Math  /  Algebra

QuestionIf $1000\$ 1000 is deposited in an account yielding an annual interest rate of 10%10 \% compounded semi-annually, how much will be in the account after 30 years? Don't include the dollar sign in your answer. Give your answer to 2 decimal places. Use the formula: A=P(1+rn)ntA=P\left(1+\frac{r}{n}\right)^{n t} where P=1000,r=0.10,n=2,t=30, A=\mathrm{P}=1000, \mathrm{r}=0.10, \mathrm{n}=2, \mathrm{t}=30, \mathrm{~A}= ?

Studdy Solution

STEP 1

What is this asking? How much money will be in the account after 30 years if you start with $1000\$1000 and it grows with interest every six months? Watch out! Don't forget to use the correct values for each part of the formula, especially the rate and time!

STEP 2

1. Identify the values
2. Plug values into the formula
3. Calculate the result

STEP 3

Alright, let's **identify** the values from the problem.
We have:
- **Initial deposit** P=1000P = 1000 - **Annual interest rate** r=0.10r = 0.10 - **Compounding frequency** n=2n = 2 (because it's semi-annually) - **Time period** t=30t = 30 years
These are the pieces we need to plug into our formula!

STEP 4

Now, let's **plug** these values into the formula A=P(1+rn)ntA = P\left(1+\frac{r}{n}\right)^{n t}.
So, we have:
\[ A = 1000\left(1+\frac{0.10}{2}\right)^{2 \cdot 30} $

STEP 5

Let's **calculate** inside the parentheses first:
1+0.102=1+0.05=1.051 + \frac{0.10}{2} = 1 + 0.05 = 1.05
This is the growth factor for each compounding period!

STEP 6

Now, let's **raise** that growth factor to the power of 2302 \cdot 30:
1.05601.05^{60}

STEP 7

Next, let's **multiply** by the initial deposit to find the final amount:
A=10001.0560A = 1000 \cdot 1.05^{60}

STEP 8

Finally, let's **compute** the result using a calculator:
\[ A \approx 1000 \cdot 18.679 $
\[ A \approx 18679.25 $

STEP 9

After 30 years, the account will have approximately **18679.25**.

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