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arlik [135]
3 years ago
13

$7600 is invested for one year, part at 4% and the remainder at 5%. The income from the two investments is $338. Find the amount

invested at 4%.
Mathematics
1 answer:
wariber [46]3 years ago
8 0
The amount invested at 4% is $4,200. This problem can be solved by using two linear equation with two different variables which represent the investment's partition and the investment's return. The first equation is x+y=$7,600 and the second equation is 0.04x + 0.05y=$338, then you must substitute the y variable in the second question with the first equation (0.04x + 0.05($7,600-x)=$338). Therefore, you will get a new equation to solve the problem which is 0.04x+380-0.05x=338 resulting x=$4,200.
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Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

<em>A</em> = video components need repair within 1 year

<em>B</em> = electronic components need repair within 1 year

<em>C</em> = audio components need repair within 1 year

The information provided is:

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P (C) = 0.001

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(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

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Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

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=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

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