Question 5 (1 point)
What interest rate, compounded daily, would be required to turn a $\$ 7,000$ investment into $\$ 9,500$ in four years?
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7.64
7.93
7.78
7.65

Step 1 ：The problem is asking for the interest rate that would turn a $7000 investment into $9500 in four years, with the interest compounded daily. This is a compound interest problem. The formula for compound interest is: \(A = P(1 + r/n)^{nt}\)

Step 2 ：Where: \(A\) is the amount of money accumulated after n years, including interest. \(P\) is the principal amount (the initial amount of money). \(r\) is the annual interest rate (in decimal). \(n\) is the number of times that interest is compounded per year. \(t\) is the time the money is invested for in years.

Step 3 ：We know that \(A = \$9500\), \(P = \$7000\), \(n = 365\) (since the interest is compounded daily), and \(t = 4\) years. We need to solve for \(r\).

Step 4 ：We can rearrange the formula to solve for \(r\): \(r = ((A/P)^{1/(nt)} - 1) * n\)

Step 5 ：Plugging in the known values and solving for \(r\), we get \(r = 0.07635339735821756\)

Step 6 ：Converting \(r\) from a decimal to a percentage, we get \(r_{percent} = 7.635339735821756\)

Step 7 ：The interest rate, compounded daily, required to turn a $7000 investment into $9500 in four years is approximately \(\boxed{7.64\%}\)