Question 703026
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Let *[tex \LARGE x] represent the amount, in pints, of 4% milk.  Then *[tex \LARGE \frac{1}{3}\ -\ x] is the amount of 1% milk.


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.04x\ +\ 0.01(\frac{1}{3}\ -\ x)\ =\ 0.02(\frac{1}{3})]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.03x\ =\ (0.02\ -\ 0.01)\,\cdot\,\frac{1}{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{0.01}{0.03}\,\cdot\,\frac{1}{3}\ =\ \frac{1}{9}]


So, *[tex \LARGE \frac{1}{9}] pint of 4% and *[tex \LARGE \frac{2}{9}] pint of 1% makes *[tex \LARGE \frac{1}{3}] pint of 2%


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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