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Sonja [21]
3 years ago
13

13+(3n) when n is 4.n=4

Mathematics
2 answers:
Art [367]3 years ago
8 0
Whoops, didn't read that right. It is 25. Thought it was 4*n=4.
scoundrel [369]3 years ago
4 0
Plug 4 in for n:
13+(3(4))
13+3(4)      (Follow PEMDAS and multiply the 3 &4 first)
13+12        
=25
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How do I compare 4/6 and 1/3 using benchmark fractions?
alex41 [277]

Steps Determine whether or not the fractions have the same denominator. This is the first step to comparing fractions. Find a common denominator. To be able to compare the fractions, you'll need to find a common denominator so you can figure out which fraction is greater. Change the numerators of the fractions.

6 0
2 years ago
Weird math question.
Aneli [31]

Answer:

It is 4/11

Step-by-step explanation:

Let x = 0.363636363636

Now as there are two digits repeating immediately after decimal point, we multiply above by 100 and get

100x = 36.363636363636

Now subtracting first equation from second we get

99x = 36

and hence

x = 36/ 99

= (9 × 4) / (9 × 11)

which then equals

= 4/11

6 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
What is the vertex of the parabola?<br><br>A. (-1,0)<br>B. (0,-3)<br>C. (1,-4)<br>D. (3,0)​
katovenus [111]

d. (1,-4) is the answer

7 0
3 years ago
Read 2 more answers
True or false !!!<br><br> T or F
trapecia [35]
Tan x = opp/adj

For angle C, opp = 4, and 3 = adj, so

tan C = 4/3

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