Abox contains the following four slips of paper, each having exactly the same dimensions. win prize 1. win prize 2. win prize 3. win prizes 1, 2, and 3. one slip will be randomly selected. let a1 = {win prize 1}, a2 = {win prize 2}, and a3 = {win prize 3}. show that a1 and a2 are independent, that a1 and a3 are independent, and that a2 and a3 are also independent (this is pairwise independence). however, show that p(a1 ∩ a2 ∩ a3) ≠ p(a1) · p(a2) · p(a3), so the three events are not mutually independent. to prove pairwise independence, we must first calculate the following probabilities.

median is the value separating the higher half of a data sample, a population or probablity distribution from the lower half.

step-by-step explanation:

say you have 3,8,7,3,1,9,8 order them from lowest to highest

if orderd it would be 1,3,3,7,8,8,9 now you take away from the lowest and the highest until you end up in the middle.

so you would remove the 1 and 9 leaving you with 5 digits remaining

when you've reached the middle that would be your answer so in this case the answer is 7

if there were two numbers left say 7 and 5 you'd and them together so 12 then divide by 2 witch would equal 6 so the answer would be 6

good luck on that project hoped i you at all

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.

thus, the first answer choice, y = sec x, is the correct one