Answer:
(2,8)
Step-by-step explanation:
Answer:
![P( A \cap B) = P(A) *P(B) = 0.34*0.32 = 0.1088](https://tex.z-dn.net/?f=%20P%28%20A%20%5Ccap%20B%29%20%3D%20P%28A%29%20%2AP%28B%29%20%3D%200.34%2A0.32%20%3D%200.1088)
And then replacing in the total probability formula we got:
![P(A \cup B) = 0.34+0.32 - 0.1088 = 0.5512](https://tex.z-dn.net/?f=%20P%28A%20%5Ccup%20B%29%20%3D%200.34%2B0.32%20-%200.1088%20%3D%200.5512)
And rounded we got ![P(A \cup B ) = 0.551](https://tex.z-dn.net/?f=%20P%28A%20%5Ccup%20B%20%29%20%3D%200.551)
That represent the probability that it rains over the weekend (either Saturday or Sunday)
Step-by-step explanation:
We can define the following notaton for the events:
A = It rains over the Saturday
B = It rains over the Sunday
We have the probabilities for these two events given:
![P(A) = 0.34 , P(B) = 0.32](https://tex.z-dn.net/?f=%20P%28A%29%20%3D%200.34%20%2C%20P%28B%29%20%3D%200.32)
And we are interested on the probability that it rains over the weekend (either Saturday or Sunday), so we want to find this probability:
![P(A \cup B)](https://tex.z-dn.net/?f=%20P%28A%20%5Ccup%20B%29)
And for this case we can use the total probability rule given by:
![P(A \cup B) = P(A) + P(B) - P(A \cap B)](https://tex.z-dn.net/?f=P%28A%20%5Ccup%20B%29%20%3D%20P%28A%29%20%2B%20P%28B%29%20-%20P%28A%20%5Ccap%20B%29)
And since we are assuming the events independent we can find the probability of intersection like this:
![P( A \cap B) = P(A) *P(B) = 0.34*0.32 = 0.1088](https://tex.z-dn.net/?f=%20P%28%20A%20%5Ccap%20B%29%20%3D%20P%28A%29%20%2AP%28B%29%20%3D%200.34%2A0.32%20%3D%200.1088)
And then replacing in the total probability formula we got:
![P(A \cup B) = 0.34+0.32 - 0.1088 = 0.5512](https://tex.z-dn.net/?f=%20P%28A%20%5Ccup%20B%29%20%3D%200.34%2B0.32%20-%200.1088%20%3D%200.5512)
And rounded we got ![P(A \cup B ) = 0.551](https://tex.z-dn.net/?f=%20P%28A%20%5Ccup%20B%20%29%20%3D%200.551)
That represent the probability that it rains over the weekend (either Saturday or Sunday)
Answer:
a^2-1
Step-by-step explanation:
(a+1)(a-1)
=a^2-1
We know:
![|a|= \left\{\begin{array}{ccc}a&if\ a \ \textgreater \ 0\\0&if\ a=0\\-a&if\ a \ \textless \ 0\end{array}\right\\\\|2|=2\\|0|=0\\|-3|=3](https://tex.z-dn.net/?f=%7Ca%7C%3D%20%20%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Da%26if%5C%20a%20%5C%20%5Ctextgreater%20%5C%20%200%5C%5C0%26if%5C%20a%3D0%5C%5C-a%26if%5C%20a%20%5C%20%5Ctextless%20%5C%20%200%5Cend%7Barray%7D%5Cright%5C%5C%5C%5C%7C2%7C%3D2%5C%5C%7C0%7C%3D0%5C%5C%7C-3%7C%3D3)
![|a+b|+|c|](https://tex.z-dn.net/?f=%7Ca%2Bb%7C%2B%7Cc%7C)
substitute a = -3; b = 7; c = -15