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:
Step-by-step explanation:
we have
Find out the domain
we know that
The radicand must be greater than or equal to zero
so
The domain is the interval -----> [0,∞)
All real numbers greater than or equal to zero
The first and the 2nd one are correct because x is 21 and y is 49
3 and 4 at wrong
For this you count rise over run aka how many up divided by how many over starting from 0. here it’s 7 up and 3.5 side, 7 divided by 3.5 is 2 and the line is going down so it’s negative 2 (D)
So what you would do is divide 988 by 26 and divide 731 by 17 and those two numbers you would add which gives you the amount of people that attended the performance
