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May 22, 2019, 04:36:46 am

Author Topic: Modelling linear growth and decay  (Read 143 times)  Share 

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yo091

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Modelling linear growth and decay
« on: April 18, 2019, 06:18:07 pm »
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The question is. Help please.
The amount which should be invested at 4.5% per annum simple interest, if you require $15 700 in three years’ time, is closest to

AlphaZero

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Re: Modelling linear growth and decay
« Reply #1 on: April 18, 2019, 09:31:34 pm »
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The question is. Help please.
The amount which should be invested at 4.5% per annum simple interest, if you require $15 700 in three years’ time, is closest to

I didn't do Further Maths in VCE, so if someone could please check if my method is what is intended by the course that would be great.

Using  \(F=P(1+rt)\implies P=\dfrac{F}{1+rt}\),  we have \[P=\frac{\$15\,700}{1+0.045\!\times\! 3}=\$13\,832.60,\ \ \text{correct to the nearest cent.}\]
2015\(-\)2017:  VCE
2018\(-\)2021:  Bachelor of Biomedicine and Concurrent Diploma in Mathematical Sciences, University of Melbourne


fish58

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Re: Modelling linear growth and decay
« Reply #2 on: April 19, 2019, 12:34:42 pm »
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A simple way to do this is to realise that 3 x 4.5 % = 13.5 %.
Therefore 15700 is 1.135 times the original amount.
Divide 15700 by 1.135 and you get the correct answer.

yo091

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Re: Modelling linear growth and decay
« Reply #3 on: April 19, 2019, 01:06:01 pm »
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What is this formula called?

I didn't do Further Maths in VCE, so if someone could please check if my method is what is intended by the course that would be great.

Using  \(F=P(1+rt)\implies P=\dfrac{F}{1+rt}\),  we have \[P=\frac{\$15\,700}{1+0.045\!\times\! 3}=\$13\,832.60,\ \ \text{correct to the nearest cent.}\]

yo091

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Re: Modelling linear growth and decay
« Reply #4 on: April 19, 2019, 02:08:12 pm »
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Also, how do i do this?
Simple interest at 12% per annum is calculated for the three years of a loan of $y. At the end of the three years the total amount owed is?