Social Security numbers are of the form XYZ - MN - ABCD The three digts XYZ are referred to as the area number. The number associated with where you registered for your social security card. X and be any number the area code 666 has never been usedThe middle two referred to as the group number. The groups be number from 01 to 99 (cannot be The last 4 digits are the serial number. The serial number can be any to 9999 be )the security number 123-45-6789 used in advertisements. How many social security numbers are possible?

step-by-step explanation:

goog be able to you with that maybe

step-by-step explanation:

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0.0076, if you meant 4.125 divided by 0.03125 it would be 132

step-by-step explanation:

first you have to move the decimal in 0.03125 over until its behind the 5 which is 5 places over. now the number is 3125

then you have to move the decimal in 4.125, 5 places over. now the number is 4,112,500.

now you divide like you would normally would and get 0.0076

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.