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K divided by 21 divided by N is equivalent to divided by?

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7mo ago

K divided by 21N.

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Q: K divided by 21 divided by N is equivalent to divided by?
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What algebraic expression can you draw from 21 less than a number divided by 4?

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How many different ways are there to choose a dozen donuts from the 21 varieties at a donut shop?

The answer depends on if you can choose the same kind of donuts more then once. Or in other words, is repetition permitted. If you can only choose the same kind of donuts only once, it is a 21 choose 12 problem: C(n,k) = n! / (k! (n - k)!) C(21, 12) = 21! / (12! (21 - 12)!) = 21! / (12! (9)!) = 293,930 If you can choose the same kind of donuts more then once, it is a combination with repetition problem. P(n+k-1,k) = (n+k-1)! / (k! (n-1)!) or put it into C(n,k) with n+k-1 as 21 + 12 - 1 = 32 and k as 12 so C(21+12-1,12) = C(32, 12) = 32! / (12! (32 - 12)!) = 32! / (12! (20)!) = 225,792,840


7 times n divided 21 equals 1 what is n?

n = 3Solution:7n/21 = 17n = 1 * 217n = 21n = 21/7n = 3


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38 plus n divided by 2 equals 21?

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Definition of a multiple of 21?

For any natural number k, 21k is a multiple of k.So for example, if k=1, 21 is the first multiple.Now if k=2, we have 42 which is the next one.In fact for any number n, and any natural number k, kn is a multiple of n.


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When a number is divided by 837 the remainder is 643 what will be the remainder if the same number is divided by 31?

23. If the remainder when a number n is divided by m is r, then it is written as n ≡ r MOD m (read as "n is equivalent to r modulo m") and n = mk + r for some integer k. if n ≡ 643 MOD 837, then n = 837k + 643 (for some k) When n is divided by 31: n ÷ 31 = (837k + 643) ÷ 31 = 837k ÷ 31 + 643 ÷ 31 = 27k + 20 + 23/31 Multiplying both sides by 31: → n = 31 x (27k + 20) + 31 x 23/31 = 31m + 23 (where m = 27k + 20) → n ≡ 23 MOD 31 Thus the remainder when the same number is divided by 31 is 23.


Is gram equivalent weight equal to molecular weight divided by two?

No, gram equivalent weight is molecular weight divided by n, where n is the number of equivalents of the substance involved in the reaction. It is not always equal to molecular weight divided by 2.


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How many possible 5 number sets can be broken down from seven numbers?

Short answer: 21 We can use combinatorics to solve this using the binomial coefficient to choose five elements out of a set of seven. The equation for this is: n! ------------ (n - k)! k! Where n is the size of the set, k is the number of elements to choose, and "------------" is a crude attempt at a division sign. In our case, n = 7, k = 5: 7! ------------ (7 - 5)! 5! 5040 ------------ (2) (120) 5040 ------------ 240 5040 / 240 = 21