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sertanlavr [38]
3 years ago
8

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

effective​ rate, and​ (c) the time to reach ​$13 comma 000.
$4700 at 4.65%
Mathematics
1 answer:
Stella [2.4K]3 years ago
3 0

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

P=4700

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

t=9 \:years

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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Answer: t-half = ln(2) / λ ≈ 0.693 / λ

Explanation:

The question is incomplete, so I did some research and found the complete question in internet.

The complete question is:

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

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Express your answer in terms of N0 and/or λ.

Answer:

1) Equation given:

N(t)=N _{0} e^{-  \alpha  t} ← I used α instead of λ just for editing facility..

Where No is the initial number of nuclei.

2) Half of the initial number of nuclei: N (t-half) =  No / 2

So, replace in the given equation:

N_{t-half} =  N_{0} /2 =  N_{0}  e^{- \alpha t}

3) Solving for α (remember α is λ)

\frac{1}{2} =  e^{- \alpha t} &#10;&#10;2 =   e^{ \alpha t} &#10;&#10; \alpha t = ln(2)

αt ≈ 0.693

⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ




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