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Black_prince [1.1K]
3 years ago
11

a person invests 9000 dollars in the back. The back pats 4.75% interest compounded quarterly. To the tenth of a year, how long m

ust a person leave the money in the back until it reaches 11800 dollars
Mathematics
1 answer:
MaRussiya [10]3 years ago
6 0

Answer:

5.7 years

Step-by-step explanation:

we know that    

The compound interest formula is equal to  

A=P(1+\frac{r}{n})^{nt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

t=?\ years\\ P=\$9,000\\A=\$11,800\\r=4.75\%=4.75/100=0.0475\\n=4

substitute in the formula above

11,800=9,000(1+\frac{0.0475}{4})^{4t}  

Solve for t

\frac{11.8}{9} =(1.011875)^{4t}  

Applying property of exponents

\frac{11.8}{9} =(1.011875^{4})^{t}  

Applying log both sides

log(\frac{11.8}{9})=log[(1.011875^{4})^{t}]  

log(\frac{11.8}{9})=(t)log[(1.011875^{4})]  

t=log(\frac{11.8}{9})/log[(1.011875^{4})]  

t=5.7\ years

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