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1. You are saving to buy a new house in 7 years. If you invest $4,500 now at 5.5% interest compounded
quarterly, how much money will you have to use for your down payment?

2. You have your heart set on buying a new car in 2 years. You invest $3,200 at 3.75% interest
compounded continuously. How much money will have to use for your down payment on your new
car?

3. You are a proud new parent of a baby girl. In eighteen years, you will want to help her pay for college.
How much do you need to invest now at 4.75% interest compounded monthly so you can help her pay
for the $40,000 expense of college?

4. After building your new home, you decide you would like to install an in-ground pool in 7 years. How
much do you need to invest now at 5.25% interest compounded continuously to have the $30,000 you
will need to build the pool?

5. You recently received an inheritance of $11,500 from your grandparents. Which option is the best way
to invest your money and how much better is the best investment? Option A: 5.6% interest
compounded semi-annually for 8 years or Option B: 3.45% interest compounded continuously for 5
years.
Part A)
Part B)
Part C)

Respuesta :

calculista

Answer:

Part 1) [tex]\$6,595.94[/tex]    

Part 2) [tex]\$3,449.23[/tex]    

Part 3) [tex]\$17,040.06[/tex]  

Part 4) [tex]\$20,773.90[/tex]  

Part 5) The Option A is the best way to invest the money by $4,223.94 than Option B

Step-by-step explanation:

Part 1)

we know that    

The compound interest formula is equal to  

[tex]A=P(1+\frac{r}{n})^{nt}[/tex]  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

[tex]t=7\ years\\ P=\$4,500\\ r=5.5\%=5.5/100=0.055\\n=4[/tex]  

substitute in the formula above  

[tex]A=4,500(1+\frac{0.055}{4})^{4*7}[/tex]  

[tex]A=4,500(1.01375)^{28}[/tex]

[tex]A=\$6,595.94[/tex]    

Part 2)

we know that

The formula to calculate continuously compounded interest is equal to

[tex]A=P(e)^{rt}[/tex]  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest in decimal  

t is Number of Time Periods  

e is the mathematical constant number

we have  

[tex]t=2\ years\\ P=\$3,200\\ r=3.75\%=3.75/100=0.0375[/tex]  

substitute in the formula above  

[tex]A=3,200(e)^{0.0375*2}[/tex]

[tex]A=\$3,449.23[/tex]    

Part 3)

we know that    

The compound interest formula is equal to  

[tex]A=P(1+\frac{r}{n})^{nt}[/tex]  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

[tex]t=18\ years\\ A=\$40,000\\ r=4.75\%=4.75/100=0.0475\\n=12[/tex]  

substitute in the formula above  

[tex]40,000=P(1+\frac{0.0475}{12})^{12*18}[/tex]  

[tex]40,000=P(\frac{12.0475}{12})^{216}[/tex]  

[tex]P=40,000/[(\frac{12.0475}{12})^{216}][/tex]  

[tex]P=\$17,040.06[/tex]  

Part 4)

we know that

The formula to calculate continuously compounded interest is equal to

[tex]A=P(e)^{rt}[/tex]  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest in decimal  

t is Number of Time Periods  

e is the mathematical constant number

we have  

[tex]t=7\ years\\ A=\$30,000\\ r=5.25\%=5.25/100=0.0525[/tex]  

substitute in the formula above  

[tex]30,000=P(e)^{0.0525*7}[/tex]  

[tex]30,000=P(e)^{0.3675}[/tex]  

[tex]P=30,000/(e)^{0.3675}[/tex]  

[tex]P=\$20,773.90[/tex]  

Part 5)

Option A

we know that    

The compound interest formula is equal to  

[tex]A=P(1+\frac{r}{n})^{nt}[/tex]  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

[tex]t=8\ years\\ P=\$11,500\\ r=5.6\%=5.6/100=0.056\\n=2[/tex]  

substitute in the formula above  

[tex]A=11,500(1+\frac{0.056}{2})^{2*8}[/tex]  

[tex]A=11,500(1.028)^{16}[/tex]

[tex]A=\$17,889.07[/tex]  

Option B

we know that

The formula to calculate continuously compounded interest is equal to

[tex]A=P(e)^{rt}[/tex]  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest in decimal  

t is Number of Time Periods  

e is the mathematical constant number

we have  

[tex]t=5\ years\\ P=\$11,500\\ r=3.45\%=3.45/100=0.0345[/tex]  

substitute in the formula above  

[tex]A=11,500(e)^{0.0345*5}[/tex]  

[tex]A=11,500(e)^{0.1725}[/tex]  

[tex]A=\$13,665.13[/tex]  

Compare the options

Option A ------> [tex]\$17,889.07[/tex]  

Option B -----> [tex]\$13,665.13[/tex]  

so

Option A > Option B

Find out the difference

[tex]\$17,889.07-$13,665.13=$4,223.94[/tex]  

therefore

The Option A is the best way to invest the money by $4,223.94 than Option B

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