2 green spaces, 2 star spaces, and 2 question mark spaces.
Suppose you rolled a twelve-sided number cube instead of a standard six-sided one. How would the probability of these events change?

The theoretical probability of landing on a star space on your first roll .

The theoretical probability of not landing on a question mark on your first roll .

The theoretical probability of reaching the end space on your first roll .

a) justify

step-by-step explanation:

justifying is just part of everyday life. justify kind of means proof so i think the answer is a) justify.

throw the computer

step-by-step explanation:

let's cover over the formula of the vertex:

y = a(x – h)^2 + k

in this formula, (h,k) are the vertex, always remember that the positive or negative value of the vertex is based on the sign it has.

for example, if the vertex is -3,-6 then both the x and y value would have the positive sign attached to it and if it were a positive x and y value, then a negative sign would be attached to it.

with that being said, for our problem the x value is a positive one so we are looking for a number with a negative sign next to it, in this case the first and second choice.

however, for the first answer choice it isn't quite part of the formula so we know that the answer choice correctly displays the x value.

to confirm, we can see if the -1 fits correctly with the second answer choice, and as mentioned above a negative number will have a positive sign attached to it. so, for our problem we can see that -1 has been changed to +1, so we can confirm that the second answer choice is correct.

hence, our final answer is y=(x - 1)^2 + 1