Search (2 results, page 1 of 1)

  • × subject_ss:"Information services"
  • × subject_ss:"Needs assessment"
  1. Nicholas, D.: Assessing information needs : tools and techniques (1996) 0.01
    0.008455964 = product of:
      0.021139909 = sum of:
        0.00851091 = weight(_text_:e in 5941) [ClassicSimilarity], result of:
          0.00851091 = score(doc=5941,freq=2.0), product of:
            0.053592302 = queryWeight, product of:
              1.43737 = idf(docFreq=28552, maxDocs=44218)
              0.037284974 = queryNorm
            0.15880844 = fieldWeight in 5941, product of:
              1.4142135 = tf(freq=2.0), with freq of:
                2.0 = termFreq=2.0
              1.43737 = idf(docFreq=28552, maxDocs=44218)
              0.078125 = fieldNorm(doc=5941)
        0.012628999 = product of:
          0.050515994 = sum of:
            0.050515994 = weight(_text_:22 in 5941) [ClassicSimilarity], result of:
              0.050515994 = score(doc=5941,freq=2.0), product of:
                0.13056563 = queryWeight, product of:
                  3.5018296 = idf(docFreq=3622, maxDocs=44218)
                  0.037284974 = queryNorm
                0.38690117 = fieldWeight in 5941, product of:
                  1.4142135 = tf(freq=2.0), with freq of:
                    2.0 = termFreq=2.0
                  3.5018296 = idf(docFreq=3622, maxDocs=44218)
                  0.078125 = fieldNorm(doc=5941)
          0.25 = coord(1/4)
      0.4 = coord(2/5)
    
    Date
    26. 2.2008 19:22:51
    Language
    e
  2. Nicholas, D.: Assessing information needs : tools, techniques and concepts for the Internet age (2000) 0.00
    0.0010213092 = product of:
      0.0051065455 = sum of:
        0.0051065455 = weight(_text_:e in 1745) [ClassicSimilarity], result of:
          0.0051065455 = score(doc=1745,freq=2.0), product of:
            0.053592302 = queryWeight, product of:
              1.43737 = idf(docFreq=28552, maxDocs=44218)
              0.037284974 = queryNorm
            0.09528506 = fieldWeight in 1745, product of:
              1.4142135 = tf(freq=2.0), with freq of:
                2.0 = termFreq=2.0
              1.43737 = idf(docFreq=28552, maxDocs=44218)
              0.046875 = fieldNorm(doc=1745)
      0.2 = coord(1/5)
    
    Language
    e