x + 6

y = -6x – 9     set both equations equal

 -x + 6  = -6x – 9      add 9, x  to both sides

15 = - 5x           divide both sides by - 5

-3  = x       sub into either equation to find y

y = - [ -3] + 6   = 9

So  {x , y}  =  {-3, 9 }

x^2 - 6x - 2016

Factor 2016  =

504 * 4

252 * 8

126 * 16

42 * 3 * 4 * 4

21 * 2 * 3 * 4 * 4

[7 * 3 * 2] * [3 * 4 * 4]

42  * 48

And we need

42 and - 48  = - 6

(x + 42) ( x - 48)