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?

Question

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.
$

Answer 1

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.

Answer 2

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.
$

Answer 3

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.

Answer 4

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

Answer 5

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.

Answer 6

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

Answer 7

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 \( y = 6x \).