Math, asked by ankitssm1362, 9 months ago

Si se invierten $5,000, con una tasa de interés de 9% anual, compuesto mensualmente, el valor futuro S en un periodo t (en meses) se obtiene por medio de = 5,000(1.0075)^t a. ¿Cuál es la cantidad después de 1 año? b. ¿Cuánto tiempo pasar antes de que se duplique la inversión?

Answers

Answered by amitnrw
0

Given :  $ 5,000 is invested, with an interest rate of 9% per year, compounded monthly, the future value S in a period t (in months) is obtained by means of = 5,000 (1.0075) ^ t

To Find :  t for doubling the amount

amount after 1 year

Solution:

P = $ 5000

R = 9 % PA  = 0.75 % per month

n = t/12 years  = t  months

A = 2P = 2 * 5000

A = P(1 + R/100)ⁿ

=> 2 * 5000 = 5000 (1 + 0.75/100)^t

=> 2 = (1.0075)^t

Taking log both sides

=> log 2 = t log (1.0075)

=> 0.3010 = t (0.003245)

=> t = 92.766  months

Approx 93 months  to double the investment

amount after 1 year = 5000(1.0075)¹² = $5,469

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