Change 0.61 into a fraction
2 answers:
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You can think of this experiment as follows: you have to select a random number between 1 and 15, where each number is equally likely to appear.
Player A wins if 3,4,5 or 6 are picked.
Player B wind is 10,11 or 12 are picked.
Two events are said to be mutually exclusive if they cannot happen at the same time. So, in our metaphor, the question is: can we pick a particular number, so that both players win simultaneously?
The question is no, because if player A wins, we have picked 3,4,5 or 6, and player B wins with none of these numbers.
Similarly, if player B wins, we have picked 10, 11 or 12, and player A wins with none of these numbers.
So, the two events cannot occur simultaneously.
If you want to be more formal, since the events are subset of the sample space, two events are mutually exclusive if their intersection is the empty set, and this is the case:
First you have to find the intercepts!
To find the y-intercept (I usually do that first), pretend x=0.
Then it becomes:
4(0)-3y=12
Which then becomes:
-3y=12.
Divide both sides, and y = -4.
Then the (x,y) coordinate would become (0, -4).
Next you would find the x-intercept, and pretend y = 0.
Which then becomes:
4x-3(0)=12
3x0 cancels out to become:
4x = 12
And x = 3 because division stuffs.
Then THIS point would then be (3,0).
So then, you would plot (0,-4) and (3,0) on a graph and connect the points! I attached a picture for reference.
hope that helped~
To find the y-intercept (I usually do that first), pretend x=0.
Then it becomes:
4(0)-3y=12
Which then becomes:
-3y=12.
Divide both sides, and y = -4.
Then the (x,y) coordinate would become (0, -4).
Next you would find the x-intercept, and pretend y = 0.
Which then becomes:
4x-3(0)=12
3x0 cancels out to become:
4x = 12
And x = 3 because division stuffs.
Then THIS point would then be (3,0).
So then, you would plot (0,-4) and (3,0) on a graph and connect the points! I attached a picture for reference.
hope that helped~