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June 16, 2024, 11:22:05 pm

Author Topic: power function/exponential function help  (Read 1445 times)  Share 

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nels

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power function/exponential function help
« on: October 24, 2009, 01:29:32 pm »
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this is a question from vcaa 02 exam 1.

x      y
1     1.6
2     2.6
3     4.3
4     7.0
5     11.3
6     18.4
7     29.9
8     48.5

the data would be best modelled using
A. a linear function
B. a power function
C. an exponential function
D. a circular function
E. a logarithmic function

i know that it is either B or C. how can i now determine which is correct? what factor separates the two?
thanks.

GoodGuys

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Re: power function/exponential function help
« Reply #1 on: October 24, 2009, 02:10:57 pm »
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C,
as x increase, so does y. Thus, the best option will be C

TrueTears

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Re: power function/exponential function help
« Reply #2 on: October 24, 2009, 02:34:56 pm »
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The best way is to type this in the TI-89, put x in list 1 and y in list 2 and then use regressions.
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bem9

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Re: power function/exponential function help
« Reply #3 on: October 24, 2009, 02:46:13 pm »
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this is a question from vcaa 02 exam 1.

x      y
1     1.6
2     2.6
3     4.3
4     7.0
5     11.3
6     18.4
7     29.9
8     48.5

the data would be best modelled using
A. a linear function
B. a power function
C. an exponential function
D. a circular function
E. a logarithmic function

i know that it is either B or C. how can i now determine which is correct? what factor separates the two?
thanks.

Power function is x^n, where n is a fraction or integer like 2/3, -1, 2 etc... i think
but exponential is in the form a^x, where a is a constant.
So its C

nels

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Re: power function/exponential function help
« Reply #4 on: October 24, 2009, 03:02:23 pm »
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thanks guys.

The best way is to type this in the TI-89, put x in list 1 and y in list 2 and then use regressions.

if i used this method, how will i know which is the correct answer?

TrueTears

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Re: power function/exponential function help
« Reply #5 on: October 24, 2009, 03:07:32 pm »
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thanks guys.

The best way is to type this in the TI-89, put x in list 1 and y in list 2 and then use regressions.

if i used this method, how will i know which is the correct answer?
Well first of all you had a hunch that it's either B or C, so go into stats list editor type your stuff in and then pick F4 9, for powerreg, if the coefficient of correlation is not close to 1 then pick F4 8 which is exponential reg, this will give a coefficient of correlation very close to 1, thus it must be C.
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Interested in asset pricing, econometrics, and social choice theory.

naved_s9994

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Re: power function/exponential function help
« Reply #6 on: October 24, 2009, 03:12:48 pm »
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Tt
wouldn't it take longer by doing that?
(statlist editor)
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TrueTears

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Re: power function/exponential function help
« Reply #7 on: October 24, 2009, 03:14:13 pm »
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Tt
wouldn't it take longer by doing that?
(statlist editor)

no, very fast if you know your stuff.
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Interested in asset pricing, econometrics, and social choice theory.

NE2000

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Re: power function/exponential function help
« Reply #8 on: October 24, 2009, 03:49:50 pm »
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thanks guys.

The best way is to type this in the TI-89, put x in list 1 and y in list 2 and then use regressions.

if i used this method, how will i know which is the correct answer?
Well first of all you had a hunch that it's either B or C, so go into stats list editor type your stuff in and then pick F4 9, for powerreg, if the coefficient of correlation is not close to 1 then pick F4 8 which is exponential reg, this will give a coefficient of correlation very close to 1, thus it must be C.

Yeah that's what I do. Won't ever fail you unless you type it in wrong. In the TI-84 you have to make sure you have DiagnosticOn to see R^2 values. The R^2 values are basically how well the correlation exists.
2009: English, Specialist Math, Mathematical Methods, Chemistry, Physics

nels

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Re: power function/exponential function help
« Reply #9 on: October 24, 2009, 04:08:42 pm »
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ahh i see. didnt have the diagnostic on before so that was why i was wondering...
thanks guys, heaps of help as usual!