Math 1300: Section 3-2 Compound Interest

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Compound InterestContinuous Compound Interest

Growth and TimeAnnual Percentage Yield

Math 1300 Finite MathematicsSection 3.2 Compound Interest

Jason Aubrey

Department of MathematicsUniversity of Missouri

Jason Aubrey Math 1300 Finite Mathematics

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If at the end of a payment period the interest due is reinvestedat the same rate, then the interest as well as the originalprincipal will earn interest at the end of the next paymentperiod. Interest payed on interest reinvested is calledcompound interest.


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Suppose you invest $1,000 in a bank that pays 8%compounded quarterly. How much will the bank owe you at theend of the year?

A = P(1 + rt)

= 1, 000[1 + 0.08

(14

)]= 1, 000(1.02) = $1, 020

This is the amount at the endof the first quarter. Then:

Second quarter:= $1,020(1.02) =$1,040.40Third quarter:= $1,040.40(1.02) =$1,061.21Fourth quarter:= $1,061.21(1.02) =$1,082.43


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A = P(1 + rt)

= 1, 000[1 + 0.08

(14

)]= 1, 000(1.02) = $1, 020




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A = P(1 + rt)

= 1, 000[1 + 0.08

(14

)]

= 1, 000(1.02) = $1, 020




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A = P(1 + rt)

= 1, 000[1 + 0.08

(14

)]= 1, 000(1.02) = $1, 020




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A = P(1 + rt)

= 1, 000[1 + 0.08

(14

)]= 1, 000(1.02) = $1, 020




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A = P(1 + rt)

= 1, 000[1 + 0.08

(14

)]= 1, 000(1.02) = $1, 020


Second quarter:= $1,020(1.02) =$1,040.40

Third quarter:= $1,040.40(1.02) =$1,061.21Fourth quarter:= $1,061.21(1.02) =$1,082.43


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A = P(1 + rt)

= 1, 000[1 + 0.08

(14

)]= 1, 000(1.02) = $1, 020


Second quarter:= $1,020(1.02) =$1,040.40Third quarter:= $1,040.40(1.02) =$1,061.21

Fourth quarter:= $1,061.21(1.02) =$1,082.43


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A = P(1 + rt)

= 1, 000[1 + 0.08

(14

)]= 1, 000(1.02) = $1, 020




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Look at the pattern:

First quarter: = $1, 000(1.02)

Second quarter: = $1, 000(1.02)2

Third quarter: = $1, 000(1.02)3

Fourth quarter: = $1, 000(1.02)4


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Third quarter: = $1, 000(1.02)3



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Third quarter: = $1, 000(1.02)3



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Third quarter: = $1, 000(1.02)3



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Third quarter: = $1, 000(1.02)3



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Definition (Amount: Compound Interest)

A = P(1 + i)n

where i = r/m andA = amount (future value) at the end of n periodsP = principal (present value)r = annual nominal ratem = number of compounding periods per yeari = rate per compounding periodn = total number of compounding periods


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A = P(1 + i)n

where i = r/m andA = amount (future value) at the end of n periods

P = principal (present value)r = annual nominal ratem = number of compounding periods per yeari = rate per compounding periodn = total number of compounding periods


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A = P(1 + i)n

where i = r/m andA = amount (future value) at the end of n periodsP = principal (present value)

r = annual nominal ratem = number of compounding periods per yeari = rate per compounding periodn = total number of compounding periods


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A = P(1 + i)n

where i = r/m andA = amount (future value) at the end of n periodsP = principal (present value)r = annual nominal rate

m = number of compounding periods per yeari = rate per compounding periodn = total number of compounding periods


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A = P(1 + i)n

where i = r/m andA = amount (future value) at the end of n periodsP = principal (present value)r = annual nominal ratem = number of compounding periods per year

i = rate per compounding periodn = total number of compounding periods


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A = P(1 + i)n

where i = r/m andA = amount (future value) at the end of n periodsP = principal (present value)r = annual nominal ratem = number of compounding periods per yeari = rate per compounding period

n = total number of compounding periods


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A = P(1 + i)n

where i = r/m andA = amount (future value) at the end of n periodsP = principal (present value)r = annual nominal ratem = number of compounding periods per yeari = rate per compounding periodn = total number of compounding periods


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Example: P = 800; i = 0.06; n = 25; A =?

A = P(1 + i)n

A = $800(1.06)25

A = $3, 433.50


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Example: P = 800; i = 0.06; n = 25; A =?

A = P(1 + i)n

A = $800(1.06)25

A = $3, 433.50


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Example: P = 800; i = 0.06; n = 25; A =?

A = P(1 + i)n

A = $800(1.06)25

A = $3, 433.50


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Example: P = 800; i = 0.06; n = 25; A =?

A = P(1 + i)n

A = $800(1.06)25

A = $3, 433.50


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Example: A = $18, 000; r = 8.12% compounded monthly;n = 90; P =?

First we compute i = rm = 0.0812

12 = 0.0068. Then,

A = P(1 + i)n

$18, 000 = P(1.0068)90

$18, 000 ≈ P(1.84)

P ≈ $9, 782.61


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12 = 0.0068. Then,

A = P(1 + i)n

$18, 000 = P(1.0068)90

$18, 000 ≈ P(1.84)

P ≈ $9, 782.61


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12 = 0.0068. Then,

A = P(1 + i)n

$18, 000 = P(1.0068)90

$18, 000 ≈ P(1.84)

P ≈ $9, 782.61


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12 = 0.0068. Then,

A = P(1 + i)n

$18, 000 = P(1.0068)90

$18, 000 ≈ P(1.84)

P ≈ $9, 782.61


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12 = 0.0068. Then,

A = P(1 + i)n

$18, 000 = P(1.0068)90

$18, 000 ≈ P(1.84)

P ≈ $9, 782.61


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12 = 0.0068. Then,

A = P(1 + i)n

$18, 000 = P(1.0068)90

$18, 000 ≈ P(1.84)

P ≈ $9, 782.61


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Definition (Amount - Continuous Compound Interest)If a principal P is invested at an annual rate r (expressed as adecimal) compounded continuously, then the amount A in theaccount at the end of t years is given by

A = Pert


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Example: P = $2, 450; r = 8.12%; t = 3 years; A =?

A = Pert

A = $2, 450e(0.0812)(3) = $3, 125.79


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A = Pert

A = $2, 450e(0.0812)(3) = $3, 125.79


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A = Pert

A = $2, 450e(0.0812)(3) = $3, 125.79


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Example: A = $15, 875; P = $12, 100; t = 48 months; r =?

A = Pert

$15, 875 = $12, 100e4r

1.311 = e4r

ln(1.311) = 4rln(1.311)

4= r

r = 0.068


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A = Pert

$15, 875 = $12, 100e4r

1.311 = e4r

ln(1.311) = 4rln(1.311)

4= r

r = 0.068


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A = Pert

$15, 875 = $12, 100e4r

1.311 = e4r

ln(1.311) = 4rln(1.311)

4= r

r = 0.068


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A = Pert

$15, 875 = $12, 100e4r

1.311 = e4r

ln(1.311) = 4rln(1.311)

4= r

r = 0.068


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A = Pert

$15, 875 = $12, 100e4r

1.311 = e4r

ln(1.311) = 4r

ln(1.311)

4= r

r = 0.068


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A = Pert

$15, 875 = $12, 100e4r

1.311 = e4r

ln(1.311) = 4rln(1.311)

4= r

r = 0.068


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A = Pert

$15, 875 = $12, 100e4r

1.311 = e4r

ln(1.311) = 4rln(1.311)

4= r

r = 0.068


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Example: If an investment company pays 6% compoundedsemiannually, how much should you deposit now to have$10,000?

5 years from now?

i = r/m = 0.06/2 = 0.03n = 5x2 = 10A = P(1 + i)n

$10, 000 = P(1.03)10

$10, 000 ≈ P(1.3439)

P ≈ $7, 441.03

10 years from now?

i = r/m = 0.06/2 = 0.03n = 10x2 = 20A = P(1 + i)n

$10, 000 = P(1.03)20

$10, 000 ≈ P(1.806)

P ≈ $5, 536.76


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5 years from now?

i = r/m = 0.06/2 = 0.03

n = 5x2 = 10A = P(1 + i)n

$10, 000 = P(1.03)10

$10, 000 ≈ P(1.3439)

P ≈ $7, 441.03

10 years from now?

i = r/m = 0.06/2 = 0.03n = 10x2 = 20A = P(1 + i)n

$10, 000 = P(1.03)20

$10, 000 ≈ P(1.806)

P ≈ $5, 536.76


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5 years from now?

i = r/m = 0.06/2 = 0.03n = 5x2 = 10

A = P(1 + i)n

$10, 000 = P(1.03)10

$10, 000 ≈ P(1.3439)

P ≈ $7, 441.03

10 years from now?

i = r/m = 0.06/2 = 0.03n = 10x2 = 20A = P(1 + i)n

$10, 000 = P(1.03)20

$10, 000 ≈ P(1.806)

P ≈ $5, 536.76


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5 years from now?

i = r/m = 0.06/2 = 0.03n = 5x2 = 10A = P(1 + i)n

$10, 000 = P(1.03)10

$10, 000 ≈ P(1.3439)

P ≈ $7, 441.03

10 years from now?

i = r/m = 0.06/2 = 0.03n = 10x2 = 20A = P(1 + i)n

$10, 000 = P(1.03)20

$10, 000 ≈ P(1.806)

P ≈ $5, 536.76


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5 years from now?

i = r/m = 0.06/2 = 0.03n = 5x2 = 10A = P(1 + i)n

$10, 000 = P(1.03)10

$10, 000 ≈ P(1.3439)

P ≈ $7, 441.03

10 years from now?

i = r/m = 0.06/2 = 0.03n = 10x2 = 20A = P(1 + i)n

$10, 000 = P(1.03)20

$10, 000 ≈ P(1.806)

P ≈ $5, 536.76


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5 years from now?

i = r/m = 0.06/2 = 0.03n = 5x2 = 10A = P(1 + i)n

$10, 000 = P(1.03)10

$10, 000 ≈ P(1.3439)

P ≈ $7, 441.03

10 years from now?

i = r/m = 0.06/2 = 0.03n = 10x2 = 20A = P(1 + i)n

$10, 000 = P(1.03)20

$10, 000 ≈ P(1.806)

P ≈ $5, 536.76


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5 years from now?

i = r/m = 0.06/2 = 0.03n = 5x2 = 10A = P(1 + i)n

$10, 000 = P(1.03)10

$10, 000 ≈ P(1.3439)

P ≈ $7, 441.03

10 years from now?

i = r/m = 0.06/2 = 0.03n = 10x2 = 20A = P(1 + i)n

$10, 000 = P(1.03)20

$10, 000 ≈ P(1.806)

P ≈ $5, 536.76


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5 years from now?

i = r/m = 0.06/2 = 0.03n = 5x2 = 10A = P(1 + i)n

$10, 000 = P(1.03)10

$10, 000 ≈ P(1.3439)

P ≈ $7, 441.03

10 years from now?

i = r/m = 0.06/2 = 0.03

n = 10x2 = 20A = P(1 + i)n

$10, 000 = P(1.03)20

$10, 000 ≈ P(1.806)

P ≈ $5, 536.76


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5 years from now?

i = r/m = 0.06/2 = 0.03n = 5x2 = 10A = P(1 + i)n

$10, 000 = P(1.03)10

$10, 000 ≈ P(1.3439)

P ≈ $7, 441.03

10 years from now?

i = r/m = 0.06/2 = 0.03n = 10x2 = 20

A = P(1 + i)n

$10, 000 = P(1.03)20

$10, 000 ≈ P(1.806)

P ≈ $5, 536.76


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5 years from now?

i = r/m = 0.06/2 = 0.03n = 5x2 = 10A = P(1 + i)n

$10, 000 = P(1.03)10

$10, 000 ≈ P(1.3439)

P ≈ $7, 441.03

10 years from now?

i = r/m = 0.06/2 = 0.03n = 10x2 = 20A = P(1 + i)n

$10, 000 = P(1.03)20

$10, 000 ≈ P(1.806)

P ≈ $5, 536.76


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5 years from now?

i = r/m = 0.06/2 = 0.03n = 5x2 = 10A = P(1 + i)n

$10, 000 = P(1.03)10

$10, 000 ≈ P(1.3439)

P ≈ $7, 441.03

10 years from now?

i = r/m = 0.06/2 = 0.03n = 10x2 = 20A = P(1 + i)n

$10, 000 = P(1.03)20

$10, 000 ≈ P(1.806)

P ≈ $5, 536.76


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5 years from now?

i = r/m = 0.06/2 = 0.03n = 5x2 = 10A = P(1 + i)n

$10, 000 = P(1.03)10

$10, 000 ≈ P(1.3439)

P ≈ $7, 441.03

10 years from now?

i = r/m = 0.06/2 = 0.03n = 10x2 = 20A = P(1 + i)n

$10, 000 = P(1.03)20

$10, 000 ≈ P(1.806)

P ≈ $5, 536.76


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5 years from now?

i = r/m = 0.06/2 = 0.03n = 5x2 = 10A = P(1 + i)n

$10, 000 = P(1.03)10

$10, 000 ≈ P(1.3439)

P ≈ $7, 441.03

10 years from now?

i = r/m = 0.06/2 = 0.03n = 10x2 = 20A = P(1 + i)n

$10, 000 = P(1.03)20

$10, 000 ≈ P(1.806)

P ≈ $5, 536.76


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Example: A zero coupon bond is a bond that is sold now at adiscount and will pay it’s face value at some time in the futurewhen it matures - no interest payments are made. Supposethat a zero coupon bond with a face value of $40,000 maturesin 20 years. What should the bond be sold for now if its rate ofreturn is to be 5.124% compounded annually.

Compounded annually means i = rm = 0.05124

1 = 0.05124. HereA = $40, 000 and we want to find P. We also know that n = 20.

A = P(1 + i)n

$40, 000 = P(1.05124)20

P =$40, 000

(1.05124)20 = $14, 723.89


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A = P(1 + i)n

$40, 000 = P(1.05124)20

P =$40, 000

(1.05124)20 = $14, 723.89


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A = P(1 + i)n

$40, 000 = P(1.05124)20

P =$40, 000

(1.05124)20 = $14, 723.89


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A = P(1 + i)n

$40, 000 = P(1.05124)20

P =$40, 000

(1.05124)20 = $14, 723.89


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A = P(1 + i)n

$40, 000 = P(1.05124)20

P =$40, 000

(1.05124)20 = $14, 723.89


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DefinitionIf a principal is invested at the annual (nominal) rate rcompounded m times a year, then the annual percentage yieldis

APY =(

1 +rm

)m− 1

If a principal is invested at the annual (nominal) rate rcompounded continuously, then the annual percentage yield is

APY = er − 1

The annual percentage yield is also referred to as the effectiverate or the true interest rate.


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Example: What is the annual nominal rate compounded dailyfor a bond that has an APY of 6.8%.

APY =(

1 +rm

)m− 1

0.068 =(

1 +r

365

)365− 1

1.068 =(

1 +r

365

)365

365√

1.068 = 1 +r

365

0.00018 =r

365r = 0.0658


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APY =(

1 +rm

)m− 1

0.068 =(

1 +r

365

)365− 1

1.068 =(

1 +r

365

)365

365√

1.068 = 1 +r

365

0.00018 =r

365r = 0.0658


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APY =(

1 +rm

)m− 1

0.068 =(

1 +r

365

)365− 1

1.068 =(

1 +r

365

)365

365√

1.068 = 1 +r

365

0.00018 =r

365r = 0.0658


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APY =(

1 +rm

)m− 1

0.068 =(

1 +r

365

)365− 1

1.068 =(

1 +r

365

)365

365√

1.068 = 1 +r

365

0.00018 =r

365r = 0.0658


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APY =(

1 +rm

)m− 1

0.068 =(

1 +r

365

)365− 1

1.068 =(

1 +r

365

)365

365√

1.068 = 1 +r

365

0.00018 =r

365

r = 0.0658


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APY =(

1 +rm

)m− 1

0.068 =(

1 +r

365

)365− 1

1.068 =(

1 +r

365

)365

365√

1.068 = 1 +r

365

0.00018 =r

365r = 0.0658


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Example: What is the annual nominal rate compoundedcontinuously has the same APY as 6% compounded monthly?

APY =(

1 +rm

)m− 1

APY =

(1 +

0.0612

)12

− 1 = 0.0617

APY = er − 10.0617 = er − 1

ln(1.0617) = rr = 0.05987


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APY =(

1 +rm

)m− 1

APY =

(1 +

0.0612

)12

− 1 = 0.0617

APY = er − 10.0617 = er − 1

ln(1.0617) = rr = 0.05987


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APY =(

1 +rm

)m− 1

APY =

(1 +

0.0612

)12

− 1 = 0.0617

APY = er − 1

0.0617 = er − 1ln(1.0617) = r

r = 0.05987


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APY =(

1 +rm

)m− 1

APY =

(1 +

0.0612

)12

− 1 = 0.0617

APY = er − 10.0617 = er − 1

ln(1.0617) = rr = 0.05987


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APY =(

1 +rm

)m− 1

APY =

(1 +

0.0612

)12

− 1 = 0.0617

APY = er − 10.0617 = er − 1

ln(1.0617) = r

r = 0.05987


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APY =(

1 +rm

)m− 1

APY =

(1 +

0.0612

)12

− 1 = 0.0617

APY = er − 10.0617 = er − 1

ln(1.0617) = rr = 0.05987


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Example: Which is the better investment and why: 9%compounded quarterly or 9.3% compounded annually?

APY1 =(

1 +rm

)m− 1 =

(1 +

0.094

)4

− 1 = 0.09308

APY2 = 0.093

The first offer is better because its APY is larger.


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APY1 =(

1 +rm

)m− 1 =

(1 +

0.094

)4

− 1 = 0.09308

APY2 = 0.093



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APY1 =(

1 +rm

)m− 1 =

(1 +

0.094

)4

− 1 = 0.09308

APY2 = 0.093



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APY1 =(

1 +rm

)m− 1 =

(1 +

0.094

)4

− 1 = 0.09308

APY2 = 0.093



Education

Math 1300: Section 3-2 Compound Interest