Find the cost of each item in 10 years, assuming an inflation rate of \( 7 \% \) (compounded continuously). (Round your answers to the nearest cent.) (a) movie admission, \( \$ 8.00 \) \( \$ \) (b) CD, \( \$ 14.95 \) \( \$ \) (c) textbook, \( \$ 110.00 \) \$ (d) electric bill, \( \$ 105 \) \( \$ \) (e) phone bill, \( \$ 35 \) \( \$ \) (f) pair of shoes, \( \$ 75 \) \( \$ \) (g) new suit, \( \$ 590 \) \( \$ \) (h) monthly rent, \( \$ 900 \) \( \$ \)

To calculate the future cost of items considering an inflation rate of \( 7\% \) compounded continuously, you can use the formula: \[ A = Pe^{rt} \] Where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial sum of money). - \( r \) is the annual interest rate (decimal). - \( t \) is the time the money is invested for in years. - \( e \) is the base of the natural logarithm (approximately equal to 2.71828). Now let’s calculate the cost for each item in 10 years: (a) Movie admission (\$8.00): \( A = 8 e^{0.07 \times 10} \) \( A \approx 8 e^{0.7} \approx 8 \times 2.0138 = 16.11 \) Cost: \( \$16.11 \) (b) CD (\$14.95): \( A = 14.95 e^{0.07 \times 10} \) \( A \approx 14.95 e^{0.7} \approx 14.95 \times 2.0138 = 30.12 \) Cost: \( \$30.12 \) (c) Textbook (\$110.00): \( A = 110 e^{0.07 \times 10} \) \( A \approx 110 e^{0.7} \approx 110 \times 2.0138 = 221.52 \) Cost: \( \$221.52 \) (d) Electric bill (\$105): \( A = 105 e^{0.07 \times 10} \) \( A \approx 105 e^{0.7} \approx 105 \times 2.0138 = 211.45 \) Cost: \( \$211.45 \) (e) Phone bill (\$35): \( A = 35 e^{0.07 \times 10} \) \( A \approx 35 e^{0.7} \approx 35 \times 2.0138 = 70.48 \) Cost: \( \$70.48 \) (f) Pair of shoes (\$75): \( A = 75 e^{0.07 \times 10} \) \( A \approx 75 e^{0.7} \approx 75 \times 2.0138 = 150.99 \) Cost: \( \$150.99 \) (g) New suit (\$590): \( A = 590 e^{0.07 \times 10} \) \( A \approx 590 e^{0.7} \approx 590 \times 2.0138 = 1186.13 \) Cost: \( \$1186.13 \) (h) Monthly rent (\$900): \( A = 900 e^{0.07 \times 10} \) \( A \approx 900 e^{0.7} \approx 900 \times 2.0138 = 1812.42 \) Cost: \( \$1812.42 \) Here’s a summary of the costs in 10 years: - (a) Movie admission: \( \$16.11 \) - (b) CD: \( \$30.12 \) - (c) Textbook: \( \$221.52 \) - (d) Electric bill: \( \$211.45 \) - (e) Phone bill: \( \$70.48 \) - (f) Pair of shoes: \( \$150.99 \) - (g) New suit: \( \$1186.13 \) - (h) Monthly rent: \( \$1812.42 \)