Bob order 151/2 pound of cherrys fron the farm, this year they want to order 1 1/4 more pound but then 2/5 of the order they want blue berrys what is the percent

Percentage of cherries    = 60 %

Percentage of berries  = 40 %

Step-by-step explanation:

He orders  15 1/2 pounds every year.

He wants more=  1 1/4 pounds

So the total pounds = 15 1/2 + 1 1/4

                                 = 15 + 1 + 1/2 + 1/4

                                 = 16 + (2+1)/4

                                  =   16 + 3/4

                                  = 16 3/4 pounds

Out of 16  3/4 pounds they want 2/5 part to be of blue berries.

So blue berries = 16 3/4 * ( 2/5)

                          = 67/4 * ( 2/5)

                            = 67/10 = 6 7/10 pounds

Amount of cherries= 16  3/4- 6 7/10

                                 = 10 3/4- 7/10

                                  = 10 (15- 14)/20

                                   = 10 1/20 pounds

Percentage of cherries = Amount of cherries/ Total amount * 100

                                      = 201/20 / 67/4 *100

                                         = 201/20 * 4/67 *100

                                         = 201/335 *100

                                          = 60 %

Percentage of berries = 100- 60 = 40 %

answer: the is put the 450

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.