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blsea [12.9K]
3 years ago
9

Please help me please

Mathematics
1 answer:
adelina 88 [10]3 years ago
5 0

Answer:

18 and 74 are true.

Step-by-step explanation:

So we want to solve this inquality. 13+4x> 75\\ We subtract 13 from the left and the right. It leaves us with 4x>62. Then we want the x by itself so we divide the 4 from 4x and 4 from 62. We get x> 15 \frac{1}{2}. So 15.5 is false since its not greater than. 0 is not a solution. 18 and 74 are true. I'm sorry if this is wrong.

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For the following amount at the given interest rate compounded​ continuously, find​ (a) the future value after 9 ​years, (b) the
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Answer:

(a) The future value after 9 ​years is $7142.49.

(b) The effective rate is r_E=4.759 \:{\%}.

(c) The time to reach ​$13,000 is 21.88 years.

Step-by-step explanation:

The definition of Continuous Compounding is

If a deposit of P dollars is invested at a rate of interest r compounded continuously for t years, the compound amount is

A=Pe^{rt}

(a) From the information given

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r=4.65\%=\frac{4.65}{100} =0.0465

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Applying the above formula we get that

A=4700e^{0.0465\cdot 9}\\A=7142.49

The future value after 9 ​years is $7142.49.

(b) The effective rate is given by

r_E=e^r-1

Therefore,

r_E=e^{0.0465}-1=0.04759\\r_E=4.759 \:{\%}

(c) To find the time to reach ​$13,000, we must solve the equation

13000=4700e^{0.0465\cdot t}

4700e^{0.0465t}=13000\\\\\frac{4700e^{0.0465t}}{4700}=\frac{13000}{4700}\\\\e^{0.0465t}=\frac{130}{47}\\\\\ln \left(e^{0.0465t}\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t\ln \left(e\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t=\ln \left(\frac{130}{47}\right)\\\\t=\frac{\ln \left(\frac{130}{47}\right)}{0.0465}\approx21.88

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