Kaylee A.
asked • 11/07/21college algebra
A woman has a total of $11,000 to invest. She invests part of the money in an account that pays 8% per year and the rest in an account that pays 9% per year. If the interest earned in the first year is $910, how much did she invest in each account?
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1 Expert Answer
Raymond B. answered • 02/01/23
Tutor
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Math, microeconomics or criminal justice
.09x + .08(11000-x) = 912
.09x -.08x = 912 -.08(11000)
.01x = 912 -880 = 32
x = 32/.01 = $3,200 invested at 9%
y = 11000-3200 = $7,800 invested at 8%
.09(3200)+ .08(7800)
= 288 + 624
= 912
912/11000= about 8.3%
9-8.3 =.7
8.3-8 =.3
3/7 = 3200/7800= .41
amount invested at 9% needs to be about 3/7 of the 8% investment. 32/78= about 3/7
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Osman A.
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