A sales manager collected the following data on annual sales for new customer accounts and the number of years of experience for a sample of 10 salespersons. Salesperson Years of Experience Annual Sales ($1000s) 1 1 80 2 3 97 3 4 92 4 4 102 5 6 103 6 8 111 7 10 119 8 10 123 9 11 117 10 13 136(a) Write an alternative hypothesis.(b). Develop an estimated regression equation that can be used to predict annual sales given the years of experience.(c) Use the estimated regression equation to predict annual sales tor a salesperson with 9 years of experience.

A. See diagram

B.y=80+4x

C.110000

Refer below.

Step-by-step explanation:

Refer to the pictures for complete illustration.

there are 25 people coming to the party

step-by-step explanation:

the text says that there will be 22 children at this party, 1 teacher, and 2 parents. so if you add them all together (22+1+2) it will equal 25. (=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.