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What letter is missing from the following sequence hnessaeas?
s
What is the missing number in sequence 98684838 8?
3. Ignore the 8's in the sequence and you get 9, 6, 4, 3,_. 9-3=6 6-2=4 4-1=3 3-0=3
What is the pattern rule for 1-1-2-6-24-120?
In this sequence s the nth term is s(n) calculated as follows where s(1) = 1: s(n) = (n - 1)(s(n - 1)) eg the 5th number (s(5)) = 4th number times 4 so next number s(7) will be 6 times s(6) ie 720
What is the next letters in the sequence O T T F F S?
s, the sequence is one two three four five six, so the next number is seven, so the answer is s
Is every cauchy sequence is convergent?
Every convergent sequence is Cauchy. Every Cauchy sequence in Rk is convergent, but this is not true in general, for example within S= {x:x€R, x>0} the Cauchy sequence (1/n) has no limit in s since 0 is not a member of S.
How is data recorded on a disk?
Using binary code. A sequence of "0's" and "1's".
How many 8's will appear in sequence of natural number from 234 to 849?
212
What are the release dates for The Roaring 20's - 1960 Among the Missing 1-27?
The Roaring 20's - 1960 Among the Missing 1-27 was released on: USA: 6 May 1961
What is the sequence if the sum of the first 10 terms is the same as the 58th term?
There are many sequences with this property: The sequence with every term equal to 0 has this property. In fact the sequence can be anything you like as long you make sure the 58th term is the sum of the first 10 terms. A more specific case: If you are dealing with an arithmetic sequence, i.e. a sequence of the form s(n)=a+bn for constants a and b, we can derive a relationship between a and b: s(1)+s(2)+...+s(10)=10a+55b and s(58)=a+58b From this, it follows that if s(1)+s(2)+...+s(10)=s(58), then we have 10a+55b=a+58b, which implies that 3a=b. Again, there are infinitely many sequences with this property, but if it is an arithmetic sequence, it will be of the general form s(n)=a+3an=a(3n+1)
What comes next in the sequence 1 11 21 1211 111221 312211?
Next in sequence is 13112221.This series describes each number that came before it:1 [One]11 [One One]21 [Two One(s)]1211 [One Two and One One(s)]111221 [One One, One Two, and Two One(s)]312211 [Three One(s), Two Two(s), and One One]The next number in the series is:13112221 [One Three, One One, Two Two(s), Two One(s)]
How do you work out the numbers in the sequence if given the mode mean median and range?
You cannot, with the information available. Probably not, but if you were given one more bit of information, the number of numbers in the sequence, then you might have a good chance if there aren't too many numbers in the sequence. If there is an odd number of numbers, then the median is the number such that half of the numbers are greater, and half are smaller. The mode is the number that occurs most often. The mean is the sum of all of the numbers, divided by the number of numbers. The range is the largest number minus the smallest number. For example, take this number sequence: 1, 2, 2. Given: mode=2, range=1, median=2, mean=5/3. Start with the mode. There must be at least two 2's, since it is the mode; so it must occur more often than any other number. The range is only 1; so it could go from 2 to 3, or from 1 to 2, assuming that only whole numbers are used. If the third number were 3, then the mean would be (2+2+3)/3=7/3. If the third number were 1, then the mean would be (1+2+2)/3=5/3, which matches the given mean; so the number sequence is 1, 2, 2. However, since we were not given the number of numbers in the sequence, could the sequence also be: 1, 1, 2, 2, 2, 2? The answer is, "Yes, it could be." So the bottom line is that if you were also given the number of numbers in the sequence, and it wasn't too many, you could have a good chance of figuring out the sequence from the mode, mean, median, and range. Another thing to think about is , if all of the numbers in the sequence are different, then you have multimodal rather than unimodal, and you might be given all of the numbers just from the mode. For example, the following number sequence 1, 3, 5, 7, 12, 21, 53, 77. Given the mode, mean, median, and range, could you figure out all of the numbers in the sequence. Answer: Yes, no problem, since it is multimodal, and no number occurs more often than any other number, the mode term would include all of the numbers in the sequence. How about this sequence: 1, 1, 2, 3, 12, 12, 17, 17? This sequence is trimodal; so the three modes are 1, 12, 17. If you were given that there were 8 numbers in the sequence, then you would know that there were only 2 numbers yet to determine, and from adding up the 6 numbers that you know from the mode, and knowing the mean, you should be able to determine that the two unknown numbers add up to 5. It can't be 1 and 4, since that would make 1 the only mode. It couldn't be 0 and 5, since you know the range, and that wouldn't fit. Any negative number wouldn't fit into the given range, which is 16. So you would be able to figure out that 2 and 3 were the remaining two numbers.
What number you s next in this sequence 2 3 10 12 13 20 21?
22
