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Elodia [21]
3 years ago
7

Simplify 18 - 2[ x + ( x - 5)]. 8 - 4 x 28 - 4 x 28 - 2 x

Mathematics
2 answers:
Alborosie3 years ago
5 0

First u have to factor the expression

2(9-(x+(x-5))x 8 - 2 x 28 - 2 x 28 - x)

next you have to remove the parenthesis and multiply

2(9 -(x + x - 5) x 8 - 2 x 28 - 2 x 28 - x) that's -2 x 28

2(9 -( x+ x - 5) x 8 - 56 - 2 x 28 - x)

 2(9 -( x + x -5) x 8 -56 -56 - x)

Check like terms

2(9 -(2x-5) x 8 -56 -56 - x)

multiply by 8

2(9 - (16x - 40) -56 -56 - x)

remove parenthesis

2(9 - 16x + 40 -56 -56 - x)

calculate  and collect like terms

2( -63 - 16x - x)

collect the like terms

2( -63 -17x)

your answer is: 2(-63 - 17x)

Arisa [49]3 years ago
3 0

The answer for this question is 28-4x

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In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the
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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:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events <em>A</em>, <em>B</em> and <em>C</em> are independent.

(a)

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

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278

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)

= P (A or B or C)

=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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