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Varvara68 [4.7K]
4 years ago
12

Seth has just decided to replace his computer. His old computer cost him $1,433 when he bought it exactly seven years ago. Seth

paid for it with his credit card, which has an APR of 11.70%, compounded monthly. He made no other purchases with the card and paid off his balance after two and a half years of making identical monthly payments. The computer consumed about $0.79 of electricity every day. In total, what percentage of the lifetime cost of the computer did the electricity make up? (Assume that two out of the seven years were leap years, and round all dollar values to the nearest cent.)
a.
82.157%
b.
58.500%
c.
21.718%
d.
54.898%


Please select the best answer from the choices provided.

A
B
C
D
Mathematics
1 answer:
puteri [66]4 years ago
3 0

Answer:

The answer is D

Step-by-step explanation:

The solution would be like this for this specific problem:

Cost of the computer:

i = 0.117 / 12 

i = 0.00975 

= monthly payment 

= (i * P * (1 + i)^n) / ((1 + i)^n - 1) 

= (0.00975 * $1433 * (1 + 0.00975)^30) / ((1 + 0.00975)^30 - 1) 

= $55.32 

Electricity cost:

= $0.79 * 365.25 * 7 

= $2019.83 

Electricity Percentage: 

= $2019.83 / ($2019.83 + $1659.60) 

= 54.89% 

Electricity: 

= ($0.79 * 365 * 5) + ($0.79 * 366 * 2) 

= $2020.03 

 

Electricity Percentage: 

= $2020.03 / ($2020.03 + $1659.60) 

= 54.898%

In total, the electricity made up 54.898% of the lifetime cost of the computer.

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3. It should be noted that there is 90% confidence that the true population proportion lies between 0.2919 and 0.4339.

Step-by-step explanation:

1)

Proportion= P = 0.8333333333333 (1/12)

Margin error= 0.06/2 = 0.03

Confidence level= 99

Significance level = α= (100 - 99)%= 1%= 0.01

α/2 = 0.01/2 = 0.0005

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= p(1 - p) (Z*/E)

=0.8333333333333 x (2.576/0.03)^2

=563.15

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=564

2) Sample size n= 124

Sample number of event x =45

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= 0.3629

Standard error =√p(1 - p) /n

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= 0.0432

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Upper limit = p+margin error= 0.3629 + 0.07103= 0.4339

The answer is (0.2919, 0.4339)

3) It should be noted that there is 90% confidence that the true population proportion lies between 0.2919 and 0.4339.

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