Francis surveyed a random sample of 707070 students at franklin high school about their favorite season. of the students surveyed, 181818 chose fall as their favorite season. there are 181618161816 students at franklin high school. based on the data, what is the most reasonable estimate for the number of students at franklin high school whose favorite season is fall?

the number of students at Franklin High School whose favorite season is fall is 467

Approx 467 students is the answer.

Step-by-step explanation:

Francis surveyed a random sample of 70 students at Franklin High School about their favorite season.

Of the students surveyed, 18 chose fall as their favorite season.

There are 1816 students at Franklin High School.

Now, students chose fall as their favorite season.

This comes to 0.257 percent.

We can say that 0.257 % of 1816 students can like fall season.

= 466.712 ≈ 467 students. (approx)

Therefore, the most reasonable estimate for the number of students at Franklin High School whose favorite season is fall is 467.

2

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.