In a certain city, the average 20- to 29-year old man is 69.8 inches tall, with a standard deviation of 3.2 inches, while the average 20- to 29-year old woman is 64.3 inches tall, with a standard deviation of 3.9 inches. who is relatively taller, a 75-inch man or a 70-inch woman?

the man

Step-by-step explanation:

The wording "relatively taller" in this context is presumed to be a reference to the corresponding percentile or z-score. By hand, the z-score is easier to calculate.

z = (x - μ)/σ

For the man,

... z = (75 -69.8)/3.2 = 1.625

For the woman,

... z = (70 -64.3)/3.9 ≈ 1.462

The man's z-score is higher, so he is taller than a larger percentage of the male population than the woman is of the female population.

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If you're doing this by calculator, you can let it show you the percentile rank of each height. The man is in the 95th percentile; the woman the 93rd percentile. So, the man is relatively taller.

mx+89b(-98)nb

step-by-step explanation:

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.