Question : float and precision

declare @t1 float, @t2 float,@t3 float

set @t1 = 2130.234
set @t2 = 3932.023
set @t3 = 2156.524

select (((@t1/@t2)*@t3)/@t1)*@t2 FloatType
go
declare @t1 float, @t2 float,@t3 float

set @t1 = 2130.23
set @t2 = 3932.02
set @t3 = 2156.52

select (((@t1/@t2)*@t3)/@t1)*@t2 FloatType

The result is based on the input variables precision? Is this how to control the precision on a float?

thanks

Answer : float and precision

Well, all this commotion, and it turns out he just had an over zealous firewall program (that I wasn't initially aware he had installed) that decided the server was trying to attack his computer, and blocked all access to it. I have configured it as a trusted site now, so this problem shouldn't happen again...