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hongkongfuiz1234
 2 years ago
Best ResponseYou've already chosen the best response.0Try reading this: http://psychassessment.com.au/PDF/Chapter%2001.pdf

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

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

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

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0so 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\[v_{av}=\frac{\sum^n_{i=1}z_i}{\sum^n_{i=1}\frac{z_i}{v_i}}\]

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0now I am wondering if I can simplify this any further?

TuringTest
 2 years ago
Best 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...

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0right that's what my thinking was.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1I 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?

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1as long as it's under the same summation sign

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0hmm 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.

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0or final answer I should say.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1you are right, that rule wouldn't make sense mathematically either

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1actually 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

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

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0right 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?

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2unless you have more info...you can't simplify it any more
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