The two dot plots below show the heights of some sixth graders and some seventh graders: Two dot plots are shown one below the other. The title for the top dot plot is Sixth Graders and the title for the bottom plot is Seventh Graders. Below the line for each dot plot is written Height followed by inches in parentheses. There are markings from 52 to 57 on the top line and the bottom line at intervals of one. For the top line there are 2 dots above the first mark, 1 dot above the second mark, 1 dot above the third mark and 2 dots above the fourth mark. For the bottom line, there is 1 dot for the first mark, there is 1 dot above the second mark, there is 1 dot above the third mark, there is 1 dot above the fourth mark and there are two dots above the sixth mark.

The mean absolute deviation (MAD) for the first set of data is 1.2 and the MAD for the second set of data is 1.7. Approximately how many times the variability in the heights of the sixth graders is the variability in the heights of the seventh graders? (Round all values to the tenths place.)

1.2
1.4
2.4
2.8

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.

139 divided by a number

step-by-step explanation:

hi,

the area of abcd is 245.

3*24=230

4*94=1039

2*12=24

add all of them up, you get 2450