Coach decaro records the cross country runners' times compared to their personal best times.

runner

gregor

hiroki

isabelle

juan

karen

3k times

time compared to

personal best

+5.6

-11.6

+3.2

0

-5.2

which list correctly orders the runners from most improved to least improved?

juan, hiroki, karen, isabelle, gregor

juan, isabelle, karen, gregor, hiroki

hiroki, karen, juan, isabelle, gregor

gregor, isabelle, juan, karen, hiroki

The 4th list: Gregor, Isabelle, Juan, Karen, Hiroki.

Step-by-step explanation:

Times compared to personal best when arranged from most imporoved to least improved are:

+5.6, +3.2, 0, -5.2, -11.6

Now +5.6 is Gregors, +3.2 is Isebelle's, 0 is Jaun's, -5.2 is Karen's, and -11.6 is Hiroki's. Therefore, the list that orders runners from most improved to least improved is

Gregor, Isabelle, Juan, Karen, Hiroki

the answer is 21%.

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.