tag:blogger.com,1999:blog-63583902967465802862017-03-19T11:03:55.960-05:00Pre-Cal 40S (Fall '06)An interactive learning ecology for students and parents in my Pre-Cal 40S class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.Darren Kuropatwahttps://plus.google.com/116565501620976083429noreply@blogger.comBlogger241125tag:blogger.com,1999:blog-6358390296746580286.post-4529172600256709452007-02-05T14:21:00.001-06:002007-02-05T14:21:28.600-06:00The Adventure Continues ...Our adventures in blogging continue....<br /><br />Watch for new blogs going live February 5, 2007 ...<br /><br /><blockquote><ul><li><a href="http://am40sw07.blogspot.com">Applied Math 40S</a> (Winter '07) (Grade 12)</li><br /><li><a href="http://pc40sw07.blogspot.com">Pre-Cal 40S</a> (Winter '07) (Grade 12)</li><br /><li><a href="http://apcalc06.blogspot.com">AP Calculus AB</a> continues ... (Grade 12)</li></ul></blockquote><p>Darren Kuropatwahttps://plus.google.com/116565501620976083429noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-43900666760488881832007-02-05T14:18:00.001-06:002007-02-05T14:18:41.583-06:00So Long ...<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://farm1.static.flickr.com/36/79811964_33754e8586_d.jpg"><img style="float:right; margin:0 0 10px 10px;cursor:pointer; cursor:hand;width: 320px;" src="http://farm1.static.flickr.com/36/79811964_33754e8586_d.jpg" border="0" alt="" /></a><br />And so we begin where we left off ... don't let the sky be your limit. ;-)<br /><br /><b><i>I'm so glad we've had this time together,<br /><br />Just to have a laugh or learn some math,<br /><br />Seems we've just got started and before you know it,<br /><br />Comes the time we have to say, "So Long!"</b></i><br /><br />So long everybody! Watch this space for pointers to new blogs for each of my classes. <br /><br />Farewell, Auf Wiedersehen, Adieu, and all those good bye things. ;-)Darren Kuropatwahttps://plus.google.com/116565501620976083429noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-90587689993729861972007-01-29T23:42:00.000-06:002007-01-30T00:05:32.999-06:00Podcapsule<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://farm1.static.flickr.com/34/123420044_107eab54fd_m_d.jpg"><img style="float:right; margin:0 0 10px 10px;cursor:pointer; cursor:hand;width: 200px;" src="http://farm1.static.flickr.com/34/123420044_107eab54fd_m_d.jpg" border="0" alt="" /></a><br />Last Thursday morning the students in this class wrote their final exam. That afternoon we wrapped up our class and recorded this podcast. It's called a podcapsule because, like a time capsule, it'll stay here as a permanent record of the students reflections and recommendations to themselves on how to guarantee their success in their next math class.<br /><br />Although the class is over and the podcast is done you can keep it going as long as you like ...<br /><br />First <a href="http://www.archive.org/download/DarrenKuropatwaPodcapsulePre-Cal40SFall2006/PodcapsulePC40sw06_64kb.mp3" target="_blank">listen to the podcast</a> (6.3 Mb, 13 minutes 11 seconds), then, if you like, you can keep it alive by <a href="http://www.vaestro.com/viewthread-119-130-196" target="_blank">adding an audio comment</a> or text comment on this post.Darren Kuropatwahttps://plus.google.com/116565501620976083429noreply@blogger.com1tag:blogger.com,1999:blog-6358390296746580286.post-13235637518410273022007-01-23T16:55:00.000-06:002008-12-09T12:39:30.815-06:00Exam Rehearsal for Identities and Counting<div>Hey people! It's JessiccaI_198 as scribe today. Today we had two exam rehearsals; one for identities in the morning and the other one was for counting in the afternoon. We talked a bit about the arc sine and about functions, but I didn't really get many notes on that. But here are the questions and solutions for the Exam Rehearsal Identities:<br /><div><br /><div><br /><br /><br /></div><br /><div><span style="color:#000099;">1) The expression (sinx + cosx)^2 - 1 is equivalent to:</span></div><br /><div><br /></div><br /><div><span style="color:#000099;">a) sin2x</span></div><br /><div><span style="color:#000099;">b) cos2x</span></div><br /><div><span style="color:#000099;">c) tan^2x</span></div><br /><div><span style="color:#000099;">d) 0</span></div><br /><div><span style="color:#000099;"></span></div><br /><div><span style="color:#ff0000;">The answer is a) sin2x.</span></div><br /><div><span style="color:#ff0000;"></span></div><br /><div><span style="color:#33cc00;"><strong>Solution:</strong></span></div><br /><div><span style="color:#33cc00;"></span></div><br /><div><span style="color:#33cc00;">sin^2x + 2sinxcosx + cos^2x - 1</span></div><br /><div><span style="color:#33cc00;">2sinxcosx</span></div><br /><div><span style="color:#33cc00;"><strong>sin2x</strong></span></div><br /><div><br /><br /></div><br /><div><span style="color:#000099;">2) If sin<em>A = </em>3/5 and sin<em>B </em>= 2/3, and <em>A </em>and<em> B</em> are acute angles, what is the value of cos(<em>A</em> - <em>B</em>)?</span></div><br /><div><br /></div><br /><div></div><a href="http://1.bp.blogspot.com/_6LbXINbIjeE/RbajE-_BbfI/AAAAAAAAAAc/kE_Sx0QZ3M0/s1600-h/wicked.JPG"></a><br /><div><span style="color:#000099;">a) -2/3</span></div><br /><div><span style="color:#000099;">b) (4√5 - 6)/15</span></div><br /><div><span style="color:#000099;">c) (4√5 + 2)/5</span></div><br /><div><span style="color:#000099;">d) (4√5 + 6)/15</span></div></div><br /><div></div><br /><div><span style="color:#ff0000;">The answer is <strong>B</strong></span></div><br /><div></div><br /><div><span style="color:#009900;"><strong>Solution:</strong></span></div><br /><div><br /></div><img id="BLOGGER_PHOTO_ID_5023384364625522194" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://4.bp.blogspot.com/_6LbXINbIjeE/Rbaldu_BbhI/AAAAAAAAAAs/Nz59lnro4w4/s320/wicked.JPG" border="0" /><br /><div><span style="color:#33cc00;">cos(A - B) = cosAcosB + sinAsinB</span></div><br /><div><span style="color:#33cc00;">= (4/5)(√5/3) + (4/5)(√5/3)</span></div><br /><div><span style="color:#33cc00;"><strong>= (4√5 +6)/15</strong></span></div><br /><div><span style="color:#000099;"></span></div><br /><div><span style="color:#000099;"></span></div><br /><div><span style="color:#000099;">3) Find all values of <em>x</em> in the interval 0 <u><</u> <em>x </em>< cos2x =" -3sinx"></div><br /><div><span style="color:#000099;"></span></div><br /><div><span style="color:#ff0000;">The answer is x = 7π/6, 11π/6</span></div><br /><div><span style="color:#ff0000;"></span></div><br /><div><strong><span style="color:#33cc00;">Solution:</span></strong></div><br /><div><strong><span style="color:#33cc00;"></span></strong></div><br /><div><span style="color:#33cc00;">cos2x = -3sinx - 1</span></div><br /><div><span style="color:#33cc00;">1 - 2sin^2x = -3sinx - 1</span></div><br /><div><span style="color:#33cc00;">0 = 2sin^2x - 3sinx - 1</span></div><br /><div><span style="color:#33cc00;">= (2sinx + 1) (sinx - 2)</span></div><br /><div><span style="color:#33cc00;">sinx = -1/2 sinx = 2</span></div><br /><div><span style="color:#33cc00;"></span></div><br /><div><span style="color:#33cc00;">2 is rejected because in the unit circle, no value can be bigger than 1. </span></div><br /><div><span style="color:#33cc00;"></span></div><br /><div><span style="color:#33cc00;"><strong>x = 7π/6, 11π/6</strong></span></div><br /><div><span style="color:#33cc00;"></span></div><br /><div></div></div><br /><div><span style="color:#000099;">4) If sinΘ = √5/3, what is the value of cos2Θ?</span></div><br /><div><span style="color:#000099;"></span></div><br /><div><span style="color:#ff0000;">The answer is -1/9</span></div><br /><div><span style="color:#000099;"></span></div><br /><div><span style="color:#33cc00;"><strong>Solution:</strong></span></div><br /><div><span style="color:#000099;"></span></div><br /><div><span style="color:#33cc00;">cos2Θ = 1 - 2sin^2Θ</span></div><br /><div><span style="color:#33cc00;">= 1 - 2(5/9)</span></div><br /><div><span style="color:#33cc00;">= 1 - 10/9</span></div><br /><div><span style="color:#33cc00;">= 9/9 - 10/9</span></div><br /><div><span style="color:#33cc00;"><strong>= -1/9</strong></span></div><br /><div><br /><br /></div><br /><div><span style="color:#000099;">5) Prove the identity: sinΘ/(1-cosΘ) + sinΘ/(1+cosΘ) = 2cscΘ</span></div><br /><div><img id="BLOGGER_PHOTO_ID_5023405723497885218" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://1.bp.blogspot.com/_6LbXINbIjeE/Rba44-_BbiI/AAAAAAAAAA0/imsteTqwhWI/s320/identity.JPG" border="0" /><br /><br /><br /><br /></div><br /><div><strong><span style="color:#000000;">Here are the problems from this afternoons class on counting:</span></strong></div><br /><div><br /><br /><br /><br /></div><br /><div><span style="color:#000099;">1) The number of different arrangements of 3 boys and 4 girls in a row, if the girls <strong>must</strong> stand together, is represented by:</span></div><br /><div></div><br /><div><span style="color:#000099;">a) 4!x4!</span></div><br /><div><span style="color:#000099;">b) 3!x4!</span></div><br /><div><span style="color:#000099;">c) 4!x4!x2!</span></div><br /><div><span style="color:#000099;">d) 3!x4!x2!</span></div><br /><div><span style="color:#ff0000;">The answer is <strong>a.</strong></span></div><strong></strong><br /><strong></strong><br /><br /><br /><br /><br /><br /><span style="color:#000099;">2) The students in a music department have practised 6 contemporary and 5 traditional choruses. For their concert, they will choose a program in which they present 4 of the contemporary and 3 of the traditional choruses. How many different programs can be presented, if the order of the choruses does <strong>not</strong> matter?</span><br /><br /><br /><span style="color:#000099;"></span><br /><span style="color:#000099;">a) 25</span><br /><br /><br /><span style="color:#000099;">b) 35</span><br /><br /><br /><span style="color:#000099;">c) 150</span><br /><br /><br /><span style="color:#000099;">d) 330</span><br /><br /><span style="color:#ff0000;">The answer is <strong>c.</strong></span><br /><p><strong><span style="color:#33cc00;">Solution:</span></strong></p><p><span style="color:#33cc00;">6C4 . 5C3</span></p><p><span style="color:#000099;"><span style="color:#33cc00;">= <strong>150</strong></span><br /></span></p><br /><br /><br /><br /><br /><span style="color:#000099;">3) All telephone numbers are preceded by a 3-digit area code. In the original Bell Telephone System of assigning area codes, the first digit could be any number from 2 to 9, the second digit was either 0 or 1, and the third digit could be any number except 0. In this system, the number of different are codes possible was:</span><br /><span style="color:#000099;"></span><br /><span style="color:#ff0000;">The answer is 144 different ways.</span><br /><span style="color:#000099;"></span><br /><strong><span style="color:#33cc00;">Solution:</span></strong><br /><strong><span style="color:#33cc00;"></span></strong><br /><span style="color:#33cc00;">8C1 . 2C1 . 9C1 = 144 different ways</span><br /><span style="color:#33cc00;"></span><br /><span style="color:#33cc00;">8C1 - there are eight numbers in the first digit - 2, 3, 4, 5, 6, 7, 8, and 9.</span><br /><span style="color:#33cc00;"></span><br /><span style="color:#33cc00;">2C1 - two number in the second digit - 0 and 1</span><br /><span style="color:#33cc00;"></span><br /><span style="color:#33cc00;">9C1 - all the numbers, excluding 0</span><br /><br /><br /><br /><br /><br /><a href="http://4.bp.blogspot.com/_6LbXINbIjeE/Rbajfu_BbgI/AAAAAAAAAAk/tasify5VIRU/s1600-h/sheet+two+problem+four.JPG"><img id="BLOGGER_PHOTO_ID_5023382199962004994" style="FLOAT: right; MARGIN: 0px 0px 10px 10px; CURSOR: hand" alt="" src="http://4.bp.blogspot.com/_6LbXINbIjeE/Rbajfu_BbgI/AAAAAAAAAAk/tasify5VIRU/s320/sheet+two+problem+four.JPG" border="0" /></a><br /><br /><span style="color:#000099;">4) A paperboy who delivers papers on his bike can travel only on the trails represented in the diagram. The number of different trails that the paperboy can take to get from house <strong>A</strong> to house <strong>B</strong> without backtracking is:</span><br /><br /><br /><br /><br /><br /><br /><span style="color:#33cc00;"><strong>Solution</strong>:</span><br /><a href="http://1.bp.blogspot.com/_6LbXINbIjeE/RbajE-_BbeI/AAAAAAAAAAU/ROuHF6IIx_k/s1600-h/sheet+two+solution+four.JPG"><img id="BLOGGER_PHOTO_ID_5023381740400504290" style="CURSOR: hand" alt="" src="http://1.bp.blogspot.com/_6LbXINbIjeE/RbajE-_BbeI/AAAAAAAAAAU/ROuHF6IIx_k/s320/sheet+two+solution+four.JPG" border="0" /></a><span style="color:#009900;">There are <u>60</u> different trails</span><br /><br /><br /><br /><br /><br /><br /><span style="color:#000099;">5. a) How many groups of 3 chairs can be chosen from 7 chairs if the chairs are all different colours?</span><br /><br /><br /><br /><span style="color:#000099;">b) How many different ways can 7 chairs be arranges in a row if 2 of the chair are blue, 3 are yellow, 1 is red and 1 is green? (Assume that all of the chairs are identical except for colour.)</span><br /><br /><br /><br /><span style="color:#000099;">c) How many different ways can the chairs in (b) above be arranged in a circle?</span><br /><span style="color:#000099;"></span><br /><span style="color:#ff0000;">5. Answers:</span><br /><span style="color:#ff0000;">a) 35</span><br /><span style="color:#ff0000;">b) 420</span><br /><span style="color:#ff0000;">c) 60</span><br /><span style="color:#000099;"></span><br /><strong><span style="color:#33cc00;">Solution:</span></strong><br /><span style="color:#33cc00;">5a) 7C3 = 35</span><br /><span style="color:#33cc00;"> - since order doesn't matter. 7 is for the total number of chairs, and 3 is for the number of chairs to be chosen.</span><br /><span style="color:#33cc00;"></span><br /><span style="color:#33cc00;">b) 7! / 3!2! = 420</span><br /><span style="color:#33cc00;"> - 7! represents the total number of arrangements, and 3! 2! represent the non-distinguishable objects.</span><br /><span style="color:#33cc00;"></span><br /><span style="color:#33cc00;">c) 6! / 3!2! = 60</span><br /><span style="color:#33cc00;">- one of the chairs is a reference point, and the rest are seated in a circle. 3!2! are the non-distinguishable objects.</span><br /><span style="color:#33cc00;"></span>10101924http://www.blogger.com/profile/09798414964359359662noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-6860827933775245312007-01-23T12:53:00.000-06:002008-12-09T12:39:30.962-06:00Autograph that provincial exam with EXCELLENCE!!<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_HiqPzpd3O9Y/RbZZyu9lVyI/AAAAAAAAAAM/7i9UVrobIB8/s1600-h/excellence.jpg"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://4.bp.blogspot.com/_HiqPzpd3O9Y/RbZZyu9lVyI/AAAAAAAAAAM/7i9UVrobIB8/s320/excellence.jpg" border="0" alt=""id="BLOGGER_PHOTO_ID_5023301162513028898" /></a>Lani Ritter Hallhttp://www.blogger.com/profile/16352862711544966770noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-26905388200144334252007-01-22T13:51:00.000-06:002008-12-09T12:39:32.080-06:00scribe post: Exam Rehearsal ProbabilityHello guys!!!!!<br />Our topics in mathematics just ended. For us it is a relief but we must always put in mind that learning in a never ending process so there are still maths stuffs out there waiting for us. Anyway, this morning we did a rehearsal examination in probability. Mr. K gave us 30 minutes to work on it then after that we had a group discussion where we talked about our answers and did some brainstorming. Afterwards, Mr. K discussed with to us the right answers with their respective solutions. The questions given are:<br /><br />(1) A certain soccer player has scored on 82% of his penalty kicks throughout his career. Given this information, the probabality that he will score on exactly 4 of his next 5 penalty kicks, correct to the nearest hundredth, is<br /><br />(a) 0.80<br />(b) 0.66<br />(c) 0.41<br />(d) 0.08<br /><br /><span style="color:#006600;">The answer is </span><span style="color:#ff0000;">(c) </span><span style="color:#ff0000;">0.41 </span><span style="color:#006600;">because the solution for this problem is </span><br /><br /><br /><br /><a href="http://3.bp.blogspot.com/_onMdYB7Msek/RbWPU7cUHhI/AAAAAAAAAAM/V5Y5v_5VNZg/s1600-h/answer+number+one(scribe).bmp"><img id="BLOGGER_PHOTO_ID_5023078549118262802" style="CURSOR: hand" alt="" src="http://3.bp.blogspot.com/_onMdYB7Msek/RbWPU7cUHhI/AAAAAAAAAAM/V5Y5v_5VNZg/s400/answer+number+one(scribe).bmp" border="0" /></a><br />(2) If P(A) = 3/4 and P(A and B)= 1/2, where A and B are dependant events, then P(BA) equals:<br /><br />(a) 1/4<br />(b) 3/8<br />(c) 2/3<br />(d) 5/4<br /><br /><span style="color:#006600;">The answer is </span><span style="color:#ff0000;">(c) 2/3</span><span style="color:#006600;">. The problem indicated that A and B occured in dependant events, which means that both have to happen. </span><br /><br /><br /><p><span style="color:#ff0000;">Solution: </span><br /></p><br /><br /><p><span style="color:#ff0000;">P(A) * P(B) = 1/2</span><br /></p><br /><br /><p><span style="color:#ff0000;">3/4 * P(B) = 1/2</span><br /></p><br /><br /><p><span style="color:#ff0000;">P(B) = 1/2 * 4/3</span><br /></p><br /><br /><p><span style="color:#ff0000;">P(B) = 2/3</span></p><p><span style="color:#ff0000;"></span><br />(3) Peter places 5 equal-sized tiles in a cloth bag. Each tile has a letter on it. The letters are P, E, T, E and R. The probability that Peter selects the 5 tiles, one at a time, in order such that they spell <strong>PETER, </strong>correct to the nearest hundredth, is<br /></p><br /><br /><br /><p><span style="color:#006600;">The answer is </span><span style="color:#ff0000;">0.02.</span><br />><a href="http://3.bp.blogspot.com/_onMdYB7Msek/RbWXD7cUHiI/AAAAAAAAAAY/Aas7fwi7Wu4/s1600-h/number+three.bmp"><img id="BLOGGER_PHOTO_ID_5023087053153508898" style="CURSOR: hand" alt="" src="http://3.bp.blogspot.com/_onMdYB7Msek/RbWXD7cUHiI/AAAAAAAAAAY/Aas7fwi7Wu4/s400/number+three.bmp" border="0" /></a><br /><br /><a href="http://4.bp.blogspot.com/_onMdYB7Msek/RbWXOLcUHjI/AAAAAAAAAAg/7_xnVpon4VY/s1600-h/number+three+(2).bmp"><img id="BLOGGER_PHOTO_ID_5023087229247168050" style="CURSOR: hand" alt="" src="http://4.bp.blogspot.com/_onMdYB7Msek/RbWXOLcUHjI/AAAAAAAAAAg/7_xnVpon4VY/s400/number+three+(2).bmp" border="0" /></a><br />(4) A child 2 quarters, 2 dimes, and 3 nickels in his pocket, but he does not understand the value of any of the coins. He puts 35cents worth of candy on the counter at a store and randomly selects two coins from his pocket. The probability that the two coins he selects will have a total at least as high as the value of the candy is:<br /><br /><a href="http://3.bp.blogspot.com/_onMdYB7Msek/RbWbh7cUHkI/AAAAAAAAAAw/RcNsCImU_yg/s1600-h/number+four.bmp"><img id="BLOGGER_PHOTO_ID_5023091966596095554" style="CURSOR: hand" alt="" src="http://3.bp.blogspot.com/_onMdYB7Msek/RbWbh7cUHkI/AAAAAAAAAAw/RcNsCImU_yg/s400/number+four.bmp" border="0" /></a><br /><br /><br />(5) It is known that 53% of graduating are boys. Three are chosen at random. Given that at least two of the three grads are boys, determine the probability that all three of the grads are boys. ( Answer accurate to at least 4 decimal places. )</p><a href="http://2.bp.blogspot.com/_onMdYB7Msek/RbWe4rcUHlI/AAAAAAAAAA8/-qyWB1vxc_8/s1600-h/number+five.bmp"><img id="BLOGGER_PHOTO_ID_5023095655973002834" style="CURSOR: hand" alt="" src="http://2.bp.blogspot.com/_onMdYB7Msek/RbWe4rcUHlI/AAAAAAAAAA8/-qyWB1vxc_8/s400/number+five.bmp" border="0" /></a><br /><br /><br />Well that's all folks.....<br />The next scribe is JESSICA!!!!!!!!<br />I already told her that she is next scribe...<br />Anyway guys, good luck to all...<br />May the good news be ours after the provincial exam....<br />God bless us all!!!!!!!!!<br /><br /><br /><br /><p></p>Dr. Jhttp://www.blogger.com/profile/06698467752880072410noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-42977691280913746102007-01-22T10:56:00.001-06:002008-12-09T12:39:32.089-06:00Sequences & Series NotesHere are the notes to add to your math dictionaries for this last unit on Sequences and Series. The <b>notes must be copied into your dictionaries by hand</b>.<br /><br />Here they are: <a href="http://1.bp.blogspot.com/_tlF3qs3v9cw/RbTrpG8CCPI/AAAAAAAAADc/JyEtBmoZJis/s1600-h/Sequences+and+Series+Notes1.jpg">Page 1</a> of 3, <a href="http://1.bp.blogspot.com/_tlF3qs3v9cw/RbTsBG8CCQI/AAAAAAAAADk/0wLDKWykUnk/s1600-h/Sequences+and+Series+Notes2.jpg">Page 2</a> of 3, <a href="http://3.bp.blogspot.com/_tlF3qs3v9cw/RbTsIm8CCRI/AAAAAAAAADs/cPBJPSlOFsc/s1600-h/Sequences+and+Series+Notes3.jpg">Page 3</a> of 3.Darren Kuropatwahttps://plus.google.com/116565501620976083429noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-43279886770921702302007-01-20T14:56:00.000-06:002008-12-09T12:39:32.340-06:00Geometric Sequences<div> </div><div> </div><div><a href="http://3.bp.blogspot.com/_kuVonSRlxSU/RbKPB-Bb46I/AAAAAAAAAAc/MV1oFMfzLbE/s1600-h/11806.jpg"><img id="BLOGGER_PHOTO_ID_5022233798463185826" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://3.bp.blogspot.com/_kuVonSRlxSU/RbKPB-Bb46I/AAAAAAAAAAc/MV1oFMfzLbE/s320/11806.jpg" border="0" /></a> sorry e, i didn't notice it was doing that..but here you go just click these links.<br /><br /><a href="http://i57.photobucket.com/albums/g215/TRILLEST06/11807.jpg">http://i57.photobucket.com/albums/g215/TRILLEST06/11807.jpg</a></div><div> </div><div><a href="http://i57.photobucket.com/albums/g215/TRILLEST06/11808.jpg">http://i57.photobucket.com/albums/g215/TRILLEST06/11808.jpg</a><br /><div></div><br /><br /><p></p><br /><br /><p>the next scribe is Daphne.</p></div>jhoannhttp://www.blogger.com/profile/11310397728134454707noreply@blogger.com3tag:blogger.com,1999:blog-6358390296746580286.post-548874891637989172007-01-18T21:54:00.000-06:002007-01-21T20:40:09.629-06:00Jefferson's Scribe postthe next scribe is <span style="color:#993399;">JHO-ANN.</span><br /><br />a) what is the sum of the integers from 1 to 5000?<br /><br /><span style="color:#ff0000;">T<sub>1</sub> = 1<br />D = 1<br />N = 5000</span><br /><br /><span style="color:#33cc00;">‘the initial term is 1<br />‘the to from getting from 1 to 5000 is 1. on account that each term increases by one. And that it doesn’t say what the difference really is so its 1<br />‘n is the rank of the rank of the term we want to find. The rank is the sum in this case and we want to find what the sum is when the sequence reaches 5000.<br /></span><br /><span style="color:#ff0000;">S<sub>n</sub> = n/2[2(t1) + (n – 1)d]<br />S<sub>5000</sub> = (5000/2)[2(1) + (5000 – 1)1]<br />S<sub>5000</sub> = 2500[2 + 4999]<br />S<sub>5000</sub> = 2500 = [5001]<br />S<sub>5000</sub> = 12 502 500</span><br /><br />b) what is the sum of all multiple of 7 between 1 & 5000?<br /><br /><span style="color:#33cc00;">T<sub>1</sub> = 7<br />D = 7<br />N = 714<br /></span><br /><span style="color:#33cc00;">Since it says multiples of 7,the initial term is 7 because you start with 7 when it comes to “multiples of 7”. The difference is 7 because it increases by 7 each time.<br />Now we want to find the sum of sequence when it reaches to 5000. but we don’t really want to find 5000. since we’re looking for the sum of the multiples 7 between 7 & 5000 then there really aren’t 5000 terms if we’re increasing by 7 each time. To find what the rank is, we need to 5000/7. and you get 714. whatever. Don’t need to worry about the decimal answers if I were you.<br /></span><br /><span style="color:#ff0000;">S<sub>n</sub> = n/2[2(t1) + (n – 1)d]<br />S<sub>714</sub> = (714/2)[2(7) + (714 – 1)7]<br />S<sub>714</sub> = 357[(14 + 7130)]<br />S<sub>714</sub> = 1 784 785</span><br /><br />c) What is the sum of all integers from 1 to 5000 inclusive that are not multiple of 7?<br /><br /><span style="color:#ff0000;">12 502 500 – 1 786 785<br />= 10 715 715</span><br /><br /><span style="color:#33cc00;">This is a no brainer but I’ll explain it to you anyways. To get the answer you take the sum of all integers from 1 to 5000 and subtracted by the sum of the multiples of 7.</span><br /><br />A super ball is dropped from a height of 200 cm. It rebounds to ¾ of the distance it fell each time it hits the ground. What is the total vertical distance traveled by the ball when it hits the ground for the fourth time?<br /><br /><span style="color:#33cc00;">We know that each time it hits the ground,the ball comes back ¾ of the last distance it traveled. So it traveled down, up, down, up, down, up, and down. So in order to calculate the total distance it traveled. We need to brake this into 2 solutions first. We need to solve the “down” distance and then the “up” distance and then add the together.<br /></span><br />Down:<br /><br /><span style="color:#33cc00;">So what are the sequences?<br /><br />Well since the first term is 200 and it decreases every time by a ¼( bounces back to ¾ of the last term) and that it hits the ground for times…..</span><br /><br /><span style="color:#ff0000;">200 + 150 + 112.5 + 84.375<br /></span><br /><span style="color:#3333ff;">(but we really didn’t need to know that only the first term)</span><br /><br /><span style="color:#3333ff;">*DID YOU KNOW: that all the things we need to know to solve the question was in the question itself? Go check it out ; - )</span><br /><br /><span style="color:#ff0000;">T<sub>1</sub> = 200<br />R = ¾<br />N = 4<br /><br />S<sub>4</sub> = 200(1 – (3/4)<sup>4</sup>)/(1 – ¾)<br />S<sub>4</sub> = 546.875</span><span style="color:#3333ff;"><br /><br />(thanks to the help of our righteous hero- the CALCULATOR!!)</span><br /><br /><span style="color:#33cc00;">The 3 term for the “down” distance is:<br />‘think everything opposite of what I said from the other part</span><br /><br /><span style="color:#ff0000;">150 + 112.5 + 84.375<br /><br />T<sub>1</sub> = 150<br />R = ¾<br />N = 3<br /><br />S<sub>3</sub> = 150( 1 – (3/4)<sup>3</sup>)/ (1 - ¾)<br />S<sub>3</sub> = 346.875<br /></span><br />Now we add the two answer to get …….<br /><br /><span style="color:#ff0000;">546.875 + 346.875 = 893.75</span> <span style="color:#3333ff;">(anybody else noticed that the is a difference of 200 for the 2 answers. Me thinks there is a shortcut for this problem….) </span><br /><br /><span style="color:#993399;">……………TAAADAAAAA!!</span><br /><br />I've voice recorded mr k's lecture today. It covered about half the class. Now i have listened to it a bit and you have to crank up the volume super high to listen on account that the mic was far from mr k. <span style="color:#ff0000;">I just wanted to test out if it worked or not.</span> <span style="color:#ff0000;">Feel free to add some comments about it</span>. Apologies to the loud scratching sounds on account i was moving the mp3 play on the desk, i assumed that it picked up the sound.<br /><br /><br /><a href="http://www.mediamax.com/jeff_boy50/Hosted/070118_01.wav">test voice lecture</a>Jeffersonhttp://www.blogger.com/profile/14519676379585208097noreply@blogger.com3tag:blogger.com,1999:blog-6358390296746580286.post-16972195285720789102007-01-17T20:03:00.000-06:002008-12-09T12:39:34.207-06:00Sum of Arithmetic / Geometric SequenceThis is <strong>RICHARD</strong> and I'm your scribe for today.<br /><br /><br />Before we started our lesson in Geometric sequences, Mr. K discussed the solutions on the on-line quiz on probability.<br /><br /><br /><strong><span style="color:#000099;">(1) A card is drawn randomly from a deck of 52 playing cards. What is the probability that it will be a spade or a face card?</span></strong><br /><br /><br /><br /><br /><img id="BLOGGER_PHOTO_ID_5021201098458819730" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 415px; CURSOR: hand; HEIGHT: 471px; TEXT-ALIGN: center" height="320" alt="" src="http://4.bp.blogspot.com/_0Ra_rrpD4Bg/Ra7jy6twbJI/AAAAAAAAAAM/JTA2ob-rvFA/s320/4scribe1.bmp" width="358" border="0" /><br /><br /><br /><strong><span style="color:#000099;">(3) A certain partial deck of cards contains 6 red cards and 4 black cards. Two cards are chosen at random. What is the probability that both cards are black?</span></strong><br /><br /><br /><span style="color:#000000;">R = 6</span><br /><br />B = 4<br /><br /><br /><br /><br /><span style="font-size:85%;">4</span><strong><span style="font-size:130%;">C</span></strong><span style="font-size:85%;">2</span><br />____ = 6/45<br /><br /><span style="font-size:85%;">10</span><strong><span style="font-size:130%;">C</span></strong><span style="font-size:85%;">2</span><br /><br /><br /><br /><strong><span style="color:#3333ff;">(6) The diagram shows a partial map of a certain city park, with walking paths located on the grid lines. A tourist starts at point A and randomly selects a path to point C, walking only to the south and east. What is the probability the tourist passes through point B?</span></strong><br /><br /><span style="font-size:85%;"><img id="BLOGGER_PHOTO_ID_5021205878757420194" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 307px; CURSOR: hand; HEIGHT: 193px; TEXT-ALIGN: center" height="157" alt="" src="http://1.bp.blogspot.com/_0Ra_rrpD4Bg/Ra7oJKtwbKI/AAAAAAAAAAU/OwlQ4SweKko/s320/4scribe2.bmp" width="307" border="0" /></span><br /><br /><br /><p>The number of ways to get to point C is RRRDD which is <span style="color:#ff0000;">5!/3!2!</span><span style="color:#000000;">. The number of ways to get to point B is RRD which is </span><span style="color:#ff0000;">3!/2!</span><span style="color:#000000;">. The number of ways to get to point B over the number of ways to get to point C is <strong>3/10</strong></span></p><br /><br /><p><strong><span style="color:#000099;">(7) It is known that about 5% of coffee cups used in a particular restaurant will be chipped. What is the probability that out of 10 cups selected at random, exactly 2 will be chipped?</span></strong><br /></p><p><span style="color:#000000;">CNCNNNNNNN</span></p><br /><br /><p>C - represents the chipped<br />N- represents not chipped</p><br /><p>10!/8!2! (o.o5)² (0.95)^8<br />= 0.0746</p><br /><p><strong><span style="color:#000099;">(8) Three coins are tossed. What is the probability that at least two heads turn up? </span></strong></p><p><span style="color:#000000;">These are the possibilities that you can get at least two heads</span></p><p>HHT = 1/8</p><p>HTH = 1/8</p><p>THH = 1/8</p><p>HHH = 1/8</p><br /><p>The sum of these events is <strong><span style="color:#ff0000;">1/2</span></strong></p><br /><p><strong><span style="color:#000099;">(10) A jar contains 5 red and 7 blue marbles. What is the probability of pulling out 2 blue marbles in a row, without replacement? </span></strong></p><br /><br /><br /><br /><br /><p><img id="BLOGGER_PHOTO_ID_5021215301915667634" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 405px; CURSOR: hand; HEIGHT: 340px; TEXT-ALIGN: center" height="268" alt="" src="http://3.bp.blogspot.com/_0Ra_rrpD4Bg/Ra7wtqtwbLI/AAAAAAAAAAc/2dk-s6htQYs/s400/4scirbe3.bmp" width="326" border="0" /></p><br /><br /><br /><br /><br /><p></p><p><strong><span style="color:#000099;">(2) A certain kind of die has four sides, labelled 1 to 4. When the die is rolled, each side has a probability of 1/4 that it will appear on the bottom. If the die is rolled 5 times, what is the probability that the side labelled “3" appears on the bottom in exactly two of the five rolls?</span></strong></p><br /><p><span style="color:#000000;">3N3NN</span></p><br /><p>= 5!/ 2!3! (1/4)² (3/4)³</p><br /><br /><br /><p><strong><span style="color:#000099;">(4) Over the course of his basketball career, Julius succeeded in making a free throw in 75% of his attempts. In a randomly selected game, he attempted to make a free throw 12 times. What is the probability that he succeeded in 8 of the 12 attempts? (You don't need to know anything about basketball to answer this problem.) </span></strong></p><p><span style="color:#000000;"><span style="font-size:85%;">12</span>C<span style="font-size:78%;">8</span> (0.75)^8 (0.25)^4</span></p><br /><p><strong><span style="color:#000099;">(5) Three different names are randomly selected from the following list of five names. Max, Kim, Codie, Lee and, Alex. Determine the probability that “Kim” is one of the three names selected.</span></strong></p><br /><p><span style="color:#000000;">(1· <span style="font-size:85%;">4</span>C<span style="font-size:85%;">2</span>)/ <span style="font-size:85%;">5</span>C<span style="font-size:85%;">3</span></span><br /></p><br /><p><span style="font-size:85%;">= </span><span style="font-size:100%;">3/10</span></p><br /><p><strong><span style="color:#000099;">(9) Assuming equally likely probabilities for male and female births, what is the probability that a four-child family will have at least one boy? </span></strong></p><br /><p><span style="color:#000000;">P(all girls) = (1/2)^4 = 1/16</span><br /></p><p>P( at least 1 boy) = 1 - 1/16 = 15/16<br /></p><br /><p></p><br /><p>After we had discussed about the on-line quiz, we proceed to our topic in Geometric sequences.</p><br /><p>A mathematician named Carl Friedrich Gauss is a brilliant student. One day his teacher got angry because the class is so noisy. She gave an activity to keep them busy. She asked them to add the numbers from 1 to 100. Gauss came out with a better solution. He instantly computed the answer 5050.<img id="BLOGGER_PHOTO_ID_5021241578525584578" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" height="209" alt="" src="http://2.bp.blogspot.com/_0Ra_rrpD4Bg/Ra8InKtwbMI/AAAAAAAAAAk/2UTAZVzbXKM/s400/4scribe4.bmp" width="400" border="0" /></p><br /><br /><br /><p><span style="color:#ff0000;">t1</span>, <span style="color:#ff0000;">t1</span> + <span style="color:#3333ff;">d</span>, <span style="color:#ff0000;">t1</span> + 2<span style="color:#3333ff;">d</span>, <span style="color:#ff0000;">t1</span> + 3<span style="color:#3333ff;">d</span>, . . . . . . <span style="color:#ff0000;">t</span><span style="color:#ff0000;">1</span> +97<span style="color:#3333ff;">d</span> , <span style="color:#ff0000;">t1</span> + 98<span style="color:#3333ff;">d</span>, <span style="color:#ff0000;">t1</span> + 99<span style="color:#3333ff;">d</span></p><br /><br /><br /><p><span style="color:#3333ff;"></span></p><img id="BLOGGER_PHOTO_ID_5021253973801200850" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://4.bp.blogspot.com/_0Ra_rrpD4Bg/Ra8T4qtwbNI/AAAAAAAAAAs/oh4GFqjBLhc/s400/4scribe5.bmp" border="0" /><br /><br /><br /><p>Find the sum of the arithmetic sequence 2, 7, 12 17, . . . . S 51</p><br /><br /><p>S51 = 51/2 [2(2) + (51-1) 5 ]</p><p>= 51/2 [ 4 + 250 ]</p><p>=51/2 (254)</p><br /><p>= 6477</p><img id="BLOGGER_PHOTO_ID_5021262052634684642" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 400px; CURSOR: hand; HEIGHT: 315px; TEXT-ALIGN: center" height="183" alt="" src="http://1.bp.blogspot.com/_0Ra_rrpD4Bg/Ra8bO6twbOI/AAAAAAAAAA0/TriqzsR6fLI/s400/4scribe6.bmp" width="400" border="0" /><br /><br /><p></p><br /><p>given the geometric sequence 16, 8, 4, 2, . . . .Find the sum of this sequence up to th 10th term.</p><br /><p>r = 1/2</p><p>t1 = 16</p><br /><p></p><img id="BLOGGER_PHOTO_ID_5021266416321457410" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://1.bp.blogspot.com/_0Ra_rrpD4Bg/Ra8fM6twbQI/AAAAAAAAABE/XxunAqh9eBk/s400/4scribe7.bmp" border="0" /> that's it for today!! the next scribe will be Jefferson..<br /><p></p>ricardonoreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-81993686306267184582007-01-17T09:54:00.000-06:002007-01-17T09:59:38.317-06:00Ashley's BoBI can't believe I totally forgot about this! Anyways Probability is alright. There are some questions that I find I'm able to do fast and other questions where I really have to think about it and I get stuck because I can't find the compliment. I actually think probablitity is a lot easier than combinations because probability uses combinations so I'm a lot more familiar with it. But overall I understand how to get some of the questions. But questions like the one on the pre-test I still need to work on. I'm just glad I made it on time to do the BoB. 2 minutes before math starts haha.... Do well guys (:Ashleyhttp://www.blogger.com/profile/04061197592082312766noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-35132264545647368502007-01-16T23:46:00.000-06:002007-01-17T19:26:07.412-06:00Richard's BOB on PROBOne thing I can say about probability is that you need to understand the situation to get the right solution...<br />we follow some basic rules such as if it is "<span style="color:#ff0000;">and</span>" it means that we multiply probabilities and if it says "<span style="color:#ff0000;">or</span>", we add probabilities. But we also need to know if it is mutually exlcusive or not, independent or dependent...<br />...but there's only one thing that we shoulld always keep in mind regarding the probabilities, it is a number between <strong>0</strong> and <strong>1</strong> and the sum of the probabilities of the events will always equal to <strong>1</strong>...<br /><br />I hope everyone will do good on the test..Good luck and God Bless!!ricardonoreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-72419879275293230902007-01-16T23:04:00.000-06:002007-01-16T23:08:52.387-06:00Jefferson's blogging on bloggingWEEEEEEELLLLLLLLLLLLL, This sort of hard to explain on my part. In probability it's easy to understand after the teacher explains it to you. But when you apply it, It's all a puzzle, and i hate puzzles! I know how to do probability questions that involve the word "given that", but I'm just a little confused with the independent and dependent and mutually exclusive and not mutually exclusive. I mean what do you do after you find that out? what sort of math is involved if these events are mutually exclusive. I have the hardest time understanding it. They rest of probability is A-OK for me! the tree diagram always helps and it gives you more than you're looking for if you know what I mean.Jeffersonhttp://www.blogger.com/profile/14519676379585208097noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-41620179487762063202007-01-16T22:57:00.000-06:002007-01-16T23:12:42.669-06:00BOB.... how about probability<span style="color: rgb(51, 153, 153);">What was the probability that after taking the day off school to sleep so that maybe i'd get my voice back, that i would remember to do my blogging on blogging?? all i can say is <span style="color: rgb(153, 51, 153);">Thanks</span> to the kids that posted them before now, cause when i went to check the scribe post to see what i missed, BOBs were posted!! Anywho enough about that, how about that probability unit?? I don't know about any of you, but I thought it was a <span style="color: rgb(153, 51, 153);">good unit</span>... I think there were only a couple times that i didn't know what i was doing, and those times were cleared up rather quickly. The other day when we did the pretest my group did so well.... minus a slip up where we were soo close. My <span style="color: rgb(153, 51, 153);">favourite</span> thing about the unit was doing the "what is the probability of A given that you already B?" those ones were no sweat... but the combinatoric ones still trip me up every once in a while... especially when i get distracted :D ... so i hope that i can <span style="color: rgb(153, 51, 153);">keep focused</span> and do well on that test tomorrow... and of course that everyone else does well too.... ok that's all for me tonight cause i'm getting some more sleep... who knows i might even get my whole voice back tomorrow.... till then, like mr. k says <span style="color: rgb(153, 51, 153);">"Learn hard!!"</span></span>Lisahttp://www.blogger.com/profile/01516225790109059190noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-66954775467473746222007-01-16T22:02:00.000-06:002007-01-16T22:08:19.099-06:00jho's blogThis unit had it's pros and cons. In the beggining i was getting confused, but since we were exposed to this unit for quite some time i was able to understand the concept, and therefore i am confident i will do well in this test. Hmmm maybe..haahha can't be too confident..but that pre-test was good, i was able to answer most of the questions without a problem. Hopefully i will do well tomorrow. Man i have been missing a lot of work lately, because of my loss of internet connection, got to earn some marks. GOOD LUCK everybody =Djhoannhttp://www.blogger.com/profile/11310397728134454707noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-40693408563860489592007-01-16T21:43:00.000-06:002007-01-16T21:48:30.470-06:00Tenny's BOBwow this unit has it's up and down. You would have to read twice ,to be able to write down your answer. sometime the answer you think its right might be even wrong.The first couple of part of unit was easy but it gets harder ,and harder. These questions you would have to do alot to be able to be familar with it. So practice,practice. Well thats all i have to say. byetennysonhttp://www.blogger.com/profile/10130195764567198314noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-81611037960323041492007-01-16T21:26:00.000-06:002007-01-16T21:31:39.769-06:00Bloggin on BloggingIn probability well I'm not so good at it and let's just leave it at that. On plenty of questions I end up making plenty of mistakes, whether it's to using the incorrect method or leaving something out when solving the problem. While practice does help improve something, I've had close to a decade of problem solving questions and I doubt another few days is going to get me ready for those questions.Jasonhttp://www.blogger.com/profile/09967057170543650076noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-43983697859768664382007-01-16T20:24:00.000-06:002007-01-16T21:34:05.221-06:00Scribe postHey everyone its Tennyson ,well today we just did some questions but i was unable to finish. i will just put down the one that we did only.The next scriber can carry out the other questions.<br /><br /><span style="color:#3366ff;">1). List the 1st four terms in each sequence</span><br /><span style="color:#3366ff;"></span><br /><span style="color:#3366ff;">a).tn= 3(n-1)</span><br /><span style="color:#3366ff;">b).tn=(n-1)2</span><br /><span style="color:#3366ff;">c).tn=2(power of)<span style="font-size:78%;">n-1</span></span><br /><span style="font-size:78%;"></span><br /><strong><span style="font-size:78%;"></span></strong><br /><span style="color:#009900;"><strong>a).</strong>tn= 3(n-1)</span><br /><span style="color:#ff0000;"></span><br /><br /><br /><ul><li><span style="color:#ff0000;">-tn=3(1-1)</span></li><li><span style="color:#ff0000;">-tn=3(0)</span></li><li><span style="color:#ff0000;">-tn=0</span></li></ul><p><span style="color:#ff0000;"></span></p><ul><li><span style="color:#ff0000;">-tn=3(2-1)</span></li><li><span style="color:#ff0000;">-tn=3(1)</span></li><li><span style="color:#ff0000;">-tn=3</span></li></ul><p><span style="color:#ff0000;"></span></p><ul><li><span style="color:#ff0000;">-tn=3(3-1)</span></li><li><span style="color:#ff0000;">-tn=3(2)</span></li><li><span style="color:#ff0000;">-tn=6</span></li></ul><span style="color:#ff0000;"></span><br /><br /><br /><ul><li><span style="color:#ff0000;">-tn=3(4-1)</span></li><li><span style="color:#ff0000;">-tn=3(3)</span></li><li><span style="color:#ff0000;">-tn=9</span></li></ul><span style="color:#ff0000;"></span><br /><span style="font-size:130%;color:#ff0000;">Answer:0,3,6,9</span><br /><br /><span style="color:#009900;"><strong>b)</strong>.tn=(n-1)2</span><br /><span style="color:#ff0000;"></span><br /><br /><br /><ul><li><span style="color:#ff0000;">-tn=(1-1)2<br />-tn=(0)2</span></li><li><span style="color:#ff0000;">-tn=0</span></li></ul><p><span style="color:#ff0000;"></span></p><ul><li><span style="color:#ff0000;">-tn=(2-1)2</span></li><li><span style="color:#ff0000;">-tn=(1)2<br />-tn=1</span></li></ul><p></p><ul><li><span style="color:#ff0000;">-tn=(3-1)2</span></li><li><span style="color:#ff0000;">-tn=(2)2<br />-tn=4</span></li></ul><p><span style="color:#ff0000;"></span></p><ul><li><span style="color:#ff0000;">-tn=(4-1)2</span></li><li><span style="color:#ff0000;">-tn=(3)2<br />-tn=9</span></li></ul><span style="color:#ff0000;"></span><br /><span style="font-size:130%;color:#ff0000;">Answer: 0,1,4,9</span><br /><span style="color:#ff0000;"></span><br /><span style="color:#009900;"><strong>c)</strong>.tn=2(power of)n-1</span><br /><br /><br /><br /><ul><li><span style="color:#ff0000;">-tn=2(power of)1-1</span></li><li><span style="color:#ff0000;">-tn=2(power of)0</span></li><li><span style="color:#ff0000;">-tn=1</span></li></ul><p><span style="color:#ff0000;"></span></p><ul><li><span style="color:#ff0000;">-tn=2(power of)2-1</span></li><li><span style="color:#ff0000;">-tn=2(power of)1<br />-tn=2</span></li></ul><p><span style="color:#ff0000;"></span></p><ul><li><span style="color:#ff0000;">-tn=2(power of)3-1</span></li><li><span style="color:#ff0000;">-tn=2(power of)2</span></li><li><span style="color:#ff0000;">-tn=4</span></li></ul><p><span style="color:#ff0000;"></span></p><ul><li><span style="color:#ff0000;">-tn=2(power of)4-1</span></li><li><span style="color:#ff0000;">-tn=2(power of)3</span></li><li><span style="color:#ff0000;">-tn=8</span></li></ul><span style="color:#ff0000;"></span><br /><span style="font-size:130%;color:#ff0000;">Answer: 1,2,4,8</span><br /><br />Later on , we talked if the question was a Geometric sequences ,Arithmetic sequences, or neither.<br /><br />a).This question is an <span style="color:#ff0000;">arithmetic sequence</span> because when the number is the same ,you add or subtract from each of the numbers to get the next number in the sequence.<br /><br />b).This question is <span style="color:#ff0000;">neither</span> because They dont involve in multiplying ,divid,adding,or subtract , it just does have the patterns like the other sequences.<br /><br />c). This question is an <span style="color:#ff0000;">geometric sequences</span>, because the common ratio is a number that you can multiply term in order to get the other next term like,Also can be divided by the previous term.<br /><br />Those question are all related to a graph,tables, and many other things<br /><br /><br />next person is ricardo , but the list was confusing so yeah , hope u didnt do 4 scribs<br /><br />by3by3by3<br /><span style="color:#ff0000;"></span><br /><span style="color:#ff0000;"></span><br /><span style="color:#ff0000;"><br /><br /><br /></span><span style="color:#ff0000;"></span><span style="color:#ff0000;"></span>tennysonhttp://www.blogger.com/profile/10130195764567198314noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-84860200439220544142007-01-16T20:11:00.000-06:002007-01-16T20:31:42.466-06:00ruschev Blogging on BloggingI <span class="blsp-spelling-corrected" id="SPELLING_ERROR_0">don't</span> really know where I am right now in the probability unit....T_T.. I get some of the answers sometimes correct and sometimes I get some of the answers wrong and sometimes I know I have the correct answer and got it wrong.....sometimes I have the right answer but get the part where you solve for the correct answer wrong............. I guess I just need a <span class="blsp-spelling-corrected" id="SPELLING_ERROR_1">Little</span> more practice!<br /><br /><br />good luck tomorrow~ruschevhttp://www.blogger.com/profile/17027604349106550491noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-12233454143300619922007-01-16T20:04:00.000-06:002007-01-16T20:16:24.798-06:00Oliver B's blogging on blogging.Well now, I sorta like probability, but the part I didn't or well, I wasn't fond of much was, the part where combinatorix was combiend with probability, only because I don't think I did that good during combinatorix, which I need to review on. I guess it is kinda neat though how they're together like that. Overall I did like this probability unit, I hope I have as much luck tomorrow during the test though! I wish everyone luck on that, okay bye bye.oliver_796http://www.blogger.com/profile/05856731575280461678noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-25414403802315665252007-01-16T19:05:00.000-06:002007-01-16T19:19:24.771-06:00daph's seventh BOBProbability for me is one of the scariest topics in mathematics. I found it hard to solve questions about cards and its decks. Moreover, thinking for the starting solution such as the tree diagram is another burden for me because I cannot really imagine things to be fitted in the diagram. On the contrary, I found it a little challenging. Some of the questions are really tough but sometimes I managed to solve it. Another is I want to acknowledge the effort that Mr. K gave to us in order to understand the probability well because he explained everything in details and still willing to explain it more if some of us was not able to understand it.<br />Honestly saying I'm not doing my assignments in probability. It is the reason why I cannot push myself in to a higher level when it comes to probability.<br />GOOD LUCK!!!!!!!!!!!!!<br />God bless us all in the upcoming test!!!!!!!!Dr. Jhttp://www.blogger.com/profile/06698467752880072410noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-49141503330611814842007-01-16T19:00:00.000-06:002007-01-16T19:10:43.144-06:00JessiccaI_198's BOB for ProbabilityI think that this unit was interesting, and I enjoyed doing it. I like making the trees for the problems which make everything easier and more organized. I just have to remember the methods to be used for certain problems. Using some of the Counting Principles helps a lot in some of the problems, and makes it easier to calculate too, so I'm glad we get to use some of that. Hopefully I won't forget to use some of the rules in the problems, but I'll probably just look at some of the problems we did in class so I would probably know which one I would need to use for the ones that will be on the test. The pre-test was a little hard, but I think I'm prepared for the unit test. Everyone Good Luck!!10101924http://www.blogger.com/profile/09798414964359359662noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-3634340356535240512007-01-16T17:34:00.000-06:002007-01-16T17:42:00.336-06:00BOBThis unit is very interesting and I get most of it. It's just that sometimes when it seems like you have the right answer, you don't..does that make any sense? What I mean by that is that a wrong answer can seem right as to the point where it looks possible so you end up getting the wrong answer. So multiple choice questions for this unit are pretty difficult. I really like the long answers where something is assumed given that an event happens, because they're a breeze. What's really hard to remember is that you always have to have the probability over the sample space and I always forget that. The pre-test was okay so I'm sure the test will be okay as well. <span style="color:#009900;">Good luck everyone.</span>Anhnoreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-22450661481931139332007-01-15T16:17:00.000-06:002007-01-15T16:31:31.489-06:00Probability Quiz<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://farm1.static.flickr.com/34/72203037_f405466d10_m_d.jpg"><img style="float:right; margin:0 0 10px 10px;cursor:pointer; cursor:hand;width: 200px;" src="http://farm1.static.flickr.com/34/72203037_f405466d10_m_d.jpg" border="0" alt="" /></a><br /><a href="http://quizstar.4teachers.org/indexs.jsp">Here</a> is it! You have until midnight tomorrow to complete it. We'll talk about in class on Wednesday morning. The test is Wednesday afternoon.<br /><br /><font size="+1"><b>Learn Hard!</b></font><br /><br />Wouldn't it be nice if the world really worked like this ... ;-)<br /><br /><center><br /><object width="425" height="350"><param name="movie" value="http://www.youtube.com/v/wP_4YWkLNFQ"></param><param name="wmode" value="transparent"></param><embed src="http://www.youtube.com/v/wP_4YWkLNFQ" type="application/x-shockwave-flash" wmode="transparent" width="425" height="350"></embed></object><br /></center>Darren Kuropatwahttps://plus.google.com/116565501620976083429noreply@blogger.com0tag:blogger.com,1999:blog-6358390296746580286.post-44780546284784375312007-01-15T14:40:00.000-06:002007-01-16T00:24:03.260-06:00scribe post; geometric sequences.today we started off with mr. k giving us questions on the board.<br><b>find the next three numbers..</b><br>1, 2, 4, <u>8</u>, <u>16</u>, <u>32</u><br>how you would get the next set of numbers was multiplying the number by two<br><b><u>OR</b></u> 1, 2, 4, <u>7</u>, <u>11</u>, <u>16</u><br>how you would get this one was to find the pattern in the sequence. so you would have to find the difference between the numbers. for example: 2 and 1, the difference is 1. and the difference between 4 and 2 is 2. so you would get the pattern. you know that the number is increasing sequentially. so the next number must be a 7 - since the next number is a 7, the difference between 7 and 4 is three. and so on..<br><br>6, 11, 16, <u>21</u>, <u>26</u>, <u>31</u><br>for this one, the difference between the numbers is 5 and it's always constant.<br><br>3, 6, 12, <u>24</u>, <u>48</u>, <u>96</u><br>this one is basically like the first one, you multiply the numbers by two.<br><br>1, 1, 2, 3, 5, <u>8</u>, <u>13</u>, <u>21</u><br>for this one.. you find a pattern by adding the numbers.. for instance: 1 plus 1 is 2. then 2 plus 3 is 5. and 3 plus 5 is 8. so you get the point..<br><br>1, 3, 6, 10, <u>15</u>, <u>21</u>, <u>28</u><br>this one is basically like the first one.. but the second answer.. but this one doesn't start at 1. it starts at two - meaning.. the difference between 3 and 1 is 2. 6 and 3 is 3. and it's going up sequentially, and so on..<br><br>x, -x<sup>2</sup>, x<sup>3</sup>, -x<sup>4</sup>, <u>x<sup>5</sup></u>, <u>-x<sup>6</sup></u>, <u>x<sup>7</sup></u><br>for this one, you would multiply by -(-x)<br><br>3, 4, 7, 11, <u>18</u>, <u>29</u>, <u>47</u><br>this one, you add the numbers together. 3 and 4 is 7. 4 and 7 is 11. 7 and 11 is 18. etc etc..<br><br>77, 49, 36, 18, <u>8</u>, <u>8</u>, <u>8</u><br>this one was kinda tricky. you would split the numbers - so you would get 7 and 7. you would then multiply them together and you would get 49. 4 times 9 is 36. 3 times 6 is 18, and soo on..<br><br>in the afternoon class, we had a pretest on probability. and here are the questions and solutions to them.<br><b>1.</b> twelve people, including you, are members of a choir. the choir director is going to choose three members to attend a workshop. the probability <i>you</i> and two other members will be chosen is:<br><b>a)</b> 1/4 <b>b)</b> 3/10 <b>c)</b> 1/12 <b>d)</b> 1/10<br><br>first of all, <b>YOU</b> have to go. so you would get 1 x 11C2 because you're part of that twelve so there are eleven people remaining to choose from - and you're going to choose two. and the denominator would be 12C3 since there are twelve people and you have to choose three. so that would look like..<br>1 x 11C2 ALL OVER 12C3<br>and that would all work out to 1/4, which would be a.<br><br><b>2.</b> rex is playing a guessing game. the probability he will guess each question correct is 0.3. what is the probability he will guess <i>exactly</i> 5 out of 10 questions correctly?<br><b>a)</b> 0.15 <b>b)</b> 0.10 <b>c)</b> 0.29 <b>d)</b> 0.80<br><br>so the P(correct)= 0.3 and the P(wrong)= 0.7<br>since there can be different ways of getting the questions wrong or right, but you still have to get <b>five</b> correct. so you would get..<br>10!/5!5!<br>the 5! represents the correct ones and the other 5! represents the wrong ones. you would then get..(C)<sup>5</sup>(W)<sup>5</sup><br>(0.3)<sup>5</sup>(0.7)<sup>5</sup><br>and when you work that all out, you would get <b>0.10</b> and that's the answer b.<br><br><b>3.</b> you choose four digits from the numbers 0-9 (ten digits in total), with no repeated digits. to win a prize, all four of your numbers need to math the four selected by a computer. the probability that you will win a prize is:<br>P(all four digits)= 4C4 ALL OVER 10C4<br>and when you work that all out, you would get 1/210 (and that's the answer)<br><br><b>4.</b> the serial nuber of a $10 bill contains 8 digits. if your $10 bill contains the digit 7 at least once, you will win a prize. what is the probability that your $10 bill will win?<br>first find the probability that you wouldn't get any 7's. so, the probability that you would get a 7 is 1/10 and the probability that you wouldn't get a 7 is 9/10. soo..<br>P(not seven)= (0.9)<sup>8</sup><br>since you now have the probability of not getting any 7's, you can now get the probability of getting at least one 7.<br>P(7 at least once)= 1-(0.9)<sup>8</sup><br>so, you would get <b>0.57</b> (and that's your answer)<br><br><b>5.</b> jeff takes a lunch to school on two days, and on the other three days he buys it. if he takes lunch - he is late for his period 4 class 15% of the time, but if he buys lunch - he is late 35% of the time.<br><b>a)</b> what is the probability that jeff arrives on time?<br><br><b>b(</b> if jeff was late, what is the probability that he took his lunch to school?<br>first make a tree diagram<br><br><a href="http://www.fileden.com/files/2006/11/28/436069/untitled.JPG"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 320px;" src="http://www.fileden.com/files/2006/11/28/436069/untitled.JPG" border="0" alt="" /></a><br><br>P(arrives ontime)= P(takes&ontime) + P(buys&ontime)<br>= (0.4)(0.85) + (0.6)(0.65)<br>= 0.34 + 0.39<br>=0.73<br>= 73%<br><br>P(late<big>|</big>took)= P(takes&late)/P(takes&late) + P(buys&late)<br>= 0.06/0.06 + 0.21<br>= 0.06/0.27<br>= 0.2222<br>= 22%<br><br>oh yeah, the test is now on wednesday.<br>umm.. the next scriber is <big><font color="FF9933">tennyson</big></font>.<br>gooday&goodnight (:jessica.noreply@blogger.com0