The Honda Accord was named the best midsized car for resale value for by the Kelley Blue Book (Kelley Blue Book website). The file AutoResale contains mileage, age, and selling price for a sample of Honda Accords. a. Estimate the selling price of a four-year-old Honda Accord with mileage of miles.
b. Develop a confidence interval for the selling price of a car with the data in part (a).
c. Develop a prediction interval for the selling price of a car with the data in part (a).

Price Mileage Age
16998 35000 3
17599 16000 3
15998 42000 3
15998 61000 4
11599 102000 7
13998 109000 4
14998 60000 5
11599 88000 9
12599 58000 8
14998 27000 6
15998 32000 4
11599 95000 8
10499 123000 8
14599 54000 5
10599 68000 11
16998 40000 4
20998 7000 1
11599 81000 9
12599 70000 6
13599 86000 5
11599 92000 7
9599 120000 10
18998 20000 2
12599 63000 7
12599 70000 6
10998 88000 10
11998 86000 6
12998 60000 6
17599 45000 2
9998 96000 11
9998 60000 11
16998 36000 2
17599 45000 2

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18

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.

1k

step-by-step explanation:

-6k+7k

(-6+7)k

1k