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Deffense [45]
3 years ago
14

What is 6/7 divided 10/3

Mathematics
1 answer:
Sidana [21]3 years ago
4 0

Answer:

6/7 divided by 10/3= 18/70 or reduced 9/35

Step-by-step explanation:

6/7 divided by 10/3

6/7*3/10= 9/35

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Can you find 2/3 of 120
polet [3.4K]
The only thing you should do is to multiply 2/3 by 120 so you will have 2*120/3 = 80 :)))
i hope this is helpful
have a nice day 
6 0
3 years ago
Suppose that your business is operating at the 4.5-Sigma quality level. If projects have an average improvement rate of 50% annu
Luda [366]

Answer:

  11.75 years

Step-by-step explanation:

If we ignore the fact that "6-sigma" quality means the error rate corresponds to about -4.5σ (3.4 ppm) and simply go with ...

  P(z ≤ -6) ≈ 9.86588×10^-10

and

  P(z ≤ -4.5) ≈ 3.39767×10^-6

the ratio of these error rates is about 0.000290372. We're multiplying the error rate by 0.5 each year, so we want to find the power of 0.50 that gives this value:

  0.50^t = 0.000290372

  t·log(0.50) = log(0.00290372) . . . . take logarithms

  t = log(0.000290372)/log(0.50) ≈ -3.537045/-0.301030

  t ≈ 11.75

It will take about 11.75 years to achieve Six Sigma quality (0.99 ppb error rate).

_____

<em>Comment on Six Sigma</em>

A 3.4 ppm error rate is customarily associated with "Six Sigma" quality. It assumes that the process may have an offset from the mean of up to 1.5 sigma, so the "six sigma" error rate is P(z ≤ (1.5 -6)) = P(z ≤ -4.5) ≈ 3.4·10^-6.

Using that same criteria for the "4.5-Sigma" quality level, we find that error rate to be P(z ≤ (1.5 -4.5)) = P(z ≤ -3) ≈ 1.35·10^-3.

Then the improvement ratio needs to be only 0.00251699, and it will take only about ...

  t ≈ log(0.00251699)/log(0.5) ≈ 8.6 . . . . years

5 0
4 years ago
Dont know how to solve for x can someone help me
vfiekz [6]
  57 + x + x + 130 = 167
  57 + 2x + 130 = 167
    187 + 2x = 167
  - 187          - 187
---------------------------
    2x = -20
  ------  -------
     2       2

x = -10
5 0
3 years ago
What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove
wolverine [178]

Answer:

The solutions are:

b=0,\:b=4

Step-by-step explanation:

Considering the expression

  • \frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}

Solving the expression

\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}

\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c

5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2

5b^3-10=6b^3-4b^2-10

\mathrm{Switch\:sides}

6b^3-4b^2-10=5b^3-10

6b^3-4b^2-10+10=5b^3-10+10

6b^3-4b^2=5b^3

\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}

6b^3-4b^2-5b^3=5b^3-5b^3

b^3-4b^2=0

Using\:the\:Zero\:Factor\:Principle: if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

So,

b=0,b-4=0

b=0,b=4

Therefore, the solutions are:

b=0,\:b=4

4 0
3 years ago
Read 2 more answers
Which of the following is a rational number?<br> 97, 98, 99, 100
Jet001 [13]

Answer:

97

Step-by-step explanation:

7 0
3 years ago
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