1. **Fit a straight line $y = a + bx$ using least squares for the data:** $x$: 2, 4, 6, 8, 10 $y$: 50, 55, 65, 70, 80 2. Calculate sums needed: $$\sum x = 2 + 4 + 6 + 8 + 10 = 30$$ $$\sum y = 50 + 55 + 65 + 70 + 80 = 320$$ $$\sum xy = 2\times50 + 4\times55 + 6\times65 + 8\times70 + 10\times80 = 100 + 220 + 390 + 560 + 800 = 2070$$ $$\sum x^2 = 2^2 + 4^2 + 6^2 + 8^2 + 10^2 = 4 + 16 + 36 + 64 + 100 = 220$$ Number of points $n=5$. 3. Use formulas for slope $b$ and intercept $a$: $$b = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} = \frac{5\times2070 - 30\times320}{5\times220 - 30^2} = \frac{10350 - 9600}{1100 - 900} = \frac{750}{200} = 3.75$$ $$a = \frac{\sum y - b \sum x}{n} = \frac{320 - 3.75 \times 30}{5} = \frac{320 - 112.5}{5} = \frac{207.5}{5} = 41.5$$ 4. The fitted line is: $$y = 41.5 + 3.75x$$ 5. Estimate $y$ when $x=7.5$: $$y = 41.5 + 3.75 \times 7.5 = 41.5 + 28.125 = 69.625$$ --- 6. **Newton’s Divided-Difference Interpolating Polynomial for:** $x$: 2, 3, 4, 5, 6, 7 $f(x)$: 16, 81, 256, 625, 1296, 2401 7. Construct divided difference table: \begin{align*} \text{Order 0 (given)}: &\quad f[2]=16, f[3]=81, f[4]=256, f[5]=625, f[6]=1296, f[7]=2401 \\ \text{First differences}: &\quad f[3,2]=\frac{81-16}{3-2}=65, \, f[4,3]=\frac{256-81}{4-3}=175, \, f[5,4]=\frac{625-256}{5-4}=369, \, f[6,5]=\frac{1296-625}{6-5}=671, \, f[7,6]=\frac{2401-1296}{7-6}=1105 \\ \text{Second differences}: &\quad f[4,3,2]=\frac{175-65}{4-2}=\frac{110}{2}=55, \, f[5,4,3]=\frac{369-175}{5-3}=\frac{194}{2}=97, \, f[6,5,4]=\frac{671-369}{6-4}=\frac{302}{2}=151, \, f[7,6,5]=\frac{1105-671}{7-5}=\frac{434}{2}=217 \\ \text{Third differences}: &\quad f[5,4,3,2]=\frac{97-55}{5-2}=\frac{42}{3}=14, \, f[6,5,4,3]=\frac{151-97}{6-3}=\frac{54}{3}=18, \, f[7,6,5,4]=\frac{217-151}{7-4}=\frac{66}{3}=22 \\ \text{Fourth differences}: &\quad f[6,5,4,3,2]=\frac{18-14}{6-2}=\frac{4}{4}=1, \, f[7,6,5,4,3]=\frac{22-18}{7-3}=\frac{4}{4}=1 \end{align*} 8. Newton polynomial form: $$P(x) = f[2] + f[3,2](x-2) + f[4,3,2](x-2)(x-3) + f[5,4,3,2](x-2)(x-3)(x-4) + f[6,5,4,3,2](x-2)(x-3)(x-4)(x-5)$$ Substitute values: $$P(x) = 16 + 65(x-2) + 55(x-2)(x-3) + 14(x-2)(x-3)(x-4) + 1(x-2)(x-3)(x-4)(x-5)$$ 9. Estimate $f(4.5)$: Calculate each term: $$x-2 = 4.5 - 2 = 2.5$$ $$x-3 = 4.5 - 3 = 1.5$$ $$x-4 = 4.5 - 4 = 0.5$$ $$x-5 = 4.5 - 5 = -0.5$$ Calculate polynomial: $$P(4.5) = 16 + 65 \times 2.5 + 55 \times 2.5 \times 1.5 + 14 \times 2.5 \times 1.5 \times 0.5 + 1 \times 2.5 \times 1.5 \times 0.5 \times (-0.5)$$ $$= 16 + 162.5 + 206.25 + 26.25 - 0.9375 = 410.0625$$ **Final answers:** - Least squares line: $y = 41.5 + 3.75x$ - Estimated $y$ at $x=7.5$ is $69.625$ - Newton interpolating polynomial estimate at $x=4.5$ is approximately $410.06$