page 27

27 / 102 Russian, pre-1918 confidence 0.96 not checked by a human

Provenance

transcript transcript: 2026-09-29T23:04:05+00:00 machine-generated
normalization normalization: 2026-10-02T18:15:38+00:00 machine-generated

Full TPME chain

[
  {
    "_schema": {
      "id": "https://doi.org/10.5281/zenodo.21287085",
      "name": "AI4LAM TPME",
      "version": "1.0"
    },
    "conventions": [
      "diplomatic",
      "https://github.com/selic/bookolot/blob/main/docs/transcription-guide.md"
    ],
    "features": {
      "bbox_space": "deskewed page, see deskew step",
      "line_bbox": true,
      "line_confidence": true,
      "word_bbox": false
    },
    "file_format": "ALTO 4",
    "human_review_level": "machine-generated",
    "media_id": "http://www.moldavica.bnrm.md/biblielmo?l=ro\u0026a=d\u0026c=carti\u0026d=JD3364",
    "media_title": "Priblijonnoe iscislenie",
    "modification_date": "2026-09-29T23:04:05+00:00",
    "originating_media_file": "27.tif",
    "processing": [
      {
        "application_name": "kraken blla",
        "application_parameters": {
          "device": "cuda",
          "model": "blla.mlmodel (встроенная)"
        },
        "application_provider": "Benjamin Kiessling",
        "application_repo": "https://github.com/mittagessen/kraken",
        "application_type": "line segmentation",
        "application_version": "7.1.1"
      },
      {
        "application_name": "kraken",
        "application_parameters": {
          "device": "cuda"
        },
        "application_provider": "Benjamin Kiessling",
        "application_repo": "https://github.com/mittagessen/kraken",
        "application_type": "OCR",
        "application_version": "7.1.1",
        "inference_model": "ppocrv6_medium (sha256:15313b51ace64cbf)"
      }
    ],
    "provider": "Bookolot",
    "transcript_checksum": "sha256:3d1cefc0a20c0419b51aeeaf639144bee8ad0853b0710c899cbcfc509d643005",
    "transcript_id": "bookolot:JD3364:27:ocr-v1",
    "transcript_language": [
      "ru-petr1708"
    ],
    "type": "transcript"
  },
  {
    "_schema": {
      "id": "https://doi.org/10.5281/zenodo.21287085",
      "name": "AI4LAM TPME",
      "version": "1.0"
    },
    "characteristics_note": "For reading and search only; quote the parent transcript.",
    "human_review_level": "machine-generated",
    "media_id": "http://www.moldavica.bnrm.md/biblielmo?l=ro\u0026a=d\u0026c=carti\u0026d=JD3364",
    "media_title": "Priblijonnoe iscislenie",
    "modification_date": "2026-10-02T18:15:38+00:00",
    "originating_media_file": "27.tif",
    "parent_transcript_id": "bookolot:JD3364:27:ocr-v1",
    "processing": [
      {
        "application_name": "bookolot.rag.normalize",
        "application_parameters": {
          "normalizer_version": 3,
          "script": "ru_prereform"
        },
        "application_provider": "Bookolot",
        "application_repo": "https://github.com/selic/bookolot",
        "application_type": "spelling normalization",
        "application_version": "fb8b81d-dirty"
      }
    ],
    "provider": "Bookolot",
    "transcript_checksum": "sha256:57c63b899d36d3ecf9a30171a3176765a09d0680730e07a82fe43fa6bcec68a0",
    "transcript_id": "bookolot:JD3364:27:modern-v3",
    "transcript_language": [
      "ru-t-ru-petr1708"
    ],
    "type": "normalization"
  }
]

confident worth a look check the scan Open the full scan

Text as printed modern spelling both

as printed − 2 + modern spelling − 2 +

0 Сложеніе и вычитаніе. 0.99
1 19. Найдемъ сумму следующихъ чиссаз стораго канониче¬ 0.94
2 скаго еида. 0.97
3 57, 47 0.89
4 32,12 0.90
5 8,34 0.99
6 9,24 0.91
7 107,17 0.98
8 Такъ какъ предѣлъ погрѣшностп въ каждомъ слагаемомъ 1.00
9 есть+0,01, то предѣлъ погрѣшности въ суммѣ будетъ+0,04. 0.97
10 Но приведеніи этой суммы къ первому канонпческому виду по¬ 0.97
11 лучшмъ 107,2-1-0,1. 0.85
12 Изъ этого примѣра заключаемъ, что сумма нѣсколькихъ 0.98
13 чиселъ втораго канонпческаго вида съ одипаковыми предѣлами 0.97
14 погрѣшности есть ирпближенное число съ нед. и предѣлъ погрѣш- 0.97
15 ности въ суммѣ равенъ единпцѣ наіінисшаго разряда суммы, 0.97
16 взятой столько разъ, сколько слагаемыхъ. 0.98
17 20. Найдемъ сумму слѣдующихъ чиселъ персаюо каноничс¬ 0.96
18 скаго вида, 0.90
19 37,42 0.94
20 2,34 0.98
21 8,75 1.00
22 0,43 0.90
23 12,87 0.98
24 14,13 0.98
25 75,94 0.99
26 Предѣлы прогрѣшности въ каждомъ слагаемомъ+0,01. 0.98
27 Если всѣ слагаемыя съ нед., то и сумча будетъ приближен¬ 0.97
28 нымъ чисхомъ съ нед. и съ точн. до+0,06; а если всѣ слага¬ 0.97
29 емыя съ изб., то и сумма будетъ съ изо. и съ точн. до —0,06. 0.96
30 Такъ какъ намъ неизвѣстно, всѣ хи слагаемыя съ нед., или 0.97
31 всѣ они съ изб., или же нѣкоторыя съ нед., а другія съ изб., 0.97
32 то мы можемъ съ увѣренностью утверждать только, что 75,94 1.00
33 есть приближенное число съ точн. до10,06. Передѣлаемъ это 0.97