Suppose \( \$ 1,200 \) dollars is deposited into an account which has an annual percentage rate of \( 5.22 \% \) per year. Assume the interest is compounded continuously. (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. \[ =\text { Number } \] (b) How many years does it take the balance to first exceed \( \$ 4,200 \) ? Solve using a graph or table and report your answer accurate to the nearest year. \( t \approx \) Number years (Round to the nearest year.)

To find the formula for the amount \( Q \) (in dollars) at year \( t \) when interest is compounded continuously, we use the formula: \[ Q = P e^{rt} \] where: - \( P = 1200 \) (the initial principal), - \( r = 0.0522 \) (the annual interest rate as a decimal), - \( t \) is the time in years, - \( e \) is the base of the natural logarithm (approximately 2.71828). So, substituting the values, the formula becomes: \[ Q = 1200 e^{0.0522t} \] To find how many years it takes for the balance to first exceed \( \$ 4,200 \), we need to solve the inequality: \[ 1200 e^{0.0522t} > 4200 \] Dividing both sides by 1200: \[ e^{0.0522t} > \frac{4200}{1200} \] Calculating the right side gives: \[ e^{0.0522t} > 3.5 \] Now, take the natural logarithm of both sides: \[ 0.0522t > \ln(3.5) \] Now, solving for \( t \): \[ t > \frac{\ln(3.5)}{0.0522} \] Using a calculator, we find \( \ln(3.5) \approx 1.25276 \). Thus, \[ t > \frac{1.25276}{0.0522} \approx 24.0 \] Rounding to the nearest year, we find: \( t \approx 24 \) years.