Question 1072621
Let n be the amount invested at 11%. Then the amount invested at 9% would be 12000-n. So:
.11n+.09(12000-n)=1200
.02n+1080=1200
.02n=120
n=$6000 to be invested at 11%, and another $6000 to be invested at 9%. &#9786;&#9786;&#9786;&#9786;