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SOVA2 [1]
2 years ago
9

A financial advisor tells you that you can make your child a millionaire if you just start saving early. You decide to put an eq

ual amount each year into an investment account that earns 7.5% interest per year, starting on the day your child is born. How much would you need to invest each year (rounded to the nearest dollar) to accumulate a million for your child by the time he is 35 years old? (Your last deposit will be made on his 34th birthday.)​
Mathematics
1 answer:
Blababa [14]2 years ago
5 0

Answer:

$12159 per year.

Step-by-step explanation:

If I invest $x each year at the simple interest of 7.5%, then the first $x will grow for 35 years, the second $x will grow for 34 years and so on.

So, the total amount that will grow after 35 years by investing $x at the start of each year at the rate of 7.5% simple interest will be given by

x( 1 + \frac{35 \times 7.5}{100}) + x( 1 + \frac{34 \times 7.5}{100}) + x( 1 + \frac{33 \times 7.5}{100}) + ......... + x( 1 + \frac{1 \times 7.5}{100})

= 35x + \frac{x \times 7.5}{100} [35 + 34 + 33 + ......... + 1]

= 35x + \frac{x \times 7.5}{100} [\frac{1}{2} (35) (35 + 1)]

{Since sum of n natural numbers is given by \frac{1}{2} (n)(n + 1)}

= 35x + 47.25x

= 82.25x

Now, given that the final amount will be i million dollars = $1000000

So, 82.25x = 1000000

⇒ x = $12,158. 05 ≈ $12159

Therefore. I have to invest $12159 per year. (Answer)

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Margin of error at 90% is 0.024

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\frac{d}{dt}(x^{2})+\frac{d}{dt}(y^{2})=\frac{d}{dt}(25)\\\\(\frac{d}{dx}(x^{2})*\frac{dx}{dt}) +(\frac{d}{dy}(y^{2})*\frac{dy}{dt})=\frac{d}{dt}(25)\\\\2x\frac{dx}{dt} +2y\frac{dy}{dt} = 0\\\\

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