The result is calculated as 10 multiplied by 8 multiplied by 5, which equals 400.
The distance from point Y to the flag post measures 38.13 m. Step-by-step explanation: Assuming point Y is located at the intersection of both lines shown. Point X is positioned 34 meters east of point Y. The flagpole at point X is observed at a bearing of N18°W, meaning it creates an angle of 18° to the west from the north at point X. Conversely, at point Y, the flagpole has a bearing of N40°E, which makes a 40° angle towards the east from the north.
Considering ∆ AXY as a right triangle, the angle FXY is established. Then, concerning ∆ BYX as another right triangle, the angle FYX is also determined. To find the third angle ∠YFX in triangle FYX, the angle sum property of triangles can be applied:
∠YFX + ∠FYX + ∠FXY = 180°
Thus, we have: ∠YFX + 50° + 72° = 180° leading to ∠YFX = 58°.
Now we can calculate the distance FY using the sine rule.
Answer:
a) 0.00019923%
b) 47.28%
Step-by-step explanation:
a) To determine the likelihood that all sockets in the sample are defective, we can use the following approach:
The first socket is among a group that has 5 defective out of 38, leading to a probability of 5/38.
The second socket is then taken from a group of 4 defective out of 37, following the selection of the first defective socket, resulting in a probability of 4/37.
Extending this logic, the chance of having all 5 defective sockets is computed as: (5/38)*(4/37)*(3/36)*(2/35)*(1/34) = 0.0000019923 = 0.00019923%.
b) Using similar reasoning as in part a, the first socket has a probability of 33/38 of not being defective as it's chosen from a set where 33 sockets are functionally sound. The next socket has a proportion of 32/37, and this continues onward.
The overall probability calculates to (33/38)*(32/37)*(31/36)*(30/35)*(29/34) = 0.4728 = 47.28%.
When there is one table (t=1), you can place 6 chairs (c=6) around it: 2 along the length of each side and 1 at each end.
With t=2, where the tables are positioned end to end (joined at the width), c=10, that means 4 chairs along each side of the joined tables and 1 chair at each end. Each additional table increases the number of chairs by 4, thus we can express this as c=4t+2, with the constant 2 representing the individual chair at each end. If the tables are spread apart, then c=6t.
