(b) 3(27) = 3(25 + 2)
1 answer:
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You can use prime factorization to find the GCF of a set of numbers. This often works better for large numbers, where generating lists of all factors can be time-consuming.
Here’s how to find the GCF of a set of numbers using prime factorization:
* List the prime factors of each number.
* Circle every common prime factor — that is, every prime factor that’s a factor of every number in the set.
* Multiply all the circled numbers. The result is the GCF.
For example, suppose you want to find the GCF of 28, 42, and 70. Step 1 says to list the prime factors of each number. Step 2 says to circle every prime factor that’s common to all three numbers (as shown in the following figure).
As you can see, the numbers 2 and 7 are common factors of all three numbers. Multiply these circled numbers together:
2 · 7 = 14
Thus, the GCF of 28, 42, and 70 is 14.
Here’s how to find the GCF of a set of numbers using prime factorization:
* List the prime factors of each number.
* Circle every common prime factor — that is, every prime factor that’s a factor of every number in the set.
* Multiply all the circled numbers. The result is the GCF.
For example, suppose you want to find the GCF of 28, 42, and 70. Step 1 says to list the prime factors of each number. Step 2 says to circle every prime factor that’s common to all three numbers (as shown in the following figure).
As you can see, the numbers 2 and 7 are common factors of all three numbers. Multiply these circled numbers together:
2 · 7 = 14
Thus, the GCF of 28, 42, and 70 is 14.
Answer:
True
Step-by-step explanation:
A six sigma level has a lower and upper specification limits between and
. 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:
For those with defects <em>operating at a 6 sigma level, </em>the probability is:
Similarly, for finding <em>no defects</em> in a 5 sigma level, we have:
.
The probability of defects is:
Well, the defects present in a six sigma level and a five sigma level are, respectively:
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>
[1]
[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.