Asked by Anonymous
It is estimated that 6,000 tons of nachos will be consumed at Super Bowl XLV viewing events. Suppose a serving of nachos is 1 1/4 oz and each person eats an average of 3 1/2 servings of nachos. If a total of 91,385 people attend the viewing parties at the 373 Champs Sports Zone locations nationwide, what percent of the total amount of nachos consumed will be eaten at all the Champs Sports Zone locations combined? Express your answer as a percent to the nearest hundredth.
Answers
Answered by
helper
6000 tons nachos
1.25 oz/serving
3.5 serving/person
3.5 serving * 1.25 oz = 4.375 oz/person
2000 lb/ton * 6000 tons = 12,000,000 lbs
12,000,000 lbs * 16 oz/lb = 192,000,000 oz
91,385 people * 4.375 oz/person = 399,809.3750 oz consumed
399,809.3750 oz consumed/192,000,000 oz
= 0.0021 = 0.2083%
Check my math and the logic of how I did this for correctness. I assumed that it was meant that the total for all viewing parties was 91,385, not 91,385 at each location.
1.25 oz/serving
3.5 serving/person
3.5 serving * 1.25 oz = 4.375 oz/person
2000 lb/ton * 6000 tons = 12,000,000 lbs
12,000,000 lbs * 16 oz/lb = 192,000,000 oz
91,385 people * 4.375 oz/person = 399,809.3750 oz consumed
399,809.3750 oz consumed/192,000,000 oz
= 0.0021 = 0.2083%
Check my math and the logic of how I did this for correctness. I assumed that it was meant that the total for all viewing parties was 91,385, not 91,385 at each location.
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