What is the volume of a rectangular prism with the dimensions 7m 8m and 9m
1 answer:
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Answer:
Given the data in the question;
Samples of size 100: 6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6, 10, 9, 2, 8, 4, 8, 10, 10, 8, 7, 7, 7, 6, 14, 18, 13, 6.
a)
For a p chart ( control chart for fraction nonconforming), the center line and upper and lower control limits are;
UCL = p" + 3√[ (p"(1-P")) / n ]
CL = p"
LCL = p" - 3√[ (p"(1-P")) / n ]
here, p" is the average fraction defective
Now, with the 30 samples of size 100
p" = [∑(6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6, 10, 9, 2, 8, 4, 8, 10, 10, 8, 7, 7, 7, 6, 14, 18, 13, 6.)] / [ 30 × 100 ]
p" = 234 / 3000
p" = 0.078
so the trial control limits for the fraction-defective control chart are;
UCL = p" + 3√[ (p"(1-P")) / n ]
UCL = 0.078 + 3√[ (0.078(1-0.078)) / 100 ]
UCL = 0.078 + ( 3 × 0.026817 )
UCL = 0.078 + 0.080451
UCL = 0.1585
LCL = p" - 3√[ (p"(1-P")) / n ]
LCL = 0.078 - 3√[ (0.078(1-0.078)) / 100 ]
LCL = 0.078 - ( 3 × 0.026817 )
LCL = 0.078 - 0.080451
LCL = 0 ( SET TO ZERO )
Diagram of the Chart uploaded below
b)
from the p chart for a) below, sample 28 violated the first western electric rule,
summary report from Minitab;
TEST 1. One point more than 3.00 standard deviations from the center line.
Test failed at points: 28
Hence, we conclude that the process is out of statistical control
Lets Assume that assignable causes can be found to eliminate out of control points.
Since 28 is out of control, we should eliminate this sample and recalculate the trial control limits for the P chart.
so
p" = 0.0745
UCL = p" + 3√[ (p"(1-P")) / n ]
UCL = 0.0745 + 3√[ (0.0745(1-0.0745)) / 100 ]
UCL = 0.0745 + ( 3 × 0.026258 )
UCL = 0.0745 + 0.078774
UCL = 0.1532
LCL = p" - 3√[ (p"(1-P")) / n ]
LCL = 0.0745 - 3√[ (0.0745(1-0.0745)) / 100 ]
LCL = 0.0745 - ( 3 × 0.026258 )
LCL = 0.0745 - 0.078774
UCL = 0 ( SET TO ZERO )
The second p chart diagram is upload below;
NOTE; the red circle symbol on 28 denotes that the point is not used in computing the control limits
The value of the angle subtended by the distance of the buoy from the
port is given by sine and cosine rule.
- The bearing of the buoy from the is approximately <u>307.35°</u>
Reasons:
Location from which the ship sails = Port
The speed of the ship = 10 mph
Direction of the ship = N35°W
Location of the warning buoy = 5 miles north of the port
Required: The bearing of the warning buoy from the ship after 7.5 hours.
Solution:
The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles
By cosine rule, we have;
a² = b² + c² - 2·b·c·cos(A)
Where;
a = The distance between the ship and the buoy
b = The distance between the ship and the port = 75 miles
c = The distance between the buoy and the port = 5 miles
Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°
Which gives;
a = √(75² + 5² - 2 × 75 × 5 × cos(35°))
By sine rule, we have;
Therefore;
Which gives;
Similarly, we can get;
The angle subtended by the distance of the buoy from the port, <em>C</em> is therefore;
C ≈ 180° - 142.68° - 35° ≈ 2.32°
By alternate interior angles, we have;
The bearing of the warning buoy as seen from the ship is therefore;
Bearing of buoy ≈ 270° + 35° + 2.32° ≈ <u>307.35°</u>
Learn more about bearing in mathematics here: