You have $1000, and a certain commodity presently sells for $2 per ounce. suppose that after one week the commodity will sell for either $1 or $4 an ounce, with these two possibilities being equally likely. (a) if your objective is to maximize the expected amount of money that you possess at the end of the week, what strategy should you employ? (b) if your objective is to maximize the expected amount of the commodity that you possess at the end of the week, what strategy should you employ?

hello i am doing good for you that it's summer in your country or state ; )

btw in my country we just have sun and rain as (because i live in the caribbean lol)

also, thx a lot for the points : )

median is the value separating the higher half of a data sample, a population or probablity distribution from the lower half.

step-by-step explanation:

say you have 3,8,7,3,1,9,8 order them from lowest to highest

if orderd it would be 1,3,3,7,8,8,9 now you take away from the lowest and the highest until you end up in the middle.

so you would remove the 1 and 9 leaving you with 5 digits remaining

when you've reached the middle that would be your answer so in this case the answer is 7

if there were two numbers left say 7 and 5 you'd and them together so 12 then divide by 2 witch would equal 6 so the answer would be 6

good luck on that project hoped i you at all

113.42

i hope that is right.

step-by-step explanation:

asnwer is (g o h o f)(x) = 2x - 6/x + 7

step-by-step explanation:

given,

f(x) = x - 3/x

g(x) = x + 3

h(x) = 2x + 1

we need to find : (g o h o f)(x)

we can write (g o h o f)(x) = g[ h{ f(x) } ]

so first we find h{f(x)}

h{f(x)} = f(x) + 3 = (x - 3/x) + 3

h{f(x)} = x - 3/x + 3

now we find g[ h{ f(x) } ].

g[ h{ f(x) } ] = g(x - 3/x + 3)       {plug h{f(x)} = x - 3/x + 3 in g[h{f(x)}] }

g(x - 3/x + 3) = 2 (x - 3/x + 3) + 1

                    = 2x - 6/x + 6 + 1

                    = 2x - 6/x + 7

hence we get (g o h o f)(x) = 2x - 6/x + 7