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question:Selene buys two instant cameras at 110 each and three digital photo frames at 120 each. She gets a 5% discount on all the items she purchased. How much does she pay in all?

answer:First, let's calculate the total cost of the instant cameras and digital photo frames before the discount. Cost of instant cameras = 2 cameras * 110/camera = 220 Cost of digital photo frames = 3 frames * 120/frame = 360 Total cost before discount = Cost of instant cameras + Cost of digital photo frames Total cost before discount = 220 + 360 = 580 Now, let's calculate the discount Selene gets. Discount = 5% of total cost before discount Discount = 5/100 * 580 Discount = 0.05 * 580 Discount = 29 Now, let's calculate the total cost after the discount. Total cost after discount = Total cost before discount - Discount Total cost after discount = 580 - 29 Total cost after discount = 551 Selene pays boxed{551} in all after the discount.

question:Alicia has two containers. Initially, the first container is tfrac{2}{3} full of water while the second container is empty. After pouring all the water from the first container into the second container, the second container becomes tfrac{1}{2} full. Determine the ratio of the volume of the first container to the volume of the second container. textbf{(A) } frac{1}{2} qquad textbf{(B) } frac{3}{5} qquad textbf{(C) } frac{3}{4} qquad textbf{(D) } frac{5}{6} qquad textbf{(E) } frac{4}{3}

answer:1. **Define Variables:** Let A be the volume of the first container and B the volume of the second container. 2. **Expression of Volume Transferred:** The volume of water transferred from the first to the second container can be represented as frac{2}{3}A, and this fills the second container to frac{1}{2}B. 3. **Set Up the Equation Derived from the Transfer:** [ frac{2}{3}A = frac{1}{2}B ] 4. **Calculation of the Volume Ratio frac{A}{B}:** Rearrange the equation to solve for frac{A}{B}: [ frac{A}{B} = frac{frac{1}{2}B}{frac{2}{3}A} = frac{frac{1}{2}}{frac{2}{3}} ] Simplifying the right-hand side, we multiply by the reciprocal of frac{2}{3}: [ frac{1}{2} times frac{3}{2} = frac{1 times 3}{2 times 2} = frac{3}{4} ] 5. **Conclusion:** The ratio of the volume of the first container to the volume of the second container is frac{3{4}}. The correct answer is boxed{textbf{(C)} frac{3}{4}}.

question:Emberly goes for morning walks on varying distances every day. Here is the summary of her walking pattern in March: Week 1: Day 1, 3, 5: 4 miles each Day 2, 4: 3 miles each Day 7: Rest Week 2: Day 8, 10, 11: 5 miles each Day 9, 12, 13: 2.5 miles each Day 14: Rest Week 3: Day 15, 16: 6 miles each Day 17 to 19: Rest Day 20, 21: 4 miles each Day 22: 3.5 miles Week 4: Day 23, 25, 27, 29: 4.5 miles each Day 24, 26, 28, 30, 31: Rest Calculate the total number of miles she's walked in March.

answer:To calculate the total number of miles Emberly walked in March, we need to sum up the miles for each day she walked. Week 1: - Day 1, 3, 5: 4 miles each, so 4 miles * 3 days = 12 miles - Day 2, 4: 3 miles each, so 3 miles * 2 days = 6 miles - Day 7: Rest Week 2: - Day 8, 10, 11: 5 miles each, so 5 miles * 3 days = 15 miles - Day 9, 12, 13: 2.5 miles each, so 2.5 miles * 3 days = 7.5 miles - Day 14: Rest Week 3: - Day 15, 16: 6 miles each, so 6 miles * 2 days = 12 miles - Day 17 to 19: Rest - Day 20, 21: 4 miles each, so 4 miles * 2 days = 8 miles - Day 22: 3.5 miles Week 4: - Day 23, 25, 27, 29: 4.5 miles each, so 4.5 miles * 4 days = 18 miles - Day 24, 26, 28, 30, 31: Rest Now, we add up all the miles from each week: Week 1: 12 miles + 6 miles = 18 miles Week 2: 15 miles + 7.5 miles = 22.5 miles Week 3: 12 miles + 8 miles + 3.5 miles = 23.5 miles Week 4: 18 miles Total miles for March: 18 miles + 22.5 miles + 23.5 miles + 18 miles = 82 miles Emberly walked a total of boxed{82} miles in March.

question:Luke wants to fence a square piece of land of at least 500 square feet. Additionally, there is an existing smaller square inside the land that is already fenced which is 10 feet on a side. What should the side of the larger square be if he wants to use the least amount of fencing?

answer:First, calculate the required side length for the larger square to have an area of at least 500 square feet. Let's denote the side of the larger square as ( S ). The area requirement gives: [ S^2 geq 500 ] Taking the square root on both sides, we find: [ S geq sqrt{500} approx 22.36 text{ feet} ] Since dimensions in real life are typically rounded to the nearest whole number, we consider ( S = 23 ) feet as the minimum possible side length that satisfies the area condition. Next, consider the existing smaller inner square fence which has a side length of 10 feet. Luke would ideally want the larger square to encompass this smaller square completely without adjusting its position. Assuming the smaller square is centrally located and must be contained entirely within the larger square, there is no additional constraint dictating a larger ( S ) than determined from the area condition. Thus, the minimal side length of the larger square fence Luke should use, if he wants to use the least amount of fencing while meeting all conditions, is ( boxed{23} text{ feet} ).

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