3 times flipped, two sides fo a coin heads and tails, 2/3 meaning most likely
First combine like terms. Add -24 and 8.
You end up with -16+w=4
Now, move 16 by doing the opposite to both sides.
w=20
you have to add 2 1/3 cups + 3 1/2 cups+ 3/4 cups together which would give you 6 7/12 cups of snack mix
This problem is a real bear. Whoever wrote it has a sense of humor, and expects
a lot from you ... probability, permutations, combinations, and geometry.
<u>First, let's talk about probability:</u>
The probability of something happening is (the number of outcomes that meet your description) divided by (the total number of all possible outcomes).
<u>#62:</u>
There are 4 points on the drawing. How many different ways could you pick
two of them ?
The first one you pick could be any one of 4 points.
For each of those, the other point could be any one of the remaining 3.
So the total number of ways to pick 2 points out of 4 is (4 x 3) = 12 ways.
But wait ! Whether you pick 'A' and then 'C', or pick 'C' and then 'A', you still wind up with the same two points. So, although there are 12 ways to pick them, there are only 6 different distinct pairs of points.
OK. How many of those pairs are collinear ? ANY two points lie on the same line, because a line can always be drawn between any two points. So out of the 6 different possible pairs of points, ALL 6 pairs are collinear. The probability of picking a pair that are collinear is 100% .
<u>#63:</u>
There are 4 points on the drawing. How many different ways could you pick
three of them ?
The first one you pick could be any one of 4 points. For each of those ...
The second point could be any one of the remaining 3. For each of those ...
The third point could be either one of the remaining 2.
So the total number of ways to pick 3 points out of 4 is (4 x 3 x 2) = 24 ways.
But
wait ! Whether you pick ABC, ACB, BAC, BCA, CAB, or CBA, you still wind up with the same three points. So, although there are 24
ways to pick them, there are only 4 different distinct sets of three points.
OK. How many sets of 3 points in this drawing are collinear ?
There is only one ! ONLY A, B, and C are collinear.
Probability = 1 out of 4 = 1/4 = 25 percent .
<u>#64:</u>
In the last problem, we saw that there are 4 distinct sets of three points.
How many of them are coplanar ?
They ALL are. A plane can be drawn through ANY three points.
So whichever three points you pick, they are coplanar.
The probability is 4 out of 4 = 100 percent.
Thank you for your 5 points. I shall cherish them.
