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Find the cost of each item in 6 years, assuming an inflation rate of \( 7 \% \) (compounded continuously). (Round your answers to the neares cent.) (a) cup of coffee, \( \$ 3.75 \) \( \$ \) (b) Sunday paper, \( \$ 2.25 \) \( \$ \) (c) Big Mac, \( \$ 4.10 \) \( \$ \) (d) gallon of gas, \( \$ 3.85 \) \( \$ \) (e) HDTV set, \( \$ 1,600 \) \( \$ \) (f) small car, \( \$ 17,000 \) \( \$ \) (g) car, \( \$ 28,000 \) \( \$ \) (h) tuition, \( \$ 27,000 \) \( \$ \)

Ask by Wells Rojas. in the United States
Feb 26,2025

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Answer

After 6 years with a 7% continuous inflation rate, the costs are: (a) \$5.70 for a cup of coffee (b) \$3.42 for a Sunday paper (c) \$6.23 for a Big Mac (d) \$5.85 for a gallon of gas (e) \$2,433.60 for an HDTV set (f) \$25,857.00 for a small car (g) \$42,588.00 for a car (h) \$41,067.00 for tuition

Solution

We use the formula for continuously compounded inflation. If the present price is P and the inflation rate is 7% per year, then after t years the future price is   Future Price = P · e^(0.07 · t). In our case, t = 6, so the multiplier is   e^(0.07·6) = e^(0.42). Using a calculator, you find   e^(0.42) ≈ 1.521. Now, multiply the current price of each item by 1.521 and round to the nearest cent. (a) For a cup of coffee at \$3.75:   3.75 × 1.521 ≈ 5.70   Answer: \$5.70 (b) For a Sunday paper at \$2.25:   2.25 × 1.521 ≈ 3.42   Answer: \$3.42 (c) For a Big Mac at \$4.10:   4.10 × 1.521 ≈ 6.23   Answer: \$6.23 (d) For a gallon of gas at \$3.85:   3.85 × 1.521 ≈ 5.85   Answer: \$5.85 (e) For an HDTV set at \$1,600:   1,600 × 1.521 ≈ 2,433.60   Answer: \$2,433.60 (f) For a small car at \$17,000:   17,000 × 1.521 ≈ 25,857.00   Answer: \$25,857.00 (g) For a car at \$28,000:   28,000 × 1.521 ≈ 42,588.00   Answer: \$42,588.00 (h) For tuition at \$27,000:   27,000 × 1.521 ≈ 41,067.00   Answer: \$41,067.00 Each of these results is rounded to the nearest cent, which completes the solution.

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To find the future cost of each item after 6 years with a continuously compounded inflation rate of 7%, we use the formula: \[ A = Pe^{rt} \] where: - \( A \) = future value - \( P \) = present value (cost of the item) - \( r \) = inflation rate (0.07) - \( t \) = time in years (6) Let's calculate the future costs for each item: (a) for the cup of coffee, \( A = 3.75 \cdot e^{(0.07)(6)} \) Calculating: \[ A \approx 3.75 \cdot e^{0.42} \approx 3.75 \cdot 1.519 \approx 5.70 \] Rounded to the nearest cent: \( \$5.70 \) (b) for the Sunday paper, \( A = 2.25 \cdot e^{(0.07)(6)} \) Calculating: \[ A \approx 2.25 \cdot 1.519 \approx 3.42 \] Rounded to the nearest cent: \( \$3.42 \) (c) for the Big Mac, \( A = 4.10 \cdot e^{(0.07)(6)} \) Calculating: \[ A \approx 4.10 \cdot 1.519 \approx 6.23 \] Rounded to the nearest cent: \( \$6.23 \) (d) for the gallon of gas, \( A = 3.85 \cdot e^{(0.07)(6)} \) Calculating: \[ A \approx 3.85 \cdot 1.519 \approx 5.85 \] Rounded to the nearest cent: \( \$5.85 \) (e) for the HDTV set, \( A = 1,600 \cdot e^{(0.07)(6)} \) Calculating: \[ A \approx 1,600 \cdot 1.519 \approx 2,430.40 \] Rounded to the nearest cent: \( \$2,430.40 \) (f) for the small car, \( A = 17,000 \cdot e^{(0.07)(6)} \) Calculating: \[ A \approx 17,000 \cdot 1.519 \approx 25,823 \] Rounded to the nearest cent: \( \$25,823.00 \) (g) for the car, \( A = 28,000 \cdot e^{(0.07)(6)} \) Calculating: \[ A \approx 28,000 \cdot 1.519 \approx 42,552 \] Rounded to the nearest cent: \( \$42,552.00 \) (h) for tuition, \( A = 27,000 \cdot e^{(0.07)(6)} \) Calculating: \[ A \approx 27,000 \cdot 1.519 \approx 40,009 \] Rounded to the nearest cent: \( \$40,009.00 \) Here are all the future costs: (a) \( \$5.70 \) (b) \( \$3.42 \) (c) \( \$6.23 \) (d) \( \$5.85 \) (e) \( \$2,430.40 \) (f) \( \$25,823.00 \) (g) \( \$42,552.00 \) (h) \( \$40,009.00 \)

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