tag:blogger.com,1999:blog-1811268622721218650.post4828559072874325949..comments2010-08-07T15:27:39.170-05:00Comments on Challenge Math - SBISD GT Book Study: Session 3 - Question 3atxteacherhttp://www.blogger.com/profile/15216583790234498239noreply@blogger.comBlogger23125tag:blogger.com,1999:blog-1811268622721218650.post-21563070822010494332010-07-16T11:05:40.523-05:002010-07-16T11:05:40.523-05:00I selected #9 on page 209. If you want to pick 3 ...I selected #9 on page 209. If you want to pick 3 of a kind, it doesn't matter what your first pick is, since that will determine what the other two need to be. So first card is 52/52 or 1, second card then would be 3/51 or 1/17, and third would be 2/50 or 1/25. Thus the probability of the three of a kind is 1/425.susanmnoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-28493196390386475662010-07-16T00:01:06.518-05:002010-07-16T00:01:06.518-05:00Problem 3 page 187. As I read the problem, I drew ...Problem 3 page 187. As I read the problem, I drew the right triangle with the storm on top and the man next to the 65 degree angle. Then found the distance of the thuderstorm =3miles.<br />Found the sine: distand off ground/3<br />distance off ground=14,344 ft <br />14,355/5280=2.72 milesRCampananoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-90095085035154430542010-07-15T21:46:20.495-05:002010-07-15T21:46:20.495-05:00Problem #3, page 245: three hours after Gabe start...Problem #3, page 245: three hours after Gabe started, his dad will catch him.<br /><br />48+48+48=144<br /><br />72+72=144<br /><br />I added using the blocks and saw that in three hours of Gabe driving, his dad would catch him.Katie Kavanaghnoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-13122585061659229502010-07-15T21:39:03.504-05:002010-07-15T21:39:03.504-05:00I did problem #9 on page 209 because these type of...I did problem #9 on page 209 because these type of problems always used to stump me in high school. By multiplying the chances of the second and third card- it made it much easier to realize your chances of getting 3 of a kind. I can't wait to share these with my class- I know they will like a challenge!Sharon G.noreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-168468313385870702010-07-15T21:06:14.004-05:002010-07-15T21:06:14.004-05:00I also did problem #9 on page 209. It took me a m...I also did problem #9 on page 209. It took me a minute that you only had to multiply the chances of the second and third card selected. You can start with any card, then multiply 3/51 by 2/50 to determine the chance of getting 3 of a kind. It was a fun problem.CKohlhttps://www.blogger.com/profile/12218582181573792567noreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-24911535488472622602010-07-15T20:55:31.919-05:002010-07-15T20:55:31.919-05:00In reponse to CynthiaM on 7/14
G/T students love t...In reponse to CynthiaM on 7/14<br />G/T students love to collaborate to come up with the correct answer. You were Einstein through collaboration! I would have never known there were answers in the back of the book... of course you picked the one problem that wasn't explained in an example :-)CKohlhttps://www.blogger.com/profile/12218582181573792567noreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-83209762638919762292010-07-15T20:18:50.984-05:002010-07-15T20:18:50.984-05:00I chose problem #6 on page 188. I loved teaching ...I chose problem #6 on page 188. I loved teaching these when I taught GT geometry, so I was reverting to the comfortable. What I like about this type of problem is you need to draw a picture to see where the triangle appears in the problem. A common mistake is to not understand that the 6 feet must be added to the height of the building as well. These types of problems were also fun to do with the students outdoors using an angle meausring device and a yardstick.Kohlerjnoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-73299684543458684922010-07-15T17:29:48.857-05:002010-07-15T17:29:48.857-05:00I chose problem #34 on page 188. I drew a diagram ...I chose problem #34 on page 188. I drew a diagram first of the tree which was 300 feet away and Greg being 5 feet tall. I had trouble working this problem at first because I didn't add in his line of site correctly.Once I found the sides using the 18 degrees I came up with 100 feet but the answer is 100.2 feet so I was still off slightly in my calculations.ndeansnoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-43754401752844222332010-07-15T16:04:11.070-05:002010-07-15T16:04:11.070-05:00I choose to complete problem #4 on pg. 245 (chapte...I choose to complete problem #4 on pg. 245 (chapter 15). Following Zaccaro's strategy, I found the sum of the blocks of time and then divided that sum by the total number of blocks. Traveling up the Mississippi I drew 3 40mph blocks and then traveling down the Mississippi I drew 2 60mph blocks. The sum of the 5 boxes equals 240 and the quotient of 240 divided by 5 equals an average boat speed of 48mph.SadloKnoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-2340546964920361892010-07-14T18:31:55.398-05:002010-07-14T18:31:55.398-05:00I worked on problem #10 on page 209.
My work was b...I worked on problem #10 on page 209.<br />My work was based on the examples that were given in this<br />chapter and then when i checked my work with the answer<br />in the back of the book I was surprised I was incorrect.<br />i tried working it another way and still didn't feel like<br />Einstein.<br />So i called a couple of friends and they helped me<br />arrive at the correct answer but also shared their<br />strategies. I think this is how we learn and I think<br />GT students would work through problems in a similar fashion.cynthiamnoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-58007296615878112132010-07-14T17:43:13.045-05:002010-07-14T17:43:13.045-05:00In response to nlopez-- Great explaination-very ea...In response to nlopez-- Great explaination-very easy to follow and very well done!!!! Drawing the picture and labeling is always a great way to start!!ReneeRnoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-11684167240289805152010-07-14T17:37:35.526-05:002010-07-14T17:37:35.526-05:00In response to ratliff--How funny, They do love to...In response to ratliff--How funny, They do love to complain about writing everything out. I am constently after them to show all possible outcomes and how they did their work. I guess we all do!!ReneeRnoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-43809826420593788702010-07-14T15:19:30.218-05:002010-07-14T15:19:30.218-05:00The Question I picked was # 2 on page 208--after a...The Question I picked was # 2 on page 208--after all I need to know my chances of winning the lotto. If you can pick 5 numbers out of 49 your problem would look like this 1/49 x 1/48 x 1/47 x 1/46 x 1/45 = 1/228,826,080. Not looking so good! The good news is we are not done, now we get to narrow the odds b/c our numbers can be in any order so we mult 1 x 2 x 3 x 4 x 5 = 120. 228,826,080 divided by 120 = 1,906,884 so your chances of winning are 1/1,906,884. YIKES!ReneeRnoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-61951022018691971902010-07-13T23:16:08.938-05:002010-07-13T23:16:08.938-05:00I solved problem #9 on page 209. Picking three car...I solved problem #9 on page 209. Picking three cards out of a deck, what is the probability that you will get three of a kind? First card can be any kind so probability is 52/52 or 1. For the second card, there are 3 more of the same kind card out of 51, so 3/51. For the thirds card, there are 2 cards out of 50.<br />I wrote the multiplication method given on page 200<br />p(three of a kind)= 52/52x3/51 x 2/50=<br /> 6/2550=1/425SDawsonnoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-53980173838029906762010-07-13T13:58:00.687-05:002010-07-13T13:58:00.687-05:00I solved problem #1 on p.208, which is probability...I solved problem #1 on p.208, which is probability. I enjoy teaching probability to my students, so I thought I would try these problems out. The first thing is you had to find all the possible outcomes, which took forever. I found that when drawing from 5 different decks, there are 1,024 possible outcomes, but only 781 of the outcomes have spades in them. It is amazing how much time you have when waiting for the a/c guy to get to your house. Now that is a probability problem in itself. What is the probability that he will come in the first hour of the allotted time period- zero!<br />This was not a difficult problem to solve, just time consuming, I find that my students don't want to take the time to write out all the possibilities, so now I can go back to school and show them this problem, so when they complain about writing it all out, I can just show them this problem.ratliffnoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-12009244718409492692010-07-13T13:54:15.391-05:002010-07-13T13:54:15.391-05:00I did Eistein Level question on page 209 #10. I go...I did Eistein Level question on page 209 #10. I got 1/5525. Because in the fraction of getting an ace the first time is 4/52 and the next time it is 3/51 and the third time is 2/50. When you multiply and simplify these fractions you get 1/5525. I enjoyed using the strategy that the book discussed in order to orgainze my thinking.sluthernoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-87144674520369663612010-07-11T16:58:23.510-05:002010-07-11T16:58:23.510-05:00nlopez in response to PKassir on July 9th- great w...nlopez in response to PKassir on July 9th- great way to model organizing data. Again, yes, you may be able to do things in your head, but using graphic organizers such as a chart actually helps to give you your proof that your thinking is correct.nlopeznoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-58660997206640964222010-07-11T16:55:40.325-05:002010-07-11T16:55:40.325-05:00nlopez in response to tatumt on July 10th- short a...nlopez in response to tatumt on July 10th- short and sweet- simple, but great, explanation!nlopeznoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-64039752876418561412010-07-10T20:27:37.423-05:002010-07-10T20:27:37.423-05:00In response to PKassir's solution:
I liked the...In response to PKassir's solution:<br />I liked the fact that you used a chart to evaluate and compare the data in the problem.<br />I encourage my students to use an organized pictorial representation to clarify information.tatumtnoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-75904766329619792932010-07-10T20:21:01.395-05:002010-07-10T20:21:01.395-05:00Einstein Level (p.245, #1):
The train travels 3.75...Einstein Level (p.245, #1):<br />The train travels 3.75 miles (75 x 0.05)<br />The car travels 3 miles (60 x 0.05)<br />The length of the train is the difference<br />(0.75 miles) (3,960 feet)tatumtnoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-29705407384657701922010-07-09T22:12:30.785-05:002010-07-09T22:12:30.785-05:00I worked on problem #3 on page 245. This problem ...I worked on problem #3 on page 245. This problem involved two people driving towards a same destination, but at different speeds, and leaving at different times. I made a chart showing both persons and the speed they were driving as well as the time they were leaving. Whenever I can, I like to make a table to show information. It helps me visualize it better. I then multiplied each person's speed as each hour passed until the secocnd person got caught up to the first person. It took 3 hours for both drivers to be caught up to each other.PKassirnoreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-50309769094684241572010-07-08T15:05:40.101-05:002010-07-08T15:05:40.101-05:00For the Einstein Level I choose Chapter 15 Distanc...For the Einstein Level I choose Chapter 15 Distance = Speed x Time <br />I do not usually teach the drawing blocks of time and adding them together then dividing by how many there are <br />Here is problem #4 on page 245<br />Going up the river you will have 3 blocks of 40 = 120 and down the river you will have 2 blocks of 60 =120 <br />Adding your block 420 divided by how many blocks 5. <br />Leads you to 420 / 5 is 48. The average speed is 48 mph.MWhttps://www.blogger.com/profile/18132644486954256919noreply@blogger.comtag:blogger.com,1999:blog-1811268622721218650.post-63965875368101750522010-07-06T20:34:35.277-05:002010-07-06T20:34:35.277-05:00I completed Einstein problem #7 on page 188. Usi...I completed Einstein problem #7 on page 188. Using the information given, you can draw a right triangle and label the information that has been provided. Side A, which represents the height of the cliff, has a value of 990 ft. (given). You are told that the ramp leading to the top of the cliff needs to have an angle of 10 degrees (given). You are asked to find how far from the cliff should the ramp begin to be built. Given the information provided, you can find the tangent for the ten degree angle which is .1763. Then to find the needed distance (side b), you need to use the ratio of opposite side/adjacent side OR (side A/side B) = (tangent/1). So (Side A = 990 ft./ side B) = (tangent= .1763/1). When you cross multiply, you'll end up with the equation of .1763b = 990 ft. Your next step is to isolate your side B (variable) so that you then have B = (990/.1763). This equation gives Side B the value of 5,615.42 ft. Round the answer (as you were told). The ramp should be built 5,615 ft away from the cliff.nlopeznoreply@blogger.com