Answered

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Suppose an investment of $1,800 doubles in value every 7 years. How much is the investment worth after 42 years?

Sagot :

x=2^n(1800)
x=2^6(1800)
x=$115,200
SociometricStar

Answer:

The investment in 42 years will become $115199.88

Step-by-step explanation:

Let the investment model is [tex]A=Pe^{rt}[/tex]

A = final amount

P = initial amount

r = rate

t = time

Now, given that

P = $1800

A = 2P = 2×1800 = $3600

t = 7 years

[tex]3600=1800e^{7r}\\\\2=e^{7r}\\ln2=7r\\r=\frac{1}{7}\ln2\\\\r=0.099021[/tex]

Now, we have to find A for t = 42 years

[tex]A=1800e^{42\times0.099021}\\\\A=115199.88[/tex]

Hence, the investment in 42 years will become $115199.88

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