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Sedaia [141]
3 years ago
15

What is the slope of a line perpendicular to the line whose equation is x-3y=-6x−3y=−6​

Mathematics
1 answer:
marissa [1.9K]3 years ago
7 0

Answer:

- 3

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Rearrange x - 3y = - 6 into this form

Subtract x from both sides

- 3y = - x - 6 ( divide all terms by - 3 )

y = \frac{1}{3} x + 2 ← in slope- intercept form

with slope m = \frac{1}{3}

Given the slope m of a line then the slope of a line perpendicular to it is

m_{perpendicular} = - \frac{1}{m} = - \frac{1}{\frac{1}{3} } = - 3

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What does product of prime numbers mean​
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product means addition. So the product of prime numbers would be all of them added together.

Step-by-step explanation:

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Tariq has a collection of 32 quarters that he wants to send to his cousin. What is the total weight of the quarters in kilograms
Mumz [18]

One quarter weighs about 5.67 grams.

5.67 x 32 = 181.44 grams

Grams to kg = .001

181.44 x .001 = .18144 kg

Answer: .18144 kg

7 0
3 years ago
James is 1 3/7 times as tall as his brother. His brother is 3 and 1/2 ft tall is James high more or less than 3 1/2 feet
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Answer:

he is higher than his brother

Step-by-step explanation:

because james states that he is 1 3/7 times as tall as his brother

3 0
3 years ago
The statistical difference between a process operating at a 5 sigma level and a process operating at a 6 sigma level is markedly
Svet_ta [14]

Answer:

True

Step-by-step explanation:

A six sigma level has a lower and upper specification limits between \\ (\mu - 6\sigma) and \\ (\mu + 6\sigma). It means that the probability of finding no defects in a process is, considering 12 significant figures, for values symmetrically covered for standard deviations from the mean of a normal distribution:

\\ p = F(\mu + 6\sigma) - F(\mu - 6\sigma) = 0.999999998027

For those with defects <em>operating at a 6 sigma level, </em>the probability is:

\\ 1 - p = 1 - 0.999999998027 = 0.000000001973

Similarly, for finding <em>no defects</em> in a 5 sigma level, we have:

\\ p = F(\mu + 5\sigma) - F(\mu - 5\sigma) = 0.999999426697.

The probability of defects is:

\\ 1 - p = 1 - 0.999999426697 = 0.000000573303

Well, the defects present in a six sigma level and a five sigma level are, respectively:

\\ {6\sigma} = 0.000000001973 = 1.973 * 10^{-9} \approx \frac{2}{10^9} \approx \frac{2}{1000000000}

\\ {5\sigma} = 0.000000573303 = 5.73303 * 10^{-7} \approx \frac{6}{10^7} \approx \frac{6}{10000000}  

Then, comparing both fractions, we can confirm that a <em>6 sigma level is markedly different when it comes to the number of defects present:</em>

\\ {6\sigma} \approx \frac{2}{10^9} [1]

\\ {5\sigma} \approx \frac{6}{10^7} = \frac{6}{10^7}*\frac{10^2}{10^2}=\frac{600}{10^9} [2]

Comparing [1] and [2], a six sigma process has <em>2 defects per billion</em> opportunities, whereas a five sigma process has <em>600 defects per billion</em> opportunities.

8 0
3 years ago
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