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Hector invests $800 in an account that earns 6.96% annual interest compounded semiannually. Rebecca invests $1,000 in an account that earns 5.44% annual interest compounded monthly. Find when the value of Rebecca's investment equals the value of Hector's investment and find the common value of the investments at that time. If necessary, enter the year to the nearest tenth and the value to the nearest cent. The value of Rebecca's investment equals the value of Hector's investment after approximately ________ years to the nearest tenth. The common value of the investments is approximately $_________ .

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Ashraf82

Answer:

The value of Rebecca's investment equals the value of Hector's investment after approximately 15.8 years to the nearest tenth

The common value of the investments is approximately $2358.05

Step-by-step explanation:

The formula of the compound interest including the principal is

[tex]A=P(1+\frac{r}{n})^{nt}[/tex] , where

  • A is the future value of the investment/loan, including interest
  • P is the principal investment amount
  • r is the annual interest rate (decimal)
  • n is the number of times that interest is compounded per unit t
  • t is the time the money is invested or borrowed for

Hector invests $800 in an account that earns 6.96% annual interest compounded semiannually

āˆµ P = 800

āˆµ r = 6.96 = [tex]\frac{6.96}{100}[/tex] = 0.0696

āˆµ n = 2 ā‡’ semiannually

- Substitute all these values in the formula to find future value

āˆ“ [tex]A=800(1+\frac{0.0696}{2})^{2t}[/tex]

āˆ“ [tex]A=800(1.0348)^{2t}[/tex]

Rebecca invests $1,000 in an account that earns 5.44% annual interest compounded monthly Ā 

āˆµ P = 1,000

āˆµ r = 5.44 = [tex]\frac{5.44}{100}[/tex] = 0.0544

āˆµ n = 12 ā‡’ monthly

- Substitute all these values in the formula to find future value

āˆ“ [tex]A=1000(1+\frac{0.0544}{12})^{12t}[/tex]

āˆ“ [tex]A=1000(1.004533333)^{12t}[/tex]

āˆµ Rebecca's investment equals the value of Hector's investment

- Equate the two equations

āˆµ Ā [tex]1000(1.004533333)^{12t}[/tex] = [tex]800(1.0348)^{2t}[/tex]

- Insert 揑 in both sides and remember 揑(ab) = 揑(a) + 揑(b)

āˆ“ 揑( [tex]1000(1.004533333)^{12t}[/tex] ) = 揑( [tex]800(1.0348)^{2t}[/tex] )

āˆ“ 揑(1000) + 揑( [tex](1.004533333)^{12t}[/tex] ) = 揑(800) + 揑( [tex](1.0348)^{2t}[/tex] )

- Remember [tex]ln(a)^{n}[/tex] = n 揑(a)

āˆ“ 揑(1000) + 12t [揑(1.004533333)] = 揑(800) + 2t [揑(1.0348)]

- Subtract 揑(800) and 12t [揑(1.004533333)] from both sides

āˆ“ 揑(1000) - 揑(800) = 2t [揑(1.0348)] - 12t [揑(1.004533333)]

- Take t as a common factor from the right hand side

āˆ“ 揑(1000) - 揑(800) = t(2[揑(1.0348)] - 12[揑(1.004533333)])

- Divide both sides by (2[揑(1.0348)] - 12[揑(1.004533333)])

āˆ“ 15.8 = t

The value of Rebecca's investment equals the value of Hector's investment after approximately 15.8 years to the nearest tenth

Let us find this value using Hector or Rebecca equations

āˆµ Ā [tex]A=800(1.0348)^{2t}[/tex]

- Substitute t by 15.8

āˆ“ Ā [tex]A=800(1.0348)^{2(15.8)}[/tex]

āˆ“ A = 2358.05

The common value of the investments is approximately $2358.05

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