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$ v. $$ v. $$$
Posted: Sun Jun 01, 2014 11:52 pm
by NoLieAbility
What does each term denote? Is it a measured by 1-9 = $, 10-99 = $$, and 100+ = $$$? Is there a set format that is used?
This feels like something that I should have been able to find for myself, but I'll be damned if I could find it.
Re: $ v. $$ v. $$$
Posted: Mon Jun 02, 2014 12:07 am
by malleus discentium
$ vs $$ vs $$$ are relative rather than absolute markers. Some people assign quartiles to them ($ is 25% tuition, $$ half etc.) but that's not standard and inferring any absolute rather than relative value from $ vs $$ vs $$$ isn't going to be useful.
Re: $ v. $$ v. $$$
Posted: Mon Jun 02, 2014 2:40 pm
by Clyde Frog
$$$$
Re: $ v. $$ v. $$$
Posted: Mon Jun 02, 2014 2:47 pm
by ManoftheHour
Re: $ v. $$ v. $$$
Posted: Mon Jun 02, 2014 3:20 pm
by Nucky
If you have to ask, you can't afford it.
Re: $ v. $$ v. $$$
Posted: Mon Jun 02, 2014 4:05 pm
by John Everyman
Asked this question a little while ago and was nearly unanimously told quartile percentages. Not sure why anyone would be against establishing a standard here.
Re: $ v. $$ v. $$$
Posted: Mon Jun 02, 2014 4:12 pm
by transferror
Nucky wrote:If you have to ask, you can't afford it.
Re: $ v. $$ v. $$$
Posted: Mon Jun 02, 2014 4:29 pm
by TheSpanishMain
$$$$ > $.
Re: $ v. $$ v. $$$
Posted: Tue Jun 10, 2014 6:03 pm
by Clyde Frog
$$+$$=$$$$
Re: $ v. $$ v. $$$
Posted: Thu Jun 12, 2014 2:50 pm
by ManoftheHour
$$$$-$$=$$
Re: $ v. $$ v. $$$
Posted: Fri Jun 13, 2014 1:11 am
by Sinatra
$x=?, if x=4?
Re: $ v. $$ v. $$$
Posted: Fri Jun 13, 2014 1:21 am
by ScottRiqui
S + I = $
Re: $ v. $$ v. $$$
Posted: Fri Jun 13, 2014 1:46 am
by Nova
Sinatra wrote:$x=?, if x=4?
$$$$
Re: $ v. $$ v. $$$
Posted: Fri Jun 13, 2014 1:51 am
by ManoftheHour
Sinatra wrote:$x=?, if x=4?
x=?/$, 4=?/$
Re: $ v. $$ v. $$$
Posted: Sat Jun 14, 2014 3:12 am
by Clyde Frog
Re: $ v. $$ v. $$$
Posted: Sat Jun 14, 2014 3:21 am
by ManoftheHour
Clyde Frog wrote:Clyde Frog wrote:$$$$
Re: $ v. $$ v. $$$
Posted: Sat Jun 14, 2014 3:07 pm
by TheSpanishMain
££££
Re: $ v. $$ v. $$$
Posted: Mon Jun 16, 2014 10:46 am
by Attax
$$^2=$$$$
Re: $ v. $$ v. $$$
Posted: Mon Jun 16, 2014 11:02 am
by ScottRiqui
Attax wrote:$$^2=$$$$
Shame on you - should be $$$.
Re: $ v. $$ v. $$$
Posted: Mon Jun 16, 2014 11:03 am
by John Everyman
√$$$$ = $$
Re: $ v. $$ v. $$$
Posted: Mon Jun 16, 2014 11:19 am
by Attax
ScottRiqui wrote:Attax wrote:$$^2=$$$$
Shame on you - should be $$$.
Why's that? $$^2 = $$*$$=$$$$ and $$$$^(1/2)=$$
But what'd d/d$ f($$)?
Re: $ v. $$ v. $$$
Posted: Mon Jun 16, 2014 11:27 am
by ScottRiqui
Attax wrote:ScottRiqui wrote:Attax wrote:$$^2=$$$$
Shame on you - should be $$$.
Why's that? $$^2 = $$*$$=$$$$ and $$$$^(1/2)=$$
But what'd d/d$ f($$)?
Exponentiation before multiplication, remember? Only the second '$' gets squared, then you multiply the result by the first '$' to give you $$$.
Re: $ v. $$ v. $$$
Posted: Mon Jun 16, 2014 11:37 am
by Attax
ScottRiqui wrote:Attax wrote:ScottRiqui wrote:Attax wrote:$$^2=$$$$
Shame on you - should be $$$.
Why's that? $$^2 = $$*$$=$$$$ and $$$$^(1/2)=$$
But what'd d/d$ f($$)?
Exponentiation before multiplication, remember? Only the second '$' gets squared, then you multiply the result by the first '$' to give you $$$.
Oh, I meant $$ as like a representation of 2 total. But in that case, since only the second $ gets squared and it is a singular unit it would be $*$^2 where $^2=$ thus $$^2=$$
Re: $ v. $$ v. $$$
Posted: Mon Jun 16, 2014 11:40 am
by ScottRiqui
Attax wrote:
Oh, I meant $$ as like a representation of 2 total. But in that case, since only the second $ gets squared and it is a singular unit it would be $*$^2 where $^2=$ thus $$^2=$$
Just do use parentheses, bro:
($$)^2 = $$$$
Re: $ v. $$ v. $$$
Posted: Mon Jun 16, 2014 11:43 am
by Attax
ScottRiqui wrote:Attax wrote:
Oh, I meant $$ as like a representation of 2 total. But in that case, since only the second $ gets squared and it is a singular unit it would be $*$^2 where $^2=$ thus $$^2=$$
Just do use parentheses, bro:
($$)^2 = $$$$
What do you think I am? A STEM person?!?!?!
So then d/d$ of f($$) where f($$)=($$)^2 would = 2($$)
The world is pure again!