1. **Situation 1: Percent error of estimated distance** Given: - Actual distance $D_a = 300$ m - Number of paces $N = 417$ - Pace length factor $L = 0.717$ m/pace Estimated distance $D_e = N \times L = 417 \times 0.717 = 299.089$ m Percent error $= \frac{|D_a - D_e|}{D_a} \times 100 = \frac{|300 - 299.089|}{300} \times 100 = 0.304\%$ 2. **Situation 2: Total area enclosed by rectangular trail** Given: - Longer side paces $N_l = 150$ - Shorter side paces $N_s = 115$ - Pace length factor $L = 0.72$ m/pace Length $= N_l \times L = 150 \times 0.72 = 108$ m Width $= N_s \times L = 115 \times 0.72 = 82.8$ m Area $= \text{Length} \times \text{Width} = 108 \times 82.8 = 8942.4$ m$^2$ 3. **Situation 3: Probable error of the mean** Measurements: 520.62, 519.92, 521.65, 520.98, 521.02 m Mean $\bar{x} = \frac{520.62 + 519.92 + 521.65 + 520.98 + 521.02}{5} = 520.838$ m Calculate deviations: $$d_i = x_i - \bar{x}$$ Squared deviations: $$(520.62 - 520.838)^2 = 0.0475$$ $$(519.92 - 520.838)^2 = 0.845$$ $$(521.65 - 520.838)^2 = 0.659$$ $$(520.98 - 520.838)^2 = 0.020$ $$(521.02 - 520.838)^2 = 0.033$$ Variance $s^2 = \frac{\sum d_i^2}{n-1} = \frac{0.0475 + 0.845 + 0.659 + 0.020 + 0.033}{4} = 0.401$ Standard deviation $s = \sqrt{0.401} = 0.633$ m Probable error $= 0.6745 \times s = 0.6745 \times 0.633 = 0.427$ m 4. **Situation 4: Area of triangle with probable errors** Given: - Base $b = 3.13567 \pm 0.06$ - Altitude $h = 105.48 \pm 0.04$ Area $A = \frac{1}{2} b h = \frac{1}{2} \times 3.13567 \times 105.48 = 165.34$ m$^2$ 5. **Situation 5: Perimeter of rectangular section with probable errors** Given: - Side 1 $a = 450.68 \pm 0.005$ m - Side 2 $b = 274.95 \pm 0.008$ m Perimeter $P = 2(a + b) = 2(450.68 + 274.95) = 2 \times 725.63 = 1451.26$ m 6. **Situation 6: Actual dimensions of rectangular lot with tape correction** Tape length $= 50$ m, found to be 15 mm (0.015 m) too short Correction factor $= \frac{50 + 0.015}{50} = 1.0003$ Recorded length $= 125.56$ m Actual length $= 125.56 \times 1.0003 = 125.598$ m Recorded width $= 100.53$ m Actual width $= 100.53 \times 1.0003 = 100.56$ m 7. **Situation 7: Corrected horizontal distance with tape correction and temperature** Given: - Measured distance $D_m = 250$ m - Tape length $L_t = 50$ m - Actual tape length $L_a = 50.025$ m - Temperature change $\Delta T = 30 - 10 = 20^\circ C$ - Weight $w = 0.02$ Kg/N/m - Pull $P = 63$ N Correction for tape length: $$C_1 = D_m \times \frac{L_a}{L_t} = 250 \times \frac{50.025}{50} = 250.125$$ Thermal expansion coefficient for steel $\alpha = 11 \times 10^{-6} / ^\circ C$ Thermal correction: $$C_2 = C_1 \times \alpha \times \Delta T = 250.125 \times 11 \times 10^{-6} \times 20 = 0.055$$ Correction for sag and pull (approximate): $$C_3 = \frac{w^2 L^2}{24 P^2} \approx 0 \text{ (negligible here)}$$ Corrected distance: $$D_c = C_1 + C_2 = 250.125 + 0.055 = 250.18 \text{ m}$$ 8. **Situation 8: Horizontal distance from slope distance and gradient** Given: - Slope distance $S = 826.75$ m - Gradient $g = 18\% = 0.18$ Horizontal distance $H = \frac{S}{\sqrt{1 + g^2}} = \frac{826.75}{\sqrt{1 + 0.18^2}} = \frac{826.75}{1.0159} = 813.68$ m 9. **Situation 9: Pull correction for steel tape** Given: - Tape length $L = 100$ m - Standard pull $P_s = 40$ Kg - Applied pull $P_a = 35$ Kg - Cross-sectional area $A = 6.18$ mm$^2 = 6.18 \times 10^{-6}$ m$^2$ - Modulus of elasticity $E = 200$ GPa $= 200 \times 10^9$ Pa Pull correction formula: $$\Delta L = \frac{(P_s - P_a) L}{A E}$$ Convert pull difference to Newtons (assuming 1 Kg = 9.81 N): $$\Delta P = (40 - 35) \times 9.81 = 49.05 \text{ N}$$ Calculate elongation: $$\Delta L = \frac{49.05 \times 100}{6.18 \times 10^{-6} \times 200 \times 10^9} = 0.00396 \text{ m} = 3.96 \text{ mm}$$ 10. **Situation 10: Correct length of tape with supports and pull** Given: - Tape length $L = 100$ m - Weight $w = 0.08$ kg/m - Supports at 0, 15, 30, 55, 85, 100 m - Pull $P = 16$ kg Calculate sag correction using standard formula for multiple supports (complex, approximate): Sag correction $\approx \frac{w^2 L^3}{24 P^2}$ Convert weight and pull to Newtons: $$w = 0.08 \times 9.81 = 0.785 \text{ N/m}$$ $$P = 16 \times 9.81 = 156.96 \text{ N}$$ Calculate sag correction: $$C = \frac{(0.785)^2 \times 100^3}{24 \times (156.96)^2} = \frac{0.616 \times 1,000,000}{24 \times 24630} = \frac{616000}{591120} = 1.042 \text{ m}$$ Corrected tape length: $$L_c = L + C = 100 + 1.042 = 101.042 \text{ m}$$ **Graph equation:** Given: $$t_f = \frac{\sqrt{2((105.98 \times 0.06)^2 + \frac{1}{2}(15.62)^2)}}{-7.062}$$ Calculate numerator inside square root: $$105.98 \times 0.06 = 6.3588$$ $$(6.3588)^2 = 40.43$$ $$(15.62)^2 = 243.94$$ Inside root: $$2(40.43 + \frac{1}{2} \times 243.94) = 2(40.43 + 121.97) = 2 \times 162.4 = 324.8$$ Square root: $$\sqrt{324.8} = 18.02$$ Divide by $-7.062$: $$t_f = \frac{18.02}{-7.062} = -2.55$$ **Final answers rounded to three decimals:** 1. Percent error = 0.304% 2. Area = 8942.400 m$^2$ 3. Probable error = 0.427 m 4. Area of triangle = 165.340 m$^2$ 5. Perimeter = 1451.260 m 6. Actual length = 125.598 m, Actual width = 100.560 m 7. Corrected distance = 250.180 m 8. Horizontal distance = 813.680 m 9. Pull correction = 3.960 mm 10. Corrected tape length = 101.042 m Graph value $t_f = -2.550$