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:
its no big deal
Step-by-step explanation:
no big deal.
9514 1404 393
Answer:
see below
Step-by-step explanation:
It is easiest to compare the equations when they are written in the same form.
The first set can be written in slope-intercept form.
y = 2x +7
y = 2x +7 . . . . add 2x
These equations are <em>identical</em>, so have infinitely many solutions.
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The second set can be written in standard form.
y +4x = -5
y +4x = -10
These equations <em>differ only in their constant</em>, so have no solutions.
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The third set is already written in slope-intercept form. The equations have <em>different slopes</em>, so have exactly one solution.
The answer is C. Hope this helps