Answered

Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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

Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.