On a single roll of a pair of dice, what are the odds against rolling a sum of 12?

Step-by-step explanation:

On a single roll of a pair of dice. When a pair of dice are rolled the possible outcomes are as follows:

(1,1)         (1,2)          (1,3)  (1,4)  (1,5)  (1,6)

(2,1)  (2,2)  (2,3)  (2,4)  (2,5)  (2,6)

(3,1)  (3,2)  (3,3)  (3,4)  (3,5)  (3,6)

(4,1)  (4,2)  (4,3)  (4,4)  (4,5)  (4,6)

(5,1)  (5,2)  (5,3)  (5,4)  (5,5)  (5,6)

(6,1)  (6,2)  (6,3)  (6,4)  (6,5)  (6,6)

The number of outcomes that gives us 12 are (6,6). There is only one outcome that gives us sum 12.

Total outcomes = 36

Odd against favor =

Number of outcomes of getting sum 12 is 1

Number of outcomes of not getting sum 12 is 36-1= 35

odds against rolling a sum of 12=

10000000000000000000000000000000

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.