The following are the grade point averages of 30 students admitted to uc berkeley's industrial engineering and operations research graduate program during a year in the 1990s. 3.46 3.72 3.95 3.55 3.62 3.80 3.86 3.71 3.56 3.49 3.96 3.90 3.70 3.61 3.72 3.65 3.48 3.87 3.82 3.91 3.67 3.72 3.66 3.79 3.75 3.93 3.74 3.50 3.83 3.69for this question, you may use a computer program but state what software and what commands you use. group the data into class intervals of width 0.05 grade points starting with the value 3.45 and draw the resulting histogram. calculate the sample mean x ¯ and sample standard deviation s. determine the proportion of the data values that lie within x ¯ ± 1.5 s. compare this proportion with the lower bound given by chebyshev's inequality. compare the proportion found in part 3 with the approximate proportion given by the empirical rule, which for 1.5 s is approximately 87%. do the data appear to be approximately normal?

(- 1, 0 )

step-by-step explanation:

an x- intercept is where the function crosses the x- axis.

to find the x- intercept/s look for the value of x when f(x) = 0

from the given table there are 3 values of x where f(x) = 0

when x = - 1, x = 2, x = 3

the one that is listed is (- 1, 0 )

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.

d. you cannot tell because there are no values shown.

it has to have an angle of 90 degrees