answer:To solve this problem, we consider placing the four indistinguishable stools among the twelve available slots. We then fill the remaining slots with indistinguishable rocking chairs. 1. Choose 4 slots for the stools out of the 12 slots. This is a combination problem (choosing 4 out of 12) and is given by the binomial coefficient: [ binom{12}{4} = frac{12 times 11 times 10 times 9}{4 times 3 times 2 times 1} ] Simplifying the above expression: [ binom{12}{4} = frac{12 times 11 times 10 times 9}{24} = frac{11880}{24} = 495 ] 2. The rocking chairs, being indistinguishable, do not add any new permutations once the stools are placed. The final number of distinct arrangements of the seats is therefore boxed{495}.

answer:Given that P_{1}(-1,y_{1}) and P_{2}(2,y_{2}) are two points on the graph of the linear function y=-x+1, we need to find the relationship between y_{1} and y_{2}. Step 1: Determine y_{1} using the coordinates of P_{1} and the equation of the line. Given y=-x+1 and P_{1}(-1,y_{1}), begin{align*} y_{1} &= -(-1) + 1 &= 1 + 1 &= 2. end{align*} Step 2: Determine y_{2} using the coordinates of P_{2} and the equation of the line. Given y=-x+1 and P_{2}(2,y_{2}), begin{align*} y_{2} &= -(2) + 1 &= -2 + 1 &= -1. end{align*} Step 3: Compare y_{1} and y_{2}. Since y_{1} = 2 and y_{2} = -1, we have 2 > -1, which means y_{1} > y_{2}. Therefore, the relationship between y_{1} and y_{2} is y_{1} > y_{2}. Hence, the correct answer is boxed{C}.

answer:To solve the problem, we will break it down into two parts (a) and (b) and give a detailed step-by-step solution for each. Part (a) We need to calculate the mathematical expectation of the number of points scored by the team of the experts over 100 games. Let's denote the number of points scored by the losing team in a single game as ( k ). 1. In each game, there are ( 6 + k ) envelopes played. Since the winning team scores 6 points, the losing team scores ( k ) points. 2. We note that the total envelopes for ( k = 0, 1, ..., ) is defined by the following sequence, and for each question, the probability of winning is ( frac{1}{2} ). 3. Therefore, the expected value (mathrm{E}(X)), the random variable representing the number of points scored by the losing team, is: [ mathrm{E}X = sum_{k=0}^{5} k cdot C_{6+k}^k cdot frac{1}{2^{6+k}} ] 4. Using ( k cdot C_{n}^k = (n - k + 1) cdot C_{n}^{k - 1} ): [ mathrm{E}X = 6 cdot sum_{k=1}^{5} C_{6+k}^{k-1} cdot frac{1}{2^{6+k}} ] 5. By extending the binomial distribution, we then sum the series considering the first part up to infinity (but it will practically converge by 10): [ mathrm{E}X = 6 left( sum_{k=0}^{5} frac{C_{6+k}^{k-1}}{2^{6+k}} right) ] 6. This series approximation will converge around: [ mathrm{E}X = 6 left( 1 - frac{11}{2^{12}} text{terms approaching 0 beyond} right) ] 7. Multiplying by 100 games, we get the total expectation: [ text{Total Expectation} = 100 cdot left( 6 - frac{6C_{12}^{6}}{2^{12}} right) ] Upon approximation: [ boxed{465} ] Part (b) In this part, we need to find the probability that envelope number 5 will be played in the next game. 1. Consider a set of 13 envelopes, and each envelope is equally likely to be selected. 2. We define ( xi_k ) to be an indicator variable that is 1 if envelope number ( k ) is played, and 0 otherwise. Therefore, the expectation (mathrm{E}(xi_k)) is the probability that envelope number ( k ) will be played. 3. The sum of all these indicators over one game is the total number of envelopes played, with an expectation of: [ sum_{k=1}^{13} xi_k = 12 cdots (12 text{ approximate number of envelopes playing per game}) ] Given all envelopes are equally likely, [ mathrm{E}(xi_k) = frac{12}{13} text{ probability is the same for each envelope being played in one set} ] Thus, [ text{Probability for envelop #5} = frac{12}{13} ] [ boxed{0.715} ] Therefore, the overall solution gives the following results: [ boxed{a) , 465} ] [ boxed{b) , 0.715} ] This completes the solution for both parts.

answer:Bill started with 50 apples. He sends each of his kids to school with 3 apples for their two favorite teachers, so that's 3 apples per teacher per child. Since he has two children, that's 3 apples x 2 teachers x 2 children = 12 apples given to the teachers. Now, subtract the 12 apples given to the teachers from the initial 50 apples: 50 apples - 12 apples = 38 apples left. Bill has 24 apples left after Jill bakes two apple pies. This means that the apples used for the pies are: 38 apples - 24 apples = 14 apples used for the pies. Since Jill bakes two apple pies with these 14 apples, the number of apples used per pie is: 14 apples ÷ 2 pies = boxed{7} apples per pie.