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Use the formula A = P( 1 + r/n)^(nt) to calculate the balance A of an investment (in dollars) when P = $4000, r = 4%, and t= 10years, and compunding is done by the day, by the hour, by the minute, and by the second. (Round your answers to the nearest cent).

a. compounding by the day: A= $

b. compounding by the hour: A= $

c. compounding by the minute: A= $

d. compounding by the second: A= $


Does increasing the number of compoundings per year result in unlimited growth of the balance? yes or no (choose one)

User Yousi
by
5.6k points

1 Answer

2 votes

Answer:

a. compounding by the day: A = $5967.17

b. compounding by the hour: A = $5967.29

c. compounding by the minute: A = $5967.30

d. compounding by the second: A = $5967.30

Explanation:


\boxed{\begin{minipage}{8.5 cm}\underline{Compound Interest Formula}\\\\$ A=P\left(1+(r)/(n)\right)^(nt)$\\\\where:\\\\ \phantom{ww}$\bullet$ $A =$ final amount \\ \phantom{ww}$\bullet$ $P =$ principal amount \\ \phantom{ww}$\bullet$ $r =$ interest rate (in decimal form) \\ \phantom{ww}$\bullet$ $n =$ number of times interest is applied per year \\ \phantom{ww}$\bullet$ $t =$ time (in years) \\ \end{minipage}}

Part (a)

If the interest is compounding by the day then n = 365.

Given:

  • P = $4000
  • r = 4% = 0.04
  • n = 365
  • t = 10 years

Substitute the values into the compound interest formula and solve for A:


\implies A=4000\left(1+(0.04)/(365)\right)^(365 * 10)


\implies A=4000\left(1.00010958...\right)^(3650)


\implies A=4000(1.49179200...)


\implies A=5967.16801...


\implies A=\$5967.17

Part (b)

If the interest is compounding by the hour then:

  • n = 365 × 24 = 8760

Given:

  • P = $4000
  • r = 4% = 0.04
  • n = 8760
  • t = 10 years

Substitute the values into the compound interest formula and solve for A:


\implies A=4000\left(1+(0.04)/(8760)\right)^(8760 * 10)


\implies A=4000\left(1.00000456...\right)^(87600)


\implies A=4000\left(1.49182333...\right)


\implies A=5967.29333...


\implies A=\$5967.29

Part (c)

If the interest is compounding by the minute then:

  • n = 365 × 24 × 60 = 525600

Given:

  • P = $4000
  • r = 4% = 0.04
  • n = 525600
  • t = 10 years

Substitute the values into the compound interest formula and solve for A:


\implies A=4000\left(1+(0.04)/(525600)\right)^(525600 * 10)


\implies A=4000\left(1.00000007...\right)^(5256000)


\implies A=4000(1.49182466...)


\implies A=5967.29867...


\implies A=\$5967.30

Part (d)

If the interest is compounding by the second then:

  • n = 365 × 24 × 60 × 60 = 31536000

Given:

  • P = $4000
  • r = 4% = 0.04
  • n = 31536000
  • t = 10 years

Substitute the values into the compound interest formula and solve for A:


\implies A=4000\left(1+(0.04)/(31536000)\right)^(31536000 * 10)


\implies A=4000\left(1.00000000...\right)^(315360000)


\implies A=4000\left(1.49182390...\right)


\implies A=5967.29562...


\implies A=\$5967.30

The more compounding periods throughout the year, the higher the future value of the investment. However, the difference between compounding by the day and compounding by the second results in a difference of 13 cents over the year, which is negligible comparatively.

User Paul Blundell
by
6.5k points
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