The ratio of boys to girls in Ella’s grade is 3 to 4. The ratio of boys to girls in Minh’s grade is 4 to 5.
Each grade has a total of 20 girls.

Choose True or False for each statement below.
Ella’s grade has more boys than Minh’s grade.
Choose...

Minh’s grade has more boys than Ella’s grade.
Choose...

Ella’s grade has 15 boys.
Choose...

Minh’s grade has 8 boys.
Choose...

answer: -8.2562234e+13

answer: (-6,3)

step-by-step explanation:

6. 104

7. 52/37

8. 2704

9. 140608

10. -52

11. |52|=52

12. 1/52

13. 156

14. 26

15. 40

16. 37

17. -364

18. 53 1/3

19. 50.96

20. 43 2/3

21. 78

22. 50.75

23. 47.63

24. 163.28

25. 10400

26. 338

27. 208

28. 2704 square inches

29. 312 square meters

30. don't know what its asking, tell me so i can tell you

31. 8490.56

32. 52 degrees would be an acute angle

33. 128 degrees

34. 20.8

35. 39

36. 208

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.