Aurora

🔍

question:a sample survey of the annual income of 1000 households in a city, the highest annual income is only one household, which is 38,000 yuan. Due to an incorrect data entry, the average annual income of these 1000 households displayed by the computer is 342 yuan higher than the actual average annual income. What is the incorrect data entered into the computer?

answer:# Problem: 某市抽样调查了1000户家庭的年收入，其中年收入最高的只有一户，是38000元。由于将这个数据输入错了，所以计算机显示的这1000户的平均年收入比实际平均年收入高出了342元，则输入计算机的那个错误数据是____元。 1. 设错输入的收入数据为 ( x ) ，正确的收入数据为 38000元。 2. 由于错误数据导致计算机显示的1000户的平均年收入比实际平均年收入高出了342元。我们可用下列公式表示计算机显示的平均年收入： [ text{错误的平均收入} = frac{text{总收入}-38000+x}{1000} ] 其中， - 正确的总收入：所有1000户家庭收入的总和，包含38000元的家庭。 - 错误的总收入：所有1000户家庭收入的总和，换成了错误输入的 ( x ) 元。 3. 实际的平均年收入为： [ text{实际的平均收入} = frac{text{总收入}}{1000} ] 4. 设置误差与实际平均收入的关系： [ text{错误的平均收入} = text{实际的平均收入} + 342 ] 5. 代入设定的错误数据公式： [ frac{text{总收入} - 38000 + x}{1000} = frac{text{总收入}}{1000} + 342 ] 6. 乘以1000消除分母： [ text{总收入} - 38000 + x = text{总收入} + 342000 ] 7. 消掉“总收入”： [ -38000 + x = 342000 ] 8. 解上述方程求 ( x ): [ x = 342000 + 38000 ] 9. 最后得出结果： [ x = 380000 ] 结论：错误的收入数据是 ( boxed{380000} ) 元。

question:What is the least integer greater than sqrt{450}?

answer:To find the least integer that is greater than sqrt{450}, we first look for perfect squares around 450. 1. Calculate squares of integers near to the value sqrt{450}: - 20^2 = 400 - 21^2 = 441 - 22^2 = 484 2. Since sqrt{450} falls between sqrt{441} and sqrt{484}, and because sqrt{441} = 21 and sqrt{484} = 22, sqrt{450} is more than 21 but less than 22. 3. Therefore, the least integer greater than sqrt{450} is boxed{22}.

question:Let the universal set be mathbb{R}, and let the sets A={x|1<x<1} and B={x|xgeq0}. Then, the complement of Acup B in mathbb{R} is ( ) A: {x|0leq x<1} B: {x|xgeq0} C: {x|xleq1} D: {x|x>1}

answer:Since the set A={x|-1<x<1} and B={x|xgeq0}, then Acup B={x|x>-1}, Therefore, the complement of Acup B in mathbb{R} is {x|xleq-1}. Hence, the correct answer is boxed{text{C}}.

question:A triangle ABC is given. Let B_1 be the reflection of B across the line AC , C_1 the reflection of C across the line AB , and O_1 the reflection of the circumcentre of ABC across the line BC . Prove that the circumcentre of AB_1C_1 lies on the line AO_1 . *Proposed by A. Akopyan*

answer:1. **Introduce New Points and Reflections:** - Let ( B_1 ) be the reflection of ( B ) across the line ( AC ). - Let ( C_1 ) be the reflection of ( C ) across the line ( AB ). - Let ( O ) be the circumcenter of ( triangle ABC ). - Let ( O_1 ) be the reflection of ( O ) across the line ( BC ). 2. **Define Midpoints:** - Let ( M ) be the midpoint of ( AC ). - Let ( N ) be the midpoint of ( AB ). - Let ( P ) be the midpoint of ( BC ). - Let ( M_1 ) be the midpoint of ( AC_1 ). - Let ( N_1 ) be the midpoint of ( AB_1 ). 3. **Reflections and Symmetry:** - Let ( O_2 ) be the reflection of ( O ) across ( AC ). - Let ( O_3 ) be the reflection of ( O ) across ( AB ). 4. **Intersection Point:** - Let ( O' ) be the intersection of lines ( O_3M_1 ) and ( O_2N_1 ). 5. **Projections:** - Let ( S ) be the projection of ( O_1 ) onto ( O'M_1 ). - Let ( T ) be the projection of ( O_1 ) onto ( O'N_1 ). 6. **Circumcenter of ( triangle AB_1C_1 ):** - It is evident that ( O' ) is the circumcenter of ( triangle AB_1C_1 ). 7. **Condition for ( O_1 ) on ( AO' ):** - We need to show that ( O_1 ) lies on the line ( AO' ). This is equivalent to: [ frac{mathrm{d}(A;O'O_3)}{mathrm{d}(A;O'O_2)} = frac{mathrm{d}(O_1;O'O_3)}{mathrm{d}(O_1;O'O_2)} ] which simplifies to: [ frac{AM_1}{AN_1} = frac{OS}{OT} ] 8. **Using Similar Triangles:** - Note that ( triangle ABC equiv triangle O_1O_2O_3 ). Therefore: [ frac{O_1O_3}{O_1O_2} = frac{AC}{AB} quad text{(1)} ] 9. **Angle Relationships:** - We also have: [ angle SO_1O_3 = angle (O_3O_1;SO_1) = angle (AC; AC_1) = angle (AB_1; AB) = angle (O_1T; O_1O_2) = angle TO_1O_2 ] Thus: [ cos widehat{SO_1O_3} = cos widehat{TO_1O_2} quad text{(2)} ] 10. **Combining Results:** - From (1) and (2), we get: [ frac{AC}{AB} = frac{O_1O_3}{O_1O_2} cdot frac{cos widehat{SO_1O_3}}{cos widehat{TO_1O_2}} ] Since ( cos widehat{SO_1O_3} = cos widehat{TO_1O_2} ), it follows that: [ frac{AC}{AB} = frac{O_1O_3}{O_1O_2} ] which confirms that ( O_1 ) lies on the line ( AO' ). (blacksquare)