Answer:
sorry cant help
Step-by-step explanation:
It would be O 945 that is the answer
The answer to your question is:
+4
-4, -3, -2, -1, 0, 1, 2, 3, 4
A single die is rolled twice. the set of 36 equally likely outcomes is {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1),
Maksim231197 [3]
<span>We need to find the rolls whose sum is greater than 10. By looking at the outcomes, we see that (5,6), (6,5), and (6,6) all have a sum greater than 10. Therefore, there are 3 chances to get a sum greater than 10. Since there are 36 chances overall, the probability of rolling greater than 10 are 3/36 = 1/12.</span>
Answer:
n=7/3
Step-by-step explanation:
1 (n – 4) – 3 = 3 - (2n+3)
1. distribute the 1 on the left side and distribute the -1 on the right side
1n-4-3=3-2n-3
2. add like terms
1n-7=-2n
3. move n to one side and by itself
-7=-3n
4. divide -3 to get n alone. Both negatives cancel out each other
7/3=n
