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3 years ago

How do we divide using long division?? (Explain plz)

Mathematics

1 answer:

Semmy [17]3 years ago

7 0

Answer:

Step-by-step explanation:

Step 1: Make sure the polynomial is written in descending order. If any terms are missing, use a zero to fill in the missing term (this will help with the spacing).

Step 2: Divide the term with the highest power inside the division symbol by the term with the highest power outside the division symbol.

Step 3: Multiply (or distribute) the answer obtained in the previous step by the polynomial in front of the division symbol.

Step 4: Subtract and bring down the next term.

Step 5: Repeat Steps 2, 3, and 4 until there are no more terms to bring down.

Step 6: Write the final answer. The term remaining after the last subtract step is the remainder and must be written as a fraction in the final answer.

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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 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

2 years ago

Hi please help thank you i’ll give brainliest

Sergio [31]

Answer:

ok so first start with 97 and 45 put together 57/97 and

Step-by-step explanation:

4 0

3 years ago

3/x^2-1 - 2(x+3) = 5/x+1

weqwewe [10]

Answer:

X=1.132589

Step-by-step explanation:

If im correct the answer would be x=1.132589

x2−1

−2(x+3)=

x+1

(x+1)(x−1)

+−2x−6=

x+1

Multiply all terms by (x+1)(x-1) and cancel:

3+(−2x−6)(x+1)(x−1)=5(x−1)

−2x3−6x2+2x+9=5x−5(Simplify both sides of the equation)

−2x3−6x2+2x+9−(5x−5)=5x−5−(5x−5)(Subtract 5x-5 from both sides)

−2x3−6x2−3x+14=0

(Use cubic formula)

x=1.132589

And dont forget to check your answer

3 0

2 years ago

A friend has offered to play a gambling game with you that involves flipping a coin that she has provided. You want to test the

Sloan [31]

Answer:

The 10% condition would not apply here

Explanation:

The 10% condition is the recommended size of sample from the population to get a non biased result. The 10% condition requires that the sample be not more than 10% of the population.

Tossing a coin is an example of a Bernoulli trial. A Bernoulli trial is one that has two possible outcomes, this face of the coin or the other face of the coin. The 10% condition does not apply to Bernoulli trials that are independent events.

Therefore the 10% condition would not apply here because tossing a coin is an an independent event. An independent event is one with replacement.

5 0

2 years ago

10 POINTS only answer if you KNOW

kenny6666 [7]

The answer is D, tony would always be behind Robert.

5 0

2 years ago