Apair of fair dice is rolled once. suppose that you lose $ 8 if the dice sum to 3 and win $ 13 if the dice sum to 10 or 12. how much should you win or lose if any other number turns up in order for the game to be fair?

You should lose $1.20 for any other outcome to make it a fair game.

There are 2 ways to get a sum of 3 out of 36 outcomes; this probability is 2/36.
There are 4 ways to get a sum of 10 or 12 out of 36 outcomes; this probability is 4/36.
There are 36-(2+4) = 36-6 = 30 other outcomes out of 36; this probability is 30/36.

We lose $8 if we get a sum of 3; this expected value is -8(2/36) = -16/36.
We win $13 if we get a sum of 10 or 12; this expected value is 13(4/36) = 52/36.
Using x as the amount we win or lose for the other outcomes, the expected value is 30x/36.

In order to be a fair game, the amount we win and/or lose should come outto( $0 total.

Together we have the equation
-16/36+52/36+30x/36 = 0

Adding the fractions, we have
(-16+52+30x)/36 = 0
(30x+36)/36 = 0

Multiplying both sides by 36,
30x+36=0*36
30x+36=0

Subtracting 36 from both sides,
30x+36-36=0-36
30x=-36

Dividing both sides by 30,
30x/30 = -36/30
x=-36/30 = -1.2

Thus we should lose $1.20 for any other roll.

Abit confused about what you mean, but i think you might be referring to slope intercept form: y = mx + bin that case b, or the y intercept is already given.(0,-1)find the slope by using the formulay2 - y1/x2 - x1the final slope will be 4/2, equivalent to 2.now your equation should look like thisy = 4/2x -1sorry if this didn’t you out!