answer:1. **Define Variables**: Let ( r ) be the number of marbles Ringo has and ( p ) be the number of marbles Paul has. 2. **Set Up Congruences**: From the problem statement: [ r equiv 5 pmod{8} ] [ p equiv 6 pmod{8} ] 3. **Add the Congruences**: Adding these two congruences gives: [ r + p equiv 5 + 6 pmod{8} ] [ r + p equiv 11 pmod{8} ] 4. **Simplify the Sum**: Since ( 11 ) is more than ( 8 ), we reduce it modulo ( 8 ): [ 11 equiv 3 pmod{8} ] Thus: [ r + p equiv 3 pmod{8} ] 5. **Interpret the Result**: This result means that when Ringo and Paul pool their marbles and try to divide them into bags of 8, they will have 3 marbles left over. Thus, the number of marbles left over when Ringo and Paul pool their marbles and place them into bags of 8 is (3). boxed{The correct answer is **C)** 3.}

answer:Proof: (1) When n=1, the left side equals frac {1}{4}, and the right side equals frac {1}{6}. The inequality holds. (2) Assume that when n=k, the original statement holds, that is frac {1}{2^{2}} + frac {1}{3^{2}} + ldots + frac {1}{(k+1)^{2}} > frac {1}{2} - frac {1}{k+2}. When n=k+1, frac {1}{2^{2}} + frac {1}{3^{2}} + ldots + frac {1}{(k+1)^{2}} + frac {1}{(k+2)^{2}} > frac {1}{2} - frac {1}{k+2} + frac {1}{(k+2)^{2}}. Since - frac {1}{k+2} + frac {1}{(k+2)^{2}} + frac {1}{k+3} = frac {1}{(k+2)^{2}(k+3)} > 0, it follows that - frac {1}{k+2} + frac {1}{(k+2)^{2}} > - frac {1}{k+3}, therefore, frac {1}{2} - frac {1}{k+2} + frac {1}{(k+2)^{2}} > frac {1}{2} - frac {1}{k+3}, which means the conclusion holds when n=k+1. Based on (1) and (2), we know the inequality holds for any positive integer n. Therefore, boxed{frac {1}{2^{2}} + frac {1}{3^{2}} + ldots + frac {1}{(n+1)^{2}} > frac {1}{2} - frac {1}{n+2}}.

answer:On each 15-sided die: - Numbers less than 10: 1 through 9, making 9 favorable outcomes. - Numbers 10 or greater: 10 through 15, making 6 favorable outcomes. Probability for one die showing a number less than 10: frac{9}{15} = frac{3}{5}. Probability for one die showing a number 10 or greater: frac{6}{15} = frac{2}{5}. To find the probability that exactly three dice show a number less than 10, and two dice show a number 10 or greater: - The probability of a specific arrangement (e.g., LLLGG, where L is less than 10 and G is 10 or greater) is left(frac{3}{5}right)^3 cdot left(frac{2}{5}right)^2. - The number of such arrangements is the number of ways to choose 3 dice out of 5 to show a number less than 10, which is binom{5}{3} = 10. The total probability is 10 cdot left(frac{3}{5}right)^3 cdot left(frac{2}{5}right)^2 = 10 cdot frac{27}{125} cdot frac{4}{25} = 10 cdot frac{108}{3125} = frac{1080}{3125} = boxed{frac{216}{625}}.

answer:Since z_1=m+i, z_2=1-2i, and frac{z_1}{z_2}=-frac{1}{2}, We have frac{m+i}{1-2i}=frac{(m+i)(1+2i)}{(1-2i)(1+2i)}=frac{m-2+(2m+1)i}{5}=-frac{1}{2}, This leads to the system of equations: begin{cases} frac{m-2}{5}=-frac{1}{2} frac{2m+1}{5}=0 end{cases}, solving which we get m=-frac{1}{2}. Therefore, the answer is: boxed{D}. The given equation frac{z_1}{z_2}=-frac{1}{2} is simplified using the algebraic form of complex number multiplication and division to obtain the answer. This problem tests basic computational skills involving complex numbers in algebraic form.