The chance and strategy unit problem was to find a strategy for the game Pig that gives you the most points in the long run. The game of pig is where you roll a die for however many rolls as you want to. Each number you roll gets added to your score, the person with the highest score wins. However if you roll a one, then all the points you won that turn are lost.

Content

Probability is any fraction or percent going from 0 to 1. There are two types of probability; theoretical probability is the probability of what should happen. The theoretical probability of getting heads when flipping a coin is ½. The other kind of probability is observed probability. This is when you take the probability of something from the results that you have gotten doing this thing. Such as if you flipped a coin 8 times and got 6 heads, the probability would be ¾. This form will be more accurate over time as the results level out.

Because of observed probability people often fall into the belief know as the Gamblers Fallacy. This when you have a string of events that lead you to believe that something different has a higher probability of happening because it hans’t happened in a while. However past events will not affect the current probability of something happening. Such as rolling a die. If you roll a bunch of sixes that doesn’t raise your probability of getting a five.

Strategy

A strategy is a set of rules or actions that you use to achieve a certain goal. Most board games are based off of strategies. If you have a good strategy that can be adapted to fit different situations, then you become a master of that game. In this Unit we will be focusing on a Strategy for the game Pig.

The best strategy that I have found so far in Pig is to keep on rolling until you get at least twenty points. I decided to go to twenty so if you do get a 1, you don’t actually loose that many points. If you keep getting a bunch of sets in the...

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The Founding Father, the Propagandist, and a Wife
Daniel Boggs
(HST201-1) – (U.S. History I)
Colorado State University – Global Campus
Dr. Bruce Ingram
August 19, 2014
The Founding Father, the Propagandist, and a Wife
Three people walked into a bar. They were a founding father, a propagandist, and a wife of a famous leader. The three introduced themselves as; Thomas Jefferson, Thomas Paine, and Abigail Adams. Ok, so they really did not meet in a bar. If they did they would have plenty of stories to share with each other about their childhood, their contributions to independence, and their influence on the United States. Maybe, they would talk about the legacy each would like to leave behind and how the world was forever changed. Regardless, they would have a lot to talk about. Back in the Revolutionary and Enlightenment era these three people overcame many obstacles in the name of independence. Each individual had a remarkable background that inspired them to be great leaders that contributed to the birth of the United States. The legacies that these three people have left behind still live on today.
Thomas Jefferson
Thomas Jefferson had many accomplishments in his life. Most notably was that he was the one who wrote the Declaration of Independence and also the Statute of Virginia for Religious Freedom. After the American Revolution, Jefferson was elected the third president of the United States. Before he died on July 4, 1826, he also established...

...Portfolio1
Personal and professional development
By
Faiza Nabil Sabita
UOG ID : 00792125
Tutor: Mr. Panchathan
Part A
Table of content
Part A ……………………………………………………………… pp. 2-5
Part B ………………………………………………………………. pp. 7-9
Introduction:
In essence, a team may be defined as two or more people who co-operate together with a common aim. A Team focuses towards common goals and clear purpose (park, 1990). The purpose of this report is to reflect on my experience on working in groups, effectiveness of group work, presentation skills, and reflect on the presentation skills.
Effectiveness of the group work:
The most popular and common model which explains the effectiveness of the team work is Tuckman (1965) the five stages group development model. According to Tuckman (1965) there are five stages of group development and these stages include: forming, storming, norming, preforming, and adjourning.
The first stage of group development is forming stage, under this stage the team members are selected, and get to know each other, objectives are well defined, and tasks are identified. Group members try to identify a group leader and the other roles, and they try to find out what behaviors are acceptable to work in group. The second stage of group development is storming, this stage often characterized as conflict stage, where member tends to disagree on leadership, objectives and the rules. In addition, some members...

...Emily Shiang
6/27/13
POW Write-up
In this POW write-up, I am trying to prove that there can be only one solution to this problem, and demonstrate and corroborate that all solutions work and are credible. What the problem of the week is asking is that the number that you put in the boxes 0-4 is the number of numbers in the whole 5-digit number. For example, if you put zero in the “one” box, you would be indicating that there is zero ones in the number. Another example is if you put a two in the “three” box. This would indicate that there are two threes in the whole 5-digit number. I was asked to find solutions where are the numbers would work in heir perspective boxes. From there I started working on the problem that would fit this criterion.
When I first saw the POW, I thought that finding the solution would be fairly easy, by just plugging different numbers into the boxes and play around with them. I soon discovered that that wasn’t the case. So using deductive reasoning, I began off by acknowledging that there couldn’t be any number over four in the boxes, because if it was above four, it wouldn’t be the right solution. If I put a five in any box, I knew that there would only be four boxes left, which wouldn’t work. So I ruled out any number over four.
From there, I began by using the largest number box, which was four. I noticed that if I used any number other than zero in the four box, the problem wouldn’t...

...1. To find my conclusions I had to think about each part of the problem. When you know that one thing means you go on to the next part. When you figure out what that means you have to see how the two statements are related. If they are related then you can deduce a conclusion that makes sense.
2. Here are my conclusions for the 6 problems on page 7.
1. a. No medicine is nice
b. Senna is a medicine
Here I deduced that Senna is not a nice medicine. I think this because the first statement says that “no medicine is nice.” That tells me that all medicines are not nice. The second statement says “Senna is a medicine”. That statement is straight forward. When you put them together you can decide that Senna is a medicine and medicines are not nice. So Senna is not nice.
2. a. All shillings are round
b. These coins are round
Here I decided that no now conclusions can be drawn. The first statement says “All shillings are round.” That statement is clear. The second statement says “These coins are round.” This tells you the coin they have are round. When you put these statements together you can see some flaws. They say these coins but you don’t know if any of these coins are shillings. They can be other coins that are round. So you cannot deduce anything.
These coins are
3. a. Some pigs are wild
b. All pigs are fat
Here I decided that there are no conclusions...

...3-by-3, 2-by-2, and 1-by-1.
So it is saying that, how many squares I can make in an 8-by-8 checkerboard?
The things I checked and figured out that are wrong is that I tried to do as much squares as
Possible but I always had less and I knew it was going to be way more. So I asked for help
From some of my classmates when I was done with my work. They seemed to actually helped
With all the struggles I was going through. They didn’t actually helped me find the answer
But they did help me through it though. I also did ask mom but she really didn’t help because
She doesn’t know how hard this could be. But the methods I tried doing was that input
Output method but then I knew I was way off. The process I went through was I was trying
A summation notation. So I started to do the process which
8*8=64
7*7=49
6*6=36
5*5=25
4*4=16
3*3=9
2*2=4
1*1=1
So I knew this has helped me in some way. The number of squares I can find in an 8 by 8 checker board is 204 possible squares. I found a summation notation that can be used as this problem which is what I put on the paper. You can use this for any checker board squares you have. (x by x)
Well an extension for my problem would be if you had a way harder problem, I would find one that would be kind of like this, like how many squares you can find on a 20 by 20 checker board. I have learned that you can find a lot of squares in any...

...“A Sticky Gum Problem” POW 4
Problem statement:
The next scenario is very similar. In this one, Ms. Hernandez passed a different gumball machine the next day with three different colors Once again her twins each want a gumball of the same color, and each gumball is still one cent. What is the most amount of money that Ms. Hernandez would have to spend in order to get each of her daughters the same color gumball?
In the last scenario, Mr. Hodges and his triplets pass the same gumball machine that Ms. Hernandez and her twins passed in scenario two. This time, each of Mr. Hodges children wants the same color gumball out of the three-color gumball machine. What is the most amount of money that Mr. Hodges would have to spend on his triplets in order to get them each the same color gumballs?
Process:
In the process of solving the first question I drew up different color gumballs.
These are both the different colored gumballs. If this was the outcome after spending two pennies on two gumballs, then the next gumball would have to be one of the previous color gumballs that already came out of the machine.
For the next scenario it was a little trickier that the first problem. Since there were still only two children involved and there were three colored gumballs it wasn’t too hard. Once again, I drew up the three different colored gumballs in the gumball machine. The gumballs were red, white, and...

...Mega POW
A very wealthy king has 8 bags of gold, which he trusts to some of his caretakers. All the bags have equal weight and contain the same amount of gold, all the gold in the kingdom. Although, the king heard a story that a woman received a gold coin. The king knew it had to be his gold so he wanted to find the lightest bag in the 3 weighing, but the mathematician thought it could be done in less, so I need to find out the least amount of weighing it takes to find the lightest bag. Also, the king used a pan balance for all of his weighing.
I started by weighing 4 bags on each side of the scale to see which side was lighter. Then from those results I thought to weigh the 4 bags that were on the lighter side by 2 and 2. After this you would find one side weighing less than another. Then you would take those results and weigh the 2 remaining bags and the lightest bag would be the bag that was taken from. However, the mathematician said it could be done in less than three steps. So throwing the answer I had just gotten to the side, I started new. This time I started with 3 bags on each side knowing that if two sides were equal than the bag with the missing gold would be one of the bags not weighed the first time. Then you would have to weigh the two remaining bags and whichever one was lighter than the other would be the bag with less gold. But, if the 3 bags from the beginning weighed different then you would weigh 2 bags of the 3 and if they are equal...

...This unit's main goal was to use similar triangles to measure the length of a shadow. While using the variables D, H, and L, we have figured out a formula to measure a shadow's length. In order to do this though, everyone had to learn the basic concepts of similarity, congruence, right triangles, and trigonometry.
Similarity and congruence were two very important factors because they helped us learn about angles and the importance of triangles. Similarity was a key to find out how to use proportions to figure out unknowns (such as in HW7). Once similarity was learned we moved on to congruence where we learned proof and how to show others what is truth by giving them accurate facts based on previous truths. If similar triangles share enough equal traits, they can be called congruent by ASA, SAS, SSS, and AAS.
Right triangles came next and with them we learned how and why they are special. As it turns out, right triangles are furtively hidden in many problems. Along with our unit problem to find out the length of a shadow. Trigonometry worked with right angles as we learned about tangent, sine and cosine. Using trigonometry, we could figure out unknowns that were considered unsolvable to us when we only knew about proportions.
All of these concepts and ideas really helped us find the final equation for finding the length of a shadow. They all built upon each other and with a little logical reasoning we were able to finally solve it. Using similarity, we were able to...