Use un diagrama de Venn para responder la pregunta. Al revisar las carteras de 240 de sus clientes, una compañía de fondos mutuos categorizó si los clientes estaban invertidos en fondos de acciones internacionales, fondos de acciones nacionales o fondos de bonos. Encontró que
16 se invirtieron en acciones nacionales, acciones internacionales y fondos de bonos;
38 fueron invertidos en acciones nacionales y fondos de bonos;
103 se invirtieron en acciones nacionales;
42 fueron invertidos en acciones internacionales y nacionales;
34 se invirtieron en acciones internacionales y fondos de bonos, pero no en acciones nacionales;
56 se invirtieron en bonos, pero no en acciones internacionales o nacionales;
73 fueron invertidos en acciones internacionales, pero no en fondos de bonos
¿Cuántos se invirtieron en fondos de bonos?

idk bring the function back   im not smart

step-by-step explanation:

step-by-step explanation:

what "following statement"?

step-by-step explanation:

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.