Let's first think about how many possible outcomes there are to a series of coin flips. One that will help us here is that coin flips are <em>independent</em> - the outcome of one flip has no effect on the outcome of the others. What this means is that there are two possible outcomes <em />for <em>each </em>flip: heads or tails. For an example with fewer coins, let's say we were flipping 2 instead of five.
- Flip 1 can either be heads or tails
- Flip 2 can either be heads or tails
So our possible outcomes are HH, HT, TH, and TT. There are two possible second flips <em />for <em>each</em> of the possible first flips, or 2 x 2 = 4 total combinations of flips. Notice that <em>only one </em>of those combinations has zero tails - the combination with all heads.
If we were to flip a coin 5 times, we'd have two possible fifth flips for each of the two possible fourth flips for each of the two possible third flips for... it gets pretty hairy to describe in words, but I've attached a diagram so you can see how quickly it grows out of control. There are 2 x 2 x 2 x 2 x 2 or
![2^5=32](https://tex.z-dn.net/?f=2%5E5%3D32)
possible combinations of heads and tails! But, in fact, we don't even need to sort through these 32 combinations to answer our question. <em>Every</em> combination will contain at least one tail, except one: the combination which contains all heads (HHHHH). Which means the rest of the 31 must contain at least one tail.
This fact will stay the same regardless of the number of coin flips you make: <em>the number of ways that contain at least one tail will always be the total number of combinations minus one (the case where all of the flips are heads).</em>
Answer:
(-5,4)
Step-by-step explanation:
If you look at A and then at AB midpoint you will see that A is 5 down from M and 4 right from M. So, if you minus 4 from -1 you get -5, and if you add 5 from -1 you get 4. Therefor the answer is (-5,4)
Answer:
The 2×2
Step-by-step explanation:
you have to solve the 2×2 first then answer the whole problem. YOUR WELCOME!!!!
Answer:
The distance between the ship at N 25°E and the lighthouse would be 7.26 miles.
Step-by-step explanation:
The question is incomplete. The complete question should be
The bearing of a lighthouse from a ship is N 37° E. The ship sails 2.5 miles further towards the south. The new bearing is N 25°E. What is the distance between the lighthouse and the ship at the new location?
Given the initial bearing of a lighthouse from the ship is N 37° E. So,
is 37°. We can see from the diagram that
would be
143°.
Also, the new bearing is N 25°E. So,
would be 25°.
Now we can find
. As the sum of the internal angle of a triangle is 180°.
![\angle ABC+\angle BCA+\angle BAC=180\\143+25+\angle BAC=180\\\angle BAC=180-143-25\\\angle BAC=12](https://tex.z-dn.net/?f=%5Cangle%20ABC%2B%5Cangle%20BCA%2B%5Cangle%20BAC%3D180%5C%5C143%2B25%2B%5Cangle%20BAC%3D180%5C%5C%5Cangle%20BAC%3D180-143-25%5C%5C%5Cangle%20BAC%3D12)
Also, it was given that ship sails 2.5 miles from N 37° E to N 25°E. We can see from the diagram that this distance would be our BC.
And let us assume the distance between the lighthouse and the ship at N 25°E is ![AC=x](https://tex.z-dn.net/?f=AC%3Dx)
We can apply the sine rule now.
![\frac{x}{sin(143)}=\frac{2.5}{sin(12)}\\ \\x=\frac{2.5}{sin(12)}\times sin(143)\\\\x=\frac{2.5}{0.207}\times 0.601\\ \\x=7.26\ miles](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7Bsin%28143%29%7D%3D%5Cfrac%7B2.5%7D%7Bsin%2812%29%7D%5C%5C%20%5C%5Cx%3D%5Cfrac%7B2.5%7D%7Bsin%2812%29%7D%5Ctimes%20sin%28143%29%5C%5C%5C%5Cx%3D%5Cfrac%7B2.5%7D%7B0.207%7D%5Ctimes%200.601%5C%5C%20%5C%5Cx%3D7.26%5C%20miles)
So, the distance between the ship at N 25°E and the lighthouse is 7.26 miles.