Suppose wesley is a marine biologist who is interested in the relationship between the age and the size of male dungeness crabs. wesley collects data on 1,000 crabs and uses the data to develop the following least-squares regression line where x x is the age of the crab in months and ˆ y y^ is the predicted value of y y , the size of the male crab in cm. ˆ y = 9.1411 + 0.4775 x y^=9.1411+0.4775x what is the value of ˆ y y^ when a male crab is 23.2417 months old? provide your answer with precision to two decimal places

20.24 cm

Step-by-step explanation:

Here, since the age of the crab is given in months and the regression line was computed using the age in months, we just have to replace the value of the age in the formula to estimate the size.

The formula obtained for the estimated size Y was

Y = 9.1411 + 0.4775X  

where X is the age in months

Replacing X with 23.2417, we get

Y = 9.1411 + 0.4775*23.2417 = 20.24 cm

mathpapa

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.