It is predicted that the population of a city will increase by 4% every year for the next many years. It is presently 60000 people.
(a) Complete the pattern below giving the predicted population as the years go by:
i) 1 year from now the population is predicted to be
4
60000 + 60000 x 60000 (1+ 60000 (1 +0.04) = 60000(1.04).
100
100
ii) 2 years from now the population is predicted to be
4
4
60000(1.04)+60000(1.04) = 60000(1.04) (1+ 60000(1.04) (1.04) = 60000(1.04).
100
4
-
(1+100)
iii) 3 years from now the population is predicted to be
60000(1.04)2 +
iv) I years from now the population is predicted to be
(b) How many years will it take the population to double? Round off your answer to the nearest
whole number. Write down the work leading to your answer.

the reporter's right

step-by-step explanation:

1/4 = .25 / 25%

c

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.

step-by-step explanation: