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Liula [17]
3 years ago
8

Question 4: What is the solution of the system? * .A .B .C .D

Mathematics
1 answer:
DedPeter [7]3 years ago
4 0

Answer:

C

Step-by-step explanation:

x+4y=-12

x+7y=-15

subtract

-3y=3

y=-1

x+4(-1)=-12

x=-12+4=-8

so point is (-8,-1)

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What value of x is in the solution set of 3(х – 4) &gt; 5х + 2?<br> 0—10<br> ОООО
MakcuM [25]
X < -7 hope this helps

4 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
Alice and Bob have two positive integers, x and y respectively, glued to their foreheads, so that each can read the other’s numb
vazorg [7]

Step-by-step explanation: |x − y| = 1, ok lets play as Alice, my number is y, and the bob number is x.

the condition says that x-y = 1 or x-y = -1.

so, if you know x, then y = 1 +y or y = y - 1. so you have two possibilities.

let's see two cases : first, let's suppose there are no code in the conversation. Then the only way of being shure of your number, is if one of them have the lowest positive number, so the other should have the next one. So if Bob have the number one, Alice knows for shure that she has the  2. Bob knows that she has a 2, but that means he could have a 1 or a 3, but when he sees that Alice is shure about her number, he knows that his number is the 1.

the second case is where the conversation may be a sort of code, saying a phrase x times and changing when x = the number of the other person, in this case, bob will have the 201 and alice the 202.

7 0
3 years ago
Order these numbers from least to greatest. 7.60, 7.228, 29/4, 7 2/9
Anarel [89]
7 2/9 < 7.228 < 29/4 < 7.60
8 0
3 years ago
Read 2 more answers
Write an equation of the line that passes through the point (5, -8) with slope 5
Oxana [17]

\Large\texttt{Answer}

equation=\bold{y=5x-33}

\overline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}

\Large\texttt{Process}

<em>⇨ Use the formula below</em>

<em />\bf{y-\hfill\stackrel{y \:co-ordinate}{y1}=\hfill\stackrel{slope}{m}(x-\hfill\stackrel{x\:co-ordinate}{x1)}}

<em>⇨ Substitute the parameters:</em>

<em />

<em />\bold{y-(-8)=5(x-5)}

\bold{y+8=5x-25}

\bold{y=5x-33}

Hope that helped

4 0
2 years ago
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