What is the prime factorization of 440?
1 answer:
Answer:
I would say the correct answer is 2^3 • 5 • 11
Step-by-step explanation:
To find the prime factors, you start by dividing the number by the first prime number, which is 2. If there is not a remainder, meaning you can divide evenly, then 2 is a factor of the number. Continue dividing by 2 until you cannot divide evenly anymore. Write down how many 2's you were able to divide by evenly. Now try dividing by the next prime factor, which is 3. The goal is to get to a quotient of 1.Here are the first several prime factors: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29...
<u>Let's start by dividing 440 by 2 </u>
440 ÷ 2 = 220 - No remainder! 2 is one of the factors!
220 ÷ 2 = 110 - No remainder! 2 is one of the factors!
110 ÷ 2 = 55 - No remainder! 2 is one of the factors!
55 ÷ 2 = 27.5 - There is a remainder. We can't divide by 2 even anymore. Let's try the next prime number
55 ÷ 3 = 18.3333 - This has a remainder. 3 is not a factor.
55 ÷ 5 = 11 - No remainder! 5 is one of the factors!
11 ÷ 5 = 2.2 -There is a remainder. We can't divide by 5 evenly anymore. Let's try the next prime number
11 ÷ 7 = 1.5714 - This has a remainder. 7 is not a factor.
11 ÷ 11 = 1 - No remainder! 11 is one of the factors!
If we put all of it together we have the factors 2 x 2 x 2 x 5 x 11 = 440. It can also be written in exponential form as 2^3 x 5^1 x 11^1.
here is also the factor tree method : -
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Answer:
And if we analyze the info provided we have 500 values
Tails: 160,165,167,171,175,...,268,269,269,269,270,270
So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.
Step-by-step explanation:
For this case we have the 5 number summary
(160,210,220,250,270)
So then we have:
minimum = 160 , Q1 = 210, Q2= Median=220, Q3 = 250, Max=270
If we find the interquartile range we got:
For this case we need to find the lower and upper limit with the following formulas:
And if we analyze the info provided we have 500 values
Tails: 160,165,167,171,175,...,268,269,269,269,270,270
So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.