What aspects of this course were most useful or valuable?
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Let's convert these inequalities to Slope Intercept Form so that we can graph them easily.
<span>−10x − 5y ≥ 15
Add 10x to both sides.
-5y </span>≥ 10x + 15
Divide both sides by -5 and flip the inequality sign.
y <span>≤ -2x - 3
</span><span>x + y ≤ − 1
Subtract x from both sides.
</span><span>y ≤ -x − 1
When graphed, the solutions lay in Quadrants II, III, and IV. You can see this in the picture I've included.</span>
<span>−10x − 5y ≥ 15
Add 10x to both sides.
-5y </span>≥ 10x + 15
Divide both sides by -5 and flip the inequality sign.
y <span>≤ -2x - 3
</span><span>x + y ≤ − 1
Subtract x from both sides.
</span><span>y ≤ -x − 1
When graphed, the solutions lay in Quadrants II, III, and IV. You can see this in the picture I've included.</span>
You made a mistake with the probability , which should be
in the last expression, so to be clear I will state the expression again.
So we want to solve the following:
Conditioned on this event, show that the probability that her paper is in drawer , is given by:
(1) and
(2)
so we can say:
is the event that you search drawer
and find nothing,
is the event that you search drawer
and find the paper,
is the event that the paper is in drawer
this gives us:
Solution to Part (1):
if , then
,
this means that
as needed so part one is solved.
Solution to Part(2):
so we have now that if =
, we get that:
remember that:
this implies that:
so we just need to combine the above relations to get:
as needed so part two is solved.