Osteoporosis is a condition where bones become weak. it affects more than 200 million people worldwide. exercise is one way to produce strong bones and to prevent osteoporosis. because we use our dominant arm (the right arm for most people) more than our nondominant arm, we expect the bone in our dominant arm to be stronger than the bone in our nondominant arm. by comparing the strengths, we can get an idea of the effect that exercise can have on bone strength. here are some data on the strength of bones, measured in newton meters divided by 1000 (nm/1000), for the arms of 15 young men.

id nondominant dominant
1 15.6 16.3
2 25.4 26.8
3 17.8 18.5
4 19.1 22.1
5 12 14.8
6 20.1 19.8
7 12.2 13.1
8 14.2 17.5
9 15.9 20.2
10 13.5 18.7
11 17.6 18.7
12 15.6 15.1
13 14.5 16.2
14 14.2 14.9
15 12.4 13

a young male who is not a baseball player has a bone strength of geteq. ashx? eqtext=16.0%20%5cc%5cm%40sup% in his nondominant arm. predict the bone strength in the dominant arm for this person.

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step-by-step explanation:

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.

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