Suppose \( \$ 1,100 \) dollars is deposited into an account which has an annual percentage rate of \( 5.7 \% \) per year. Assume the interest is compounded quarterly. (a) Complete the boxes below to write a formua for the amount \( Q \) (in dollars) at year \( t \). You can enter arithmetic expressions (containing \( +,-,{ }^{*}, / \), or \( \wedge \) ) in any of these boxes. Do not type any rounded values in these boxes. \( Q= \) Number (b) How many years does it take the balance to first exceed \( \$ 1,500 \) ? Solve with a graph or table, and report your answer accurate to the nearest 0.01 years. \( t \approx \) Number years (Round to the nearest 0.01 year.)

To determine the amount \( Q \) in the account after \( t \) years with quarterly compounding, we can use the formula: \[ Q = P \left(1 + \frac{r}{n}\right)^{nt} \] Here, \( P \) is the principal amount (initial deposit), \( r \) is the annual interest rate (as a decimal), \( n \) is the number of times interest is compounded per year, and \( t \) is the number of years. Plugging in the values, we get: - \( P = 1100 \) - \( r = 0.057 \) - \( n = 4 \) Thus, the formula for \( Q \) will be: \[ Q = 1100 \left(1 + \frac{0.057}{4}\right)^{4t} \] Now let's calculate \( t \) to find out when the balance first exceeds \( \$ 1,500 \). To find the number of years it takes, we will calculate \( Q \) for different values of \( t \) using a table. | \( t \) (years) | \( Q \) (dollars) | |------------------|-------------------| | 0.00 | 1100 | | 0.25 | \(1100 \left(1 + \frac{0.057}{4}\right)^{1} \approx 1114.74\) | | 0.50 | \(1100 \left(1 + \frac{0.057}{4}\right)^{2} \approx 1129.71\) | | 1.00 | \(1100 \left(1 + \frac{0.057}{4}\right)^{4} \approx 1157.11\) | | 2.00 | \(1100 \left(1 + \frac{0.057}{4}\right)^{8} \approx 1216.89\) | | 3.00 | \(1100 \left(1 + \frac{0.057}{4}\right)^{12} \approx 1283.66\) | | 4.00 | \(1100 \left(1 + \frac{0.057}{4}\right)^{16} \approx 1357.85\) | | 5.00 | \(1100 \left(1 + \frac{0.057}{4}\right)^{20} \approx 1439.92\) | | 6.00 | \(1100 \left(1 + \frac{0.057}{4}\right)^{24} \approx 1529.29\) | Now, since at \( t = 6.00 \) years the amount goes over \( 1500 \), we can refine our search between \( 5.00 \) and \( 6.00 \) years. Continuing from the previous values, we can calculate for \( t = 5.50 \) to get a closer approximation. \( Q \text{ at } t = 5.50 \approx 1100 \left(1 + \frac{0.057}{4}\right)^{22} \approx 1489.97 \) This still does not exceed \( 1500 \), so we continue with \( t = 5.75 \): \( Q \text{ at } t = 5.75 \approx 1100 \left(1 + \frac{0.057}{4}\right)^{23} \approx 1502.46 \) Thus, we find: \( t \approx 5.75 \) years (accurate to the nearest 0.01 year).