Find the required annual interest rate to the nearest tenth of a percent for $1100 to grow to $1900 if interest is compounded quarterly for 10yr. The required annual interest rate is _%?

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1 + r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = 1100

A = 1900

n = 4 because it was compounded 3 times in a year and n = 12/3 = 4

t = 10 years

Therefore,.

1900 = 1100(1 + r/4)^4 × 10

1900/1100 = (1+ r/4)^40

1.73 = (1+ r/4)^40

Taking log to base 10 of both sides, it becomes

Log 1.73 = 40log(1 + 0.25r)

0.238 = 40log(1 + 0.25r)

Log(1 + 0.25r) = 0.238/40 = 0.00595

Take exponent of both sides, it becomes

10^log(1 + 0.25r) = 10^0.00595

1 + 0.25r = 1.0138

0.25r = 1.0138 - 1 = 0.0138

r = 0.0138/0.25

r = 0.0552

The The required annual interest rate is

0.0552 × 100 = 5.5%

the answer is 200

step-by-step explanation:

the way the question is worded, it seems to imply the $264 interest is simple interest, not compound interest.

the simple interest for 2 years is $264; the simple interest for 1 year is $132.

$132 interest on a $600 investment means an interest rate of (132/600)*100 = 22%.

the value of $200 at 22% interest compounded annually for three years is

200%281.22%29%5e3

the amount of interest is that value, minus the original $200.

good luck on your studies! : )

david edward.