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nothing more nothing else [May. 8th, 2009|04:58 am]
[music |the dmb]

stumbling on memories, waiting for new ones oh at the idle lounge .


One of my students, a freshman in high school, said this to me the other day:

" You know I've had many tutors. A lot of tutors. And I think I remember working with you once before and I've been going here for three weeks. And I think you teach math better than anyone I've ever had."

there's nothing left for me to do now. this is destiny.

i need to find a place to live this fall in old town davis.

fact of the day: once what you know what you gotta do, you've gotta do it.

no turning back-

From: bigmach
2009-05-08 07:44 pm (UTC)
" You know I've had many tutors. A lot of tutors. And I think I remember working with you once before and I've been going here for three weeks. And I think you teach math better than anyone I've ever had."

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[User Picture]From: blo_lobo
2009-05-10 05:15 pm (UTC)
It reminds me of a similar memory I've had about my place.

"Matt, it's worthless to call IT when we have problems here in the system, we should just ask you from now on," said my old boss Peter.

"Matt the CBORD guru." DTR Annie L.

I think it's destiny, time to be a Dietitian with all this computer knowledge...
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[User Picture]From: dazz04
2009-05-10 09:44 pm (UTC)
can you tutor me? is fridays still possible?
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[User Picture]From: screamingryan
2009-05-11 05:46 am (UTC)
sorry steve i totally forgot about you and your test.

this friday isn't good i have an orientation, but next friday (may 22nd) will be definitely fine. where do you want to meet? give me a call this week or next week.
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[User Picture]From: dazz04
2009-05-12 12:10 am (UTC)
that'll work. im down to meet up in berkely or where ever.
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[User Picture]From: mythicalmenace
2009-05-12 05:49 am (UTC)
The only thing I find truly annoying about math is how these mathematical geniuses are so terrible at explaining all the good things they're doing and the joys of math. Or it could just be that certain people get math and certain people don't.

I texted you a problem that a middle school teacher gave me, and I'm wondering if you can solve it.

50 students take a test and their average score is 54%. The students were then separated into two groups. One group had an average score of 50% while the other group had an average score of 60%. How many students are in each group?

It's been bugging me all weekend! Help, Mr Math Guy!
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[User Picture]From: screamingryan
2009-05-12 10:53 am (UTC)
It's tricky but after 15 minutes I figured it out. This would only be solvable by middle school students who truly understand what average means. Anyway recall what average means:

It's taking a certain set of data points, divided by the number of data points. For instance, we have 50 students each with individual scores let's call them s_1, s_2, s_3...,s_50. We don't care about the individual scores we only care about the total of these scores. so we're going to let this sum = X. (i.e s_1 + s_2 + ... + s_50 = X). Remember that X is the total sum of the individual scores.

So look back at one of the first things we know. We know what the average of all 50 students' scores are which is .54 = 54%. It doesn't matter what the test is out of we only care about what the average is. However recall what the average is...the sum of all the scores divided by 50 (the number of students). which then means:

X/50 = .54

However we know our students are going to be split up into two groups but we want to know how many are in each group. We don't know these two different numbers, let's call them A and B.

So this means that

1) A + B = 50.

Which changes our equation, by plugging this in for 50, above to

2) X+Y/A+B = .54

And for each group, there are different sums of scores. For instance (this of course isn't the answer) the two groups have both 25 students. This will then give us two different averages. We want these two different sums of scores but we don't know them, let's call these sum of the 1st group with A students Z, and the sum of the scores of the group with B students, Y. So for the two groups we are given one average is .5 and the other is .6. So this means

3) Z/A = .5
4) Y/B = .6

We are trying to find A and B, remember.

In algebra, the number of variables you are trying to find can only be found with the same number of equations. As you can see we have 4 equations but luckily enough for us we need to only find A and B. What we're going to use is substitution.

and we turn #1 by subtracting A in both sides

1) B = 50 - A

we can rewrite each other equation in 3) and 4) to look like these via cross-multiplication.
3) Z = (.5)(A)
4) Y = (.6)(B)

but since from 1), 4) can be rewritten as
4) Y = (.6)(50-A)

So we have B,Z,Y all in terms of A. So let's plug these in into our equation in 2)

Z+Y/A+B =

(.5)(A) + (.6)(.50-A) / A + (50 - A) which we know this is equal to = .54

so we can solve this question for A
(.5)(A) + (.6)(.50-A) / A + (50 - A) = .54 >>
(.5)(A) + 30 - .6(A) / 50 = .54 >> combine like terms in the

numerator, and distributive property on the numerator.
(-.1)(A) + 30 / 50 = .54 >> combine like terms in numerator (.5 - .6 = -.1
(-.1)(A) + 30 = 27 >> multiply both sides by 50

(-.1)(A) = -3
A = 30.
That is our number of kids in the group with the average .5 = 50%. We can go back and find B, but that's easy for us because we know B = 50-A = 20.

So the group with the average .6 = 60% has 20 students.

It is a lot of work but it is easy to do. The hardest and most time consuming part of these problem is figuring out what to do. It's a lot but i tried to be detailed enough as possible.


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[User Picture]From: screamingryan
2009-05-12 10:56 am (UTC)
ahh type, the first line that has a 2) should say
X/A+B = .54

there shouldn't be an X there.
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[User Picture]From: blo_lobo
2009-05-12 11:41 pm (UTC)
dude the
(.5)(A) + (.6)(.50-A) / A + (50 - A) = .54 >>
(.5)(A) + 30 - .6(A) / 50 = .54 >>

(.6)(.50-A) doesn't = 30-.6(A). Typo
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[User Picture]From: mythicalmenace
2009-05-13 07:50 pm (UTC)
Mental overload! Brain will explode in T-minus ten seconds!
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