You are depositing $1000 in a savings account at 6% annual interest rate, compounded continuously. What will the balance be after 5 years? Round your answer to the nearest hundredth.
amount = 1000 e^(.06*5)
= 1000 e^.3 = 1349.86
Ignore the gibberish of the previous reply.
To calculate the balance after 5 years with continuous compounding, we can use the formula:
B = P * e^(rt)
where:
B is the balance after the specified time (in this case, 5 years),
P is the principal amount (initial deposit),
e is the mathematical constant e (approximately 2.71828),
r is the annual interest rate (as a decimal),
and t is the time period (in years).
In this case, the principal amount (P) is $1000, the annual interest rate (r) is 6% or 0.06 as a decimal, and the time period (t) is 5 years.
Plugging in these values into the formula, we get:
B = 1000 * e^(0.06 * 5)
Calculating this using a calculator:
B ≈ 1000 * 2.71828^(0.06 * 5)
B ≈ 1000 * 2.71828^(0.3)
B ≈ 1000 * 1.3498588
B ≈ 1349.86
Therefore, the balance after 5 years, with continuous compounding, will be approximately $1349.86. Rounded to the nearest hundredth, the answer is $1349.86.
Sorry, i missed the fact that it was continuously compounded, but Reiny there was no need to make me feel incompetent.
First you multiply your starting balance (A) by .06 (or if your percentage[p] was larger like 43%, it would be .43 because of the way math works.) you will now have x% of your starting balance, multiply that number by the number of Z number of years. Now you have Y ( The percentage of your starting balance over z years), Add Y to your starting balance (A).
A*p = x% of starting number
x*z= y% for the number of years
y+A= new balance