4Number 4Explanationx0x_0x0 is the average of data x1,x2,...,x10x_1, x_2, ..., x_{10}x1,x2,...,x10. If the data changes following the pattern x12+3,x22+5,x32+7\frac{x_1}{2} + 3, \frac{x_2}{2} + 5, \frac{x_3}{2} + 72x1+3,2x2+5,2x3+7, and so on, then the average value becomes...x0+11x_0 + 11x0+11x0+12x_0 + 12x0+1212x0+11\frac{1}{2}x_0 + 1121x0+1112x0+12\frac{1}{2}x_0 + 1221x0+1212x0+21\frac{1}{2}x_0 + 2121x0+21ExplanationThe new data follows the pattern: x12+3,x22+5,x32+7,...\frac{x_1}{2} + 3, \frac{x_2}{2} + 5, \frac{x_3}{2} + 7, ...2x1+3,2x2+5,2x3+7,... The new average can be obtained by: xˉ=(x12+3)+(x22+5)+(x32+7)+...+(x102+(2(10)+1))10\bar{x} = \frac{\left(\frac{x_1}{2} + 3\right) + \left(\frac{x_2}{2} + 5\right) + \left(\frac{x_3}{2} + 7\right) + ... + \left(\frac{x_{10}}{2} + (2(10)+1)\right)}{10}xˉ=10(2x1+3)+(2x2+5)+(2x3+7)+...+(2x10+(2(10)+1)) First, let's find the sum of the arithmetic series 3,5,7,...3, 5, 7, ...3,5,7,.... This sequence has the first term a=3a = 3a=3 and difference b=2b = 2b=2. The 10th term is: U10=3+9(2)=21U_{10} = 3 + 9(2) = 21U10=3+9(2)=21 The sum of the first 10 terms is: S10=102(3+21)=5(24)=120S_{10} = \frac{10}{2}(3 + 21) = 5(24) = 120S10=210(3+21)=5(24)=120 The new average can be written as: xˉ=12(x1+x2+...+x10)+12010=12(x1+x2+...+x10)10+12010=12x0+12\begin{aligned} \bar{x} &= \frac{\frac{1}{2}(x_1 + x_2 + ... + x_{10}) + 120}{10} \\ &= \frac{1}{2}\frac{(x_1 + x_2 + ... + x_{10})}{10} + \frac{120}{10} \\ &= \frac{1}{2}x_0 + 12 \end{aligned}xˉ=1021(x1+x2+...+x10)+120=2110(x1+x2+...+x10)+10120=21x0+12 Thus, the new average is 12x0+12\frac{1}{2}x_0 + 1221x0+12.
4Number 4Explanationx0x_0x0 is the average of data x1,x2,...,x10x_1, x_2, ..., x_{10}x1,x2,...,x10. If the data changes following the pattern x12+3,x22+5,x32+7\frac{x_1}{2} + 3, \frac{x_2}{2} + 5, \frac{x_3}{2} + 72x1+3,2x2+5,2x3+7, and so on, then the average value becomes...x0+11x_0 + 11x0+11x0+12x_0 + 12x0+1212x0+11\frac{1}{2}x_0 + 1121x0+1112x0+12\frac{1}{2}x_0 + 1221x0+1212x0+21\frac{1}{2}x_0 + 2121x0+21ExplanationThe new data follows the pattern: x12+3,x22+5,x32+7,...\frac{x_1}{2} + 3, \frac{x_2}{2} + 5, \frac{x_3}{2} + 7, ...2x1+3,2x2+5,2x3+7,... The new average can be obtained by: xˉ=(x12+3)+(x22+5)+(x32+7)+...+(x102+(2(10)+1))10\bar{x} = \frac{\left(\frac{x_1}{2} + 3\right) + \left(\frac{x_2}{2} + 5\right) + \left(\frac{x_3}{2} + 7\right) + ... + \left(\frac{x_{10}}{2} + (2(10)+1)\right)}{10}xˉ=10(2x1+3)+(2x2+5)+(2x3+7)+...+(2x10+(2(10)+1)) First, let's find the sum of the arithmetic series 3,5,7,...3, 5, 7, ...3,5,7,.... This sequence has the first term a=3a = 3a=3 and difference b=2b = 2b=2. The 10th term is: U10=3+9(2)=21U_{10} = 3 + 9(2) = 21U10=3+9(2)=21 The sum of the first 10 terms is: S10=102(3+21)=5(24)=120S_{10} = \frac{10}{2}(3 + 21) = 5(24) = 120S10=210(3+21)=5(24)=120 The new average can be written as: xˉ=12(x1+x2+...+x10)+12010=12(x1+x2+...+x10)10+12010=12x0+12\begin{aligned} \bar{x} &= \frac{\frac{1}{2}(x_1 + x_2 + ... + x_{10}) + 120}{10} \\ &= \frac{1}{2}\frac{(x_1 + x_2 + ... + x_{10})}{10} + \frac{120}{10} \\ &= \frac{1}{2}x_0 + 12 \end{aligned}xˉ=1021(x1+x2+...+x10)+120=2110(x1+x2+...+x10)+10120=21x0+12 Thus, the new average is 12x0+12\frac{1}{2}x_0 + 1221x0+12.
