Question

A. You deposit $1000 in a savings account that earns 6% interest per year.
Complete each table.
Simple Interest
𝑡𝑡 Principal Annual
Interest
End of
year
Balance
1 $1000.00 $60.00 $1060.00
Compound Interest
𝑡𝑡
Principal
and
Interest
Annual
Interest
End of
year
Balance
1 $1000.00 $60.00 $1060.00
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2 $1000.00 $60.00 $1120.00
3
4
5
6
7
8
9
10
2 $1060.00 $63.60 $1123.60
3
4
5
6
7
8
9
10
Which type of interest gives the greater balance?
B. Graph the end-of-year balances for each type of interest in B on the same
graph. Label each.
C. 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?
D. 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?
E. Find the missing variable for compound interest compounded annually.
a. 𝑃𝑃 = $2000, 𝑦𝑦 = 5%, 𝑡𝑡 = 5. Find I.
b. 𝐼𝐼 = $2566.72, 𝑦𝑦 = 12%, 𝑡𝑡 = 3. Find P.
c. 𝐼𝐼 = $842.96, 𝑃𝑃 = 500, 𝑡𝑡 = 2. Find R.
d. 𝐼𝐼 = $1738.91, 𝑃𝑃 = 1500, 𝑦𝑦 = 3%. Find t.

Answer 1

for the graph in part b can you give me an equation to help

Answer 2

give me a simple equation nothing els

Answer 3

Let's go through each part of your question step by step.

### A. Completing the Tables

**1. Simple Interest Table:**
The formula for calculating simple interest is:
\[ I = P \times r \times t \]
Where:
- \( I \) = Interest
- \( P \) = Principal
- \( r \) = Rate (as a decimal)
- \( t \) = Time (in years)

For 1 year:
- \( P = 1000 \)
- \( r = 0.06 \)
- \( t = 1 \)

Calculating interest:
\[ I = 1000 \times 0.06 \times 1 = 60 \]

End of Year Balance:
\[ \text{Balance} = P + I = 1000 + 60 = 1060 \]

Continuing with the simple interest for subsequent years, simply add \( I = 60 \) for each year:
- 2: Balance = 1060 + 60 = 1120
- 3: Balance = 1120 + 60 = 1180
- 4: Balance = 1180 + 60 = 1240
- 5: Balance = 1240 + 60 = 1300
- 6: Balance = 1300 + 60 = 1360
- 7: Balance = 1360 + 60 = 1420
- 8: Balance = 1420 + 60 = 1480
- 9: Balance = 1480 + 60 = 1540
- 10: Balance = 1540 + 60 = 1600

### Simple Interest Table
| t | Principal | Annual Interest | End of year Balance |
|----|-----------|----------------|----------------------|
| 1 | $1000.00 | $60.00 | $1060.00 |
| 2 | $1000.00 | $60.00 | $1120.00 |
| 3 | $1000.00 | $60.00 | $1180.00 |
| 4 | $1000.00 | $60.00 | $1240.00 |
| 5 | $1000.00 | $60.00 | $1300.00 |
| 6 | $1000.00 | $60.00 | $1360.00 |
| 7 | $1000.00 | $60.00 | $1420.00 |
| 8 | $1000.00 | $60.00 | $1480.00 |
| 9 | $1000.00 | $60.00 | $1540.00 |
| 10 | $1000.00 | $60.00 | $1600.00 |

**2. Compound Interest Table:**
The formula for compound interest is:
\[ A = P(1 + r)^t \]
Where:
- \( A \) = Amount (or balance after t years)
- \( P \) = Principal
- \( r \) = Rate (as a decimal)
- \( t \) = Time (in years)

Calculating for each year:
- After 1 year:
\[ A = 1000(1 + 0.06)^1 = 1000(1.06) = 1060 \]
- After 2 years:
\[ A = 1000(1 + 0.06)^2 = 1000(1.06)^2 = 1000(1.1236) = 1123.60 \]
- After 3 years:
\[ A = 1000(1 + 0.06)^3 = 1000(1.06)^3 = 1000(1.191016) = 1191.02 \]
- After 4 years:
\[ A = 1000(1 + 0.06)^4 = 1000(1.26247696) = 1262.48 \]
- After 5 years:
\[ A = 1000(1 + 0.06)^5 = 1000(1.33822594) = 1338.23 \]
- After 6 years:
\[ A = 1000(1 + 0.06)^6 = 1000(1.41851917) = 1418.52 \]
- After 7 years:
\[ A = 1000(1 + 0.06)^7 = 1000(1.50363033) = 1503.63 \]
- After 8 years:
\[ A = 1000(1 + 0.06)^8 = 1000(1.59385336) = 1593.85 \]
- After 9 years:
\[ A = 1000(1 + 0.06)^9 = 1000(1.68947818) = 1689.48 \]
- After 10 years:
\[ A = 1000(1 + 0.06)^{10} = 1000(1.79084769) = 1790.85 \]

### Compound Interest Table
| t | Principal | Annual Interest | End of year Balance |
|----|-----------|----------------|----------------------|
| 1 | $1000.00 | $60.00 | $1060.00 |
| 2 | $1060.00 | $63.60 | $1123.60 |
| 3 | $1123.60 | $67.42 | $1191.02 |
| 4 | $1191.02 | $71.46 | $1262.48 |
| 5 | $1262.48 | $75.75 | $1338.23 |
| 6 | $1338.23 | $80.29 | $1418.52 |
| 7 | $1418.52 | $86.11 | $1503.63 |
| 8 | $1503.63 | $92.22 | $1593.85 |
| 9 | $1593.85 | $98.59 | $1689.48 |
| 10 | $1689.48 | $106.37 | $1790.85 |

### Which type of interest gives the greater balance?
At the end of 10 years:
- Simple Interest Balance: $1600.00
- Compound Interest Balance: $1790.85

So, compound interest gives a greater balance.

### B. Graphing the Balances
To graph, you will plot the end-of-year balances for both simple and compound interest on the same graph with the x-axis representing time (years) and the y-axis representing account balance. The simple interest line will be straight, while the compound interest line will be a curve that goes up more sharply as time increases.

### C. Calculate Future Balances for $400 with 3.5% Interest Compounded Annually

Using the compound interest formula:
\[ A = P(1 + r)^t \]

Where:
- \( P = 400 \)
- \( r = 0.035 \)

1. After 2 years:
\[ A = 400(1 + 0.035)^2 = 400(1.035)^2 = 400(1.071225) \approx 428.49 \]

2. After 10 years:
\[ A = 400(1 + 0.035)^{10} = 400(1.035)^{10} = 400(1.4107) \approx 564.28 \]

3. After 25 years:
\[ A = 400(1 + 0.035)^{25} = 400(1.035)^{25} = 400(2.2082) \approx 883.28 \]

### D. Final Balance After 2 Years for $40,000 with 6% Compounded Monthly
Using the formula:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( P = 40000 \)
- \( r = 0.06 \)
- \( n = 12 \) (compounded monthly)
- \( t = 2 \)

Calculating:
\[ A = 40000 \left(1 + \frac{0.06}{12}\right)^{12 \times 2} = 40000 \left(1 + 0.005\right)^{24} = 40000(1.005)^{24} \]

Calculating \( (1.005)^{24} \):
\[ (1.005)^{24} \approx 1.12749 \]

So,
\[ A \approx 40000 \times 1.12749 \approx 45099.60 \]

Final Balance ≈ $45,099.60

### E. Find the Missing Variables for Compound Interest

a. Find \( I \)
\[ P = 2000, \, y = 5\% = 0.05, \, t = 5 \]
Using the formula:
\[ A = P(1 + r)^t \]
\[ A = 2000(1 + 0.05)^5 = 2000(1.27628) \approx 2552.56 \]
Now, find \( I \):
\[ I = A - P = 2552.56 - 2000 = 552.56 \]

b. Find \( P \)
\[ I = 2566.72, \, y = 12\% = 0.12, \, t = 3 \]
Using the formula
\[ A = P(1 + r)^t \]
First, find \( A \):
\[ A = I + P \]
Assuming \( r = 0.12 \) and \( t = 3 \):
\[ 2566.72 = P \cdot (1 + 0.12)^3 \]
\[ 2566.72 = P \cdot 1.404928 \]
Solving for \( P \):
\[ P = \frac{2566.72}{1.404928} \approx 1824.15 \]

c. Find \( R \)
\[ I = 842.96, \, P = 500, \, t = 2 \]
Using the rearranged formula:
\[ I = P \cdot r \cdot t \]
\[ 842.96 = 500 \cdot r \cdot 2 \]
\[ r = \frac{842.96}{500 \cdot 2} = \frac{842.96}{1000} \approx 0.84296 \text{ or } 84.296\% \]

d. Find \( t \)
\[ I = 1738.91, \, P = 1500, \, y = 3\% = 0.03 \]
Using the formula:
\[ 1738.91 = 1500 \cdot 0.03 \cdot t \]
Solving for \( t \):
\[ t = \frac{1738.91}{1500 \cdot 0.03} = \frac{1738.91}{45} \approx 38.86 \]
So \( t \approx 39 \) years.

Feel free to ask if you need further clarification or assistance!

Answer 4

Certainly! To graph the end-of-year balances for both simple and compound interest, you'll want to convert your calculated balances from both methods into equations that can be plotted.

### 1. Simple Interest Equation

For simple interest, the balance at the end of year \( t \) can be expressed with the following equation:

\[
B_{\text{SI}}(t) = P + (P \cdot r \cdot t)
\]

Where:
- \( B_{\text{SI}}(t) \) is the balance at the end of year \( t \).
- \( P = 1000 \) (initial principal)
- \( r = 0.06 \) (annual interest rate)
- \( t \) is the number of years.

Replacing \( P \) and \( r \):

\[
B_{\text{SI}}(t) = 1000 + (1000 \cdot 0.06 \cdot t)
\]

This simplifies to:

\[
B_{\text{SI}}(t) = 1000 + 60t
\]

### 2. Compound Interest Equation

For compound interest, the balance at the end of year \( t \) can be expressed with the following compound interest formula:

\[
B_{\text{CI}}(t) = P(1 + r)^t
\]

Where:
- \( B_{\text{CI}}(t) \) is the balance at the end of year \( t \).
- \( P = 1000 \) (initial principal)
- \( r = 0.06 \) (annual interest rate)

Substituting \( P \) and \( r \):

\[
B_{\text{CI}}(t) = 1000(1 + 0.06)^t
\]

This simplifies to:

\[
B_{\text{CI}}(t) = 1000(1.06)^t
\]

### Summary of Equations to Plot

1. **Simple Interest**:
\[ B_{\text{SI}}(t) = 1000 + 60t \]

2. **Compound Interest**:
\[ B_{\text{CI}}(t) = 1000(1.06)^t \]

### Points to Plot

You can use integer values for \( t \) from 0 to 10 to calculate the values of \( B_{\text{SI}}(t) \) and \( B_{\text{CI}}(t) \).

- For \( t = 0 \) to \( t = 10 \):
- Use the equations to calculate the balance for each year.

### Example Calculation

For example, if \( t = 0 \):
- Simple Interest:
\[ B_{\text{SI}}(0) = 1000 + 60(0) = 1000 \]
- Compound Interest:
\[ B_{\text{CI}}(0) = 1000(1.06)^0 = 1000 \]

If \( t = 1 \):
- Simple Interest:
\[ B_{\text{SI}}(1) = 1000 + 60(1) = 1060 \]
- Compound Interest:
\[ B_{\text{CI}}(1) = 1000(1.06)^1 = 1060 \]

Continue this for each year up to \( t = 10 \).

Now, you can plot these values on a graph, with the x-axis as \( t \) (years) and the y-axis as the balance. This will give you two lines: one linear for simple interest and one exponential for compound interest.

Answer 5

Sure! Here are the simple equations for both interest types:

### Simple Interest
\[ B_{\text{SI}}(t) = 1000 + 60t \]

### Compound Interest
\[ B_{\text{CI}}(t) = 1000(1.06)^t \]