Answer:
TU = 19 cm
coz both the triangles are same.
Since the definition of a parallelogram literally says that it's a quadrilateral with 2 parallel sides, we have AB || CD so the 2 other sides (AC and BD) need to be either similar or parallel (either works, due to that if they're similar we can prove that since AB || CD that they're parallel). B is our answer
The answer is 0.6616
Hope it helped :)
Answer: 270 minutes or 4 hours and 30 minutes
Step-by-step explanation: 3x6 =18 and 45 x 6= 270
Answer:
![P(B' \cup A') = P((A \cap B)') = 1-P(A \cap B)= 1-0.32=0.68](https://tex.z-dn.net/?f=P%28B%27%20%5Ccup%20A%27%29%20%3D%20P%28%28A%20%5Ccap%20B%29%27%29%20%3D%201-P%28A%20%5Ccap%20B%29%3D%201-0.32%3D0.68)
See explanation below.
Step-by-step explanation:
For this case we define first some notation:
A= A new training program will increase customer satisfaction ratings
B= The training program can be kept within the original budget allocation
And for these two events we have defined the following probabilities
![P(A) = 0.8, P(B) = 0.2](https://tex.z-dn.net/?f=%20P%28A%29%20%3D%200.8%2C%20P%28B%29%20%3D%200.2)
We are assuming that the two events are independent so then we have the following propert:
![P(A \cap B ) = P(A) * P(B)](https://tex.z-dn.net/?f=%20P%28A%20%5Ccap%20B%20%29%20%3D%20P%28A%29%20%2A%20P%28B%29)
And we want to find the probability that the cost of the training program is not kept within budget or the training program will not increase the customer ratings so then if we use symbols we want to find:
![P(B' \cup A')](https://tex.z-dn.net/?f=%20P%28B%27%20%5Ccup%20A%27%29%20)
And using the De Morgan laws we know that:
![(A \cap B)' = A' \cup B'](https://tex.z-dn.net/?f=%20%28A%20%5Ccap%20B%29%27%20%3D%20A%27%20%5Ccup%20B%27)
So then we can write the probability like this:
![P(B' \cup A') = P((A \cap B)')](https://tex.z-dn.net/?f=%20P%28B%27%20%5Ccup%20A%27%29%20%3D%20P%28%28A%20%5Ccap%20B%29%27%29)
And using the complement rule we can do this:
![P(B' \cup A') = P((A \cap B)')= 1-P(A \cap B)](https://tex.z-dn.net/?f=%20P%28B%27%20%5Ccup%20A%27%29%20%3D%20P%28%28A%20%5Ccap%20B%29%27%29%3D%201-P%28A%20%5Ccap%20B%29)
Since A and B are independent we have:
![P(A \cap B )=P(A)*P(B) =(0.8*0.4) =0.32](https://tex.z-dn.net/?f=%20P%28A%20%5Ccap%20B%20%29%3DP%28A%29%2AP%28B%29%20%3D%280.8%2A0.4%29%20%3D0.32)
And then our final answer would be:
![P(B' \cup A') = P((A \cap B)') = 1-P(A \cap B)= 1-0.32=0.68](https://tex.z-dn.net/?f=P%28B%27%20%5Ccup%20A%27%29%20%3D%20P%28%28A%20%5Ccap%20B%29%27%29%20%3D%201-P%28A%20%5Ccap%20B%29%3D%201-0.32%3D0.68)