Question
Erica opened a savings account and deposited $200.00 as principal. The account earns 10% interest, compounded annually. What is the balance after 6 years?
Use the formula A=P1+
r
n
nt, where A is the balance (final amount), P is the principal (starting amount), r is the interest rate expressed as a decimal, n is the number of times per year that the interest is compounded, and t is the time in years.
Round your answer to the nearest cent.
$
Use the formula A=P1+
r
n
nt, where A is the balance (final amount), P is the principal (starting amount), r is the interest rate expressed as a decimal, n is the number of times per year that the interest is compounded, and t is the time in years.
Round your answer to the nearest cent.
$
Answers
GPT-4o mini
It seems like there's a small error in the provided formula. The correct formula for compound interest is:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the final amount (balance)
- \( P \) is the principal (starting amount)
- \( r \) is the annual interest rate (as a decimal)
- \( n \) is the number of times interest is compounded per year
- \( t \) is the number of years
In this case:
- \( P = 200 \)
- \( r = 0.10 \) (10% expressed as a decimal)
- \( n = 1 \) (since it is compounded annually)
- \( t = 6 \)
Now we can plug these values into the formula:
\[
A = 200 \left(1 + \frac{0.10}{1}\right)^{1 \cdot 6}
\]
\[
A = 200 \left(1 + 0.10\right)^{6}
\]
\[
A = 200 \left(1.10\right)^{6}
\]
Now we need to calculate \( (1.10)^{6} \):
\[
(1.10)^{6} \approx 1.771561
\]
Now, multiply by the principal:
\[
A \approx 200 \times 1.771561 \approx 354.3122
\]
Rounding to the nearest cent gives:
\[
A \approx 354.31
\]
Thus, the balance after 6 years is **$354.31**.
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) is the final amount (balance)
- \( P \) is the principal (starting amount)
- \( r \) is the annual interest rate (as a decimal)
- \( n \) is the number of times interest is compounded per year
- \( t \) is the number of years
In this case:
- \( P = 200 \)
- \( r = 0.10 \) (10% expressed as a decimal)
- \( n = 1 \) (since it is compounded annually)
- \( t = 6 \)
Now we can plug these values into the formula:
\[
A = 200 \left(1 + \frac{0.10}{1}\right)^{1 \cdot 6}
\]
\[
A = 200 \left(1 + 0.10\right)^{6}
\]
\[
A = 200 \left(1.10\right)^{6}
\]
Now we need to calculate \( (1.10)^{6} \):
\[
(1.10)^{6} \approx 1.771561
\]
Now, multiply by the principal:
\[
A \approx 200 \times 1.771561 \approx 354.3122
\]
Rounding to the nearest cent gives:
\[
A \approx 354.31
\]
Thus, the balance after 6 years is **$354.31**.
Eliana puts $5,619.00 into an account to use for school expenses. The account earns 4.04% interest, compounded quarterly. How much will be in the account after 10 years?
Use the formula A=P1+
r
n
nt, where A is the balance (final amount), P is the principal (starting amount), r is the interest rate expressed as a decimal, n is the number of times per year that the interest is compounded, and t is the time in years.
Round your answer to the nearest cent.
$
Use the formula A=P1+
r
n
nt, where A is the balance (final amount), P is the principal (starting amount), r is the interest rate expressed as a decimal, n is the number of times per year that the interest is compounded, and t is the time in years.
Round your answer to the nearest cent.
$
GPT-4o mini
To calculate the final amount in the account after 10 years with a principal amount that earns compounded interest, we will use the following formula:
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \( A \) = final amount (balance)
- \( P \) = principal (starting amount)
- \( r \) = annual interest rate (as a decimal)
- \( n \) = number of times interest is compounded per year
- \( t \) = number of years
In your case:
- \( P = 5,619.00 \)
- \( r = 4.04\% = 0.0404 \) (as a decimal)
- \( n = 4 \) (since the interest is compounded quarterly)
- \( t = 10 \)
Now, we can plug in these values into the formula:
\[
A = 5619 \left(1 + \frac{0.0404}{4}\right)^{4 \times 10}
\]
Calculate \( \frac{0.0404}{4} \):
\[
\frac{0.0404}{4} = 0.0101
\]
Now we can continue with the calculation:
\[
A = 5619 \left(1 + 0.0101\right)^{40}
\]
\[
A = 5619 \left(1.0101\right)^{40}
\]
Next, we need to calculate \( (1.0101)^{40} \):
\[
(1.0101)^{40} \approx 1.488864
\]
Now, multiply by the principal:
\[
A \approx 5619 \times 1.488864 \approx 8367.35956
\]
Rounding to the nearest cent gives:
\[
A \approx 8367.36
\]
Thus, the amount in the account after 10 years will be **$8,367.36**.
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \( A \) = final amount (balance)
- \( P \) = principal (starting amount)
- \( r \) = annual interest rate (as a decimal)
- \( n \) = number of times interest is compounded per year
- \( t \) = number of years
In your case:
- \( P = 5,619.00 \)
- \( r = 4.04\% = 0.0404 \) (as a decimal)
- \( n = 4 \) (since the interest is compounded quarterly)
- \( t = 10 \)
Now, we can plug in these values into the formula:
\[
A = 5619 \left(1 + \frac{0.0404}{4}\right)^{4 \times 10}
\]
Calculate \( \frac{0.0404}{4} \):
\[
\frac{0.0404}{4} = 0.0101
\]
Now we can continue with the calculation:
\[
A = 5619 \left(1 + 0.0101\right)^{40}
\]
\[
A = 5619 \left(1.0101\right)^{40}
\]
Next, we need to calculate \( (1.0101)^{40} \):
\[
(1.0101)^{40} \approx 1.488864
\]
Now, multiply by the principal:
\[
A \approx 5619 \times 1.488864 \approx 8367.35956
\]
Rounding to the nearest cent gives:
\[
A \approx 8367.36
\]
Thus, the amount in the account after 10 years will be **$8,367.36**.
Eliana puts $5,619.00 into an account to use for school expenses. The account earns 4.04% interest, compounded quarterly. How much will be in the account after 10 years?
Use the formula A=P1+
r
n
nt, where A is the balance (final amount), P is the principal (starting amount), r is the interest rate expressed as a decimal, n is the number of times per year that the interest is compounded, and t is the time in years.
Round your answer to the nearest cent
Use the formula A=P1+
r
n
nt, where A is the balance (final amount), P is the principal (starting amount), r is the interest rate expressed as a decimal, n is the number of times per year that the interest is compounded, and t is the time in years.
Round your answer to the nearest cent
GPT-4o mini
To calculate the amount in Eliana's account after 10 years with a principal amount that earns compounded interest, we will use the correct formula for compound interest:
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \( A \) is the final amount (balance)
- \( P \) is the principal (starting amount)
- \( r \) is the annual interest rate (as a decimal)
- \( n \) is the number of times interest is compounded per year
- \( t \) is the number of years
For Eliana's account:
- \( P = 5619.00 \)
- \( r = 4.04 \% = 0.0404 \)
- \( n = 4 \) (since interest is compounded quarterly)
- \( t = 10 \)
Now we can plug in these values into the formula:
\[
A = 5619 \left(1 + \frac{0.0404}{4}\right)^{4 \cdot 10}
\]
First, calculate \( \frac{0.0404}{4} \):
\[
\frac{0.0404}{4} = 0.0101
\]
Now we substitute back into the equation:
\[
A = 5619 \left(1 + 0.0101\right)^{40}
\]
\[
A = 5619 \left(1.0101\right)^{40}
\]
Next, calculate \( (1.0101)^{40} \):
Using a calculator:
\[
(1.0101)^{40} \approx 1.48985
\]
Now multiply by the principal:
\[
A \approx 5619 \times 1.48985 \approx 8369.10015
\]
Finally, rounding to the nearest cent:
\[
A \approx 8369.10
\]
Thus, the amount in the account after 10 years will be **$8,369.10**.
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \( A \) is the final amount (balance)
- \( P \) is the principal (starting amount)
- \( r \) is the annual interest rate (as a decimal)
- \( n \) is the number of times interest is compounded per year
- \( t \) is the number of years
For Eliana's account:
- \( P = 5619.00 \)
- \( r = 4.04 \% = 0.0404 \)
- \( n = 4 \) (since interest is compounded quarterly)
- \( t = 10 \)
Now we can plug in these values into the formula:
\[
A = 5619 \left(1 + \frac{0.0404}{4}\right)^{4 \cdot 10}
\]
First, calculate \( \frac{0.0404}{4} \):
\[
\frac{0.0404}{4} = 0.0101
\]
Now we substitute back into the equation:
\[
A = 5619 \left(1 + 0.0101\right)^{40}
\]
\[
A = 5619 \left(1.0101\right)^{40}
\]
Next, calculate \( (1.0101)^{40} \):
Using a calculator:
\[
(1.0101)^{40} \approx 1.48985
\]
Now multiply by the principal:
\[
A \approx 5619 \times 1.48985 \approx 8369.10015
\]
Finally, rounding to the nearest cent:
\[
A \approx 8369.10
\]
Thus, the amount in the account after 10 years will be **$8,369.10**.
Complete the table for the function y=6x.
x y
–
1
0
1
2
Now, graph the function.
Plot two points to graph the function. an 0 to 8 axis
x y
–
1
0
1
2
Now, graph the function.
Plot two points to graph the function. an 0 to 8 axis
GPT-4o mini
To complete the table for the function \( y = 6x \), we will calculate the value of \( y \) for each given value of \( x \).
### Completing the Table
1. **For \( x = -1 \)**:
\[
y = 6(-1) = -6
\]
2. **For \( x = 0 \)**:
\[
y = 6(0) = 0
\]
3. **For \( x = 1 \)**:
\[
y = 6(1) = 6
\]
4. **For \( x = 2 \)**:
\[
y = 6(2) = 12
\]
### Completed Table
| \( x \) | \( y \) |
|---------|---------|
| -1 | -6 |
| 0 | 0 |
| 1 | 6 |
| 2 | 12 |
### Graphing the Function
To graph the function \( y = 6x \), we will plot the points calculated from the table.
**Points to Plot:**
1. \((-1, -6)\)
2. \((0, 0)\)
3. \((1, 6)\)
4. \((2, 12)\) — Note that this point is outside the 0 to 8 range on the graph, so we can focus on plotting the ones within your specified axis.
#### Selected Points for Graphing:
- \((0, 0)\)
- \((1, 6)\)
### Graphing Steps:
1. Draw an x-axis and a y-axis.
2. Mark the points \((0, 0)\) and \((1, 6)\).
3. Draw a straight line through these points since this is a linear function.
### Visual Representation
```
8 | *
7 | *
6 | *
5 | *
4 | *
3 | *
2 | *
1 | *
|_________________________
-1 0 1 2 3 4 5 6 7 8
```
### Overview:
- The line passing through points \((0, 0)\) and \((1, 6)\) extends infinitely in both directions, representing the function \( y = 6x \).
### Completing the Table
1. **For \( x = -1 \)**:
\[
y = 6(-1) = -6
\]
2. **For \( x = 0 \)**:
\[
y = 6(0) = 0
\]
3. **For \( x = 1 \)**:
\[
y = 6(1) = 6
\]
4. **For \( x = 2 \)**:
\[
y = 6(2) = 12
\]
### Completed Table
| \( x \) | \( y \) |
|---------|---------|
| -1 | -6 |
| 0 | 0 |
| 1 | 6 |
| 2 | 12 |
### Graphing the Function
To graph the function \( y = 6x \), we will plot the points calculated from the table.
**Points to Plot:**
1. \((-1, -6)\)
2. \((0, 0)\)
3. \((1, 6)\)
4. \((2, 12)\) — Note that this point is outside the 0 to 8 range on the graph, so we can focus on plotting the ones within your specified axis.
#### Selected Points for Graphing:
- \((0, 0)\)
- \((1, 6)\)
### Graphing Steps:
1. Draw an x-axis and a y-axis.
2. Mark the points \((0, 0)\) and \((1, 6)\).
3. Draw a straight line through these points since this is a linear function.
### Visual Representation
```
8 | *
7 | *
6 | *
5 | *
4 | *
3 | *
2 | *
1 | *
|_________________________
-1 0 1 2 3 4 5 6 7 8
```
### Overview:
- The line passing through points \((0, 0)\) and \((1, 6)\) extends infinitely in both directions, representing the function \( y = 6x \).
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