Use the simple interest formula.How can $42,000 be invested, part at 4% annual simple interest and the remainder at 10% annual simple interest, so that the interest earned by the two accounts is equal at the end of the year?
A. $30,000 invested at 4%; $12,000 invested at 10%
B. $22,000 invested at 4%; $20,000 invested at 10%
C. $12,000 invested at 4%; $30,000 invested at 10%
D. $20,000 invested at 4%; $22,000 invested at 10%
Answer: A
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Solve the problem.Linda invests $25,000 for one year. Part is invested at 5%, another part at 6%, and the rest at 8%. The total income from all 3 investments is $1,600. The income from the 5% and 6% investments is the same as the income from the 8% investment. Find the amount invested at each rate.
A. $10,000 at 5%, $10,000 at 6%, $5,000 at 8% B. $8,000 at 5%, $10,000 at 6%, $7,000 at 8% C. $10,000 at 5%, $5,000 at 6%, $10,000 at 8% D. $5,000 at 5%, $10,000 at 6%, $10,000 at 8%
Solve the problem.Suppose $2500 is deposited into an account earning 5% annual interest, compounded daily. Find the balance in the account at the end of 6 years.
A. $3374.58 B. $337.46 C. $3372.54 D. $50,110.86
Factor completely. If the polynomial cannot be factored, say it is prime.27y3 - 1
A. (3y - 1)(9y2 + 3y + 1) B. (27y - 1)(y2 + 3y + 1) C. (3y - 1)(9y2 + 1) D. (3y + 1)(9y2 - 3y + 1)
Change to a percent.5.1
A. 510% B. 0.51% C. 51% D. 0.0051%