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Non-Cattle Specific Topics
Every Thing Else Board
I should've listened to Ms Johnson.........
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<blockquote data-quote="cfpinz" data-source="post: 573463" data-attributes="member: 2383"><p>You've already got .079oz of active ingredient in your one ounce sample (assuming we're using ounces as weight rather than volume). So if you set it up .079/x=.0005 and solve for x, you wind up with 158 ounces. 158 is the total number of ounces in the final product, so you must subtract the ounce you already have which supplied the .079oz of active ingredient, leaving 157oz of water to add. Not that any of that really amounts to a hill of beans, but if I'm wrong please explain why.</p></blockquote><p></p>
[QUOTE="cfpinz, post: 573463, member: 2383"] You've already got .079oz of active ingredient in your one ounce sample (assuming we're using ounces as weight rather than volume). So if you set it up .079/x=.0005 and solve for x, you wind up with 158 ounces. 158 is the total number of ounces in the final product, so you must subtract the ounce you already have which supplied the .079oz of active ingredient, leaving 157oz of water to add. Not that any of that really amounts to a hill of beans, but if I'm wrong please explain why. [/QUOTE]
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I should've listened to Ms Johnson.........
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