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anonymous
 5 years ago
Tian opened a saving account on January 1st with a deposite of $150. The following July 1st, January 1st, and July 1st, he made 3 more deposits of $150 each. The account paid interest at 3.75% per annum, compounded semiannually. a) What amount did Tian have in his account at the end of the second year? b) How much interest did Tian earn by the end of the second year?
anonymous
 5 years ago
Tian opened a saving account on January 1st with a deposite of $150. The following July 1st, January 1st, and July 1st, he made 3 more deposits of $150 each. The account paid interest at 3.75% per annum, compounded semiannually. a) What amount did Tian have in his account at the end of the second year? b) How much interest did Tian earn by the end of the second year?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do it in segments remember that there are 12 months in a year. n=2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I tried R, refular deposits (payment) = 150, n, number of deposits or payment = 3, and i, interest rate per compounding period: 0.03765/2 = 0.01875. I got $458. 50 in the end using this formula: A = R [(1 + i)^n  1 / i. However, the correct answers are: a) $617.09 and b) $17.09

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0n = 2 because it is compounded twice a year so every six months 3.75% of the money present in the account is added to the original amount invested.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh.. so what's the function of the 'he made 3 more deposits of $150.'?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We know the original amount was compounded once before he made his second investment, then compounded again when he made his third investment than compounded again when he made his fourth investment

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To answer b) The first compound can be expressed as (0.0375)*150 + 150 The second compound he added 150 so for the second compound (first year we have the expression) ((((0.0375)*150 + 150) + 150)0.03 + ((((0.0375)*150 + 150) + 150)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would probably just try answering this manually :L

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohhh, okay. So do I still use the formula or.. no?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well you could use it for every compound but make sure you change the principle accordingly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm sure there is a way to modify the equation so that ever time there is a compound 150 is added plus the interest

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh.. okay. ohhhh.. like this: R + R(1 + i)^1 + R(1 + i)^2 + R(1+i)^3 +... + R (1+i)^n1? And calculate the sum of the series: which is basically the formula?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, that one is for compounding interests isn't it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah but I was wrong the formula has to take in account the added 150 after each compound
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