Isle Royale, an island in Lake Superior, has provided an important study site of wolves and their prey. Of special interest is the study of the number of moose killed by wolves. In the period from 1958 to 1974, there were 296 moose deaths identified as wolf kills. The age distribution of the kills is as follows.
Age of Moose in Years Number Killed by Wolves
Calf (0.5 yr)
1-5
6-10
11-15
16-20 113
51
71
56
5
(a) For each age group, compute the probability that a moose in that age group is killed by a wolf. (Round your answers to three decimal places.)
0.5
.382

Correct: Your answer is correct.
1-5
.172

Correct: Your answer is correct.
6-10
.240

Correct: Your answer is correct.
11-15
.189

Correct: Your answer is correct.
16-20
.017

Correct: Your answer is correct.

(b) Consider all ages in a class equal to the class midpoint. Find the expected age of a moose killed by a wolf and the standard deviation of the ages. (Round your answers to two decimal places.)
μ =
5
σ =
Ask for details Edit by Pplinke 3 minutes ago
i new how to cal;cu;ate the probabilities i need the expected age and standard deviation

18a

step-by-step explanation:

simply because it is. it's the highest common factor

answer: 230 miles ( vote ! )

step-by-step explanation:

180.39 minus 17.99 equals 162.4

then divide 162.4 by .80 to get 230

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.