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max2010maxim [7]
2 years ago
8

What is the solution to the equation?

Mathematics
1 answer:
UkoKoshka [18]2 years ago
7 0
\frac{1}{3}x+\frac{1}{3}(2x-15)=3\frac{1}{2}\\\\
\frac{1}{3}x+\frac{1}{3}(2x-15)=\frac{7}{2}\ \ \ |Multiply\ by\ 6\\\\
2x+2(2x-15)=21\\\\
2x+4x-30=21\\\\
6x-30=21\ \ |Add\ 30\\\\
6x=51\ \ \ |Divide\ by\ 6\\\\
\boxed{x=\frac{51}{6}=8\frac{1}{2}}
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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
2 years ago
Find compound interest on12600rupees for 2years at10% per annum compounded anually<br>​
vekshin1

Answer:

₹2520

Step-by-step explanation:

First, converting R percent to r a decimal

r = R/100 = 10%/100 = 0.1 per year,

then, solving our equation

I = 12600 × 0.1 × 2 = 2520

I = ₹ 2,520.00

The simple interest accumulated

on a principal of ₹ 12,600.00

at a rate of 10% per year

for 2 years is ₹ 2,520.00.

3 0
2 years ago
Simplify the following expression: 5(y + 12) - 34<br><br> Help me please! (15 points)
hram777 [196]
The simplified expression would be as follows: 5y + 26
4 0
2 years ago
Billy makes $5 a week in allowance plus $4 for each lawn that he mows.
Ad libitum [116K]

Answer:

$15 < $4n + $5

Step-by-step explanation:

We know that Billy needs to make more than $15 between his allowance and the lawns that he mows. This means our inequality should include $15<. Also, since Billy will make $4 per lawn, that means we need to multiply $4 by the number of lawns he needs to mow, n: $4n. So far we have the following: $15<$4n. Next, we know that he makes $5 each week, on top of what he makes mowing each law. This means we need to add the $5 to the $4n. When we put all of these pieces together, we will get the following inequality: $15<$4n+$5

7 0
3 years ago
Write many equations that are equal to the equation 2x+4=x-4.
Jobisdone [24]

Answer:

<u>x = -8</u>

4x + 8 = 2x - 8

3x + 4 = 2x - 4

2x + 16 = x + 8

x - 1 = 2x + 7

17x + 36 = 16x + 28

13x + 2 = 12x + 11

.......

(this is infinite)

As long as it cancels out to x = -8

4 0
2 years ago
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