Date: Tue Oct 29 14:52:30 PST 2002
Student's Name: Mattie
Answer(or question to Mentor): PROBLEM SOLUTION: A. The storm surged 9 feet in 2002.
B. None of Andy’s stilt was left dry in 2002.
RATIONALE: A. First I had to find out was how many feet high the pole was and how much of it was not in the ground. I knew that ¼ was in the ground and 1/3 was covered with water, so I had to find the least common denominator. I found that it was 12 and ¼ became 3/12 and 1/3 became 4/12. So I added them together and got that 7/12 of the pole was either in the ground of covered in water. So then I subtracted 7/12 from 12/12 and found that 5/12 of the pole was dry, and that that is the 5 feet that is dry. So to find how many feet the water surged in 2002 I took ¼(the part in the ground) and ¾(the part that was covered in water) and again found that the LCD was 12; making these fractions become 3/12 of the pole was in the ground and 9/12 was covered by water. Since before when 5/12=5 feet, then the 9/12 of the pole in the water would be that the water rose 9 feet.
B. Referring back to part A, 3/12 of the pole was in the water and 9/12 was covered in water. I added those two together and got 12/12 of the pole was either under ground or covered in water. Therefore, none of the pole was left dry.
Date: Sat Nov 2 22:20:43 PST 2002
Student's Name: Mattie
Instructor Comments: Mimi,
Please respond to this student as soon as possible. Please know that timing is critical in this project, and if you don’t respond on time, then it throws off my schedule and the schedule for the student. Therefore, timeliness is a significant part of your grade.
This student should be congratulated for his (her?) excellent description of his thinking, which makes it easier for you to find out where the error might have occurred. There are actually two problems here: a) the student is adding fractions as if they were actual measurements (instead of fractions of lengths), and b) the fractions do not refer to the same whole, and thus they cannot be added. Please try to elaborate in detail on a) why her answer is incorrect, and b) a more specific hint about what to do next. If you want me to look over your response before you send it, let me know by writing "PLEASE READ DR. BOWERS" at the top.
Dr. Bowers
Date: Tue Nov 5 09:04:47 PST 2002
Student's Name: Mattie
Mentor Reply: Hi Mattie!
Once again thanking you for sending a response to the problem of the week.
You've done a nice job of describing your thinking.
Let me help you to make a few changes so you can try this problem again okay?
The two fractions you are working with are not part of the same whole. In other words you don't have to add the two fractions. These are two different storms on different years.
You can begin by drawing a diagram and compare the two different storms and seperate the measurements of the stilts. (HINT: Don't worry about the 1/4 that is underground). You know that there is 1/3 of the pole was covered leaving 2/3 of it uncovered which are the 5 ft. This information will help you to now figure out questions A and B for the 2002 storm. If you still need more help, don't hesitate to ask.
I'd like to hear from you soon.
Bye for now, Noemi
Date: Tue Nov 5 14:54:10 PST 2002
Student's Name: Mattie
Answer(or question to Mentor): I’m still not sure I understand how to do this problem. I did what you said thought I drew a diagram. I’m pretty sure this is incorrect but this is what I did and if I did miss it again, I want to figure out what I am doing wrong.
PROBLEM SOLUTION: A. The storm surged 5.625 feet in 2002. B. There was 1.875 feet of the pole left dry in 2002.
RATIONALE: A. Like you suggested I drew a diagram of the pole for 2000. Since 1/3 was covered in water I divided it into 3 sections to make it easier for me to work with. The 1/3 of the pole that was covered in water I shaded in. Beside the part left blank I wrote 2/3=5 feet, again just to make it easier for me. Since 2/3=5 feet I divided 5 feet into 2 parts and found that each of the 2 parts that was dry was 2.5 feet. And since each of those was 2.5 feet I knew the 1/3 of the pole that was underground had to be 2.5 also. Since there were 3 “parts” of the pole, each 2.5 feet, I multiplied 2.5 times 3 and found that the entire pole was 7.5 feet.
Date: Tue Nov 5 14:55:48 PST 2002
Student's Name: Mattie
Answer(or question to Mentor): B. Since it was the same pole in 2002 as it was in 2000, I knew that the pole was 7.5 feet high. I also knew that ¾ was covered in water, so obviously ¼ was dry. This time I divided the pole I drew into 4 parts, since ¼ of the pole was dry. Again I shaded the ¾ that was in water and left the ¼ that was dry blank. Since I knew the pole was 7.5 feet I divided it by 4, since there were 4 “sections,” and found that each “section” was 1.875 feet. Since only ¼ of the pole was covered in water, and ¼ of the pole was 1.875 feet, then 1.875 feet of the pole was left dry.
I couldnt send it all at 1 time so i sent it in 2 parts agian. Thanks for helping me and explaining what i should have done differently, it helped. Talk to you later, Mattie
Date: Tue Nov 5 15:09:43 PST 2002
Student's Name: Mattie
Answer(or question to Mentor): CALLENGE SOLUTION: No, the house did not get washed away.
RATIONALE: For Andy’s pole he had ¼ in the ground, and since it was 7.5 feet above the ground I added another 2.5 feet to that and found that the whole pole was 10 feet. Since 40% of Bob’s pole was under ground, 4 is 40% of 10, then 4 feet would be under ground so 6 feet would be above the ground. I found that 4 was 40% of 10 by multiplying 10 times .40 to get 4. And since the storm surged 5.625 feet in 2002 and that is less than 6, since there is 6 feet of the pole above the ground, I knew that the house was not washed away.
Date: Fri Nov 8 09:01:45 PST 2002
Student's Name: Mattie
Mentor Reply: Mattie, guess what?
You absolutely have the right idea of coming up with the way to do it. That is GREAT! You did it, that is the correct answer and you explained it to me just perfectly. I'm glad I was able to assist you making it more clear. Don't be discouraged about coming up with an answer even if you may think it's not right. Give it a try and you'll see that there is always someone willing to help :-)
Hope you keep up the great work Mattie and thanks again for submitting an answer. Take care.
Noemi
P.S. I almost forgot to mention that the challenge problem, was it kind of hard? You have done a great job at it once again. Congratulations!
Date: Mon Nov 11 15:42:17 PST 2002
Student's Name: Mattie
Answer(or question to Mentor): It was a little challenging but once i got the oringinal problem right it made it easier. Thanks agian for all the help! Mattie
Date: Mon Nov 11 16:29:50 PST 2002
Student's Name: Mattie
Answer(or question to Mentor): PROBLEM SOLUTION: 1. Charle’s cheeseburger has 330 calories. 2. She has ingested 16.5% of the daily intake. 3. 38% of the cheeseburger is fat.
RATIONALE: 1. Since 1 gram of fat has 9 calories and there were 14 grams of fat I multiplied 14 times 9 and found that there were 126 calories from fat. Then since there were 36 carbohydrates (carbs) and each carb was 4 calories I multiplied 36 times 4 and got that there were 144 calories from carbs. Then since there were 15 grams of protein and each had 4 calories I multiplied 15 times 4 and got that were 60 calories from proteins. 2. Since I know the cheeseburger was 330 calories and her RDA for calories is 2000 then I divided 330 by 2000 to find that she had eaten 16.5% of the RDA. 3. Since there were 330 calories in all and 126 of those were from fat I divided 126 by 330 and found that 38% of the cheeseburger was fat.
Date: Mon Nov 11 17:09:40 PST 2002
Student's Name: Mattie
Answer(or question to Mentor): CHALLENGE SOLUTION: She originally bought 24 chicken nuggets.
RATIONALE: For me to find this answer I had to work the problem backward. I started with the 3 nuggets that were left. I knew that 3 was half of 6, so there must have been 6 in the bucket when Miles got it. Then I had to find out what 6 was ¾ of. To do that I wrote ¾ and changed it to where the numerator was 6 and so that made the denominator become 8. Therefore, there where 8 pieces in the bucket when Juan got it. Then I had to find was 8 was 1/3 of. So I changed the numerator to 8 which made the denominator 24. Therefore, there were 24 pieces of chicken when she first bought it.
~I hope i did a good job of explaining it. If I didn't i'll try to explain it agian more clearly. Thanks agian for all your help. Mattie
Date: Fri Nov 15 08:34:42 PST 2002
Student's Name: Mattie
Mentor Reply: READ DR. BOWERS!!!
This students has a great way of explaining I think and has the correct solutions too. Yet this is the only one that has responded, I understand the problem too. :-). But this one has been saved and just wanted to let you know he or she has done a great job. Another thing is that I don't know if I should respond since they have submitted and hasn't been lost.
Thank You
Mimi
Date: Mon Nov 18 10:37:10 PST 2002
Student's Name: Mattie
Mentor Reply: Hello Mattie!
Thank you so much for submitting a response so quickly. Sorry I hadn't sent you anything, but you might know that there have been a few problems with the system.
In regards to your submission, I must say I am really proud you have come up with this great solution. Your explanation has been absolutely nicely done. Congratulations!
And the challenge problem, you have given me the correct answer. Thank you for that great explanation too. Keep doing a wonderful job.
Talk to you soon, Noemi
Date: Wed Nov 20 22:51:57 PST 2002
Student's Name: Mattie
Instructor Comments: Noemi,
First you don't have to apologize because you are not late: all students will be getting their responses at the same time, which will be when you send them some time on Thursday. Second, I agree that his answers are very well stated. HOWEVER, his answer to the challenge problem is incorrect! The problem stems from his thinking when he states, 'Then I had to find out what 6 was ¾ of'. Ask him to consider that 6 represents what is left after 3/4 is taken out, so the question to ask is 6 is 1/4 of what.--DR. Bowers
