Let x be the original position.
After the first play they gain 9 yards. The position can be represented by
x + 9
After the second play they lose 22 yards. The position can be represented by
x + 9 - 22 = x - 13
Therefore, in total they lost 13 yards.
Answer:
1/12
Step-by-step explanation:
first we have to find all the possibilities of getting a sum of 3 or less: 1+1 or 1+2 and we count the second combination 2 times because the numbers can be on either of the dices so we have a total of 3 possibilities. all the possible pairimg of dice are 6*6=36 because each dice has 6 sides and we can get either of them. so the probability would be the chance of getting a sum of 3 or less divided by all the diff combination which equals 3/36 or 1/12 which is roughly around 8.3%
The second one that you did
In the first 10 -> 10 - 9 = 1 (contain a 3).
In the first 100 -> 100 - 9 * 9 = 19 (contain a 3).
In the first 10^n ->
10^n - 9^n (contain a 3).
<u>The answer is 3.439 numbers contain a 3 in the first 10.000</u>