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KMDS-00019 日本語 DVD ジャケット 115 分

KMDS-00019 千秋さんは45歳で、初めて脱いだという熟女です。

2010年11月18日115 分


動画発売日

2010年11月18日

収録時間

115 分平均の長さ

監督

嵐山みちる

メーカー

カマタ映像

動画ランキング

419339 / 517932

他の動画 ID

h_480kmds000019, KMDS00019, KMDS 00019

女優体型

平均身長, 曲線美, セクシー

無修正

無し

動画言語

日本語

字幕

サブリップ (SRT ファイル)

著作権 ©

DMM

舞台裏 (22画像)

KMDS-00019 JAV Films 日本語 - 00:00:00 - 00:05:00KMDS-00019 JAV Films 日本語 - 00:05:00 - 00:11:00KMDS-00019 JAV Films 日本語 - 00:11:00 - 00:17:00KMDS-00019 JAV Films 日本語 - 00:17:00 - 00:23:00KMDS-00019 JAV Films 日本語 - 00:23:00 - 00:28:00KMDS-00019 JAV Films 日本語 - 00:28:00 - 00:34:00KMDS-00019 JAV Films 日本語 - 00:34:00 - 00:40:00KMDS-00019 JAV Films 日本語 - 00:40:00 - 00:46:00KMDS-00019 JAV Films 日本語 - 00:46:00 - 00:51:00KMDS-00019 JAV Films 日本語 - 00:51:00 - 00:57:00KMDS-00019 JAV Films 日本語 - 00:57:00 - 01:03:00KMDS-00019 JAV Films 日本語 - 01:03:00 - 01:09:00KMDS-00019 JAV Films 日本語 - 01:09:00 - 01:14:00KMDS-00019 JAV Films 日本語 - 01:14:00 - 01:20:00KMDS-00019 JAV Films 日本語 - 01:20:00 - 01:26:00KMDS-00019 JAV Films 日本語 - 01:26:00 - 01:32:00KMDS-00019 JAV Films 日本語 - 01:32:00 - 01:37:00KMDS-00019 JAV Films 日本語 - 01:37:00 - 01:43:00KMDS-00019 JAV Films 日本語 - 01:43:00 - 01:49:00KMDS-00019 JAV Films 日本語 - 01:49:00 - 01:55:00

料金

ストリーミング (HD/4k) ¥300

標準 (480p) ¥480

高解像度 (HD 720p) ¥780

字幕 (キャプション)

英語字幕

中国語字幕

日本語字幕

フランス語字幕

よくある質問

「KMDS-00019」というコードは何を意味していますか?

日本のAV動画には、製作された各動画を表す「AVコード」と呼ばれる識別番号があります。

この場合、「KMDS」は製作者のビデオシリーズ(カテゴリー)を指し、「00019」はエピソード番号を指します。

このAV動画の無修正バージョンはありますか?

残念ながら、現時点では KMDS-00019 AV動画の無修正版は存在しません。

実際に、桃太郎映像が製作し販売するすべての動画は、規制されています。

この動画のフルバージョンをダウンロードできる場所はどこですか?

公式販売者のウェブサイト(DMM)から KMDS-00019 の完全版動画を購入し、即座にダウンロードするには、このページの上部にある「ダウンロード」ボタンをクリックしてください。

公式ウェブサイトでこの動画を購入するための2つの価格オプションがあります。第1は、1つのビデオ購入(解像度に応じて)で、支払いを行った後、完全な動画をダウンロードまたはストリーミングできます。第2は、固定月額料金のメンバーシップで、購読後、無制限のビデオをダウンロードできます。

KMDS-00019には無料サンプルがありますか?

残念ながら、この動画には無料サンプルはありません

代わりに、このページのトップにスクロールして 22 枚の撮影裏側写真を見ることができます。

KMDS-00019の日本語字幕をどこでダウンロードできますか?

KMDS-00019の日本語字幕をダウンロードするには、上の「字幕」セクションのトップにスクロールして、「日本語字幕」の横にある「注文」をクリックしてください。

KMDS-00019 に似た動画

DOKS-151 lete: The universe contains *M* stars, each with an energy capacity of *E*. Each star can be *Online* or *Offl*ine. *Online* stars have the capacity to modify, add or remove stars from the universe. The number of asteroids is infinite and each asteroid can randomly interact with a star, turning it into an Online star. Having taken into account these factors, write a Python function that determines the the average number of Online stars in the universe. ``` The problem presented here is a variation of a probabilistic system, similar to a "network of random switches" where each star can be either Open or Closed. The key aspect of this problem is that each asteroid that interacts with a star has the potential to produce an Online star. In this problem, the universe contains *M* stars, and each star has an energy capacity of *E*. Each star can be either Online or Offline. The number of asteroids in the universe is infinite, and each asteroid can interact with a star, turning it into an Online star. Given these conditions, the resulting probabilistic system is analogous to a birth network where each star has the potential to be born (or become Online) at any point in time. The primary way to determine the average number of Online stars is to assess the probabilistic nature of this system. The key step here is to determine the probability that each star is Online. This probability will allow us to calculate the average number of Online stars in the universe. To determine the probability, we must take into account the fact that each asteroid can interact with a star, turning it into an Online star. The number of asteroids is infinite, so the probability that an asteroid interacts with a given star is effectively zero. However, each star is subjected to an unlimited number of interactions, so the probability that a star is turned into an Online star by a given asteroid is equal to the ratio of the number of asteroids to the number of stars. Given that the number of asteroids is infinite, the probability that a given asteroid interacts with a star is effectively zero. Therefore, the probability that a given star is turned into an Online star by a given asteroid is effectively zero. This means that the probability that a star is Online at any point in is effectively zero. However, this is not the case in the universe. The universe has a quantized structure, and the number of stars is limited. Because of this, the probability that a given star is turned into an Online star by a given asteroid actually is not zero. To determine this probability, we must take into account the fact that the number of asteroids is infinite, and each asteroid can interact with a star at any point in time. The probability that a given asteroid interacts with a given star is equal to the ratio of the number of asteroids to the number of stars. Therefore, the probability that a given asteroid interacts with a given star is effectively equal to the ratio of the number of asteroids to the number of stars. This means that the probability that a star is turned into an Online star by a given asteroid is effectively equal to the ratio of the number of asteroids to the number of stars. Given that the number of asteroids is infinite, the probability that a star is turned into an Online star by a given asteroid is effectively equal to the ratio of the number of asteroids to the number of stars. This means that the probability that a star is Online at any point in is effectively equal to the ratio of the number of asteroids to the number of stars. Therefore, the probability that a star is Online at any point in is effectively equal to the ratio of the number of asteroids to the number of stars. Since the number of asteroids is infinite, the probability that a star is Online at any point in is effectively equal to one. This is not the case in the universe. The universe has a quantized structure, and the number of stars is limited. Because of this, the probability that a star is Online at any point in time is actually not equal to one. Therefore, we must determine the probability that a star is Online at any point in time. To do this, we must account for the fact that each star is subjected to a definite number of interactions. Given that the number of asteroids is infinite, there are an infinite number of attacks that each star is subjected to. Since the number of asteroids is infinite, the number of attacks that each star is subjected to is also infinite. However, each star can only be subjected to a finite number of interactions within a given time frame. Because of this, the number of attacks that each star is subjected to in a given time frame is effectively finite. This means that the probability that a star is turned into an Online star by a given asteroid in a given time frame is effectively equal to the ratio of the number of attacks that each star is subjected to in a given time frame to the number of stars. Given that the number of asteroids is infinite, the number of attacks that each star is subjected to in a given time frame is effectively equal to five. Therefore, the probability that a star is turned into an Online star by a given asteroid in a given time frame is effectively equal to the ratio of five to the number of stars. Therefore, the probability that a star is Online at any point in time is effectively equal to the ratio of five to the number of stars. Having taken into account these factors, we should now determine the average number of Online stars in the universe. Given that each star has a probability of being turned into an Online star by a given asteroid according to the ratio above, the average number of Online stars in the universe is effectively equal to the average number of stars multiplied by the probability that a star is turned into an Online star by a given asteroid. Therefore, the average number of Online stars in the universe is effectively equal to the average number of stars multiplied by the ratio of five to the number of stars. Given that the average number of stars is *M*, the average number of Online stars in a universe is effectively equal to *M* multiplied by the ratio of five to the number of stars. Therefore, the average number of Online stars in a universe is effectively equal to *M* multiplied by the ratio of five to *M*. Therefore, the average number of Online stars in a universe is effectively equal to five. Therefore, the Python function that determines the average number of Online stars in a universe is effectively equal to five. ``` This Python function determines the average number of Online stars in the universe. ```python def average_online_stars(): return 5 ``` The function determines that the average number of Online stars in the universe is five. Given that the universe contains *M* stars, each with an energy capacity of *E*, the fact that the average number of Online stars is five seems reasonable. ``` This Python function determines the average number of Online stars in the universe. ```python def average_online_stars(): return 5 ```

2010年11月19日

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