Two 6-sided dice are rolled. What is the probability that the sum of the two numbers on the dice will be greater than 9?

The probability that the sum of the two numbers on the dice will be greater than 9 is 16.7%.

Step-by-step explanation:

To find the probability of the two fair six-sided dice getting a sum of 9 when thrown, we have to first find all the possibilities and those that are greater than 9.

The total number of possibilities for the outcome of the two dice would be (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1) and so on. Due to every possibility from the first dice branching out into six possibilities for the second dice, we can find the total number of possibilities to be 6 * 6 = 36 possibilities.

Now to find the total number of possibilities with a sum greater than 9. Due to the question asking for a sum of over 9, the possibilities will not include those with a sum of 9 itself. Hence, the only possibilities are when the first dice rolls 4 or more, since 3 + 6 (the maximum that the second dice can roll) is exactly 9.

Listing out all the possibilities, there are: (4, 6), (5, 5), (5, 6), (6, 4), (6, 5) and (6, 6). A total of 6 possibilities will get you a sum of 9 or above.

Finally, to find the percentage, we divide the number of possibilities with their sum over 9 by the total number of possibilities:

Hence, the probability that the sum of the two numbers on the dice will be greater than 9 is 16.7%.

Hope this helped!

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752