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tiny-mole [99]
3 years ago
15

Carlos bought a $235 water heater with his credit card. He used the water heater for five years before replacing it. He paid off

the water heater after two years, making monthly payments. The water heater cost him an average of $1.56 per week in electricity, and $0.78 per week in water. If Carlos’s credit card has an APR of 14.15%, compounded monthly, and he made no other purchases with it, what percentage of the lifetime cost of the water heater was interest? (Round all dollar values to the nearest cent.)
Mathematics
2 answers:
Sergio [31]3 years ago
7 0

Answer:

C. 4.12%

Step-by-step explanation:

n200080 [17]3 years ago
6 0

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%

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3 years ago
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What is the value of the expression i 0 × i 1 × i 2 × i 3 × i 4?
astraxan [27]
ANSWER


The value of the expression is
- 1


EXPLANATION

Method 1: Rewrite as product of
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The expression given to us is,

{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}


We use the fact that
{i}^{2}  =  - 1
to simplify the above expression.



{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  {i}^{0}  \times {i}^{1}  \times {i}^{3}   \times {i}^{2}   \times {i}^{4}


This implies,


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  {i}^{0}  \times {i}^{2}  \times {i}^{2}   \times {i}^{2}   \times {i}^{2} \times {i}^{2}


We substitute to obtain,

{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  1\times  - 1 \times  - 1  \times  - 1\times  - 1 \times  - 1


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  1\times  1 \times   1  \times  - 1 =  - 1


Method 2: Use indices to solve.



{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  = {i}^{0 + 1 + 2 + 3 + 4}



This implies that,


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  = {i}^{10}




{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  (  {{i}^{2}} )^{5}


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  (   - 1 )^{5}   =  - 1


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