answer:For totaling to 17 with number cards from 2 through 10, we consider the following cases: bullet~ Case 1: The first card is 8, and the second card is 9. - There are 4 eights and 4 nines in a deck. - Probability of choosing an 8 first: dfrac{4}{52}. - Probability of then choosing a 9: dfrac{4}{51}. bullet~ Case 2: The first card is 9, and the second card is 8. - Probability of choosing a 9 first: dfrac{4}{52}. - Probability of then choosing an 8: dfrac{4}{51}. Calculating the probabilities for each case: - Probability for Case 1: dfrac{4}{52} times dfrac{4}{51} = dfrac{16}{2652}. - Probability for Case 2: dfrac{4}{52} times dfrac{4}{51} = dfrac{16}{2652}. Therefore, the overall probability is dfrac{16}{2652} + dfrac{16}{2652} = dfrac{32}{2652} = dfrac{16}{1326} = boxed{dfrac{8}{663}}.

answer:First, using Pascal’s identity, find binom{25}{5} and binom{25}{6}: - To calculate binom{25}{5}, we need binom{24}{4} which is not provided, instead, using binom{24}{5}=42504 directly. - To calculate binom{25}{6}: [ binom{25}{6} = binom{24}{5} + binom{24}{6} = 42504 + 134596 = 177100 ] Then, using binom{25}{6} and binom{25}{5} to find binom{26}{6}: [ binom{26}{6} = binom{25}{5} + binom{25}{6} ] However, binom{25}{5} calculation is not supported by previous steps directly and would need the value of binom{24}{4} which is missing. The calculation needs revising due to the absence of required values. Let's directly use Pascal's Identity for the next step with provided values: - To calculate binom{26}{6}: [ binom{26}{6} = binom{25}{5} + binom{25}{6} = 53130 + 177100 = boxed{230230} ]

answer:1. **Given**: ( A_1A_2A_3 ) is an acute-angled triangle inscribed in a unit circle centered at ( O ). The cevians from ( A_i ) passing through ( O ) meet the opposite sides at points ( B_i ) ((i = 1, 2, 3)) respectively. 2. **Objective**: - (a) Find the minimal possible length of the longest of three segments ( B_iO ). - (b) Find the maximal possible length of the shortest of three segments ( B_iO ). # Part (a): Minimal possible length of the longest of three segments ( B_iO ) 3. **Using Gergonne-Euler Theorem**: [ frac{OB_1}{1 + OB_1} + frac{OB_2}{1 + OB_2} + frac{OB_3}{1 + OB_3} = 1 ] This can be rewritten as: [ sum_{i=1}^{3} left( 1 - frac{1}{1 + OB_i} right) = 1 implies sum_{i=1}^3 frac{1}{1 + OB_i} = 2 ] 4. **Analyzing the inequality**: [ 2 = sum_{i=1}^3 frac{1}{1 + OB_i} geq frac{3}{1 + OB_1} ] This implies: [ 1 + OB_1 geq frac{3}{2} implies OB_1 geq frac{1}{2} ] 5. **Conclusion for part (a)**: The minimal possible length of the longest segment ( OB_1 ) is ( boxed{frac{1}{2}} ). # Part (b): Maximal possible length of the shortest of three segments ( B_iO ) 6. **Analyzing the inequality**: [ 2 = sum_{i=1}^3 frac{1}{1 + OB_i} leq frac{3}{1 + OB_3} ] This implies: [ 1 + OB_3 leq frac{3}{2} implies OB_3 leq frac{1}{2} ] 7. **Conclusion for part (b)**: The maximal possible length of the shortest segment ( OB_3 ) is ( boxed{frac{1}{2}} ). 8. **Verification**: It can be easily seen that one case of equality is when ( triangle A_1A_2A_3 ) is equilateral. In this case, all ( OB_i ) are equal and each ( OB_i = frac{1}{2} ).

answer:Solution: (Ⅰ) ( frac {25}{9})^{0.5}+( frac {27}{64})^{- frac {2}{3}}+(0.1)^{-2}-100cdot π^{0} = ( frac {5}{3})^{2×0.5} + ( frac {3}{4})^{3×(- frac {2}{3})} + 100 - 100 = frac {5}{3} + frac {16}{9} = frac {31}{9}; (Ⅱ) lg frac {1}{2}-lg frac {5}{8}+lg12.5-log_{8}9cdot log_{27}8+e^{2ln2} = lg( frac {1}{2} times frac {8}{5} times frac {25}{2}) - frac {lg9}{lg8} cdot frac {lg8}{lg27} + 4 = 1 - frac {2}{3} + 4 = frac {13}{3}. Therefore, the answers are boxed{frac {31}{9}} for (Ⅰ) and boxed{frac {13}{3}} for (Ⅱ).