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marta [7]
3 years ago
12

In the ________ method, the distance between groups is defined as the distance between the closest pair of objects, where only p

airs consisting of one object from each group are considered.
Mathematics
1 answer:
andrew-mc [135]3 years ago
8 0

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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consecutive odd integers, like 1, 3, 5, 7

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Express the following quotient in scientific notation
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1.98 x 10^-12       1.98

------------------ = ---------- x 10^-10

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4 years ago
Write an expression for the 12th partial sum of the series 3/2+7/3+19/6+... using summation notation
lapo4ka [179]

Answer:

S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]

S_{12}=73

Step-by-step explanation:

First\ term\ of\ the\ series(a_1)=\frac{3}{2}\\\\Second\ term\ of\ the\ series(a_2)=\frac{7}{3}\\\\Third\ term\ of\ the\ series(a_3)=\frac{19}{6}\\\\a_2-a_1=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\\\\a_3-a_2=\frac{19}{6}-\frac{7}{3}=\frac{5}{6}\\\\Hence\ it\ is\ an\ Arithmetic\ Series\ with\ first\ term=\frac{3}{2}\ and\ constant\ difference=\frac{5}{6}

a_1=\frac{3}{2}+0\times \frac{5}{6}\\\\a_2=\frac{3}{2}+1\times \frac{5}{6}\\\\a_3=\frac{3}{2}+2\times \frac{5}{6}\\\\.\\.\\.\\a_n=\frac{3}{2}+(n-1)\times \frac{5}{6}\\\\S_n=a_1+a_2+a_3+......+a_n\\\\S_n=(\frac{3}{2}+0\times \frac{5}{6})+(\frac{3}{2}+1\times \frac{5}{6})+(\frac{3}{2}+2\times \frac{5}{6})+....+(\frac{3}{2}+[n-1]\times \frac{5}{6})\\\\S_n=\sum_{i=1}^n [\frac{3}{2}+(i-1)\times \frac{5}{6}]\\\\S_n=(\frac{3}{2}+\frac{3}{2}+\frac{3}{2}+...n\ times)+\frac{5}{6}(1+2+3+4+...+(n-1))\\\\

S_n=\frac{3}{2}\times n+\frac{5}{6}\times \frac{n(n-1)}{2}\\\\

S_{12}=\sum_{i=1}^{12} [\frac{3}{2}+(i-1)\times \frac{5}{6}]

S_{12}=\frac{3}{2}\times 12+\frac{5}{6}\times \frac{(12)(12-1)}{2}\\\\S_{12}=18+55\\\\S_{12}=73

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