pencils come in packages of 10. erasers come in packages of 12. phillip wants to purchase the smallest number of pencils and era
1 answer:
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Start by setting up a factor tree for each number.
Factor in anyway that you'd like.
At the end, circle all prime numbers.
Remember, prime numbers are numbers
with only 2 factors, 1 and the number itself.
Now write these factors in least to greatest
multiplying them separate rows.
Circle factors they have in common.
Since 3 pairs of two's match up, multiply by 2 three times to get 8.
So the GCF is 8.
Answer:
We have been given a unit circle which is cut at k different points to produce k different arcs. Now we can see firstly that the sum of lengths of all k arks is equal to the circumference:
Now consider the largest arc to have length \small l . And we represent all the other arcs to be some constant times this length.
we get :
where C(i) is a constant coefficient obviously between 0 and 1.
All that I want to say by using this step is that after we choose the largest length (or any length for that matter) the other fractions appear according to the above summation constraint. [This step may even be avoided depending on how much precaution you wanna take when deriving a relation.]
So since there is no bias, and \small l may come out to be any value from [0 , 2π] with equal probability, the expected value is then defined as just the average value of all the samples.
We already know the sum so it is easy to compute the average :