George gathered samples for the average monthly precipitation, in inches, for two cities and recorded the results in the box-and-whisker plots shown. What conclusion can you make from the box-and-whisker plots?

The figure shows two box-and-whisker plot on a number line for City Upper A and City Upper B and is labeled as Average Monthly Precipitation. The numbers on the number line ranges from 0 to 5 in increments of 1. For City Upper A, the lower extreme is labeled at 0 decimal point 5, the upper extreme is labeled at 4 decimal point 5, the lower quartile is labeled between 1 and 2 but near 2, the upper quartile is labeled between 3 and 4 but near 4, and the median is labeled between 2 and 3 but near 3. For City Upper B, the lower extreme is labeled at 0, the upper extreme is labeled between 4 and 5 but near 4, the lower quartile is labeled at 4, the upper quartile is labeled at same value as it is labeled for City Upper A, and the median is also labeled at the same value as it is labeled for City Upper A.

A.
The median amount of precipitation for City A is greather than the median for City B.

B.
The median amount of precipitation for City A is less than the median for City B.

C.
The amount of precipitation in City B is more spread out than in City A.

D.
The amount of precipitation in City A is more spread out than in City B.

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.