answer:Let the side length of the cube be ( s ). The volume of the cube is ( s^3 ). For a regular tetrahedron inside the cube, each edge of the tetrahedron equals the space diagonal of the cube. The space diagonal ( d ) of the cube can be calculated using the Pythagorean theorem in three dimensions: ( d = sqrt{s^2 + s^2 + s^2} = ssqrt{3} ). The volume ( V ) of a regular tetrahedron with edge length ( a ) is given by the formula: [ V = frac{a^3 sqrt{2}}{12} ] Substitute ( a = ssqrt{3} ) into the volume formula: [ V = frac{(ssqrt{3})^3 sqrt{2}}{12} = frac{s^3 3sqrt{3} sqrt{2}}{12} = frac{s^3 3sqrt{6}}{12} = frac{s^3sqrt{6}}{4} ] Thus, the volume of the inscribed tetrahedron is ( frac{s^3sqrt{6}}{4} ). The ratio of the volume of the tetrahedron to the volume of the cube is: [ frac{frac{s^3sqrt{6}}{4}}{s^3} = frac{sqrt{6}}{4} ] Hence, the required ratio is ( boxed{frac{sqrt{6}}{4}} ).

answer:To find the least positive integral value of ( n ) for which the equation [ x_1^3 + x_2^3 + cdots + x_n^3 = 2002^{2002} ] has integer solutions ((x_1, x_2, x_3, cdots, x_n)), we follow these steps: 1. **Step 1: Consider the properties of cubes modulo 9.** The possible residues of ( x^3 ) modulo 9 when ( x ) is an integer are: [ 0, pm 1 ] This is because for any integer ( x ), ( x^3 equiv 0 pmod{9} ) if ( x equiv 0 pmod{9} ), and ( x^3 equiv 1 pmod{9} ) or ( x^3 equiv -1 pmod{9} ) for other values modulo 9. 2. **Step 2: Compute ( 2002^{2002} mod 9 ).** First, reduce 2002 modulo 9: [ 2002 equiv 4 pmod{9} ] Now calculate the residue of ( 4^{2002} mod 9 ). Begin by understanding the powers of 4 modulo 9: [ 4^1 equiv 4 pmod{9} ] [ 4^2 equiv 16 equiv 7 pmod{9} ] [ 4^3 equiv 28 equiv 1 pmod{9} ] From the above cycles, ( 4^3 equiv 1 pmod{9} ). Therefore: [ 4^{2002} equiv 4^{667 times 3 + 1} = (4^3)^{667} times 4 equiv 1^{667} times 4 equiv 4 pmod{9} ] 3. **Step 3: Determine the minimum ( n ) such that the sum of ( n ) cubes can be congruent to 4 modulo 9.** Given that the residues of ( x^3 mod 9 ) are 0, 1, or -1, observe that: - For ( n = 1 ): ( x_1^3 equiv 0, pm 1 not equiv 4 pmod{9} ) - For ( n = 2 ): ( x_1^3 + x_2^3 equiv (0, pm 1) + (0, pm 1) not equiv 4 pmod{9} ) - For ( n = 3 ): ( x_1^3 + x_2^3 + x_3^3 equiv (0, pm 1) + (0, pm 1) + (0, pm 1) not equiv 4 pmod{9} ) Let's try ( n = 4 ). We need to see if four cubes can sum to 4 modulo 9: [ 1^3 + 1^3 + 1^3 + 1^3 = 1 + 1 + 1 + 1 = 4 pmod{9} ] This shows that ( n = 4 ) is indeed a solution since the sum of four ones cubed gives a residue of 4 modulo 9. # Conclusion: Therefore, the least positive integral value of ( n ) for which the equation has integer solutions is: [ boxed{4} ]

answer:To solve the problem, let's denote the number of female students among the 200 student volunteers as x. According to the stratified sampling by gender, the ratio of selected female students to the total selected students should be equal to the ratio of female students in the entire pool of 200 students. Given that 12 male students are selected out of the 30, the number of female students selected is 30 - 12 = 18. Therefore, the ratio of selected female students to the total selected students is frac{18}{30}. This ratio should be equal to the ratio of female students in the entire pool, which is frac{x}{200}. Setting these ratios equal to each other gives us the equation: [ frac{x}{200} = frac{18}{30} ] Solving for x involves cross-multiplying and simplifying the equation: [ x = frac{18}{30} times 200 = frac{18 times 200}{30} ] Further simplifying: [ x = frac{3600}{30} = 120 ] Therefore, the number of female students among the 200 student volunteers could be boxed{120}.

answer:Let's denote the speed of ferry P as ( v_P ) km/h and the speed of ferry Q as ( v_Q ) km/h. According to the problem, we have: 1. ( v_Q = v_P + 3 ) (since ferry Q is 3 km/h faster than ferry P) 2. The distance covered by ferry P in 3 hours is ( 3v_P ) km. 3. The distance covered by ferry Q is twice the distance of ferry P, so it is ( 2 times 3v_P = 6v_P ) km. 4. The time taken by ferry Q to cover its distance is 1 hour more than the time taken by ferry P, so it is ( 3 + 1 = 4 ) hours. Now, we can use the relationship between speed, distance, and time for ferry Q: [ v_Q = frac{Distance}{Time} ] [ v_Q = frac{6v_P}{4} ] [ v_Q = frac{3}{2}v_P ] Now we can substitute ( v_Q ) from equation 1 into this equation: [ v_P + 3 = frac{3}{2}v_P ] To solve for ( v_P ), we rearrange the equation: [ 3 = frac{3}{2}v_P - v_P ] [ 3 = frac{1}{2}v_P ] [ v_P = 2 times 3 ] [ v_P = 6 ] km/h So, the speed of ferry P is boxed{6} km/h.