16L of antifreeze is 30% antifreeze how much mixture should be drained and replaces with pure antifreeze so final mix is 50% anitfreeze

Question

anonymous · Answer

Let x be amount removed to make way for pure, then

30(16-x)/100  + 100x/100 =50*16/100 ->

480 + 70x = 800 ->

x = 320/70 = 32/7 litres to  remove

anonymous · Answer

thanks

anonymous · Answer

ur welcome

anonymous · Answer

are you familiar with the greek letter "mu"$$\mu$$? it look like a population problem but its got that thing in it
I=I$$I=I_{0}e ^{-mux}                             \mu=1.4$$
percentage of light intensity at 1m,3m,5m,10m, 50m?

anonymous · Answer

I don't really know, scouting about a bit seems to indicate it is used in descriptions of light intensity as u suggest.

Eg 86 mu mol m^-2 s^-1 which I would be inclined to read as micro mol but that's just a guess I'm afraid.