Find the interest rate to the nearest hundredth of a percent that will produce $\$ 3000$, if $\$ 2500$ is left at interest compounded quarterly for $3.5 \mathrm{yr}$. Use the formula

Step 1 ：We are given that the amount of money accumulated after n years, including interest (A) is $3000, the principal amount (P) is $2500, the number of times that interest is compounded per year (n) is 4 (since interest is compounded quarterly), and the time the money is invested for in years (t) is 3.5 years. We need to find the annual interest rate (r).

Step 2 ：We can rearrange the formula for compound interest to solve for r: \(r = n[(A/P)^(1/nt) - 1]\)

Step 3 ：Substituting the given values into the formula, we get \(r = 4[(3000/2500)^(1/(4*3.5)) - 1]\)

Step 4 ：Solving the equation, we find that \(r = 0.0524325460286974\)

Step 5 ：The interest rate r is in decimal form. To convert it to a percentage, we multiply by 100. Also, we need to round it to the nearest hundredth of a percent. So, \(r_{percent} = 0.0524325460286974 * 100 = 5.24\)

Step 6 ：Final Answer: The interest rate that will produce $3000, if $2500 is left at interest compounded quarterly for 3.5 years is \(\boxed{5.24\%}\)