Here are the 30 best lifetime baseball batting averages of all time, arranged in order from lowest to highest: 0.330, 0.331, 0.331, 0.333, 0.333, 0.333, 0.334, 0.334, 0.334, 0.336, 0.337, 0.338, 0.338, 0.338, 0.340, 0.340, 0.341, 0.341, 0.342, 0.342, 0.342, 0.344, 0.344, 0.345, 0.346, 0.349, 0.356, 0.358, 0.366, 0.371

If you were to build a histogram of these data, which of the following would be the best choice of classes?
A.
0.320 - 0.330, 0.330 - 0.340, 0.340 - 0.350, 0.350 - 0.360, 0.360 - 0.370
B.
0.300 - 0.349, 0.350 - 0.399
C.
0.320 - 0.329, 0.330 - 0.339, 0.340 - 0.349, 0.350 - 0.359, 0.360 - 0.369, 0.370 - 0.379
D.
0.330 - 0.333, 0.334 - 0.337, 0.338 - 0.341, ..., 0.364 - 0.367
E.
0.329 - 0.330, 0.331 - 0.332, 0.333 - 0.334, ..., 0.365 - 0.366​

3x3 = 9 and 2x2 = 4

step-by-step explanation:

a: inscribed angles

step-by-step explanation:

i reviewed this question with a peer and teacher, seems to be correct, me later

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.