Vanessa invested $2,500 into an account that will increase in value by 3.5% each year. Write an exponential function to model this situation, then find the value of the investment after 20 years.
Solution:
To model the exponential equation for Vanessa's investment, we use the compound interest formula:
\[ A = P \times (1 + r)^t \]
Where:
- \( A \) is the final amount
- \( P \) is the principal amount (initial investment)
- \( r \) is the annual interest rate (as a decimal)
- \( t \) is the time in years
- \( P = \$2,500 \)
- \( r = 0.035 \) (3.5% expressed as a decimal)
- \( t = 20 \) years
\[ A = 2500 \times (1 + 0.035)^{20} \]
This equation models the situation. To find the value of the investment after 20 years, you can use a calculator to evaluate this expression.
The average price of a movie ticket in 1990 was $4.22. Since then, the price has increased by approximately 3.1% each year. Write an exponential function to model this situation, then find the price of a ticket in 2016.
Solution:
To model the exponential equation for the movie ticket price, we use the compound interest formula:
\[ P(t) = P_0 \times (1 + r)^t \]
Where:
- \( P(t) \) is the price of the ticket at time \( t \)
- \( P_0 \) is the initial price of the ticket (in 1990)
- \( r \) is the annual growth rate (as a decimal)
- \( t \) is the time in years
- \( P_0 = \$4.22 \)
- \( r = 0.031 \) (3.1% expressed as a decimal)
- \( t = 2016 - 1990 = 26 \) years
\[ P(26) = 4.22 \times (1 + 0.031)^{26} \]
This equation models the situation. To find the price of a ticket in 2016, you can use a calculator to evaluate this expression.
A virus has infected 400 people in the town and is spreading to 25% more people each day. Write an exponential function to model this situation, then find the number of people that will be infected in 10 days.
Solution:
To model the exponential equation for the virus spread, we use the compound interest formula:
\[ N(t) = N_0 \times (1 + r)^t \]
Where:
- \( N(t) \) is the number of infected people at time \( t \)
- \( N_0 \) is the initial number of infected people
- \( r \) is the daily growth rate (as a decimal)
- \( t \) is the time in days
- \( N_0 = 400 \)
- \( r = 0.25 \) (25% expressed as a decimal)
- \( t = 10 \) days
\[ N(10) = 400 \times (1 + 0.25)^{10} \]
This equation models the situation. To find the number of people infected in 10 days, you can use a calculator to evaluate this expression.
The population of a small town was 10,800 in 2002. Since then, the population has decreased at a rate of 2.5% each year. Write an exponential function to model the situation, then find the population of the town in 2020.
Solution:
To model the exponential equation for the population decline, we use the compound interest formula:
\[ P(t) = P_0 \times (1 - r)^t \]
Where:
- \( P(t) \) is the population at time \( t \)
- \( P_0 \) is the initial population
- \( r \) is the annual decline rate (as a decimal)
- \( t \) is the time in years
- \( P_0 = 10,800 \)
- \( r = 0.025 \) (2.5% decline expressed as a decimal)
- \( t = 2020 - 2002 = 18 \) years
\[ P(18) = 10,800 \times (1 - 0.025)^{18} \]
This equation models the situation. To find the population of the town in 2020, you can use a calculator to evaluate this expression.
Manny bought a brand new car in 2012 for $28,750. If the car depreciates by 12% each year, write an exponential function to model the situation, then find the value of the car in 2018.
Solution:
To model the exponential equation for the car depreciation, we use the compound interest formula:
\[ V(t) = V_0 \times (1 - r)^t \]
Where:
- \( V(t) \) is the value of the car at time \( t \)
- \( V_0 \) is the initial value of the car (in 2012)
- \( r \) is the annual depreciation rate (as a decimal)
- \( t \) is the time in years
- \( V_0 = \$28,750 \)
- \( r = 0.12 \) (12% depreciation expressed as a decimal)
- \( t = 2018 - 2012 = 6 \) years
\[ V(6) = 28750 \times (1 - 0.12)^6 \]
This equation models the situation. To find the value of the car in 2018, you can use a calculator to evaluate this expression.
Anisha invested $8,000 in an account that earns 10% interest. How much money will she have in 15 years if the interest is compounded quarterly?
Solution:
To model the exponential equation for Anisha's investment growth, we use the compound interest formula:
\[ A = P \times \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the final amount
- \( P \) is the principal amount (initial investment)
- \( r \) is the annual interest rate (as a decimal)
- \( n \) is the number of times interest is compounded per year
- \( t \) is the time in years
- \( P = \$8,000 \)
- \( r = 0.10 \) (10% interest rate)
- \( n = 4 \) (compounded quarterly)
- \( t = 15 \) years
\[ A = 8000 \times \left(1 + \frac{0.10}{4}\right)^{4 \times 15} \]
This equation models the situation. To find the final amount in 15 years, you can use a calculator to evaluate this expression.
Kevin borrowed $32,500 to purchase a new car. If the rate on the loan is 6% compounded annually, how much will he pay in total over the course of the 5 year loan?
Solution:
To model the exponential equation for Kevin's loan repayment, we use the compound interest formula:
\[ A = P \times \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the total amount repaid
- \( P \) is the principal amount (loan)
- \( r \) is the annual interest rate (as a decimal)
- \( n \) is the number of times interest is compounded per year
- \( t \) is the time in years
- \( P = \$32,500 \)
- \( r = 0.06 \) (6% interest rate)
- \( n = 1 \) (compounded annually)
- \( t = 5 \) years
\[ A = 32500 \times \left(1 + \frac{0.06}{1}\right)^{1 \times 5} \]
This equation models the situation. To find the total amount repaid over 5 years, you can use a calculator to evaluate this expression.
Scott invested $1,600 into a retirement account that earns 2.4% interest compounded monthly. What will the balance of the account be after 30 years?
Solution:
To model the exponential equation for Scott's retirement account, we use the compound interest formula:
\[ A = P \times \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the final balance of the account
- \( P \) is the principal amount (initial investment)
- \( r \) is the annual interest rate (as a decimal)
- \( n \) is the number of times interest is compounded per year
- \( t \) is the time in years
- \( P = \$1,600 \)
- \( r = 0.024 \) (2.4% interest rate)
- \( n = 12 \) (compounded monthly)
- \( t = 30 \) years
\[ A = 1600 \times \left(1 + \frac{0.024}{12}\right)^{12 \times 30} \]
This equation models the situation. To find the final balance after 30 years, you can use a calculator to evaluate this expression.
Kaylee used her graduation money to set up a savings account that earns 3.4% interest compounded weekly. If the original amount deposited was $500, how much interest will she have earned after 10 years?
Solution:
To model the exponential equation for Kaylee's savings account, we use the compound interest formula:
\[ A = P \times \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the final amount
- \( P \) is the principal amount (initial deposit)
- \( r \) is the annual interest rate (as a decimal)
- \( n \) is the number of times interest is compounded per year
- \( t \) is the time in years
- \( P = \$500 \)
- \( r = 0.034 \) (3.4% interest rate)
- \( n = 52 \) (compounded weekly)
- \( t = 10 \) years
\[ A = 500 \times \left(1 + \frac{0.034}{52}\right)^{52 \times 10} \]
This equation models the situation. To find the final amount after 10 years, you can use a calculator to evaluate this expression.
Mr. and Mrs. Rainer took out a $240,000 loan to purchase their home. If the interest rate on the loan is 1.2% compounded bimonthly, how much interest will they have paid after 30 years?
Solution:
To model the exponential equation for the loan repayment, we use the compound interest formula:
\[ A = P \times \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the total amount repaid
- \( P \) is the principal amount (loan)
- \( r \) is the annual interest rate (as a decimal)
- \( n \) is the number of times interest is compounded per year
- \( t \) is the time in years
- \( P = \$240,000 \)
- \( r = 0.012 \) (1.2% interest rate)
- \( n = 6 \) (compounded bimonthly)
- \( t = 30 \) years
\[ A = 240000 \times \left(1 + \frac{0.012}{6}\right)^{6 \times 30} \]
This equation models the situation. To find the total amount repaid after 30 years, you can use a calculator to evaluate this expression.