Mobon oil company owns a lease that entitles it to explore for oil on a parcel of offshore land in california. since the lease is about to expire mobon must now decide whether to drill for oil at the site or to sell the lease to excel oil company, which has just offered mobon $50,000. mobon estimates it would cost $100,000 to drill at the site. if the well were dry, all this cost would be lost. if the well were successful, its value to mobon would depend on the extent of the oil discovered. for simplicity, mobon assumes there are only two types of successful wells: minor and major success. mobon estimates that a minor success would result in revenues of $200,000 in excess of the drilling cost, whereas a major success would result in revenue of $500,000 in excess of the drilling cost. mobon has assessed the following probabilities: if the well is dry then the probability is 0.7, if it is a minor success the probability is 0.2, and if it is a major success the probability is 0.1. a) construct an appropriate payoff matrix, b) using the maximax criterion, identify the optimal decision c) using the maximin criterion, identify the optimal decision d) using the minimax criterion, identify the optimal decision e) using the emv criterion, identify the optimal decision

69.64%

step-by-step explanation:

t

step-by-step explanation:

ever thing time 1 is a multiply by 1 it get  

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.

x^3-4x^2+3x-2/x^2+1

step-by-step explanation:

factor the numerator and denominator and cancel the common factors