1. Investment Problem:

- $5000 are invested in a bank account at an interest rate of 6 percent per year.
1. Find the amount in the bank after 7 years if interest is compounded annually.
2. Find the amount in the bank after 7 years if interest is compounded quarterly.
3. Find the amount in the bank after 7 years if interest is compounded monthly.
4. Find the amount in the bank after 7 years if interest is compounded continuously.

2. Car Depreciation Problem:

A car was valued at $25,000 in the year 1995. The value depreciated to $13,000 by the year 2004.

A) What was the annual rate of change between 1995 and 2004? Round the rate of decrease to 4 decimal places.

B) What is the correct answer to part A written in percentage form?

C) Assume that the car value continues to drop by the same percentage. What will the value be in the year 2007? Round to the nearest 50 dollars.

3. Radioactive Substance Decay:

A scientist begins with 180 milligrams of a radioactive substance. After 25 hours, 90 mg of the substance remains. How many milligrams will remain after 41 hours? Give your answer accurate to at least one decimal place.

4. Doctor's Office Problem:

You go to the doctor and he gives you 10 milligrams of radioactive dye. After 20 minutes, 7.5 milligrams of dye remain in your system. To leave the doctor's office, you must pass through a radiation detector without sounding the alarm. If the detector will sound the alarm if more than 2 milligrams of the dye are in your system, how long will your visit to the doctor take, assuming you were given the dye as soon as you arrived? Give your answer to the nearest minute. You will spend minutes at the doctor's office.

5. Palladium-100 Decay:

The half-life of Palladium-100 is 4 days. After 12 days, a sample of Palladium-100 has been reduced to a mass of 7 mg.

- What was the initial mass (in mg) of the sample?
- What is the mass (in mg) 5 weeks after the start? You may enter the exact value or round to 4 decimal places.

6. Carbon-14 Dating Problem:

A wooden artifact from an ancient tomb contains 30 percent of the carbon-14 that is present in living trees. How long ago, to the nearest year, was the artifact made? (The half-life of carbon-14 is 5730 years.)

7. Bacterial Population Growth:

The doubling period of a bacterial population is 10 minutes. At time t = 90 minutes, the bacterial population was 50,000.

- What was the initial population at time t = 0?
- Find the size of the bacterial population after 5 hours.

Additionally, the count in a bacterial culture was 400 after 15 minutes and 1600 after 35 minutes. Assuming the count grows exponentially:

- What was the initial size of the culture?
- Find the doubling period.
- Find the population after 85 minutes.
- When will the population reach 11,000? You may enter the exact value or round to 2 decimal places.

8. Fox Population Growth:

The fox population in a certain region has a continuous growth rate of 9 percent per year. It is estimated that the population in the year 2000 was 23,700.

(a) Find a function that models the population t years after 2000 (t = 0 for 2000). Hint: Use an exponential function with base e. Your answer is P(t) =

(b) Use the function from part (a) to estimate the fox population in the year 2008.

The amounts in the bank after 7 years for each compounding period are: Annually compounded: $7926.64; Quarterly compounded: $7951.62; Monthly compounded: $7959.45; Continuously compounded: $7968.64.

To calculate the amount in the bank after 7 years with different compounding periods, we can use the formula for compound interest:

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

Where:
A = the final amount
P = the principal amount (initial investment)
r = the annual interest rate (as a decimal)
n = the number of times the interest is compounded per year
t = the number of years

For the first case, where interest is compounded annually:
P = $5000
r = 0.06 (6% as a decimal)
n = 1 (compounded annually)
t = 7

Substituting these values into the formula, we get:
A = 5000(1 + 0.06/1)^(1*7)

Calculating this, we find that the amount in the bank after 7 years with annual compounding is approximately $7926.64.

For the second case, where interest is compounded quarterly:
P = $5000
r = 0.06
n = 4 (compounded quarterly)
t = 7

Substituting these values into the formula:
A = 5000(1 + 0.06/4)^(4*7)

Calculating this, we find that the amount in the bank after 7 years with quarterly compounding is approximately $7951.62.

For the third case, where interest is compounded monthly:
P = $5000
r = 0.06
n = 12 (compounded monthly)
t = 7

Substituting these values into the formula:
A = 5000(1 + 0.06/12)^(12*7)

Calculating this, we find that the amount in the bank after 7 years with monthly compounding is approximately $7959.45.

Finally, for the case of continuous compounding:
P = $5000
r = 0.06
n = infinity (continuous compounding)
t = 7

Substituting these values into the formula:
A = 5000 * e^(0.06*7)

Calculating this, we find that the amount in the bank after 7 years with continuous compounding is approximately $7968.64.

To know more about compound interest:

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