Yu and Elena each bought between 7 and 9 yards of ribbon. Yu bought 3 pieces of ribbon. Elena bought 6 pieces of ribbon. The lengths of ribbons are given below. Drag ribbons to the box below each persons name to show what they callus have bought.

Given: Yu, Nailah, and Elena each bought between 7 and 9 yards of ribbon  Yu bought 3 pieces of ribbon. Nailah bought 5 pieces of ribbon. Elena bought 6 pieces of ribbon.

To find: Who can buy which ribbon

Solution:

Ribbon Sizes

1 2/3 = 5/3  yard

4/5  yard

3 1/2  = 7/2 Yard

Yu, Nailah, and Elena each bought between 7 and 9 yards of ribbon

Yu bought 3 pieces of ribbon

=> 3 * 5/3  = 5

  3 * 4/5  = 2.4

  3 * 7/2   = 10.5

Nailah  bought 5 pieces of ribbon

=> 5 * 5/3  = 8.33

   5 * 4/5  =  4

   5 * 7 /2   = 17.5

Elena bought 6 pieces of ribbon

=> 6 * 5/3  = 10

  6 * 4/5  = 4.8

  6 * 7 /2   = 21

Only value between 7 & 9 is 5 * 5/3  = 8.33

hence Nailah only can buy 1 2/3 = 5/3  yard ribbon

or there is some mistake in the data

7

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.