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tamaranim1 [39]
4 years ago
11

3/15divied by 4/9

Mathematics
2 answers:
pochemuha4 years ago
4 0

Answer:

2 2/9

Step-by-step explanation:

(i simplified the 15 and 9 because they both can be divisible by 3.) P.s. to get a better look at the picture use the arrow keys. Look at picture.

Novay_Z [31]4 years ago
3 0
Some things you can do is put them in decimals but the answer is 0.45
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Step-by-step explanation:

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xxTIMURxx [149]

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Find two fractions with the difference of 1/5 but with neither denominator equal to 5? I know the answer from Google but how is
Bas_tet [7]
Suppose we try to write out this problem symbolically:

a     c      1
-- - --- = --- , b not equal to 5, d not equal to 5
b     d      5

Unfortunately, this equation contains 4 unknowns (too many)


We could do a bit of guesswork here:

a     c      1
-- - --- = --- , b not equal to 5, d not equal to 5
b     d      5

  8     6        2
--- - ----- = -----
10    10      10

This is true, but not all that wonderful, because all 3 fractions could be reduced and would then have denominator 5.
6 0
4 years ago
In the ________ method, the distance between groups is defined as the distance between the closest pair of objects, where only p
andrew-mc [135]

Answer: single linkage clustering

Step-by-step explanation: because here

defining feature of the method is that distance between groups is defined as the distance between the closest pair of objects, where only pairs consisting of one object from each group are considered.

In the single linkage method, D(r,s) is computed as

D(r,s) = Min { d(i,j) : Where object i is in cluster r and object j is cluster s }

The distance between every possible object pair (i,j) is computed, where object i is in cluster r and object j is in cluster s. The minimum value of these distances is said to be the distance between clusters r and s. In other words, the distance between two clusters is given by the value of the shortest link between the clusters.

At each stage of hierarchical clustering, the clusters r and s , for which D(r,s) is minimum, are merged. In this case, those two clusters are merged such that the newly formed cluster, on average, will have minimum pairwise distances between the points.

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