The answer: $1,000 was invested at %5. $2,500 was invested @ 8% It is done with a system of equations when "x" is a portion of $3,500 and "y" is a portion of $3,500: x + y = 3500AND .05x +.08y = 250 Use algebra and solve #2 for "x" as follows: .05x + .08y = 250 .05x = 250 - .08y x = 5000 - 1.6y plug the value of "x" into the first equation and solve for "y": 5000 - 1.6y + y = 3500 5000 - 3500 - 1.6y + y = 0 1500 - 1.6y + y = 0 1500 = 1.6y -y 1500 = .6y 2500 = y plug the value of "y" into the original equation and solve for "x" x + 2500 = 3500 x = 3500 - 2500 x = 1000 You can plug them both into the 2nd equation in the system to check it.
The annual equivalent rate is 15.5625%. The amount invested is irrelevant to calculation of the equivalent rate.
Let P be the amount of invested money. Then, .08P = 336 P = 336/.08 = 4,200
To find 8% of an amount we multiply by 0.08 So we have an amount $x: $x x 0.08 = $250 Rearranging gives $x = 250/1.08 = $3125
x = amount of money invested at 5% y = amount of money invested at 4% x=2y .05x+.04y=350 .05(2y)+.04y=350 .1y+.04y=350 .14y=350 y=$2500 x=$5000
$14,693.28
The annual equivalent rate is 15.5625%. The amount invested is irrelevant to calculation of the equivalent rate.
Let P be the amount of invested money. Then, .08P = 336 P = 336/.08 = 4,200
To find 8% of an amount we multiply by 0.08 So we have an amount $x: $x x 0.08 = $250 Rearranging gives $x = 250/1.08 = $3125
x = amount of money invested at 5% y = amount of money invested at 4% x=2y .05x+.04y=350 .05(2y)+.04y=350 .1y+.04y=350 .14y=350 y=$2500 x=$5000
y = ln(3)/ln(1.0575) = 19.65 years, approx.
$14,693.28
Find the annual amount of FICA at a 7.51% rate by computing his annual salary
1/12th of 5% because there are 12 months in a year. ANSWER:- 1/60th per cent, which is the same as 0.01667 of the amount invested.
2500
2.5 years
2500
To find the annual yield of an investment, you can calculate it by dividing the annual income generated by the investment by the initial amount invested, and then multiplying by 100 to get a percentage.