X=4 or X=-4 because 4(4)^2 does =64
2/5(x - 1) < 3/5(1 + x)
To find the solution, we can use the distributive property to simplify.
2/5x - 2/5 < 3/5 + 3/5x
Multiply all terms by 5.
2x - 2 < 3 + 3x
Subtract 2x from both sides.
-2 < 3 + x
Subtract 3 from both sides.
-5 < x
<h3><u>The value of x is greater than the value of -5.</u></h3>
Answer: -1
The negative value indicates a loss
============================================================
Explanation:
Define the three events
A = rolling a 7
B = rolling an 11
C = roll any other total (don't roll 7, don't roll 11)
There are 6 ways to roll a 7. They are
1+6 = 7
2+5 = 7
3+4 = 7
4+3 = 7
5+2 = 7
6+1 = 7
Use this to compute the probability of rolling a 7
P(A) = (number of ways to roll 7)/(number total rolls) = 6/36 = 1/6
Note: the 36 comes from 6*6 = 36 since there are 6 sides per die
There are only 2 ways to roll an 11. Those 2 ways are:
5+6 = 11
6+5 = 11
The probability for event B is P(B) = 2/36 = 1/18
Since there are 6 ways to roll a "7" and 2 ways to roll "11", there are 6+2 = 8 ways to roll either event.
This leaves 36-8 = 28 ways to roll anything else
P(C) = 28/36 = 7/9
-----------------------------
In summary so far,
P(A) = 1/6
P(B) = 1/18
P(C) = 7/9
The winnings for each event, let's call it W(X), represents the prize amounts.
Any losses are negative values
W(A) = amount of winnings if event A happens
W(B) = amount of winnings if event B happens
W(C) = amount of winnings if event C happens
W(A) = 18
W(B) = 54
W(C) = -9
Multiply the probability P(X) values with the corresponding W(X) values
P(A)*W(A) = (1/6)*(18) = 3
P(B)*W(B) = (1/18)*(54) = 3
P(C)*W(C) = (7/9)*(-9) = -7
Add up those results
3+3+(-7) = -1
The expected value for this game is -1.
The player is expected to lose on average 1 dollar per game played.
Note: because the expected value is not 0, this is not a fair game.
