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If XYZ invested $5,800 today in an account that is expected to earn 3.2 percent per year, and she expects to make another investment in the same account in 3 years from today, then how much money does XYZ expect to invest in 3 years if she expects to have $15,000 in her account in 4 years from today

Sagot :

Answer:

The answer is "$8,160.08".

Explanation:

[tex]A= \text{future value} = \$ 15,000 \\\\P= \text{present value}= \$ 5,800 \\\\r=\tex{rate} =3.2 \%\\\\n= \text{time in years} = 4[/tex]

Using formula:

[tex]A=P(1+ \frac{r}{100})^n + \text{Investment in 3 years} \times (1.032)\\\\15,000=5,800 (1+ \frac{3.2}{100})^4 + \text{Investment in 3 years} \times (1.032)\\\\15,000=5,800 (1+ 0.032)^4 + \text{Investment in 3 years} \times (1.032)\\\\15,000=5,800 (1.032)^4 + \text{Investment in 3 years} \times (1.032)\\\\[/tex]

[tex]15,000 = 5,800 \times 1.13427612 +\text{Investment in 3 years} \times (1.032)\\\\15,000=6,578.8015 + \text{Investment in 3 years} \times (1.032)\\\\\text{Investment in 3 years} = \frac{(15000-6578.8015)}{1.032}\\\\\text{Investment in 3 years} = \frac{8,421.1985}{1.032}\\\\\text{Investment in 3 years} = 8,160.07607\\\\ \text{Investment in 3 years} = 8,160. 08[/tex]

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