There are the combinations that result in a total less than 7 and at least one die showing a 3:
[3, 3] [3,2] [2,1] [1,3] [2,3]
The probability of each of these is 1/6 * 1/6 = 1/36
There is a little ambiguity here about whether or not we should count [3,3] as the problem says "and one die shows a 3." Does this mean that only one die shows a 3 or at least one die shows a 3? Assuming the latter, the total probability is the sum of the individual probabilities:
1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 5/36
Therefore, the required probability is: 5/36
Answer:
Option H
Step-by-step explanation:
Perimeter of rectangle= 2(length) +2(width)
100= 2(5a -22) +2[½(a +1)]
Expand:
100= 2(5a) +2(-22) +a +1
100= 10a -44 +a +1
Simplify:
100= 11a -43
+43 on both sides:
11a= 100 +43
11a= 143
Divide both sides by 11:
a= 143 ÷11
a= 13
Thus, option H is correct.
Answer:
Step-by-step explanation:
Y=18 and x=9 redical 3
