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hongkongfuiz1234 Group TitleBest ResponseYou've already chosen the best response.0
Try reading this: http://psychassessment.com.au/PDF/Chapter%2001.pdf
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
ok so I have this formula \[v_{av}=\frac{\sum^n_{i=1}v_i\delta t_i}{\sum^n_{i=1}\delta t_i}\]
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
yeah I have looked it up. I would just like some clarification on a problem I am working on
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
I also have that \[\delta t_i=\frac{2z_i}{v_i}\]
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
so plugging this in I get \[v_{av}=\frac{\sum^n_{i=1}v_i\frac{2z_i}{v_i}}{\sum^n_{i=1}\frac{2z_i}{v_i}}\]factoring out constants\[v_{av}=\frac{\sum^n_{i=1}v_i\frac{z_i}{v_i}}{\sum^n_{i=1}\frac{z_i}{v_i}}\]canceling\[v_{av}=\frac{\sum^n_{i=1}z_i}{\sum^n_{i=1}\frac{z_i}{v_i}}\]
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
now I am wondering if I can simplify this any further?
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
@TuringTest
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\(v_i\) and \(z_i\) are all different terms depending on index, so I can't see how you could simplify that. I could be wrong...
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
right that's what my thinking was.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
I mean, what about this?\[v_{av}=\frac{\sum^n_{i=1}\cancel v_i\frac{z_i}{\cancel v_i}}{\sum^n_{i=1}\frac{z_i}{v_i}}\]or am I misreading your notation?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[\sum_i(v_i)\cdot(\frac{z_i}{v_i})=\sum_iz_i\]is valid
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
as long as it's under the same summation sign
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
hmm no I don't think that is correct. This is a physics problem and the answer does not make sense. v is a velocity and z is a distance so the final sum is a velocity.
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
or final answer I should say.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
you are right, that rule wouldn't make sense mathematically either
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
actually the more I thinking about it the more I am thinking it should make sense, but I am obviously unsure... @Zarkon @satellite73 series rules question
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
this is valid... \[\sum_i\left[(v_i)\cdot\left(\frac{z_i}{v_i}\right)\right]=\sum_iz_i\]
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
right and I did that in the numerator of the fraction to get. \[v_{av}=\frac{\sum^n_{i=1}z_i}{\sum^n_{i=1}\frac{z_i}{v_i}}\] Is there any way to express this as a product of \(z_i\) and \(v_i\) in the numerator?
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
unless you have more info...you can't simplify it any more
 one year ago
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