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USPshnik [31]
2 years ago
11

Solve for x: 5x + one third(3x + 6) > 14

Mathematics
2 answers:
Masteriza [31]2 years ago
6 0

Answer:

x > 2

Step-by-step explanation:

Given

5x + \frac{1}{3}(3x + 6) > 14 ← distribute and simplify left side

5x + x + 2 > 14

6x + 2 > 14 ( subtract 2 from both sides )

6x > 12 ( divide both sides by 6 )

x > 2

ahrayia [7]2 years ago
6 0

Answer: x > 2  

Step-by-step explanation:

5x + 1/3(3x +6) > 14     Distribute

5x + 1x + 2 > 14     Combine like terms on the left side

  6x + 2 > 14      Subtract 2 from both sides

         -2    -2

   6x > 12      Divde both sides by 6

 x > 2

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10*10=100

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Step-by-step explanation:


3 0
2 years ago
A quality control engineer tests the quality of produced computers. suppose that 5% of computers have defects, and defects occur
11111nata11111 [884]
(a) 0.059582148 probability of exactly 3 defective out of 20

 (b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.

  (a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So

 0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)

 = 0.05^3 * 0.95^17 * 20! / (3!17!)

 = 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)

 = 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)

 = 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)

 = 0.05^3 * 0.95^17 * 20*19*3

 = 0.000125* 0.418120335 * 1140

 = 0.059582148

  (b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.

 The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)

 = 0.05*0.95^3*24/(1!3!)

 = 0.05*0.857375*24/6

 = 0.171475

 
 So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
3 0
2 years ago
A department store purchases logo shirts at a cost of $10 per shirt. They increase the price 100% and put them on the sales floo
Andrei [34K]

The original selling price is found by multiplying 100% by $10 and adding that to the original $10. The logo shirt goes on the sale floor for $20. Each of the next three months, the reduced price is found by multiplying the current sale price by 75%. The tax is found by multiplying 5% by the final discounted price and adding the tax to the final sale price.

7 0
2 years ago
Read 2 more answers
ou have two parents, four grandparents, eight great-grandparents, and so forth. (a) If all your ancestors were distinct, what wo
Maslowich

Answer:

Part A:

2^{40}-2\\=1.0995*10^{12} ancestors

Part B:

Generations Age=25*39=975 years

Part C:

  1. Some ancestors on different branches of the family tree must be the same.
  2. There could not have been 39 generations in my line of ancestry.

Step-by-step explanation:

Given Data:

Two Parents, Four Grand Parents, Eight great Grand Parents.

Generation=39

Solution:

Part A:

From given data Following series is made:

2+2^2+2^3+.........+2^{39}

Now, Above series will become:

2[1+2+2^2+......+2^{38}]

From geometric Sequence:

2[\frac{2^{38+1}-1}{2-1} ]

2^{40}-2\\=1.0995*10^{12} ancestors

Part B:

Generations Age=years*Number of generations

Generations Age=25*39=975 years

Part C:

Number of people lived= 10^11

Ancestors=1.0995*10^{12} ancestors

10^{11}

It means ancestors are not distinct, it means:

  1. Some ancestors on different branches of the family tree must be the same.
  2. There could not have been 39 generations in my line of ancestry.
5 0
2 years ago
The quantities xxx and yyy are proportional.
Valentin [98]

Answer:

The answer is 0.5

Step-by-step explanation:

If you do 4.5 divided by 9 you get 0.5

5 0
2 years ago
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