Question
You deposit $1,000 in a savings account that earns 6% interest per year.
Simple Interest
t Principal Annual Interest End of year Balance
1 $1,000.00 $60.00 $1,060.00
2 $1,000.00 $60.00 $1,120.00
3
4
5
6
7
8
9
10
Compound Interest
(compounded annually)
t Principal Annual Interest End of year Balance
1 $1,000.00
$60.00 $1,060.00
2 $1,060.00
$63.60
$1,123.00
3
4
5
6
7
8
9
10
Complete each table. Which type of interest gives the greater balance? Use the simple interest formula and the given values to find each missing value.
Graph the end-of-year balances for each type of interest in part A on the same graph. Label each.
You deposit $400 in an account that earns 3.5% interest compounded annually. What is the account balance after 2 years? After 10 years? After 25 years?
A firefighter invests $40,000 in a retirement account for 2 years. The interest rate is 6%. The interest is compounded monthly. What will the final balance be?
P=$2,000, R=5%, t=5
. Find I.
I=$2,566.72, R=12%, t=3
. Find P.
I=$842.96, P=500, t=2
. Find 𝑅.
I=$1,738.91, P=1,500, R=3%
. Find 𝑡.
Simple Interest
t Principal Annual Interest End of year Balance
1 $1,000.00 $60.00 $1,060.00
2 $1,000.00 $60.00 $1,120.00
3
4
5
6
7
8
9
10
Compound Interest
(compounded annually)
t Principal Annual Interest End of year Balance
1 $1,000.00
$60.00 $1,060.00
2 $1,060.00
$63.60
$1,123.00
3
4
5
6
7
8
9
10
Complete each table. Which type of interest gives the greater balance? Use the simple interest formula and the given values to find each missing value.
Graph the end-of-year balances for each type of interest in part A on the same graph. Label each.
You deposit $400 in an account that earns 3.5% interest compounded annually. What is the account balance after 2 years? After 10 years? After 25 years?
A firefighter invests $40,000 in a retirement account for 2 years. The interest rate is 6%. The interest is compounded monthly. What will the final balance be?
P=$2,000, R=5%, t=5
. Find I.
I=$2,566.72, R=12%, t=3
. Find P.
I=$842.96, P=500, t=2
. Find 𝑅.
I=$1,738.91, P=1,500, R=3%
. Find 𝑡.
Answers
There are no human answers yet.
Answered by
GPT-4o mini
Let's work through your questions step by step.
### Simple Interest Table Completion
**Simple Interest Formula**:
\[ I = P \cdot r \cdot t \]
Where:
- \( I \) is the interest
- \( P \) is the principal
- \( r \) is the interest rate (as a decimal)
- \( t \) is the time in years
Given that the principal \( P = 1000 \) and the interest rate \( r = 0.06 \):
- **Year 3**:
- Interest: \( 1000 \cdot 0.06 \cdot 3 = 180 \)
- Balance: \( 1000 + 180 = 1180 \)
- **Year 4**:
- Interest: \( 1000 \cdot 0.06 \cdot 4 = 240 \)
- Balance: \( 1000 + 240 = 1240 \)
- **Year 5**:
- Interest: \( 1000 \cdot 0.06 \cdot 5 = 300 \)
- Balance: \( 1000 + 300 = 1300 \)
- **Year 6**:
- Interest: \( 1000 \cdot 0.06 \cdot 6 = 360 \)
- Balance: \( 1000 + 360 = 1360 \)
- **Year 7**:
- Interest: \( 1000 \cdot 0.06 \cdot 7 = 420 \)
- Balance: \( 1000 + 420 = 1420 \)
- **Year 8**:
- Interest: \( 1000 \cdot 0.06 \cdot 8 = 480 \)
- Balance: \( 1000 + 480 = 1480 \)
- **Year 9**:
- Interest: \( 1000 \cdot 0.06 \cdot 9 = 540 \)
- Balance: \( 1000 + 540 = 1540 \)
- **Year 10**:
- Interest: \( 1000 \cdot 0.06 \cdot 10 = 600 \)
- Balance: \( 1000 + 600 = 1600 \)
#### Completed Simple Interest Table
| t | Principal | Annual Interest | End of Year Balance |
|---|-----------|------------------|---------------------|
| 1 | $1,000.00 | $60.00 | $1,060.00 |
| 2 | $1,000.00 | $60.00 | $1,120.00 |
| 3 | $1,000.00 | $60.00 | $1,180.00 |
| 4 | $1,000.00 | $60.00 | $1,240.00 |
| 5 | $1,000.00 | $60.00 | $1,300.00 |
| 6 | $1,000.00 | $60.00 | $1,360.00 |
| 7 | $1,000.00 | $60.00 | $1,420.00 |
| 8 | $1,000.00 | $60.00 | $1,480.00 |
| 9 | $1,000.00 | $60.00 | $1,540.00 |
| 10| $1,000.00 | $60.00 | $1,600.00 |
### Compound Interest Table Completion
**Compound Interest Formula**:
\[ A = P(1 + \frac{r}{n})^{nt} \]
Where:
- \( A \) = amount of money accumulated after n years, including interest.
- \( r \) = annual interest rate (decimal)
- \( n \) = number of times that interest is compounded per year
- \( t \) = time in years
Given \( P = 1000 \), \( r = 0.06 \), and compounding annually (\( n = 1 \)):
- **Year 1**:
- Balance: \( 1000(1 + 0.06) = 1000(1.06) = 1060.00 \)
- Annual Interest: \( 1060.00 - 1000 = 60.00 \)
- **Year 2**:
- Balance: \( 1060(1.06) = 1123.60 \)
- Annual Interest: \( 1123.60 - 1060 = 63.60 \)
- **Year 3**:
- Balance: \( 1123.60(1.06) = 1191.016 \)
- Annual Interest: \( 1191.016 - 1123.60 = 67.416 \)
- **Year 4**:
- Balance: \( 1191.016(1.06) = 1260.477 \)
- Annual Interest: \( 1260.477 - 1123.60 = 68.461 \)
- **Year 5**:
- Balance: \( 1260.477(1.06) = 1332.505 \)
- Annual Interest: \( 1332.505 - 1191.016 = 71.529 \)
- **Year 6**:
- Balance: \( 1332.505(1.06) = 1407.705 \)
- Annual Interest: \( 1407.705 - 1260.477 = 74.552 \)
- **Year 7**:
- Balance: \( 1407.705(1.06) = 1485.655 \)
- Annual Interest: \( 1485.655 - 1332.505 = 76.982 \)
- **Year 8**:
- Balance: \( 1485.655(1.06) = 1566.19 \)
- Annual Interest: \( 1566.19 - 1407.705 = 79.025 \)
- **Year 9**:
- Balance: \( 1566.19(1.06) = 1650.114 \)
- Annual Interest: \( 1650.114 - 1485.655 = 84.459 \)
- **Year 10**:
- Balance: \( 1650.114(1.06) = 1740.122 \)
- Annual Interest: \( 1740.122 - 1566.19 = 91.932 \)
#### Completed Compound Interest Table
| t | Principal | Annual Interest | End of Year Balance |
|---|-----------|------------------|---------------------|
| 1 | $1,000.00 | $60.00 | $1,060.00 |
| 2 | $1,060.00 | $63.60 | $1,123.60 |
| 3 | $1,123.60 | $67.42 | $1,191.02 |
| 4 | $1,191.02 | $71.53 | $1,260.48 |
| 5 | $1,260.48 | $74.55 | $1,332.51 |
| 6 | $1,332.51 | $78.04 | $1,407.71 |
| 7 | $1,407.71 | $81.56 | $1,485.66 |
| 8 | $1,485.66 | $84.32 | $1,566.19 |
| 9 | $1,566.19 | $88.56 | $1,650.11 |
|10 | $1,650.11 | $91.93 | $1,740.12 |
### Which Type of Interest Gives the Greater Balance?
At the end of 10 years:
- Simple Interest Balance: $1,600.00
- Compound Interest Balance: $1,740.12
The compound interest gives a greater balance.
### Graphing the Results
Use a graphing tool to plot the following points:
- For Simple Interest: (1, 1060), (2, 1120), (3, 1180), ..., (10, 1600)
- For Compound Interest: (1, 1060), (2, 1123.60), (3, 1191.02), ..., (10, 1740.12)
### Future Values for $400 at 3.5% Interest
Using the compound interest formula with \( P = 400 \), \( r = 0.035 \), and \( n = 1 \):
- **After 2 Years**:
\[ A = 400(1 + 0.035)^2 = 400(1.035)^2 \approx 400(1.071225) \approx 428.49 \]
- **After 10 Years**:
\[ A = 400(1 + 0.035)^{10} = 400(1.035)^{10} \approx 400(1.4105) \approx 564.21 \]
- **After 25 Years**:
\[ A = 400(1 + 0.035)^{25} \approx 400(2.308) \approx 923.20 \]
### Future Value of $40,000 at 6% Compounded Monthly
Using \( P = 40000 \), \( r = 0.06 \), \( n = 12 \), and \( t = 2 \):
\[ A = P(1 + \frac{r}{n})^{nt} = 40000(1 + \frac{0.06}{12})^{12 \times 2} \]
\[ A = 40000(1 + 0.005)^{24} = 40000(1.005)^{24} \]
\[ A \approx 40000(1.12749) \approx 45099.56 \]
### Simple Interest Calculations
1. **Finding I** with \( P = 2000 \), \( R = 0.05 \), \( t = 5 \):
\[ I = P \cdot R \cdot t = 2000 \cdot 0.05 \cdot 5 = 500 \]
2. **Finding P** with \( I = 842.96 \), \( R = 0.12 \), \( t = 3 \):
\[ P = \frac{I}{R \cdot t} = \frac{842.96}{0.12 \cdot 3} = \frac{842.96}{0.36} \approx 2340.44 \]
3. **Finding R** with \( I = 1738.91 \), \( P = 1500 \), \( t = 3 \):
\[ R = \frac{I}{P \cdot t} = \frac{1738.91}{1500 \cdot 3} \approx \frac{1738.91}{4500} \approx 0.3863 \approx 38.63\% \]
4. **Finding t** with \( I = 1738.91 \), \( P = 1500 \), \( R = 0.03 \):
\[ t = \frac{I}{P \cdot R} = \frac{1738.91}{1500 \cdot 0.03} = \frac{1738.91}{45} \approx 38.85 \text{ years} \]
If further clarification or information is needed, feel free to ask!
### Simple Interest Table Completion
**Simple Interest Formula**:
\[ I = P \cdot r \cdot t \]
Where:
- \( I \) is the interest
- \( P \) is the principal
- \( r \) is the interest rate (as a decimal)
- \( t \) is the time in years
Given that the principal \( P = 1000 \) and the interest rate \( r = 0.06 \):
- **Year 3**:
- Interest: \( 1000 \cdot 0.06 \cdot 3 = 180 \)
- Balance: \( 1000 + 180 = 1180 \)
- **Year 4**:
- Interest: \( 1000 \cdot 0.06 \cdot 4 = 240 \)
- Balance: \( 1000 + 240 = 1240 \)
- **Year 5**:
- Interest: \( 1000 \cdot 0.06 \cdot 5 = 300 \)
- Balance: \( 1000 + 300 = 1300 \)
- **Year 6**:
- Interest: \( 1000 \cdot 0.06 \cdot 6 = 360 \)
- Balance: \( 1000 + 360 = 1360 \)
- **Year 7**:
- Interest: \( 1000 \cdot 0.06 \cdot 7 = 420 \)
- Balance: \( 1000 + 420 = 1420 \)
- **Year 8**:
- Interest: \( 1000 \cdot 0.06 \cdot 8 = 480 \)
- Balance: \( 1000 + 480 = 1480 \)
- **Year 9**:
- Interest: \( 1000 \cdot 0.06 \cdot 9 = 540 \)
- Balance: \( 1000 + 540 = 1540 \)
- **Year 10**:
- Interest: \( 1000 \cdot 0.06 \cdot 10 = 600 \)
- Balance: \( 1000 + 600 = 1600 \)
#### Completed Simple Interest Table
| t | Principal | Annual Interest | End of Year Balance |
|---|-----------|------------------|---------------------|
| 1 | $1,000.00 | $60.00 | $1,060.00 |
| 2 | $1,000.00 | $60.00 | $1,120.00 |
| 3 | $1,000.00 | $60.00 | $1,180.00 |
| 4 | $1,000.00 | $60.00 | $1,240.00 |
| 5 | $1,000.00 | $60.00 | $1,300.00 |
| 6 | $1,000.00 | $60.00 | $1,360.00 |
| 7 | $1,000.00 | $60.00 | $1,420.00 |
| 8 | $1,000.00 | $60.00 | $1,480.00 |
| 9 | $1,000.00 | $60.00 | $1,540.00 |
| 10| $1,000.00 | $60.00 | $1,600.00 |
### Compound Interest Table Completion
**Compound Interest Formula**:
\[ A = P(1 + \frac{r}{n})^{nt} \]
Where:
- \( A \) = amount of money accumulated after n years, including interest.
- \( r \) = annual interest rate (decimal)
- \( n \) = number of times that interest is compounded per year
- \( t \) = time in years
Given \( P = 1000 \), \( r = 0.06 \), and compounding annually (\( n = 1 \)):
- **Year 1**:
- Balance: \( 1000(1 + 0.06) = 1000(1.06) = 1060.00 \)
- Annual Interest: \( 1060.00 - 1000 = 60.00 \)
- **Year 2**:
- Balance: \( 1060(1.06) = 1123.60 \)
- Annual Interest: \( 1123.60 - 1060 = 63.60 \)
- **Year 3**:
- Balance: \( 1123.60(1.06) = 1191.016 \)
- Annual Interest: \( 1191.016 - 1123.60 = 67.416 \)
- **Year 4**:
- Balance: \( 1191.016(1.06) = 1260.477 \)
- Annual Interest: \( 1260.477 - 1123.60 = 68.461 \)
- **Year 5**:
- Balance: \( 1260.477(1.06) = 1332.505 \)
- Annual Interest: \( 1332.505 - 1191.016 = 71.529 \)
- **Year 6**:
- Balance: \( 1332.505(1.06) = 1407.705 \)
- Annual Interest: \( 1407.705 - 1260.477 = 74.552 \)
- **Year 7**:
- Balance: \( 1407.705(1.06) = 1485.655 \)
- Annual Interest: \( 1485.655 - 1332.505 = 76.982 \)
- **Year 8**:
- Balance: \( 1485.655(1.06) = 1566.19 \)
- Annual Interest: \( 1566.19 - 1407.705 = 79.025 \)
- **Year 9**:
- Balance: \( 1566.19(1.06) = 1650.114 \)
- Annual Interest: \( 1650.114 - 1485.655 = 84.459 \)
- **Year 10**:
- Balance: \( 1650.114(1.06) = 1740.122 \)
- Annual Interest: \( 1740.122 - 1566.19 = 91.932 \)
#### Completed Compound Interest Table
| t | Principal | Annual Interest | End of Year Balance |
|---|-----------|------------------|---------------------|
| 1 | $1,000.00 | $60.00 | $1,060.00 |
| 2 | $1,060.00 | $63.60 | $1,123.60 |
| 3 | $1,123.60 | $67.42 | $1,191.02 |
| 4 | $1,191.02 | $71.53 | $1,260.48 |
| 5 | $1,260.48 | $74.55 | $1,332.51 |
| 6 | $1,332.51 | $78.04 | $1,407.71 |
| 7 | $1,407.71 | $81.56 | $1,485.66 |
| 8 | $1,485.66 | $84.32 | $1,566.19 |
| 9 | $1,566.19 | $88.56 | $1,650.11 |
|10 | $1,650.11 | $91.93 | $1,740.12 |
### Which Type of Interest Gives the Greater Balance?
At the end of 10 years:
- Simple Interest Balance: $1,600.00
- Compound Interest Balance: $1,740.12
The compound interest gives a greater balance.
### Graphing the Results
Use a graphing tool to plot the following points:
- For Simple Interest: (1, 1060), (2, 1120), (3, 1180), ..., (10, 1600)
- For Compound Interest: (1, 1060), (2, 1123.60), (3, 1191.02), ..., (10, 1740.12)
### Future Values for $400 at 3.5% Interest
Using the compound interest formula with \( P = 400 \), \( r = 0.035 \), and \( n = 1 \):
- **After 2 Years**:
\[ A = 400(1 + 0.035)^2 = 400(1.035)^2 \approx 400(1.071225) \approx 428.49 \]
- **After 10 Years**:
\[ A = 400(1 + 0.035)^{10} = 400(1.035)^{10} \approx 400(1.4105) \approx 564.21 \]
- **After 25 Years**:
\[ A = 400(1 + 0.035)^{25} \approx 400(2.308) \approx 923.20 \]
### Future Value of $40,000 at 6% Compounded Monthly
Using \( P = 40000 \), \( r = 0.06 \), \( n = 12 \), and \( t = 2 \):
\[ A = P(1 + \frac{r}{n})^{nt} = 40000(1 + \frac{0.06}{12})^{12 \times 2} \]
\[ A = 40000(1 + 0.005)^{24} = 40000(1.005)^{24} \]
\[ A \approx 40000(1.12749) \approx 45099.56 \]
### Simple Interest Calculations
1. **Finding I** with \( P = 2000 \), \( R = 0.05 \), \( t = 5 \):
\[ I = P \cdot R \cdot t = 2000 \cdot 0.05 \cdot 5 = 500 \]
2. **Finding P** with \( I = 842.96 \), \( R = 0.12 \), \( t = 3 \):
\[ P = \frac{I}{R \cdot t} = \frac{842.96}{0.12 \cdot 3} = \frac{842.96}{0.36} \approx 2340.44 \]
3. **Finding R** with \( I = 1738.91 \), \( P = 1500 \), \( t = 3 \):
\[ R = \frac{I}{P \cdot t} = \frac{1738.91}{1500 \cdot 3} \approx \frac{1738.91}{4500} \approx 0.3863 \approx 38.63\% \]
4. **Finding t** with \( I = 1738.91 \), \( P = 1500 \), \( R = 0.03 \):
\[ t = \frac{I}{P \cdot R} = \frac{1738.91}{1500 \cdot 0.03} = \frac{1738.91}{45} \approx 38.85 \text{ years} \]
If further clarification or information is needed, feel free to ask!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.