Based on the data in the two-way table, the probability of being older than 25 years and having a hemoglobin level above 11 is a: 0.29 b: 0.34 c: 0.54 the probability of having a hemoglobin level above 11 is . a: 0.25 b: 0.3 c: 0.36 being older than 25 years and having a hemoglobin level above 11 a. are b. are not dependent on each other. 20 points : 3

A) 0.29; C) 0.36; A) are

Step-by-step explanation:

There are a total of 128+162 = 290 people that are over 25 years old.  Out of these, there are 84 people with a hemoglobin level above 11; this is a probability of 84/290 = 0.29.

There are 154 people with a hemoglobin level over 11 out of 429 people total; 154/429 = 0.36.

In order for two events to be independent,

P(A∩B) = P(A)·P(B)

The probability that someone is over 25 and has a hemoglobin level above 11 was 0.29.

The probability that someone is over 25 is 290/429; the probability that someone has a hemoglobin level above 11 is 154/429.  This makes

P(A)·P(B) = (290/429)·(154/429) = 0.24

The probabilities are not equal, so the events are not independent; this means they are dependent.

describing distributions is one of the key skills you’ll need to earn a high score on the ap statistics exam. if you need proof of this, just flip through some past exam questions, which can be found at the collegeboard website. you’ll notice that the first free response question is almost always a question requiring you to look at a graph and describe it.

step-by-step explanation: just trying to get points sorry : (

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.