Cattle currently weighing 528 pounds will be fed a diet at 2.3% of their body we

Question

Cattle currently weighing 528 pounds will be fed a diet at 2.3% of their body weight until they weigh 755 pounds. The diet consists of corn grain, corn silage, and supplement, and supplies 1.74 Mcal/kg NEm and 1.11 Mcal/kg NEg on a DM basis. If the maintenance energy requirements of the cattle are 5.4 Mcal/day NEm, determine the daily amount of dietary energy remaining for gain (NEg). Note, express your answer as Mcal/d NEg.

Answer: 4.0 Mcal/d NEg

I have the answer but I need the steps to get the correct answer

Explanation / Answer

HERE WEIGHT OF CATTLE IS IN POUND AND FEED IN KG . WEIGHT OF 528 POUNDS = 528 X 0.4536 KG

= 239.5008 KG

AGAIN 755 POUNDS = 755 X 0.4536 KG =342.468 KG, AERAGE OF THESE WEIGHT =( 342.468 + 239.5008)/2 = 290.98, NOW 2.3% OF BODY SHOULD

BE GIVEN AS FEED . HENCE AVERAGE FEED WEIGHT = 290.98 X 2.3 % = 6.69254 KG

NOW FOR 5.4 Mcal/DAY TOTAL NEm required = 5.4/1.74 = 3.1034 kg HENCE NEg required 6.69254-3.1034

= 3.58914 kg

NOW 3.58914 KG WILL GIVE 3.58914 X 1.11 NEg = 3.98395 Mcal/DAY ~ 4 Mcal/ DAY

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