answer:Let the rectangle's length be 5k and its width be 2k. Given the perimeter of the rectangle, which is the sum of twice the length and twice the width, we have: [ 2(5k) + 2(2k) = 60 ] [ 10k + 4k = 60 ] [ 14k = 60 ] [ k = frac{60}{14} = frac{30}{7} ] Next, we calculate the rectangle's diagonal using the Pythagorean theorem: [ text{Diagonal} = sqrt{(5k)^2 + (2k)^2} ] [ = sqrt{25k^2 + 4k^2} ] [ = sqrt{29k^2} ] [ = ksqrt{29} ] [ = frac{30}{7}sqrt{29} ] Therefore, the length of the diagonal is boxed{frac{30sqrt{29}}{7}} meters.

answer:Given the following counts of apples for seven traders: (20, 40, 60, 80, 100, 120,) and (140), let's determine the price at which they sold apples to ensure that each trader received the same total revenue. 1. **Assumption of Uniform Price with Residuals**: - Each trader sells apples primarily at the same bulk rate, let's say (1 text{ cent per } n) apples, where (n) is a certain number of apples. - Remaining apples, if any, are sold at a higher price, let's say (b text{ cents per apple}). 2. **Calculation for Uniform Revenue**: - Let's denote the batch price (p) per (n) apples as (1 text{ cent per } n) apples and the higher price per apple as (3) cents for any remaining apples. 3. **Determine the Batch Size ((n)) and Price ((p))**: - It's given in the solution that (p = 1 text{ cent per } 7 text{ apples}), so (n = 7). - Any remaining apples are sold at (3) cents per apple. To check this systematically and understand how this leads to equal revenue, calculate the revenue for each trader: 4. **Revenue for Trader**: - Compute first for each trader with total apple counts dividing and obtaining remainders: - For (20) apples: (20 div 7 = 2 text{ batches} text{ (14 apples)}), and (6 text{ apples leftover}). [ text{Revenue} = 2 times 1 text{ cent} + 6 times 3 text{ cents} = 2 + 18 = 20 text{ cents} ] - For (40) apples: (40 div 7 = 5 text{ batches (35 apples)}, and 5 text{ apples leftover}). [ text{Revenue} = 5 times 1 text{ cent} + 5 times 3 text{ cents} = 5 + 15 = 20 text{ cents} ] - For (60) apples: (60 div 7 = 8 text{ batches (56 apples)}, and 4 apples leftover). [ text{Revenue} = 8 times 1 text{ cent} + 4 times 3 text{ cents} = 8 + 12 = 20 text{ cents} ] - For (80) apples: (80 div 7 = 11 text{ batches (77 apples)}, and 3 apples leftover). [ text{Revenue} = 11 times 1 text{ cent} + 3 times 3 text{ cents} = 11 + 9 = 20 text{ cents} ] - For (100) apples: (100 div 7 = 14 text{ batches (98 apples)}, and 2 apples leftover). [ text{Revenue} = 14 times 1 text{ cent} + 2 times 3 text{ cents} = 14 + 6 = 20 text{ cents} ] - For (120) apples: (120 div 7 = 17 text{ batches (119 apples)}, and 1 apple leftover). [ text{Revenue} = 17 times 1 text{ cent} + 1 times 3.cent} = 17 + 3 = 20 text{ cents} ] - For (140) apples: (140 div 7 = 20 text{ batches (140 apples)}}, and 0 apples leftover). [ text{Revenue} = 20 times 1 text{ cent} + 0 times 3 (no leftover) = 20 + 0 = 20 text{ cents} ] Each trader gets exactly (20) cents revenue. Conclusion: [ boxed{1 text{ cent per 7 apples; 3 cents per leftover apple}} ]

answer:To solve this problem, we proceed in a step-by-step manner as follows: **Step 1: Calculate the total number of outcomes.** Since each coin flip has 2 possible outcomes (heads or tails), and there are 7 flips, the total number of outcomes is calculated by raising 2 to the power of 7: [2^7 = 128] **Step 2: Use casework to count the number of successful outcomes.** We need to consider all cases where we have at least 5 heads. This includes having exactly 5 heads, exactly 6 heads, and exactly 7 heads. - **Case 1: Exactly 5 heads.** To find the number of ways to get exactly 5 heads out of 7 flips, we choose 5 flips to be heads. The number of ways to do this is given by the combination formula binom{7}{5}, which calculates to: [binom{7}{5} = 21] - **Case 2: Exactly 6 heads.** Similarly, to find the number of ways to get exactly 6 heads out of 7 flips, we choose 6 flips to be heads. The number of ways to do this is given by the combination formula binom{7}{6}, which calculates to: [binom{7}{6} = 7] - **Case 3: Exactly 7 heads.** For all 7 flips to be heads, there is only 1 way this can happen, as all flips must be heads. **Step 3: Calculate the total number of successful outcomes.** Adding the successful outcomes from each case together: [21 + 7 + 1 = 29] **Step 4: Calculate the probability.** The probability of getting at least 5 heads is the ratio of the number of successful outcomes to the total number of outcomes: [frac{29}{128}] Therefore, the probability that at least 5 of the flips come up heads is boxed{frac{29}{128}}.

answer:Let's denote the cost of one pencil as P and the cost of one notebook as N. According to the given information, we have two equations: 1) The cost of 3 pencils and 4 notebooks is 60 rupees: 3P + 4N = 60 (Equation 1) 2) The sum of the cost of 1 pencil and 1 notebook is 15.512820512820513 rupees: P + N = 15.512820512820513 (Equation 2) We can solve these two equations to find the values of P and N. From Equation 2, we can express N in terms of P: N = 15.512820512820513 - P (Equation 3) Now, we can substitute Equation 3 into Equation 1 to find the value of P: 3P + 4(15.512820512820513 - P) = 60 Expanding the equation: 3P + 62.05128205128205 - 4P = 60 Combining like terms: -P + 62.05128205128205 = 60 Now, we solve for P: P = 62.05128205128205 - 60 P = 2.05128205128205 Now that we have the value of P, we can find the value of N using Equation 3: N = 15.512820512820513 - 2.05128205128205 N = 13.461538461538463 Now we have the cost of one pencil (P) and one notebook (N). We need to find the total cost of 8 dozen pencils and 2 dozen notebooks. Since one dozen is equal to 12, we have: Total cost = (8 dozen * 12 pencils/dozen * P) + (2 dozen * 12 notebooks/dozen * N) Total cost = (96 * 2.05128205128205) + (24 * 13.461538461538463) Calculating the total cost: Total cost = (96 * 2.05128205128205) + (24 * 13.461538461538463) Total cost = 196.9230769230769 + 323.0769230769231 Total cost = 520 rupees Therefore, the total cost of the 8 dozen pencils and 2 dozen notebooks is boxed{520} rupees.