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Old 04-20-2008, 04:29 AM   #1
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Anybody here good at Math?

So I'm teaching Algebra II this year, which I haven't taught in seven years (and then only for one year). It's tough going compared to Pre-Algebra and Algebra I which I've teaching for years.

I hit this problem I just cannot figure out and I was wondering if any Interference math whizzes could help me out (so that I can, in turn, help my students out).

"Most of todays languages are thought to be descended from a few common ancestral languages. The longer the time since the languages split from the ancestral language the fewer common words there are in the descendant languages. Let c = the number of centuries since two languages split from an ancestral language. Let w = the percentage of words from the ancestral language which are common to the the two descendant languages. In linguistics, the equation below (in which r = .86 is the index of retention) has been used to relate c and w.

10/c = 2 log r/log w

If about 15% of the words in an ancestral language are common to two different languages, about how many centuries ago did they split from the ancestral language?"

The book provides the answer of about 63 centuries. Try as I might I cannot seem to get this answer or anything close.

Can anyone help me? Thanks!
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Old 04-20-2008, 06:03 AM   #2
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I think I'd rather have a urinary tract infection than do maths again any time soon.

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Old 04-20-2008, 10:32 AM   #3
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10/c = 2 log r/log w

r = 0.86
w = 15%. which is the same as 0.15 (which is probably where you go wrong)

10/c = 2 log 0.86/log 0.15

10/c = 0.159

10/0.159 = approx. 62.89
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Old 04-20-2008, 10:35 AM   #4
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^agreed! I would do it the same way.
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Old 04-20-2008, 11:37 AM   #5
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Quote:
Originally posted by DrTeeth
10/c = 2 log r/log w

r = 0.86
w = 15%. which is the same as 0.15 (which is probably where you go wrong)

10/c = 2 log 0.86/log 0.15

10/c = 0.159

10/0.159 = approx. 62.89
QFT


That's how I would have worked it out.



On an unrelated note, as it is not true in this case, surely after teaching for so many years you are familiar with how often textbooks get the answers wrong?
I think they do it on purpose to mess with people's minds.
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Old 04-20-2008, 11:50 AM   #6
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Quote:
Originally posted by DrTeeth
10/c = 2 log r/log w

r = 0.86
w = 15%. which is the same as 0.15 (which is probably where you go wrong)

10/c = 2 log 0.86/log 0.15

10/c = 0.159

10/0.159 = approx. 62.89
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Old 04-20-2008, 11:54 AM   #7
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I used to be good at math. Then I went to law school. Now I only need to know how to bill. In 6 minute increments. Which really, is an art unto itself.
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Old 04-20-2008, 12:48 PM   #8
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Quote:
Originally posted by ~BrightestStar~

how often textbooks get the answers wrong?
I've found that. I prefer to do everything myself first anyway. What's annoying is when something in the pupils' book is wrong.
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Old 04-20-2008, 03:50 PM   #9
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I haven't taken math since high school. Dr. Teeth's answer looks like an alien language to me.
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Old 04-20-2008, 04:37 PM   #10
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^

Maddy had algebra homework the other night, and needed help (she's in 8th grade). Her math teacher has been out 2 weeks due to illness, so she's had 2 substitutes during that time. The 2nd sub wasn't very helpful she said, and I thought I could try & offer my assistance. She asked if I knew something (monomial? - a word I've never heard of). So I said, let me look at it to see if anything made sense. I made a suggestion, which turned out to be helpful to her. But I haven't had math since high school, which was over 20 years ago I was shocked, and pleased with myself
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Old 04-20-2008, 09:53 PM   #11
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Quote:
Originally posted by DrTeeth
10/c = 2 log r/log w

r = 0.86
w = 15%. which is the same as 0.15 (which is probably where you go wrong)

10/c = 2 log 0.86/log 0.15

10/c = 0.159

10/0.159 = approx. 62.89
Okay, but how do you end up dividing 10 by 0.159? Shouldn't it be the other way around, dividing 0.159 by 10 (i.e. multiplying by one tenth?).

by the way, thanks for the help and commentary everyone! I teach my Algebra II class in about 15 minutes. We'll see how it goes.
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Old 04-20-2008, 10:16 PM   #12
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Quote:
Originally posted by Diemen
I haven't taken math since high school. Dr. Teeth's answer looks like an alien language to me.
Me too! I definitely had the face going when I read his answer.
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Old 04-20-2008, 10:21 PM   #13
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It's easier to figure out the mechanism if you substitute the numbers for numbers you're more familiar with.

Instead of 10/c = 0.159, you could write 6/3 =2. It's easy now to see how what you have to do with the 6 and the 2, to get 3 as an answer. Once you've figured that out, you only have to change back those numbers to the numbers in your formula.
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Old 04-20-2008, 10:38 PM   #14
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Shoot me now!
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Old 04-21-2008, 01:35 AM   #15
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Quote:
Originally posted by maycocksean


Okay, but how do you end up dividing 10 by 0.159? Shouldn't it be the other way around, dividing 0.159 by 10 (i.e. multiplying by one tenth?).

I take it today has been one of those horrible days where the longer you look at something that should be simple, the more confusing it gets?


I suppose even math teachers can have their bad days



And I mean no offense.


Anyway, just remember it's a simple equation of:

number1 = number2
variable

(or alternatively)
1 = 2
x


how would you solve that on any other day?
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