Bo deposits $490 every year into an account earning an annual interest rate of 5.3% compounded annually. How much would he have in the account after 3 years, to the nearest dollar? Use the following formula to determine your answer. A=d(((1+i)^(n)-1)/(i)) A= the future value of the account after n periods d= the amount invested at the end of each period i= the interest rate per period n= the number of periods Answer:

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Bo deposits  $490 every year into an account earning an annual interest rate of  5.3% compounded annually. How much would he have in the account after 3 years, to the nearest dollar? Use the following formula to determine your answer.  A=d(((1+i)^(n)-1)/(i))  A= the future value of the account after  n periods  d= the amount invested at the end of each period  i= the interest rate per period  n= the number of periods Answer:

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Identify variables and amounts: Identify the variables from the problem to use in the formula. We have: d = $490 (the amount invested at the end of each period) i = 5.3% or 0.053 (the interest rate per period) n = 3 (the number of periods)Convert interest rate to decimal: Convert the interest rate from a percentage to a decimal. i = 5.3% = 5.3/100 = 0.053Plug values into formula: Plug the values into the formula to calculate the future value of the account. A = d * (((1 + i)^n - 1) / i) A = 490 * (((1 + 0.053)^3 - 1) / 0.053)Calculate value inside parentheses: Calculate the value inside the parentheses. (1 + i)^n = (1 + 0.053)^3 (1 + i)^n = 1.053^3 (1 + i)^n ≈ 1.053 * 1.053 * 1.053 (1 + i)^n ≈ 1.162727Subtract one from result: Subtract 1 from the result obtained in Step 4. (1 + i)^n - 1 ≈ 1.162727 - 1 (1 + i)^n - 1 ≈ 0.162727Divide by interest rate: Divide the result from Step 5 by the interest rate i. ((1 + i)^n - 1) / i ≈ 0.162727 / 0.053 ((1 + i)^n - 1) / i ≈ 3.071075Multiply by amount deposited: Multiply the result from Step 6 by the amount deposited at the end of each period d. A ≈ 490 * 3.071075 A ≈ 1504.82675Round to nearest dollar: Round the result to the nearest dollar. A ≈ $1505

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