Respuesta :
Answer:
FYAB gives a better deal.
Step-by-step explanation:
Compound interest:
[tex] A = P(1 + \dfrac{r}{n})^{nt} [/tex]
Continuously compounded interest:
[tex] A = Pe^{rt} [/tex]
For the quarterly compounded interest, r = 0.25%, and n = 4.
[tex] 2P = P(1 + \dfrac{0.0025}{4})^{4t} [/tex]
[tex] 1.000625^{4t} = 2 [/tex]
[tex] \log (1.000625^{4t}) = \log 2 [/tex]
[tex] 4t(\log 1.000625) = \log 2 [/tex]
[tex] t = \dfrac{\log 2}{4\log 1.000625} [/tex]
[tex] t = 277 [/tex]
For the continuously compounded interest, r =0.23%
[tex] A = Pe^{rt} [/tex]
[tex] 2P = Pe^{0.0023t} [/tex]
[tex] 2 = e^{0.0023t} [/tex]
[tex] \ln 2 = \ln e^{0.0023t} [/tex]
[tex] \ln 2 = 0.0023t [/tex]
[tex] t = \dfrac {\ln 2}{0.0023} [/tex]
[tex] t = 301 [/tex]
The quarterly compounded doubles in 277 years.
The continuously compounded doubles in 301 years.
Answer: FYAB gives a better deal.
Applying the interest formulas, it is found that due to the smaller doubling time, the FYAB bank is giving the better deal.
The amount of money is continuous compounding, with a interest rate of r, as a decimal, is given by:
[tex]A(t) = A(0)e^{rt}[/tex]
Interest rate of 0.23%, thus [tex]r = 0.0023[/tex].
The doubling time is t for which:
[tex]A(t) = 2A(0)[/tex]
Hence:
[tex]2A(0) = A(0)e^{0.0023t}[/tex]
[tex]e^{0.0023t} = 2[/tex]
[tex]\ln{e^{0.0023t}} = \ln{2}[/tex]
[tex]0.0023t = \ln{2}[/tex]
[tex]t = \frac{\ln{2}}{0.0023}[/tex]
[tex]t = 301.37[/tex]
Using continuous compounding, the doubling time is of 301.37 years.
Compound interest:
[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]
- A(t) is the amount of money after t years.
- P is the principal(the initial sum of money).
- r is the interest rate(as a decimal value).
- n is the number of times that interest is compounded per year.
- t is the time in years for which the money is invested or borrowed.
In this problem, 0.25% compounded quarterly, hence [tex]r = 0.0025, n = 4[/tex].
[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]
[tex]2P = P\left(1 + \frac{0.0025}{4}\right)^{4t}[/tex]
[tex](1.000625)^{4t} = 2[/tex]
[tex]\log{(1.000625)^{4t}} = \log{2}[/tex]
[tex]4t\log{1.000625} = \log{2}[/tex]
[tex]t = \frac{\log{2}}{4\log{1.000625}}[/tex]
[tex]t = 277.35[/tex]
Due to the smaller doubling time, the FYAB bank is giving the better deal.
A similar problem is given at https://brainly.com/question/14098039
