The question is finding the set for the next dates. It is obvious that the dates have an alternating 3 and 4 difference of interval. You can guess if this problem is all about an arithmetic series or a geometric series.
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As for an arithmetic series, you need to use this equation:
where a1 is the first number of the sequence, d is the common difference, n is the number of the term to find. But before using this equation, you need to evaluate if the series you have has constant d (difference).
Let us assume that the allowable series to be solved is the in the column and the row. Let us take a look first at the column series. Take column 1;
1, 14, 7, 25, 23, 8, 15, 19, 10, 10
Let us take d if it has a constant value:
14-1= 1325-7= 188-23=-15 19-15= 4
It is evident that d has no constant value therefore you cannot use the equation for the arithmetic series for column 1, Solve the succeeding columns, and it will also give a fluctuating value for d.
Now let us take row 1;
1, 10, 13, 18, 19, 27
Evidently, it also has a fluctuating value for d. And you cannot also use the rows in predicting the next set of numbers.
Now let us consider a geometric sequence formula;
Let us use column number 1:
14/1= 1423/25= 0. 928/23= 0. 35
7/14= 0. 5
25/7= 3. 57
This makes the sequence unidentified. It could not be a geometric or an arithmetic sequence. Therefore you could only use the probability formula with this problem, but the only thing you could get is a percentage of a number to come out in a set.
You can actually have many combinations of numbers for the following dates.
Take this equation as a reference for probability problems;
the probability of winning the lottery= the number of winning lottery numbersthe total number of possible lottery numbers
nr= n! r! n-r!
You can use that equation for probability.