A class quiz has 5 True/False questions. The correct answers are T, T, E, F, and T, in that order. John answered the questions randomly and is now wondering if he will pass the quiz. He needs to have at least 3 correct answers. John
used a calculator to generate random numbers from 1 to 100. He let odd numbers stand for True and even numbers
stand for False. He recorded the results in the table. What is the experimental probability that John passed the quiz?
Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 Trial 6 Trial 7 Trial 8
Random 22, 33, 11, 3, 72, 28, 36, 43, 27, 56, 41, 12, 34, 20, 11, 7,50, 17, 44, 76,2, 93, 53,
Numbers 18, 45, 80 98,67
85, 71 90, 55 59, 31 28,82 39,20 38, 29
Result Fail Pass Fail Pass Fail Pass Fail Pass
Experimental nrnhahility

a reflection across the x-axis , followed by the by a reflection across the y-axis , and then a scale dilation of 0.5 .

answer: y=4000x+14000

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.

answer: 5

step-by-step explanation:

since, the range is the difference between the lowest and highest values of the given sample space.

and, here the sample space = { 3, 1, 0, 4, 1, 4, 2, 3, 0, 5, 0, 1, 1, 2, 0, 4, 1,4, 4 ,2 ,3 ,2 ,0 ,1 ,4 }

here, the highest value of the given sample = 5

and, the lowest value = 0

therefore, the range of the sample space = 5 - 0 = 5