PLEASE WILL MARK BRAINLIEST The bearing of town x from town y is S50°E. What is the bearing of y from x ?.

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

3 years ago

Maria plans to use fencing to build an enclosure or enclosures for her two horses. A single enclosure would be square shaped and

Gala2k [10]

Step 1
find the perimeter of a single enclosure
perimeter of a square=4*b
where b is the long side of a square
area square=b²
area square=2025 ft²
b²=2025-------> b=√2025-----> b=45 ft
so
perimeter=4*45-------> 180 ft

step 2
find the perimeter of a two individual enclosure
perimeter=4*20+3*40------> 200 ft
area=20*40*2------> 1600 ft²

therefore
fencing singular enclosure < fencing two individual enclosure
180 ft < 200 ft

area singular enclosure > area two individual enclosure
2025 ft² > 1600 ft²

the answer is the option
a The singular enclosure would minimize cost because it requires 180 feet of fencing.

5 0

3 years ago

Read 2 more answers

Don’t understand could I get some help pls :)

Lemur [1.5K]

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the concept of Parallel lines.

We must first understand that, Parallel Lines always have a same Slope, hence the 'm' value in y=mx+c equation will same, here it is '1/2' in the above equation,

so the points here are (-6,-17)

==>

(y-(-17)) = m(x-(-6))

==>

here m = 1/2 ,hence

y+17 = 1/2(x+6)

==> y+17 = 1/2(x) + 3

==> y = 1/2(x) + 3 - 17

==> y = 1/2(x) - 14

hence the Option 4.) is the correct answer!!

5 0

2 years ago

Divide. −1/9÷3/4 −4/27 −1/12 1/12 4/27 PLZZZZ HELP FAST!

ivanzaharov [21]

$\frac{ - 1}{9} \div \frac{3}{4} \\ \\ \frac{ - 1}{9} \times \frac{4}{3} \\ \\ \frac{( - 1) \times 4}{9 \times 3} \\ \\ \frac{ - 4}{27}$

7 0

3 years ago

It is given that D is inversely proportional to v m

Darina [25.2K]

Answer:

Step-by-step explanation:

$D=\frac{k}{\sqrt{m}}\\\\64=\frac{k}{\sqrt{9}}\\\\64=\frac{k}{3}\\\\ 64*3=k\\\\k=192\\a)D=\frac{192}{\sqrt{4}}\\\\=\frac{192}{2}\\\\D=96\\\\b)32=\frac{192}{\sqrt{m}}\\\\ 32*\sqrt{m}=192\\\\\sqrt{m}=\frac{192}{32}=6\\\\m=36$

8 0

3 years ago

PLEASE WILL MARK BRAINLIEST The bearing of town x from town y is S50°E. What is the bearing of y from x ?​.

PLEASE WILL MARK BRAINLIEST The bearing of town x from town y is S50°E. What is the bearing of y from x ?.