Answered Essay: Call the error in the measurement of the edge of the cube Delta X = 0.02 cm. Then the maximum

Call the error in the measurement of the edge of the cube AX -0.02 cm Then the maximum possible diameter is dma,-dact + Δ And the minimum possible diameter is dmin dact AX estion 3 A) Using dmax and dmin calculate both Vmax and Vmin B) Calculate the possible range in volume: Vmax- Vmin Vact was your answer to question 1 Range as % Actual = Vmax-Vinn Vminx 100 Vact C) Calculate the Range as % Actual of the volume of the cube. Enter you answer for C) Enter a numeric answer only, do not include the % sign Enter your answer using TWO decimal places Enter Your Answer 6 3 5 6

Call the error in the measurement of the edge of the cube Delta X = 0.02 cm. Then the maximum possible diameter is d_max = d_act + Delta X And the minimum possible diameter is d_min = d_act – Delta X A) Using d_max and d_min calculate both V_max and V_min B) Calculate the possible range in volume: V_max – V_min V_act was your answer to question 1 Range as % Actual = V_max – V_min/V_act times 100 C) Calculate the Range as % Actual of the volume of the cube. Enter you answer for C) Enter a numeric answer only, do not include the % sign Enter your answer using TWO decimal places

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