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**addyta.org** · 03-27-2007, 06:39 PM

That works great, thank you. I knew it would be simple - I just could not see the answer....

Huh? That's like saying you need a formula where 1+1=3 and solving it by writing it as 1+1=3 - or am I missing something?

What you need, IMO, is something like (100*(75-45)/45)k=40%, in this case k=0.6 (I think). However it may not be the same in other cases, it may be progressive, proportional, exponentially increasing etc - hence the request for other samples.

[edit] I mean (100(b-a)/a)k=x%, where a is initial value, b is new value, k is the modifier (term?) and x is the resultant.

03-27-2007, 06:39 PM	#7
addyta.org Join Date Oct 2005 Posts 557 Senior Member	That works great, thank you. I knew it would be simple - I just could not see the answer.... Huh? That's like saying you need a formula where 1+1=3 and solving it by writing it as 1+1=3 - or am I missing something? What you need, IMO, is something like (100*(75-45)/45)k=40%, in this case k=0.6 (I think). However it may not be the same in other cases, it may be progressive, proportional, exponentially increasing etc - hence the request for other samples. [edit] I mean (100(b-a)/a)k=x%, where a is initial value, b is new value, k is the modifier (term?) and x is the resultant.
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