answer:Let's denote the wrongly noted mark as x. The sum of the marks of all 30 students with the wrong mark is: 30 * 100 = 3000 If we correct the student's mark from x to 10, the new sum of the marks will be: 3000 - x + 10 The correct average marks for the 30 students is 98, so the correct sum of the marks should be: 30 * 98 = 2940 Now we can set up the equation: 3000 - x + 10 = 2940 Solving for x: 3010 - x = 2940 x = 3010 - 2940 x = 70 The wrongly noted mark of the student was boxed{70} .

answer:The first eight positive composite numbers are 4, 6, 8, 9, 10, 12, 14, 15. The product of these numbers is ( P_1 = 4 cdot 6 cdot 8 cdot 9 cdot 10 cdot 12 cdot 14 cdot 15 ). The next eight positive composite numbers are 16, 18, 20, 21, 22, 24, 25, 26. The product of these numbers is ( P_2 = 16 cdot 18 cdot 20 cdot 21 cdot 22 cdot 24 cdot 25 cdot 26 ). The first prime number is 2 and the second prime number is 3. The modified products are: [ P_1 + 2 = 4 cdot 6 cdot 8 cdot 9 cdot 10 cdot 12 cdot 14 cdot 15 + 2 ] [ P_2 + 3 = 16 cdot 18 cdot 20 cdot 21 cdot 22 cdot 24 cdot 25 cdot 26 + 3 ] Thus, the modified quotient is: [ frac{P_1 + 2}{P_2 + 3} = frac{4 cdot 6 cdot 8 cdot 9 cdot 10 cdot 12 cdot 14 cdot 15 + 2}{16 cdot 18 cdot 20 cdot 21 cdot 22 cdot 24 cdot 25 cdot 26 + 3} ] Performing the exact calculation of these products is complex and typically not feasible without computational aid, but the expression is valid and would yield a unique fraction upon computation. Conclusion: The result of the division of these modified products is a unique fraction, represented by: [ boxed{frac{4 cdot 6 cdot 8 cdot 9 cdot 10 cdot 12 cdot 14 cdot 15 + 2}{16 cdot 18 cdot 20 cdot 21 cdot 22 cdot 24 cdot 25 cdot 26 + 3}} ]

answer:To determine the number of Young tableaux of a given weight ( s ), we need to list all possible ways to partition ( s ) into sums of natural numbers in a non-decreasing order. Let's break down each case step-by-step. a) For ( s = 4 ): We consider all partitions of ( 4 ): 1. ( 4 ) = ( 4 ) 2. ( 4 ) = ( 3 + 1 ) 3. ( 4 ) = ( 2 + 2 ) 4. ( 4 ) = ( 2 + 1 + 1 ) 5. ( 4 ) = ( 1 + 1 + 1 + 1 ) There are a total of 5 partitions. b) For ( s = 5 ): We consider all partitions of ( 5 ): 1. ( 5 ) = ( 5 ) 2. ( 5 ) = ( 4 + 1 ) 3. ( 5 ) = ( 3 + 2 ) 4. ( 5 ) = ( 3 + 1 + 1 ) 5. ( 5 ) = ( 2 + 2 + 1 ) 6. ( 5 ) = ( 2 + 1 + 1 + 1 ) 7. ( 5 ) = ( 1 + 1 + 1 + 1 + 1 ) There are a total of 7 partitions. c) For ( s = 6 ): We consider all partitions of ( 6 ): 1. ( 6 ) = ( 6 ) 2. ( 6 ) = ( 5 + 1 ) 3. ( 6 ) = ( 4 + 2 ) 4. ( 6 ) = ( 4 + 1 + 1 ) 5. ( 6 ) = ( 3 + 3 ) 6. ( 6 ) = ( 3 + 2 + 1 ) 7. ( 6 ) = ( 3 + 1 + 1 + 1 ) 8. ( 6 ) = ( 2 + 2 + 2 ) 9. ( 6 ) = ( 2 + 2 + 1 + 1 ) 10. ( 6 ) = ( 2 + 1 + 1 + 1 + 1 ) 11. ( 6 ) = ( 1 + 1 + 1 + 1 + 1 + 1 ) There are a total of 11 partitions. d) For ( s = 7 ): We consider all partitions of ( 7 ): 1. ( 7 ) = ( 7 ) 2. ( 7 ) = ( 6 + 1 ) 3. ( 7 ) = ( 5 + 2 ) 4. ( 7 ) = ( 5 + 1 + 1 ) 5. ( 7 ) = ( 4 + 3 ) 6. ( 7 ) = ( 4 + 2 + 1 ) 7. ( 7 ) = ( 4 + 1 + 1 + 1 ) 8. ( 7 ) = ( 3 + 3 + 1 ) 9. ( 7 ) = ( 3 + 2 + 2 ) 10. ( 7 ) = ( 3 + 2 + 1 + 1 ) 11. ( 7 ) = ( 3 + 1 + 1 + 1 + 1 ) 12. ( 7 ) = ( 2 + 2 + 2 + 1 ) 13. ( 7 ) = ( 2 + 2 + 1 + 1 + 1 ) 14. ( 7 ) = ( 2 + 1 + 1 + 1 + 1 + 1 ) 15. ( 7 ) = ( 1 + 1 + 1 + 1 + 1 + 1 + 1 ) There are a total of 15 partitions. # Conclusion: - Number of Young tableaux for ( s = 4 ): ( boxed{5} ) - Number of Young tableaux for ( s = 5 ): ( boxed{7} ) - Number of Young tableaux for ( s = 6 ): ( boxed{11} ) - Number of Young tableaux for ( s = 7 ): ( boxed{15} )

answer:Solution: (Ⅰ) From the given condition, |x-4| leq 2, which implies -2 leq x-4 leq 2, thus 2 leq x leq 6. Therefore, the range of x is boxed{[2, 6]}. (Ⅱ) Given 2 leq x leq 6, we have g(x) = 2sqrt{x-2} + sqrt{6-x}. By the Cauchy-Schwarz inequality, we get g(x) leq sqrt{(4+1)(x-2+6-x)} = 2sqrt{5}. The equality holds if and only if frac{sqrt{x-2}}{2} = frac{sqrt{6-x}}{1}, which happens when x = frac{26}{5}. Therefore, the maximum value of g(x) is boxed{2sqrt{5}}.