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So in mathematics you can divide 100% by 3 without having 0.1% left. 100% / 3 = 1 / 3 = 1 3. Imagine an apple which was cloned two times, so the other 2 are completely equal in 'quality'. The totality of the 3 apples is 100%. Now, you can divide those 3 apples for 3 persons and you will get 100% divided by 3 and none left.
There are $366$ possible birthdays; if we have $366\times 2=732$ people, then it is possible for each day, exactly two people call it their birthday. This is the maximum number of people where we can have at most $2$ per birthday. If we have one more person, then at least three people must share a birthday.
So the bonification of each employee will be: bi bi bi = αxi = 100000 300A − Wxi = 100000 300A − W(A −ai) b i = α x i b i = 100000 300 A − W x i b i = 100000 300 A − W (A − a i) If you prefer an inverse proportional distribution, the distribution variable would be xi = 1 ai x i = 1 a i. However in this distribution the top earner ...
Here (30 3) is the number of triples, and 1 / 3652 is the chance that any particular triple is a success. The probability of getting at least one success is obtained from the Poisson distribution: P( at least one triple birthday with 30 people) ≈ 1 − exp(− (30 3) / 3652) =.0300. You can modify this formula for other values, changing ...
2. Hint 1: Firstly, solve the problem "considering the order of the groups". You would be looking for the number of surjective functions. f: {1, …, n} → {1, …, k} f: {1, …, n} → {1, …, k} The j j -th group would be the inverse image f−1(j) f − 1 (j). Think about it: Firstly, try to solve other problems about the number of such ...
1. When the rhs is changed, we have 2 state : 1- all constraints whose rhs are being changed are nonbinding constraint so the current basis remain optimal off each rhs remains within the allowable range, so the value of decision variable and z remains unchanged. 2- at least one constraint whose rhs is being binding, the 100% rule is helpful.
That leaves $49$ coins to be distributed among the five people, including Alicia. This can be done in $\binom{49+5-1}{5-1}$ ways. This can be done in $\binom{49+5-1}{5-1}$ ways. So the total number of bad distributions is $5\binom{53}{4}$.
First divide the number and forget the fractions. 660/7 = 94.28 (forget 0.28) it becomes 94. and 94 * 7 = 658, 660 - 658 = 2 (you have this much extras) Now you can add 2 to one part or add them to two parts. i.e 5x94, 2x95 or 6x94, 1x96.
Per cent means for every 100. So we multiply by 100 to see how much we get for one group of 100. Because "percent" means 1 100 1 100. 50/200 is not representing 50 out of 100 because 50/200 is not 50%. It is 25 % and it absolutely does indeed represent 25 out of 100. In response to suggested edits...
Thus, there are $365 \times 3 \times 364$ ways for exactly $2$ people to share a birthday. We can uniquely construct every possibility where all three people share a birthday by selecting that common birthday, so there are $365$ choices. So, the number on top is $3 \times 365 \times 364 + 365$.