Answer:
I'm pretty sure that they are 4, 16, 64, and 256.
Step-by-step explanation:
Answer:
A)
Step-by-step explanation:
Given expression:
To factor the given expression completely.
Solution:
In order to factor the expression, we will factor in pairs.
We will factor the G.C.F of the terms in the pairs.
G.C.F. of and can be given as:
Thus, G.C.F. =
G.C.F. of and can be given as:
Thus, G.C.F. =
The expression after factoring the G.C.F. pairs is given as:
Taking G.C.F. of the whole expression as is a common term.
The expression is completely factored.
Step-by-step explanation:
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)
Answer: This statement isn't true.
Step-by-step explanation: By finding like denominators, you can add the fractions on each side. Then, compare by cross multiplying.