Y-5=3-9 (y+2)
Solve for y
Distribute the 9 to (y+2)
Y-5=3-9y-18
Y-5=-15-9y
+9y to both sides
10y-5=-15
+5 to both sides
10y=-10
÷10 both sides
Y= -1
2 (x-7)-10=12-4x
Solve for X
Distribute 2 to (x-7)
2x-14-10=12-4x
2x-24=12-4x
+4x to both sides
6x-24=12
+24 to both sides
6x=36
÷6 to both sides
X=6
When atleast one dice shows a 6 the possible outcomes will be:
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
(1,6), (2,6), (3,6), (4,6), (5,6)
Thus there are 11 total possible outcomes.
Among these outcomes, the sum of numbers greater than or equal to 9 can be obtained from:
(6,3), (6,4), (6,5), (6,6), (3,6), (4,6), (5,6)
This means there are 7 outcomes with sum greater than or equal to 9.
Thus, Probability of rolling a number greater than or equal to 9 with atleast one dice showing a 6 = 9/11
So, option A gives the correct answer
Answer:
No
Step-by-step explanation:
Flipping two coins is independent because the outcome of the first coin doesn't affect the outcome of the second coin. There is still a 1/2 chance of of getting heads or tails.
If this answer is correct, please make me Brainliest!
Answer:
I think it would be D.
I apologize if that is wrong