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Answer :
Final answer:
To sustain themselves, 7 gobblers would consume 152 oz of insects daily.
Explanation:
To calculate the amount of insects that 7 gobblers would consume daily, we first need to find the difference in weight between a male and a female turkey. The male turkey weighs 17.5 lbs and the female turkey weighs 8.0 lbs, so the difference is 17.5 - 8.0 = 9.5 lbs.
Next, we need to convert the weight from pounds to ounces. Since 1 lb is equivalent to 16 oz, we multiply 9.5 lbs by 16 to get 9.5 * 16 = 152 oz. This is the extra weight that the male turkey has compared to the female turkey.
Therefore, 7 gobblers would consume 152 oz of insects daily to sustain themselves.
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Answer:
The answer is below
Explanation:
It have been shown that domestic turkeys require about 13.6 grams of food per pound of body weight each day to sustain weight. A gobbler is a male turkey.
The food consumed by a male turkey that is 17.5 lbs = 13.6 grams of food per pound * 17.5 pound = 238 grams.
7 gobblers = 7 male turkeys, hence:
The food consumed by 7 gobblers = 238 grams * 7 = 1666 grams
