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Sedbober [7]
3 years ago
10

Two perpendicular lines have opposite y-intercepts. The equation of one of these lines is y = mx + b. Express the x-coordinate o

f the intersection point of the lines in terms of m and b.
Mathematics
1 answer:
Fofino [41]3 years ago
6 0

Answer:

x = -2bm/(m²+1)

Step-by-step explanation:

One line has equation

... y = mx + b

The slope of the perpendicular line is the negative reciprocal of the slope of the original line. The perpendicular line with the opposite y-intercept has equation

... y = (-1/m)x - b

The point of intersection is where the x- and y-values are equal, so ...

... mx + b = (-1/m)x - b

... (m +1/m)x = -2b . . . . . . . add 1/m - b to both sides

... (m²+1)x = -2bm . . . . . . . multiply by m

... x = -2bm/(m²+1) . . . . . . divide by the coefficient of x

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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
The sum of the first ten terms of a linear sequence is 145. the sum of the next ten term is 445. find the sum of the first four
chubhunter [2.5K]

The sum of the first four terms of the sequence is 22.

In this question,

The formula of sum of linear sequence is

S_n =\frac{n}{2}(2a+(n-1)d)

The sum of the first ten terms of a linear sequence is 145

⇒ S_{10} =\frac{10}{2}(2a+(10-1)d)

⇒ 145 = 5 (2a+9d)

⇒ \frac{145}{5} =2a+9d

⇒ 29 = 2a + 9d  ------- (1)

The sum of the next ten term is 445, so the sum of first twenty terms is

⇒ 145 + 445

⇒ S_{20} =\frac{20}{2}(2a+(20-1)d)

⇒ 590 = 10 (2a + 19d)

⇒ \frac{590}{10}=2a+19d

⇒ 59 = 2a + 19d -------- (2)

Now subtract (2) from (1),

⇒ 30 = 10d

⇒ d = \frac{30}{10}

⇒ d = 3

Substitute d in (1), we get

⇒ 29 = 2a + 9(3)

⇒ 29 = 2a + 27

⇒ 29 - 27 = 2a

⇒ 2 = 2a

⇒ a = \frac{2}{2}

⇒ a = 1

Thus, sum of first four terms is

⇒ S_4 =\frac{4}{2}(2(1)+(4-1)(3))

⇒ S_4 =2(2+(3)(3))

⇒ S₄ = 2(2+9)

⇒ S₄ = 2(11)

⇒ S₄ = 22.

Hence we can conclude that the sum of the first four terms of the sequence is 22.

Learn more about sum of sequence of n terms here

brainly.com/question/20385181

#SPJ4

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