Respuesta :
First we will find the interest on:
P = $235 principal
t = 2 years
r = 0.1415 annual rate
A = future value
I = A - P the interest
A = P(1 + r)^t
A = 235(1 + 0.1415)^2
A = $306.21
I = A - P
I = $306.21 - $235
I = $71.21
the interest was $71.21.
Next lets find the lifetime cost value:
Lifetime cost value = 306.21 + 5*1.56*52 + 5*0.78*52 = $914.61 (considering that 1 year = 52 weeks)
Now lets find the percentage what percentage the interest is of the lifetime cost:
(71.21/914.61)*100 = 7.79%
← Previous ProblemNext Problem →The percentage of the lifetime cost of the water heater was [tex]7.78\%[/tex] .
What is compound interest ?
Compound interest is the interest on principal which is charged on both the initial principal and the interest from previous periods.
[tex]Amount = P(1+\frac{R}{100} )^t[/tex],
Here, [tex]P= Principal,[/tex]
[tex]R= Rate[/tex] and [tex]t=time[/tex]
We have,
Cost of Water Heater [tex]=\$ 235[/tex]
Water heater cost an average in electricity [tex]=\$1.56[/tex] per week
Water heater cost an average in water [tex]=\$0.78[/tex] per week
APR of credit card [tex]=14.15\%[/tex] compounded monthly
Now,
Find Amount which was compounded,
[tex]Amount = P(1+\frac{R}{100} )^t[/tex]
[tex]A=235(1+\frac{14.15}{100} )^2[/tex]
[tex]A=306.20[/tex]
Now,
So, Total Interest was charged ;
[tex]Interest=Amount - Principal[/tex]
[tex]I = 306.20-235[/tex]
[tex]I=71.2[/tex],
To find the lifetime cost;
Heater was used for [tex]5[/tex] years,
[tex]1\ Year =52\ weeks[/tex]
So,
Cost of electricity [tex]= 5\ year * 52\ week\ *\ \$ 1.56=\$405.6[/tex]
Cost of water [tex]= 5\ year * 52\ week\ *\ \$ 0.78=\$202.8[/tex]
So,
Total lifetime cost [tex]= 306.2+405.6+202.8=\$914.6[/tex]
Now,
The percentage of lifetime cost as interest;
Percentage [tex]=\frac{Percentage}{Total\ cost} *\ 100[/tex]
[tex]=\frac{71.2}{914.6} *100[/tex]
Percentage [tex]=7.78\%[/tex]
Hence, we can say that the percentage of the lifetime cost of the water heater was [tex]7.78\%[/tex] .
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