Suppose \( \$ 1,100 \) dollars is deposited into an account which has an annual percentage rate of \( 6.4 \% \) per year. Assume the interest is compounded annually. (a) Complete the boxes below to write a formua for the amount \( Q \) (in dollars) at year \( t \). Do not type any rounded values in these boxes. \( Q= \) Number (b) How many years does it take the balance to first exceed \( \$ 1,300 \) ? Solve with a graph or table and report your answer accurate to the nearest 0.1 years. \( t \approx \) Number years (Round to the nearest 0.1 year.)

The formula for calculating the amount \( Q \) in an account with compound interest is given by: \( Q = P(1 + r)^t \) In this case, \( P = 1100 \) (the principal amount), and \( r = 0.064 \) (the interest rate as a decimal). So the formula can be written as: \( Q = 1100(1 + 0.064)^t \) or \( Q = 1100(1.064)^t \) To find out how many years \( t \) it takes for the balance to first exceed \( 1300 \), we need to solve the equation: \( 1100(1.064)^t > 1300 \) First, simplify the inequality: \( (1.064)^t > \frac{1300}{1100} \) \( (1.064)^t > 1.1818 \) Now, take the logarithm of both sides to solve for \( t \): \( t \cdot \log(1.064) > \log(1.1818) \) Then, solve for \( t \): \( t > \frac{\log(1.1818)}{\log(1.064)} \) Calculating the logs: \( t > \frac{0.0834}{0.0270} \) \( t > 3.09 \) So, to the nearest tenth, \( t \approx 3.1 \) years. If we create a table to visualize: - Year 0: \( Q = 1100(1.064)^0 = 1100 \) - Year 1: \( Q = 1100(1.064)^1 \approx 1169.4 \) - Year 2: \( Q = 1100(1.064)^2 \approx 1242.0 \) - Year 3: \( Q = 1100(1.064)^3 \approx 1318.4 \) By year 3, the amount first exceeds 1,300, confirming \( t \approx 3.1 \) years.