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A livestock operation is planning to use feed additives to improve growth performance of ruminants. A...

A livestock operation is planning to use feed additives to improve growth performance of ruminants. A commercial product containing the active ingredient monensin was selected. Animals are currently consuming 19.2 lb per day of diet DM. Current commercial product contains 90g of active ingredient (monensin) for every pound as-is (as-fed) of commercial product. The DM content of such commercial product is 95.76 %. After discussion with nutritionists, the dose recommended of monensin for the diet was 30 g/short ton of diet DM. Assuming 1 lb = 453.6 g, 1 short ton = 2,000 lb, and 1 g = 1,000 mg, how many grams (DM) of commercial product will have to be included in each short ton of diet DM? (Provide 2 decimals for your answer, and perform rounding only at the end of your calculation).

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Answer #1

Recommended dose of monesin = 30g/ short ton of diet DM

1lb = 453.6g

1short ton = 2000lb

1gm = 1000mg

2000 lb = 2000× 453.6 g

= 907200 gm (1 short ton )

Recommended dose = 30 × 907200 gm

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