A box is measured to have a length of L=(10.0+_0.01)FT A WIDTH OF W=(1.534+_0.002)FT AND A HEIGHT OF H=(0.013+_0.001)FT. FIND THE VOLUME OF THE BOX .WRIE THE EXPRESSION FOR V IN THE FOR OF EQUATION 2-3 AND FIND TTHE EXPRESSION FOR SIGN STANDARD DEVIATION V. COMPUTE STANDARD DEVIATION V?
To find the volume of the box, you multiply the length, width, and height. Let's go step by step to find the volume.
Step 1: Calculate the lower and upper bounds for each dimension:
- Length (L): Lower bound = 10.0 - 0.01 = 9.99 ft, Upper bound = 10.0 + 0.01 = 10.01 ft
- Width (W): Lower bound = 1.534 - 0.002 = 1.532 ft, Upper bound = 1.534 + 0.002 = 1.536 ft
- Height (H): Lower bound = 0.013 - 0.001 = 0.012 ft, Upper bound = 0.013 + 0.001 = 0.014 ft
Step 2: Calculate the volume using the upper and lower bounds:
- Lower bound volume = (9.99 ft) * (1.532 ft) * (0.012 ft) = 0.186 ft³
- Upper bound volume = (10.01 ft) * (1.536 ft) * (0.014 ft) = 0.225 ft³
Step 3: Express the volume as an equation:
V = (10.0 ± 0.01) ft * (1.534 ± 0.002) ft * (0.013 ± 0.001) ft
Step 4: Find the expression for the standard deviation of V:
To find the standard deviation, you need to find the range of values for the volume. The range is the difference between the upper bound and lower bound volumes.
Range = Upper bound volume - Lower bound volume
Range = 0.225 ft³ - 0.186 ft³ = 0.039 ft³
Thus, the expression for the standard deviation of V is ±0.039 ft³.
Step 5: Compute the standard deviation of V:
The standard deviation is half of the range. Divide the range by 2.
Standard deviation V = Range / 2
Standard deviation V = 0.039 ft³ / 2 = 0.0195 ft³
Therefore, the standard deviation of V is ±0.0195 ft³.