Sage Wentzell-Brehme
Problem Statement
In this problem I am looking at strategies for a game called What’s on Back? This game has 3 cards. One card has an X on both sides, one card has an O on both sides, and one has an O on one side and a X on the other. On your turn you choose one of the cards at random from a bag. You hold it up with one side facing you. You are not allowed to look at the other side or the other cards. You have to predict whether it is an O or an X on the other side of the card. My task is to look at a bunch of different strategies for playing and find the probability of success for each one. For each strategy I look at I have to find the probability using both an experimental and theoretical model. My real goal is to find the strategy with the highest probability of success or the one most likely to help me win.

Strategy # 1
a. Always choose the same thing the card says. So if it is an O choose O, if it is an X choose X. b. 30 trials
1. yes 6.no 11.no 16.yes 21.yes 26. yes 2. yes 7.yes 12.yes 17.yes 22.no 27. yes 3. yes 8.no 13.yes 18.no 23.yes 28. yes 4. no 9.no 14.yes 19.yes 24.no 29. yes 5.no 10.yes 15.yes 20.no 25.no 30. Yes

P (right) – 18/30 or 6/10 or .6
Success rate = .6
c. (XX)-works-1/6
(XX)-works-1/6
(OO)-works-1/6
(OO)-works-1/6
(OX)-doesn’t work
(XO)-doesn’t work total successful – 4/6 or 2/3 or .66 Success rate = .66
Strategy #2
a. Always choose O no matter what.
b. 30 trials
1. yes 6.no 11.yes 16.yes 21.no 26.yes
2. yes 7.no 12.yes 17.yes 22.no 27.no
3. no 8.yes 13.no 18.yes 23.no 28.yes
4. no 9.no 14.yes 19.yes 24.yes 29.yes
5.no 10.yes 15.yes 20.yes 25.no 30.no
P (right) – 17/30 or .56
Success rate = .56
c. (XX)-doesn’t work
(XX)-doesn’t work
(OX)-doesn’t work...

...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 that can be made. The first...

...“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...

...POW Problem Statement
A. A farmer is going to sell her eggs at the market when along the way she hits a pot hole causing all of her eggs to spill and break. She meets an insurance agent to talk about the incident, and during the conversation he asks, how many eggs did you have? The farmer did not know any exact number, but proceeded to explain to the insurance agent that when she was packing the eggs, she remembered that when she put the eggs in groups of 2-6 she had even groups with 1 left over, but when she put them in groups of 7 she had even groups of 7 with none left over.
B. Why does groups of 2,3,4,5 or 6 results in 1 left over egg, but groups of 7 has an equal amount of eggs with none left over. What # of eggs has equal groups of 2,3,4,5, or 6 with one left over and 7 goes into the number evenly.
C. They think the answer is 49 eggs because 7 goes into 49 eggs evenly with none left over.
D. It cannot be 49 eggs because if it were 49 then 2-6 would need to go into 48 evenly to have a left over egg, but 5 does not go into 48 evenly which is why 49 wouldn’t work
POW Process
A. My initial ideas concerning the task is we need to find a number that has 2,3,4,5 and 6 go into that number evenly with 1 left over and have 7 go into it evenly with none left over. Making a chart with multiples of all those numbers out on paper will help us track a number down and...

...IMPPOW1: The Broken Eggs
Problem Statement:
A farmer’s cart hits a pothole, causing all her eggs to fall out and break. Luckily, she is unhurt. To cover the cost of the eggs, her insurance agent needs to know how many she had. She can’t remember the number, but can remember some problems she had when packing the eggs. When she put the eggs in groups of two to six eggs, there was always one left over. However, in groups of seven, there were none left over. From what she knows, how can she figure out how many eggs she had?
Process:
First, I thought the answer would be forty-nine. Then, I realized my mistake and tried to think of different ways to do it. I decided to make a chart showing the remainders of the numbers two to six into multiples of seven. I then started to find a pattern. I noticed that the number two has a remainder of one every other multiple of seven, the number three has a remainder of one every two multiples, the number four has a remainder of one every three multiples, and so on. I marked a dot every time there was a remainder of one. I knew that when I had dots marked from two through seven, that would be the answer. So I set out on the long journey of calculating when there was a remainder of one until I reached a number that was all filled up.
Solution:
Using my process, I found that the number of eggs the farmer had was three-hundred one. I know this because it was the first number to fit my...

...ranges of data sets, summation notation, consecutive numbers, factorials, arithmetic sequence and order of operations. On top of that, we investigated triangular numbers, finding formulas, the number of diagonals in a polygon, the sum of interior angles in a polygon, measure of angles in regular polygons, exponents, and squaring negative numbers. As you can see, this was a productive unit!
The “In and Out” tables were a vital piece of this unit. Not only were they an effective way to show information, we were also able to solve patterns using them and find missing terms in a sequence with them. In the example “In and Out” table below:
“To find the out value, multiply the in value by itself and then subtracts three.”
In Out
0 -3
1 -2
2 1
“In and Out” tables taught us to look at problems differently and to find multiple ways as to how to address a problem.
When we solved the “In and Out” tables, we came up with functions to express a rule derived from the table that could describe how to find the out value from the in value and would work for all numbers being applied to it. By solving the functions, we were able to come up with formulas. Upon coming up with formulas, we were able to also solve for missing domains and ranges in the tables. Domains are the inputs in data sets and ranges are the outputs.
Another useful way to display data sets is through summation notation. Summation notation uses the “Sigma” sign and can...

...POW 16: Spiralaterals
Problem Statement: Spiralaterals-a spiralateral is a sequence of numbers that forms a pattern or a spiral like shape. Spiralaterals can form a complete spiral-like shape or it could form an open spiral that never recrosses itself or return to it's original starting point.
To make a spiralateral: Each spiralateral is based on a sequence of numbers.To draw the spiralateral, you need to choose a starting point. The starting point is always "up" on the paper. Next take your first number and draw a line as many squares long as your number is. Now turn your paper 90° and take your next number and do the same thing as the first. Each line will be drawn 90° clockwise from the previous line drawn. Stop this whenever the shape has come to its starting point or if you realize the shape will continue in one direction and will probably not return.
Begin by drawing odd and even number sequences,shorter sequences, longer ones,and eventually more elaborate and random sequences to get different products.
Conclusion: Spiralaterals containing 4 numbers will not return to their starting point but ones with 3,2, or 5 numbers will almost always make a complete spiral.
Spiralaterals with 2 numbers will make a rectangle or a parallelogram. Any set of numbers that is a repetition of one number (I.e. 1,1,1,1,1,1,1,1. 7,7,7,7,7,7,7. e.t.c) will also make rectangles. Also numbers that repeat two of the numbers but not the...

...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...