If the 4 states have to be in a specific order say ABCD,
then the total number of different possible routes is:
43P4 = 2,961,840
So the probability is:
1 / 2,961,840 = 3.38 x 10^-7
But if the 4 states can be in any order such as DBAC,
ACBD etc, then the total number of different possible routes is:
43C4 = 123,410
So the probability is:
1 / 123,410 = 8.1 x 10^-6
No I don’t think it is practical to list all the
different possible routes to select the one that is best. We can simply use
mathematical models to solve for that one.
<span> </span>
ABCD is a square . if the coordinates of three of its vertices are A(-1,2a), B(a,2a), C(a,0), find the coordinates of D. BC is obviously 2a (it's a vertical line) then AB must be 2a (notice AB is horizontal) so BC = AC because it is a square 2a = a+1 a=1
√8 = 2√2 = 2√2 = 2 x 1.414... =2.828....(irrational number between 2 and 3)
2.π = 6.283...........................................(irrational number between 6 and 7)
7/3 = 2.33333.. (mind you a repeating decimal is a rational number), then:
7/3 = 2.33333 ......................................(rational number between 2 and 3)
e/2 = 1.359...........................................(irrational number between 1 and 2)
3/2 = 1.5 ..............................................(rational number between 1 and 2)
3 in the first question 4 on the last question!!!
