HEL 1. Shown are the final scores for Norm Duke in the past eight bowling tournaments.
299, 281, 285, 269, 280, 269, 286, 287

Part A: What is the mean absolute deviation of Norm Duke’s scores?

A.
7.50

B.
7.00

C.
7.25

D.
6.75

Part B: Based on the mean absolute deviation from Part A, who has been the more consistent bowler if Ryan Shafer has a mean absolute deviation of 7?

A.
Norm Duke has been more consistent because his mean absolute deviation is greater than Ryan Shafer’s.

B.
Ryan Shafer has been more consistent because his mean absolute deviation is less than Norm Duke’s.

C.
Ryan Shafer has been more consistent because his mean absolute deviation is greater than Norm Duke’s.

D.
Norm Duke has been more consistent because his mean absolute deviation is less than Ryan Shafer’s.

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.

i’m here for points.

answer: 0.26

i used the calculator on google

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.