A person invest $10,000 into a bank the bank pays 4.75% interest compounded semi annually. To the nearest 10th of a year, how long was the person leave the money in the bank until it reaches $19,200 dollars

T is 13.9 years to the nearest 10th of a year

Step-by-step explanation:

In this question, we are to calculate the number of years at which someone who invests a particular amount will have a particular amount based on compound interest.

To calculate the number of years, what we do is to use the compound interest formula.

Mathematically,

A = P(1+ r/n) ^nt

Where A is the final amount after compounding all interests which is $19,200 according to the question

P is the initial amount invested which is $10,000 according to the question

r is the rate which is 4.75% according to the question = 4.75/100 = 0.0475

n is the number of times per year in which interest is compounded. This is 2 as interest is compounded semi-annually

t= ?

we plug these values;

19200 = 10,000(1+0.0475/2)^2t

divide through by 10,000

1.92 = (1+0.02375)^2t

1.92 = (1.02375)^2t

We find the log of both sides

log 1.92 = log [(1.02375)^2t)

log 1.92 = 2tlog 1.02375

2t = log 1.92/log 1.02375

2t = 27.79

t = 27.79/2

t = 13.89 years

The question asks to give answer to the nearest tenth of a year and thus t = 13.9 years

step-by-step explanation: 4 cause 2+2=4

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step-by-step explanation: