Baseball Cards

Third grade students at Hanover Street School made this awesome video as a Free Scenario based on the Math Forum problem called “Baseball Cards.” We are so excited to share their video!

 

 

 

The students’ video is based on the Math Forum Baseball Cards Scenario [PDF]

I’ve been reading a lot lately about the idea of a “modeling curriculum.” Not as in America’s Next Top Model and also not as in the teacher models the thinking and the student learns from watching and trying it themselves.  A modeling approach to teaching science and math means that the students work together to develop better and better conceptual models to explain situations. So in physics, you might roll two objects down a ramp and try to make a mathematical model to describe what was going on. At first you might include the weight of the balls in your model, but then you might observe that two objects with different weights behave the same, and so your model would change based on new data and new understanding.

Some of the studies of this kind of teaching show us that students come into situations with models already in their heads — they already have ideas about how balls fall, for example. Their models might not be the most accurate or easiest to use, and so as they encounter new situations and new demands, they change their models. While that’s happening, students might use lots of different competing ideas at once. One minute the same kid will go from making really accurate predictions about two balls of different weights rolling down an incline, but then say that gravity will make a bowling ball fall faster than a beach ball.

This week’s AlgPoW, Filling Glasses, asked students to match graphs of water level vs. time of glasses being filled at a steady rate, to pictures of the glasses. Students used many different models for thinking about the problem:

• Try to match the shape of the graph to the shape of the glass (e.g. count the wavy parts, look for straight graphs for straight glasses).
• Relate the skinniness of the glass to speed.
• Relate the skinniness of the glass to the steepness of the graph.
• Relate the height of the glass as a whole to the maximum height reached in the graph.
• Relate the skinniness of the graph to speed at which the glass fills and the speed at which the glass fills to the steepness of the graph.

What was most interesting, though, was the students who used different strategies at different moments. Students who are in the middle of learning often switch models based on small details or when a problem seems easier or harder for some reason.

Like this:

For this problem, you have to really visualize the glasses and their shape.

First, I looked at glass A. It starts out skinny for a tiny bit, then there is a huge bulge before it is a little skinnier. So the height would rise quickly for the shortest amount of time, then go slower, then finally go a little faster. I visualized the graph to be a slightly zigzaggy line that was not too tall. Graph 4 did not have any zigzags, and graphs 2 and 3 went too high. So, graph 1 matched with glass A.

Glass B is like a funnel, starting skinny and getting wider and wider as the top draws nearer. So the height would rise quickly at first and get slower and slower. Since there are no bulges in glass 2, the graph it matched up to would have to be zigzag-free. And the only graph without zigzags is graph 4.

Finally, glass C starts skinny, gets wider, gets skinnier, and then gets wider. The water will go fast at first, then slower, then faster, then slower. Graphs 2 and 3 are very similar, but only graph 3 starts out fast.

Glass A= First of all glass A is the shortest so the line on the graph would be less steep. Also, since the glass is kind of round, at first the water would pour fast then gradually pour slower then after you get to the middle the water would gradually pour faster.

i figured this out becauause if you look at the glasses and the graphs. the arches in the graphs are like the glasses when get bigger because you need to have more water and then it would fill it up.

Last year brought us the unveiling of the Primary Problem of the Week, a series of PoWs geared towards the youngest problem solvers. This year, we’re focusing on older students, students who have studied math beyond Algebra I and Geometry. We will be publishing 20 problems from our Trig & Calc library to make them available to all teachers with a Current PoW membership. Each problem will have links to enhanced teacher materials (strategy alignments, Online Resource Pages, Scenario-Only versions of the problems, and Teacher Packets including Common Core alignments).

We’re excited to be able to offer these problems to our Current PoW members so that teachers of higher-level math can be part of the Current PoW community. As the name suggests, most of the problems can be solved using techniques from Trigonometry or Calculus. However, many of the problems can be solved in multiple ways: with right-triangle trig that students may have learned in early grades; with algebraic techniques and software; or they draw on content areas like Discrete Math or Probability. So these problems can also be extra challenges for students who aren’t yet in the Trigonometry or Calculus class.

This year, we don’t plan to feature highlighted solutions for the Trig & Calculus PoW on the PoW site (teachers can always see sample successful solutions with different strategies in the Teacher Packet). If, however, we do get interesting submissions, we’ll certainly be blogging about them here! And if we get to a consistent level of submissions, we’ll be excited to have highlighted solutions from the Trig & Calculus PoW next year!

So we’d love to have you check out the current Trig & Calculus PoW, “Building Boxes,” a PoW that can be approached by older students using derivatives or by middle-school students through careful tables and virtual manipulatives. View the Online Resources Page in the “blue box” on the PoW for a link to an applet that will help students from middle school through calculus make sense of this problem.

This week, we asked Pre-Algebra students to solve a problem that turned out to be really tough. Students were asked to think about scoring in a game used to practice place value skills in the context of scientific notation. We used some examples to illustrate how the scoring worked, and many students used the examples to start teaching themselves place value in scientific notation.

It was pretty awesome to have a window into how students interpreted the examples we gave. Some students were able to make sense of what was happening. Other students got correct answers, but some of their words made me wonder if they were getting the right answer without really understanding what was going on. And some students saw things in the examples that I never would have thought of… that might lead to problems down the road! The whole experience could be a microcosm of what students see when teachers work out examples on the board.

Here are the examples we gave:

If a card contains the number 3.38 x 10^14 and Mrs. Nasamy picks an 8, the card is worth 8 x 10^12 points. If she picks a 3, it’s worth 3.3 x 10^14 points since 3 appears in two places, you must add both values:

(3 x 10^14) + (3 x 10^13) = (3.3 x 10^14) points

We include this example so you can think about how the game is played when the digit pulled appears in more than one place; note that the process would be the same if the digit appeared on two different cards.

And here are some ways students interpreted those examples:

Student 1 thinks about how the exponent and place value of the target digit are manipulated together:

When the 3.38×10^14 card was there the next time they showed it had gone from 3.38×10^14 to 8×10^12. It was like this because you cross out everything but the number pulled. If the number pulled is the first in back of the decimal the exponent needs to take 1 away. If it is the second back you do the same thing but take two away from the exponent.

Student 2 has an AHA! moment relating to the distributive property:

first off, i had to do a lot of time figuring out what the question is. it took me a while, but an AHA! was when i realized that these cards were like three #’s in sci. notation mashed together for example, (1.92×10^9) is (1×10^9)+(9×10^8)+(2×10^7). and if a 2 is pulled, your points are (2×10^7).

Student 3 provides a third example to check his/her understanding:

my interpretation of the problem is this. The teacher pick a card between 1 and 9. Whatever number she draws, you find the number in your scientific notation. for example, if she drew a 7, and one of your notations was 7.37 X 10^³, it would turn into 7.07 X 10^³. then you would solve the exponent, which is 10^³, and that is 1,000. Now your problem is simply 7.07 X 1,000. When you solve that simple multiplication problem, you get 7,070. That is your answer for that card.

Student 4 applies the examples to the main problem, finding the point score if Ms. Nasamy pulls an 9 and you have cards with 1.92 x 10^9 and 6.59 x 10^8. I am not sure if Student 4 has a misconception or not, because they do eventually come to the correct answer, however, I suspect something’s wrong after very careful reading:

I…started with 1.92*10^9. I then need to cross out the place values that aren’t 9, like the example showed us. Now the problem is 0.9*10^10. The exponent changed because I took the 0.02 place value away. Now, since there are 10 zeros instead of 9, the exponent changes to 10. To solve the problem, I first calculated 10^10, which is 10,000,000,000. Then I did 10,000,000,000*9, which is 90,000,000,000. Then I did the last one which is 6.59*10^8. Next I used the same process for the last card and changed the problem to 0.09*10^8. The exponent didn’t change because the 0.09 is the last place value there. After that, I did the same process as the first card and and solved the problem. I ended up with 900,000,000. Then I added the points together and got 90,900,000,000. I then converted that number into a decimal. I did that by counting the number of place value zeros and the turning that number into the exponent next to ten. I got 9.09*10^8.

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A middle-school math teacher, Ms. Alcala, shared the activity, “My Favorite No” on the Teacher Channel. I highly recommend watching the video. The idea is to collect work from students and then share, anonymously, one incorrect solution that you really like. You might like it because it’s a different approach or because of how much was correct about it or because it’s a common mistake. The idea is to share some work that’s not perfect in a very positive, non-judgmental way, and engage students in thoughtfully critiquing the work. Students get to work on the mathematical practice, “Construct Viable Arguments and Critique the Reasoning of Others” while also sharing and strengthening their content knowledge.

This week, reading student submissions to “Teeter Trio” I was struck by this awesome, but ultimately incorrect, submission:

Seesaw Balance POTW
A seesaw can balance with more than two people on it. The product of each person’s weight and distance from the fulcrum contributes to the balancing. If the sum of those products on one side equals the sum of the products on the other side, balance is achieved.
-As far as I know balance can be achieved when both sides are equal when the products are added in other words, wd=wd, when w is weight and d is distance from the fulcrum.
Shareef and his two little sisters, Marshay and Janeka, are playing on a seesaw. Shareef weighs 30 pounds more than Marshay and 35 pounds more than Janeka, so Shareef sits on one side to balance the two girls on the other.
-If Shareef is trying to balance the other two girls on the other side than we need to know everyone’s weights to fill in the equation of wd=wd.
Shareef=s
Marshay=m
Janeka=j
-Since Shareef is 30 pounds more than Marshay and 35 pounds more than Janeka than we have two different equations for her.
s=30+m
s=35+j
-Since we know this information we can find Marshay’s weight by putting these two equations together like so.
35+j=30+m     Subtract 30 on both sides
5+j=m
-For Marshay’s weight we get m=5+j and Janeka’s weight can’t be found so it will just be represented by the letter j.
Shareef is sitting 6 feet from the fulcrum and Janeka is sitting 4 feet from the fulcrum. If the seesaw is balanced, find a function that expresses Marshay’s distance from the fulcrum in terms of her weight.
-Now that we know Janeka’s distance and Shareef’s distance from the fulcrum but we need to know Marshay’s distance from the fulcrum. Her distance will be represented by the letter d. now that we have all the pieces of information than we can fill in the equation of wd=wd. (The right side will be Shareef and the left will be Marshay and Janeka)
wd=wd
6(35+j)=d(5+j)+4j       subtract 4j on both sides
6(35+j)-4j=d(5+j)        divide (5+j) on both sides
6(35+j)-4j/5+j=d         distribute the 6 and combine like terms
210+2j/5+j=d              Simplify
2(105+j)/j+5=d
Marshay’s distance from the fulcrum is 2(105+j)/j+5=d.

To help you think about what might be incorrect, it’s important to know that in the follow-up you will be given Marshay’s weight and asked to help calculate how far from the fulcrum she should sit.

My questions to you are:

1) What do you find awesome about this submission?

2) How would you fix it without redoing the problem? Is there a way to tweak this answer to make it better match what was asked?

Some Teeter Trio links in case you are interested:

• The problem [requires a Math Forum PoW Membership].
• Information about accessing Teeter Trio (and a selection of all our PoWs) for 21 days with a free Math Forum trial account.
• Information about becoming a Math Forum Problems of the Week Member. Consider starting with a $25 membership, which gives you access to all of this year’s Current PoWs — and now you can create 36 student logins as well!

Join the John Ehret Patriots in New Orleans, Lousiana, USA, in Noticing and Wondering about this Problem of the Week!

The Hopi Indians invented Totolospi, a game of chance. The game is played with three cane dice, a counting board, and a counter for each player. Each cane die can land round side up (r) or flat side up (f).

The moves of the game are determined by tossing the three cane dice with these rules:

Toss	Move
three round sides up (rrr)	player advances 2 lines
three flat sides up (fff)	player advances 1 line
any other toss of the three cane dice	player doesn't advance