{"id":38692,"date":"2022-05-15T10:05:41","date_gmt":"2022-05-15T08:05:41","guid":{"rendered":"https:\/\/www.sssh.ch\/?post_type=forum&p=38692"},"modified":"2022-05-16T10:07:43","modified_gmt":"2022-05-16T08:07:43","slug":"propositions-de-solutions","status":"publish","type":"forum","link":"https:\/\/sssh.ch\/fr\/forum\/propositions-de-solutions\/","title":{"rendered":"Propositions de solutions"},"content":{"rendered":"\t\t
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(Il existe d\u2019autres possibilit\u00e9s de trouver les r\u00e9ponses)<\/strong><\/h5>
R\u00e9ponse 1
<\/strong><\/h5>

Non<\/p>

Pour conserver le b\u00e9b\u00e9 mammouth st\u00e9rile, il faudrait qu\u2019il soit emball\u00e9 dans un emballage de st\u00e9rilisation.<\/p>

La contamination initiale du b\u00e9b\u00e9 mammouth est inconnue, mais tr\u00e8s certainement sup\u00e9rieure \u00e0 102<\/sup>.<\/p>

25’000 Gy = 25 kGy permettent une r\u00e9duction de 8 log (25\/3.1), ce qui est insuffisant pour atteindre le NAS de 10-6<\/sup>.<\/p>

\u00a0<\/h5>
R\u00e9ponse 2
<\/strong><\/h5>

Soit x le prix de l\u2019emballage, alors les instruments co\u00fbtent x + 10<\/p>

L\u2019\u00e9quation devient\u00a0: x + x + 10 = 11 => 2 x = 1 => x = 0.5<\/p>

\u00a0<\/h5>
R\u00e9ponse 3
<\/strong><\/h5>

De tous les quadrilat\u00e8res et \u00e0 m\u00eame p\u00e9rim\u00e8tre, c’est le carr\u00e9 qui couvre la plus grande surface. Par cons\u00e9quent, c\u2019est un carr\u00e9 de 15 m x 15 m.<\/p>

\u00a0<\/h5>
R\u00e9ponse 4
<\/strong><\/h5>

Il lui faudra 1 heure, soit 60 minutes.<\/p>

Temps 0\u00a0: 6 plateaux<\/p>

Temps 30\u00a0: 6 + 6<\/p>

Temps 60\u00a0: 6 + 6 + 6<\/p>

\u00a0<\/h5>
R\u00e9ponse 5
<\/strong><\/h5>

La \"\"= 9\u00a0; la \"\" = 3 ; par cons\u00e9quent la \"\" = 3.<\/p>

Si 3 repr\u00e9sente le 6 % de l\u2019\u00e9quipe, alors le 100 % est 3\/6*100 = 50<\/p>

\u00a0<\/h5>
R\u00e9ponse 6
<\/strong><\/h5>

Un panier d\u2019instrument p\u00e8se 9’800 \u2013 6’400 = 3’400 g<\/p>

Deux paniers p\u00e8sent\u00a0: 2*3’400 = 6’800 g, donc le conteneur p\u00e8se 9’800 \u2013 6’800 = 3\u2019000 g soit 3 Kg<\/p>

\u00a0<\/h5>
R\u00e9ponse 7
<\/strong><\/h5>

Le volume d\u2019acier est de 2’496’000’000 mg \/ 8 mg = 312\u2019000’000 mm3<\/sup><\/p>

La base du pav\u00e9 est de 1\u2019500 x 2\u2019600 = 3\u2019900’000 mm3<\/sup><\/p>

La hauteur est de 312’000\u2019000 \/ 3\u2019900\u2019000 = 80 mm<\/p>

\u00a0<\/h5>
R\u00e9ponse 8
<\/strong><\/h5>

Tc = (5\/9) (270 \u2013 32) = 132.22<\/p>

Votre cycle de st\u00e9rilisation est programm\u00e9 \u00e0 134\u00b0 C. La phase plateau doit se d\u00e9rouler entre 134\u00b0 C et 137\u00b0 C. Par cons\u00e9quent, un dispositif qui ne supporte que 132.2 pourrait \u00eatre abim\u00e9.<\/p>

Dans un tel cas, il est utile de contacter le fabricant et de lui demander par \u00e9crit si son DM supporte un cycle de 134\u00b0 C pendant 18 minutes. SI ce n\u2019est pas le cas, un cycle de st\u00e9rilisation \u00e0 121 \u00b0 C pourrait \u00eatre une solution.<\/p>

\u00a0<\/h5>
R\u00e9ponse 9
<\/strong><\/h5>

Soit x le volume de la cuve du bac \u00e0 ultra-sons.<\/p>

L\u2019\u00e9quation devient :\u00a0 (x\/2) + 2 =\u00a0 (3x\/4)<\/p>

En mettant tout sur le m\u00eame d\u00e9nominateur :\u00a0 (4x\/8) + (16\/8)\u00a0 =\u00a0 (6x\/8) => 2x = 16, x = 8 litres<\/p>

\u00a0<\/h5>
R\u00e9ponse 10<\/strong><\/h5>

La question est\u00a0: En quelle ann\u00e9e a \u00e9t\u00e9 fond\u00e9e la SSSH ?<\/p>

C\u2019est en 1983.<\/p>

\u00a0<\/h5>
R\u00e9ponse 11<\/strong><\/h5>

Soit R le poids du conteneur rouge, B le bleu et J le jaune.<\/p>

L\u2019\u00e9quation devient R + B + 2J = 15<\/p>

R = 3 B<\/p>

B = 2J<\/p>

En mettant tout dans l\u2019\u00e9quation, cela donne\u00a0:<\/p>

3B + B + B = 15 => B = 15\/5 = 3<\/p>

R = 3*3 = 9<\/p>

J = 3\/2 = 1.5<\/p>

\u00a0<\/h5>
R\u00e9ponse 12<\/strong><\/h5>

Il a une faute d\u2019orthographe dans le mot erreur.<\/p>

\u00a0<\/h5>
R\u00e9ponse 13<\/strong><\/h5>

Soit X la distance totale et Y la vitesse sur la deuxi\u00e8me partie du parcours.<\/p>

En sachant qu\u2019une vitesse est une distance divis\u00e9e par un temps, l\u2019\u00e9quation devient\u00a0:<\/p>

\"\"<\/p>

60X\/100 + 60 X\/2Y = X<\/p>

En mettant tout sous le m\u00eame d\u00e9nominateur\u00a0:<\/p>

60 XY\/100 Y + 3000 X\/100 Y = 100 XY\/100 Y<\/p>

60 XY + 3000 X = 100 XY<\/p>

En divisant tout par X, l\u2019\u00e9quation devient\u00a0:\u00a0 60 Y + 3000 = 100 Y => 40 Y = 3000 => Y = 75<\/p>

Il y a certainement une fa\u00e7on plus simple de r\u00e9soudre le probl\u00e8me, puisqu\u2019il n\u2019y a pas besoin du X\u00a0!<\/p>

\u00a0<\/h5>
R\u00e9ponse 14<\/strong><\/h5>

A partir de la premi\u00e8re lettre dans le tableau, il faut suivre les fl\u00e8ches (\u2191 = monter d\u2019une case, etc.)<\/p>

Les pr\u00e9noms sont\u00a0: Nicole et Markus<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<\/div>\n\t\t","protected":false},"featured_media":0,"template":"","meta":{"_acf_changed":false,"site-sidebar-layout":"no-sidebar","site-content-layout":"page-builder","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"disabled","ast-breadcrumbs-content":"","ast-featured-img":"disabled","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}}},"forum_cat":[1030],"parution":[],"page_sommaire":[],"class_list":["post-38692","forum","type-forum","status-publish","hentry","forum_cat-edition-1-2022"],"acf":[],"yoast_head":"\nPropositions de solutions - 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