Storm

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question:6 successive games, a baseball team scored a certain number of runs once, 4 runs twice, and 5 runs 3 times. The average (arithmetic mean) number of runs the team scored per game over the 6game period was 4. How many runs did the team score in the game with the lowest number of runs?

answer:Let's denote the number of runs scored in the game with the lowest number of runs as x. The total number of runs scored in the 6 games can be calculated as follows: - The team scored x runs once. - The team scored 4 runs twice, which is 4 * 2 = 8 runs. - The team scored 5 runs three times, which is 5 * 3 = 15 runs. The sum of all the runs scored in the 6 games is x + 8 + 15. The average number of runs scored per game is the total number of runs divided by the number of games. We know the average is 4 runs per game over 6 games, so the total number of runs scored over the 6 games is 4 * 6 = 24 runs. Now we can set up the equation: x + 8 + 15 = 24 Combine the constant terms: x + 23 = 24 Subtract 23 from both sides to solve for x: x = 24 - 23 x = 1 Therefore, the team scored boxed{1} run in the game with the lowest number of runs.

question:Given the function f(x)=left{ begin{array}{ll} 2^{x+1}, & xleq 0, 1log_{2}x, & x > 0, end{array} right. then f(f(3))= ( ) A: dfrac{4}{3} B: dfrac{2}{3} C: dfrac{4}{3} D: 3

answer:**Analysis** This question tests the evaluation of a piecewise function. According to the given function expression, it can be solved by evaluating each piece separately. **Solution** Solution: f(f(3))=f(1-log_{2}3)=2^{1-log_{2}3+1}= dfrac{2^{2}}{3}= dfrac{4}{3}, Therefore, the correct choice is boxed{A}.

question:Jo adds up all the positive integers from 1 to 50. Kate does a similar thing with the first 50 positive integers; however, she first rounds every integer to its nearest multiple of 10 (rounding 5s up) and then adds the 50 values. What is the positive difference between Jo's sum and Kate's sum?

answer:To solve this problem, let's break it down step by step, closely following the logic provided in the standard solution: 1. **Understanding the Rounding Process for Each Group of Ten Numbers:** For each group of ten numbers, such as 1, 2, 3, ldots, 10, we analyze the rounding process as Kate would do it: - The first four numbers (1, 2, 3, 4) are rounded down to 0, which decreases the sum by 1 + 2 + 3 + 4 = 10. - The next six numbers (5, 6, 7, 8, 9, 10) are rounded up to 10. The increase in the sum due to this rounding is calculated by considering how much each number is increased to reach 10: 10 - 5 = 5, 10 - 6 = 4, 10 - 7 = 3, 10 - 8 = 2, 10 - 9 = 1, and 10 - 10 = 0. Adding these differences gives us 5 + 4 + 3 + 2 + 1 + 0 = 15. 2. **Calculating the Net Change for Each Group of Ten Numbers:** The net change in Kate's sum for each group of ten numbers, compared to Jo's sum, is the sum of the decrease and the increase: -10 + 15 = 5. 3. **Applying the Logic to All Groups of Ten Numbers from 1 to 50:** Since the numbers 1 to 50 can be divided into five groups of ten numbers (i.e., 1-10, 11-20, 21-30, 31-40, 41-50), and each group contributes a net change of 5 to Kate's sum compared to Jo's sum, we multiply this net change by the number of groups: 5 times 5 = 25. 4. **Conclusion:** Therefore, the positive difference between Jo's sum and Kate's sum is boxed{25}.

question:Calculate: 1+i^2+i^3+ldots+i^{10} (where i represents the imaginary unit).

answer:We have i+i^2+i^3+ldots+i^{10} = (i+i^2+i^3+i^4)+(i^5+i^6+i^7+i^8)+i^9+i^{10}. This simplifies to (i-1-i+1)+(i-1-i+1)+i-1. Therefore, the result is -1+i. Hence, the answer is boxed{-1+i}.