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:
93
Step-by-step explanation:
by remainder theorem p(3) is the remainder
and therefore it is 93
Answer:
true
Step-by-step explanation:
because CD divided AB into equal halves, so they are of course perpendicular.
Linear positive because you can see, it’s moving up not down.
None of the offered choices is correct.
The correct equation is
.. a = (2s -2ut)/t^2 . . . . . . . . parentheses are required