No arguments on this end, Bob. I may not be the sharpest tool in the shed, but I'm smart enough to realize that I was lucky, and I've got a pretty good thing going on here!
Thanks to everyone for wishing us well! It was a good day.
"Forget pounds and ounces, I'm figuring displacement!"
If we accept that: MBG(+)FGSF(=)HBG(F1) And we surmise that: BG(>)HBG(F1) while GSF(<)HBG(F1) Would it hold true that: HBG(F1)(+)AM500(x)q.d.(=)1.5lbGRWT? PB answer: It depends.
