Question

pre-algebra
assignment
use simple interest to find the ending balance.

1. $6,300 at 6% for 2 years
2. $56,600 at 5% for 2 years
3. $12,300 at 5% for 5 years
4. $20,600 at 15% for 2 years
5. $29,900 at 6% for 7 years
6. $1,090 at 7% for 3 years
7. $45,000 at 6% for 7 years
8. $10,600 at 4% for 7 years
9. $1,870 at 4% for 2 years
10. $27,000 at 15% for 8 years
11. $620 at 1% for 4 years
12. $52,000 at 9% for 2 years
13. $55,500 at 12% for 2 years
14. $225 at 7% for 4 years
15. $28,400 at 7% for 6 years
16. $480 at 14% for 3 years
17. $19,600 at 5% for 2 years
18. $45,500 at 15% for 2 years
19. $7,000 at 9% for 7 years
20. $1,380 at 11% for 4 years

Explanation:

Let's solve problem 1) as an example. The formula for simple interest is $I = P \times r \times t$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal), and $t$ is the time in years. The ending balance $A$ is the principal plus the interest, so $A = P + I = P(1 + rt)$.

Step 1: Identify the values

For problem 1): $P = \$6,300$, $r = 6\% = 0.06$, $t = 2$ years.

Step 2: Calculate the ending balance

Using the formula $A = P(1 + rt)$, substitute the values:

$$\begin{align*} A &= 6300 \times (1 + 0.06 \times 2)\\ &= 6300 \times (1 + 0.12)\\ &= 6300 \times 1.12\\ &= 7056 \end{align*}$$

Step 1: Identify the values

For problem 2): $P = \$56,600$, $r = 5\% = 0.05$, $t = 2$ years.

Step 2: Calculate the ending balance

Using the formula $A = P(1 + rt)$:

$$\begin{align*} A &= 56600 \times (1 + 0.05 \times 2)\\ &= 56600 \times (1 + 0.1)\\ &= 56600 \times 1.1\\ &= 62260 \end{align*}$$

Answer:

The ending balance for problem 1) is $\$7,056$.

If you want solutions for other problems, you can follow the same steps: identify $P$, $r$ (convert percentage to decimal), $t$, then use $A = P(1 + rt)$ to calculate the ending balance. For example, let's solve problem 2):