The student body at a high school was surveyed about sports participation. the table below shows the data by grade level.
the number of students who are in 9th grade and play two or more sports is a
'marginal frequency'
'joint frequency'
the number of students who are in the 11th grade is a
'marginal frequency'
'joint frequency'
the percentage of the student body that plays one sport is a
'marginal relative frequency'
'joint relative frequency'
the percentage of the student body that are in 12th grade and do not play a sport is a
'marginal relative frequency'
'joint relative frequency'

Marginal frequency:

The number in the end column or row - "Total".

Joint frequency:

The ratio of the frequency in a particular category and the total number of data values.

Basically, the values that aren't in the final row or column - "Total".

The number of students who are in 9th grade and play two or more sports is a  joint frequency.

This is only asking for the students in 9th grade who play two or more sports.

The number of students who are in the 11th grade is a  marginal frequency.

This is asking for the total number of students in 11th grade.

The percentage of the student body that plays one sport is a  marginal relative frequency'.

This is asking for the total number of students who play 1 sport.

The percentage of the student body that are in 12th grade and do not play a sport is a  joint relative frequency.

This is only asking for the students in 12th grade and don't play sports.

50 min

step-by-step explanation:

they will be decreasing both their times in half

60 / 2 = 30

40/2 = 20

30 + 20 = 50

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.