Answer:
a) 66
b) 45
Step-by-step explanation:
We have 12 horses and after the first race 2 of them have to be eliminated
that condition is equal to consider that 10 horses will be available for the second race.
The number of possiblities we have for the first race is the combination of 12 horses in group of 10
C¹²₁₀ = 12!/ (2!)* (12-2)! = 12*11/2 = 66
For the second race
C¹⁰₈ = 10! / 8! * (10-8)! = 10*9/ 2 = 45
So we are given two points, say P1(4,7), P2(x,19).
Slope is given by
m=3=(y2-y1)/(x2-x1)=(19-7)/(x-4)
solve for x
3=(19-7)/(x-4)
cross multiply
3(x-4)=12
3x-12=12
3x=12+12=24
x=8
