(a) At 10% interest compounded yearly:

The formula for compound interest is A = P(1 + r/n)^(nt), where:
- A is the amount of money accumulated after n years, including interest.
- P is the principal amount (the initial amount of money).
- r is the annual interest rate (in decimal).
- n is the number of times that interest is compounded per year.
- t is the time the money is invested for in years.

In this case, we want to find out when the amount of money will double, so A/P = 2. The interest rate r is 10% or 0.10, and since the interest is compounded yearly, n = 1. So we have:

2 = (1 + 0.10/1)^(1*t)
2 = 1.10^t

Taking the natural logarithm of both sides to solve for t:

ln(2) = t * ln(1.10)
t = ln(2) / ln(1.10) ≈ 7.27 years

So it would take approximately 7.27 years for your money to double at 10% interest compounded yearly.

(b) At 10% interest compounded weekly:

Now, n = 52 because there are 52 weeks in a year. So we have:

2 = (1 + 0.10/52)^(52*t)
2 = (1.001923)^(52t)

Taking the natural logarithm of both sides to solve for t:

ln(2) = 52t * ln(1.001923)
t = ln(2) / (52 * ln(1.001923)) ≈ 7.21 years

To find out the number of weeks, we take the decimal part of the year and multiply it by 52:

0.21 * 52 ≈ 11 weeks

So it would take approximately 7 years and 11 weeks for your money to double at 10% interest compounded weekly.

(c) At 10% interest compounded continuously:

The formula for continuously compounded interest is A = Pe^(rt). Again, we want to find out when the amount of money will double, so A/P = 2. The interest rate r is 10% or 0.10. So we have:

2 = e^(0.10*t)

Taking the natural logarithm of both sides to solve for t:

ln(2) = 0.10 * t
t = ln(2) / 0.10 ≈ 6.93 years

So it would take approximately 6.93 years for your money to double at 10% interest compounded continuously.

Question

(a) At 10% interest compounded yearly:

The formula for compound interest is A = P(1 + r/n)^(nt), where:
- A is the amount of money accumulated after n years, including interest.
- P is the principal amount (the initial amount of money).
- r is the annual interest rate (in decimal).
- n is the number of times that interest is compounded per year.
- t is the time the money is invested for in years.

In this case, we want to find out when the amount of money will double, so A/P = 2. The interest rate r is 10% or 0.10, and since the interest is compounded yearly, n = 1. So we have:

2 = (1 + 0.10/1)^(1*t)
2 = 1.10^t

Taking the natural logarithm of both sides to solve for t:

ln(2) = t * ln(1.10)
t = ln(2) / ln(1.10) ≈ 7.27 years

So it would take approximately 7.27 years for your money to double at 10% interest compounded yearly.

(b) At 10% interest compounded weekly:

Now, n = 52 because there are 52 weeks in a year. So we have:

2 = (1 + 0.10/52)^(52*t)
2 = (1.001923)^(52t)

Taking the natural logarithm of both sides to solve for t:

ln(2) = 52t * ln(1.001923)
t = ln(2) / (52 * ln(1.001923)) ≈ 7.21 years

To find out the number of weeks, we take the decimal part of the year and multiply it by 52:

0.21 * 52 ≈ 11 weeks

So it would take approximately 7 years and 11 weeks for your money to double at 10% interest compounded weekly.

(c) At 10% interest compounded continuously:

The formula for continuously compounded interest is A = Pe^(rt). Again, we want to find out when the amount of money will double, so A/P = 2. The interest rate r is 10% or 0.10. So we have:

2 = e^(0.10*t)

Taking the natural logarithm of both sides to solve for t:

ln(2) = 0.10 * t
t = ln(2) / 0.10 ≈ 6.93 years

So it would take approximately 6.93 years for your money to double at 10% interest compounded continuously.

Knowee AI · Accepted Answer

(a) At 10% interest compounded yearly:

The formula for compound interest is A = P(1 + r/n)^(nt), where:
- A is the amount of money accumulated after n years, including interest.
- P is the principal amount (the initial amount of money).
- r is the annual interest rate (in decimal).
- n is the number of times that interest is compounded per year.
- t is the time the money is invested for in years.

In this case, we want to find out when the amount of money will double, so A/P = 2. The interest rate r is 10% or 0.10, and since the interest is compounded yearly, n = 1. So we have:

2 = (1 + 0.10/1)^(1*t)
2 = 1.10^t

Taking the natural logarithm of both sides to solve for t:

ln(2) = t * ln(1.10)
t = ln(2) / ln(1.10) ≈ 7.27 years

So it would take approximately 7.27 years for your money to double at 10% interest compounded yearly.

(b) At 10% interest compounded weekly:

Now, n = 52 because there are 52 weeks in a year. So we have:

2 = (1 + 0.10/52)^(52*t)
2 = (1.001923)^(52t)

Taking the natural logarithm of both sides to solve for t:

ln(2) = 52t * ln(1.001923)
t = ln(2) / (52 * ln(1.001923)) ≈ 7.21 years

To find out the number of weeks, we take the decimal part of the year and multiply it by 52:

0.21 * 52 ≈ 11 weeks

So it would take approximately 7 years and 11 weeks for your money to double at 10% interest compounded weekly.

(c) At 10% interest compounded continuously:

The formula for continuously compounded interest is A = Pe^(rt). Again, we want to find out when the amount of money will double, so A/P = 2. The interest rate r is 10% or 0.10. So we have:

2 = e^(0.10*t)

Taking the natural logarithm of both sides to solve for t:

ln(2) = 0.10 * t
t = ln(2) / 0.10 ≈ 6.93 years

So it would take approximately 6.93 years for your money to double at 10% interest compounded continuously.

(1 point) How many years would it take your money to double:(a) At 10% interest compounded yearly. years.(b) At 10% interest compounded weekly. years and weeks.(c) At 10% interest compounded continuously. years.Note: You can earn partial credit on this problem.

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