
Shaina H.
asked 03/02/15I think this is missing info?
Regina owns 150 shares of a company's stock. When the stock market opened, each share traded for $23.65. When the stock market closes, Regina will calculate how much money she has gained or lost from the stock that day. How much can Regina reasonably expect her stock to be worth?
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1 Expert Answer
Matthew H. answered 03/02/15
Tutor
New to Wyzant
Math - Algebra to Calculus: Middle and high school, some college.
I will take this problem as far as I can with what information we're given. It helps that you've added "They want us to make an equation and then put a domain of the least and most it could be" since that is all we can really do.
Let us denote the number of shares with N and the trading value at market opening of V.
So at the time of opening, her stock is worth N*V or more simply NV.
But over the course of the day, the trading value changes by d. This can be positive (she's earning money) or negative (she's losing money). So at the end of the day, the trading value of each share is:
V + d
And worth of her stock will be:
N(V + d)
Now to address "put a domain of the least and most it could be", we will use an inequality. Obviously the least value a share can have is 0 (unless there's some feature of stock markets that prevents stock from becoming completely worthless or somehow allows negative values). This gives us:
N(V + d) >= 0
We will use another inequality to express the most she could have. We'll use M to represent the arbitrary "most that Regina can reasonably expect".
N(V + d) < M
Combined we have the inequality:
0 < N(V + d) < M
We know of N and V, which are 150 and $23.65 respectively. The possible values of d are apparently where we must make some wild guesses. So let's go with:
-V < d < V
I use -V because this would result in a share worth $0. Using V for the upper limit assumes a share can do no more than double in value. We can now "build up" the earlier inequality, 0 < N(V + d) < M, around the above.
-V < d < V
V + -V < V + d < V + V
0 < V + d < 2V
N*0 < N(V + d) < N*2V
0 < N(V + d) < 2NV
Again, this assumes that the most a value of a share can do in a day is double. We can replace the 2 with a 1.5 if we instead want to assume a share cannot increase to more than 150% of its value at the start or with a 3 if we say it can triple. Unless special knowledge of stock markets provides missing information, this is all we can do.
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Jon P.
03/02/15