Invested, r is the interest rate as a decimal, n is the number of times compounded annually, and t is the number of years. A person is investing $1000 at an interest rate of 12% interest for 25 years, and is curious how much difference the number of compounds (n) increases the value of the account after 25 years.

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 = $1000

r = 12% = 12/100 = 0.12

n = 1 because it was compounded once in a year.

t = 25 years

Therefore,.

A = 1000(1 + 0.12/1)^1 × 25

A = 1000(1.12)^25

A = $17000

If the number of compounding periods increases, the amount of compound interest would be greater.

send it over

step-by-step explanation:

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)