Question

6. If you invest $4700 into an account that pays 4.3% interest rate compounded continuously, find (a) the future value after 9 years, (b) the effective rate, and (c) the time to reach $10,000. Round all values to the nearest hundredth. 7. June made an initial deposit of $ 2100 in an account for her son. Assuming an interest rate of 2% compounded quarterly, how much will the account be worth in 19 years? 8. A small company borrows $ 81.000 at 6 % compounded monthly. The loan is due in 5 years. How much interest will the company pay? 9. Barbara knows that she will need to buy a new car in 4 years. The car will cost $15,000 by then. How much should she invest now at 2%, compounded quarterly, so that she will have enough to buy a new car?

please help to do 6 8 9 thank u soooo much

Answer 1

Question - 6

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for continuous compound formula is

$A=Pe^{rt}$

P=4700

r=4.3% = 0.043

t= 9 years

$A=4700e^{0.043\cdot 9}$

$A= 4700\cdot \:1.4725$

${\color{Red} A \approx 6921.02}$

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effective rate is

$R=\left[e^r-1\right]\cdot 100\%$

$R=\left[e^{0.043}-1\right]\cdot 100\%$

$R=\left[1.04393-1\right]\cdot 100\%$

$R=\left[0.04393\right]\cdot 100\%$

${\color{Red} R= 4.393\%}$

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when A=10000

$10000=4700e^{0.043t}$

$\frac{10000}{4700}=e^{0.043t}$

$2.12765=e^{0.043t}$

$\ln \left(2.12765\right)=\ln \left(e^{0.043t}\right)$

$\ln \left(2.12765\right)=0.043t$

$t=\frac{\ln \left(2.12765\right)}{0.043}$

$t=17.5585$

${\color{Red} t \approx 17.56}$ years

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Question - 7

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$A=P\left(1+\frac{r}{n}\right)^{nt}$

P=2100

r=2% = 0.02

n=4 for quarterly compound

t= 19 years

$A=2100\left(1+\frac{0.02}{4}\right)^{4\cdot 19}$

$A= 2100\left(1+0.005\right)^{76}$

$A= 2100\left(1.005\right)^{76}$

$A= 2100\cdot \:1.4609$

A = 3067.89144

${\color{Red} A \approx 3067.89 }$

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Question - 8

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$A=P\left(1+\frac{r}{n}\right)^{nt}$

P=81000

r=6% = 0.06

n=12 for monthly compound

t= 5 years

$A=81000\left(1+\frac{0.06}{12}\right)^{12\cdot 5}$

$A= 81000\left(1+0.005\right)^{60}$

$A= 81000\left(1.005\right)^{60}$

$A= 81000\cdot \:1.34885$

${\color{Blue} A \approx 109256.86 }$

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interest is

$I=A-P$

$I=109256.86-81000$

${\color{Red} I= 28256.86}$

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Question - 9

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$A=P\left(1+\frac{r}{n}\right)^{nt}$

A=15000

r=2% = 0.02

n=4 for quartely compound

t=4 years

$15000=P\left(1+\frac{0.02}{4}\right)^{4\cdot 4}$

$15000=P\left(1+0.005\right)^{16}$

15000 = P(1.005)16

$15000=P\cdot 1.08307$

$P=\frac{15000}{1.08307}$

${\color{Red} P \approx 13849.5 1}$