Respuesta :
Answer:
a) $3,571.02
b) $3,572.9
c) $3,573.74
Step-by-step explanation:
Data provided in the question:
p = $3000,
r = 3.5%,
t = 5 years
a) quarterly
number of periods in a year, n = 4
Interest rate per period = 3.5% ÷ 4 = 0.875%
Now,
[tex]A = p\times \left( 1 + \frac{r}{n} \right)^{\Large{n \cdot t}}[/tex]
A = total amount
n = number of times compounded per year
on substituting the respective values, we get
A = 3000 × [tex]\left( 1 + \frac{ 0.035 }{ 4 } \right)^{\Large{ 4 \cdot 5 }}[/tex]
A = 3000 × [/tex]\cdot { 1.00875 } ^ { 20 }[/tex]
A = 3000 × 1.19034
A = $3,571.02
b) monthly
number of periods in a year, n = 12
Now,
A = [tex]p\times \left( 1 + \frac{r}{n} \right)^{\Large{n \cdot t}}[/tex]
on substituting the respective values, we get
A = 3000 × [tex]\left( 1 + \frac{ 0.035 }{ 12 } \right)^{\Large{ 12 \cdot 5 }}[/tex]
A = 3000 × [/tex]\cdot { 1.002917} ^ { 60 }[/tex]
A = 3000 × 1.190967
A = $3,572.9
c) continuously
A = [tex]pe^{r\times t}[/tex]
on substituting the respective values, we get
A = 3,000 × [tex]e^{0.035\times 5}[/tex]
or
A = 3,000 × [tex]e^{0.175}[/tex]
or
A = 3,000 × 1.1912
or
A = $3,573.74
Answer:
Step-by-step explanation:
a) P = 3000
It was compounded quarterly. This means that it was compounded 4 times in a year. So
n = 4
The rate at which the principal was compounded is 3.5%. So
r = 3.5/100 = 0.035
It was compounded for just 5 years. So
t = 5
The formula for compound interest is
A = P(1+r/n)^nt
A = total amount in the account at the end of t years. Therefore
A = 3000 (1+0.035/4)^4×5
A = $3571
b) It was compounded monthly. This means that it was compounded 12 times in a year. So
n = 12
Therefore,
A = 3000 (1+0.035/12)^12×5
A = $3572.88
c) The formula for continuously compounded interest is
A = Pe (r x t)
where
e = 2.7183.
Therefore
A = 3000 × 2.7183^(0.035 × 5)
A = $3573.7
