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The Water Cooler
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Math Problem!
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<blockquote data-quote="turkeyrun" data-source="post: 4103436" data-attributes="member: 27991"><p>Chicks are easily sexed, 100% can be achieved.</p><p></p><p>Mixed will always be mostly roosters.</p><p></p><p>With bushes and getting, 3 bushes has 8 possibilities, with 6 having opportunities for berries.6/8 = 75% POSSIBILITY of berries.</p><p></p><p>With chicks, assume 50 - 50 mix on 100 chicks. An order for 100 chicks, 25 straight run and 75 mix pulled. Results in 25 and 50 - 25 mix. So, 2 in 3 are rooster. Order comes in for 35 and 65, nets 50 - 15.</p><p>I bought 48 mixed, at TSC and got 4 hens, with 42 roosters. 2 didn't survive.</p><p>Luckily, I was wanting meat birds, not layers. IF you want hens, pay the extra for straight run, unless you can get a helpful employee that will sex them as you go.</p><p></p><p>Son ordered 100 fertilized eggs. Hatched 54 hens and 45 roosters.</p><p></p><p>Dingleberries is a whole nuther subject, I WILL NOT get into.</p><p></p><p>Time for coffee, my head is starting to hurt.</p></blockquote><p></p>
[QUOTE="turkeyrun, post: 4103436, member: 27991"] Chicks are easily sexed, 100% can be achieved. Mixed will always be mostly roosters. With bushes and getting, 3 bushes has 8 possibilities, with 6 having opportunities for berries.6/8 = 75% POSSIBILITY of berries. With chicks, assume 50 - 50 mix on 100 chicks. An order for 100 chicks, 25 straight run and 75 mix pulled. Results in 25 and 50 - 25 mix. So, 2 in 3 are rooster. Order comes in for 35 and 65, nets 50 - 15. I bought 48 mixed, at TSC and got 4 hens, with 42 roosters. 2 didn't survive. Luckily, I was wanting meat birds, not layers. IF you want hens, pay the extra for straight run, unless you can get a helpful employee that will sex them as you go. Son ordered 100 fertilized eggs. Hatched 54 hens and 45 roosters. Dingleberries is a whole nuther subject, I WILL NOT get into. Time for coffee, my head is starting to hurt. [/QUOTE]
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