Q:

A worker at the zoo calculates the amount of fish, in pounds, needed in the weekly diet of an eagle and a bear.·The eagle eats 6 pounds of food each week, and 60% of that weight must be fish.·The bear eats 105 pounds of food each week, and 25% of that weight must be fish.PART AWhat is the total number of fish, in pounds, that the eagle and bear should eat each week? Round your answer to the nearest hundredth of a pound.PART BThe zoo increases the amount of food that the bear eats each week to 115 pounds. What is the percent increase in the amount of food that the bear eats each week? Round your answer to the nearest tenth of a percent.

Accepted Solution

A:
Answer: Part A: The two will consume a total of 29.85 pounds per weekPart B: The percentage increase in the amount of food the bear eats per week is 9.5%( to the nearest tenth)Step-by-step explanation: For part A: since the eagle eats 6 pounds of food per week, and 60% of this must be fish, we will need to determine the exact amount of fish (in pounds) that the eagle eats: (60/100) × 6 = 360/100 = 3.60 pounds This means that the eagle eats 3.60 pounds of fish per week. Similarly, since the bear eats 105 pounds of food each week and 25% of this 105 pounds must be fish which the bear consumes per week, we will also find the exact amount of fish (in pounds) that the bear consumes: (25/100) × (105/1) = 2625/100 = 26.25 pounds. Therefore, the bear eats 26.25 pounds of fish every week. In order to determine the amount of fishes (in pounds) that the eagle and the bear consumes per wey, we will add up the amount consumed by the eagle with the amount consumed by the bear: = 26.25 + 3.60 = 29.85 pounds altogether (to the nearest hundredth) Part B: if the zoo increases the amount of food the bear eats to 115 pounds from 105 pounds, the increment is therefore: 115 pounds - 105 pounds = 10 pounds. The percentage increase in the amount of food the bear consumes each week is then : (10 pounds/105 pounds) × (100/1) = 1000/105 = 9.524% = 9.5% (rounding to the nearest tenth)