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2 years ago

What is the fractional equivalent of the repeating decimal n = 0.5151...?

Mathematics

2 answers:

slava [35]2 years ago

5 0

Answer:

I believe that it is 2, I'm not 100% but ill take the guess

Step-by-step explanation:

Arte-miy333 [17]2 years ago

4 0

Answer:

It is 2

Step-by-step explanation:

You might be interested in

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 operating at a 6 sigma level, the probability is:

$\\ 1 - p = 1 - 0.999999998027 = 0.000000001973$

Similarly, for finding no defects 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 6 sigma level is markedly different when it comes to the number of defects present:

$\\ {6\sigma} \approx \frac{2}{10^9}$ [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 2 defects per billion opportunities, whereas a five sigma process has 600 defects per billion opportunities.

8 0

2 years ago

Somebody help pls Complete the table by filling in the multiplicative inverse of the numbers below

Harlamova29_29 [7]

Answer:

2---> 1/2

1/5 ----> 5

-4 ----> 1/-4

1 2/5 = 7/5 ----> 1/(7/5) = 5/7

Step-by-step explanation:

8 0

2 years ago

Find the remaining trigonometric functions of θ based on the given information.csc θ = 257 and cos θ < 0

katrin [286]

Cos(θ) < 0, so we know it would be in Quadrant 2 or 3
then csc(θ) = 257, but csc(θ) = $\frac{1}{sin(θ)}$ = 257
==> sin(θ) = $\frac{1}{257}$
it is positive, so now we can determine that is in Quadrant 2

sin(θ) = opp./hyp both of opp and hyp are positive but adj suppose to negative because that way it leads the cos(θ) < 0
cos(θ) = adj/hyp
Pythereom to find the adj: $257^{2} - 1^{2} = adj ==\ \textgreater \ adj \sqrt{ 257^{2} - 1^{2} } ==\ \textgreater \ adj = 16 \sqrt{258}$

cosθ = $\frac{16 \sqrt{258} }{257} 
$
tanθ = $\frac{1}{16 \sqrt{258} }$
cosθ = $16 \sqrt{258}$