Rachel has just completed a study of precipitation in three different locations over the course of a year. The data she wishes to present in graph form is shown in the table below, with precipitation listed in inches. Boston Phoenix Minneapolis January 3.27 0.45 1.04 February 3.44 0.48 .76 March 3.10 0.71 1.12 April 3.89 0.31 2.10 May 3.75 0.17 3.15 June 3.12 0.09 5.11 July 2.98 0.11 5.78 August 3.07 0.33 4.65 September 3.15 0.52 3.61 October 3.55 0.40 3.22 November 3.42 0.42 2.34 December 3.44 0.29 1.84 Rachel wants her graph to emphasize the relation of each city’s precipitation to each other city’s at each point in time. What kind of graph would be appropriate? a. bar graph b. line graph c. pictograph d. histogram

6.5

step-by-step explanation:

you said it was in the question

~courtney

57.09 minus 7.0435 equals

step-by-step explanation:

just subtrach them both and u get the answer of 50.0465

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.