Scores on the GRE. A college senior who took the Graduate Record Examination exam scored 610 on the Verbal Reasoning section and 660 on the Quantitative Reasoning section. The mean score for Verbal Reasoning section was 459 with a standard deviation of 113, and the mean score for the Quantitative Reasoning was 447 with a standard deviation of 155. Suppose that both distributions are nearly normal. A) Write down the short-hand for these two normal distributions.
B) What is her Z score on the Verbal Reasoning section?
C) What do these Z scores tell you?
D) Relative to others, which section did she do better on?
E) Find her percentile scores for the two exams.
F) What percent of the test takers did better than her on the Verbal Reasoning section? On the Quantitative Reasoning section?
G) Explain why simply comparing her raw scores from the two sections would lead to the incorrect conclusion that she did better on the Quantitative Reasoning section.
H) If the distributions of the scores on these exams are not nearly normal, would your answers to parts (b) - (f) change? Explain.

7.25

step-by-step explanation:

we know that 7 equals to 7.0. we also know that 1/4=0.25. so we can add the two together and we get 7.25

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.

well

44-3-3-3=35

35+4+4+4+4=51

so the ansewr would be 51 degreesat 7: 00

the answer is 64 cm

me l8t3r