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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)

[tex]\boxed{\begin{minipage}{8.5 cm}\underline{Compound Interest Formula}\\\\$ A=P\left(1+\frac{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}}[/tex]

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:

[tex]\implies A=4000\left(1+\dfrac{0.04}{365}\right)^{365 \times 10}[/tex]

[tex]\implies A=4000\left(1.00010958...\right)^{3650}[/tex]

[tex]\implies A=4000(1.49179200...)[/tex]

[tex]\implies A=5967.16801...[/tex]

[tex]\implies A=\$5967.17[/tex]

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:

[tex]\implies A=4000\left(1+\dfrac{0.04}{8760}\right)^{8760 \times 10}[/tex]

[tex]\implies A=4000\left(1.00000456...\right)^{87600}[/tex]

[tex]\implies A=4000\left(1.49182333...\right)[/tex]

[tex]\implies A=5967.29333...[/tex]

[tex]\implies A=\$5967.29[/tex]

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:

[tex]\implies A=4000\left(1+\dfrac{0.04}{525600}\right)^{525600 \times 10}[/tex]

[tex]\implies A=4000\left(1.00000007...\right)^{5256000}[/tex]

[tex]\implies A=4000(1.49182466...)[/tex]

[tex]\implies A=5967.29867...[/tex]

[tex]\implies A=\$5967.30[/tex]

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:

[tex]\implies A=4000\left(1+\dfrac{0.04}{31536000}\right)^{31536000 \times 10}[/tex]

[tex]\implies A=4000\left(1.00000000...\right)^{315360000}[/tex]

[tex]\implies A=4000\left(1.49182390...\right)[/tex]

[tex]\implies A=5967.29562...[/tex]

[tex]\implies A=\$5967.30[/tex]

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.