a. The events are mutually exclusive, so
![P(A\cup B\cup C)=P(A)+P(B)+P(C)=\boxed{0.9}](https://tex.z-dn.net/?f=P%28A%5Ccup%20B%5Ccup%20C%29%3DP%28A%29%2BP%28B%29%2BP%28C%29%3D%5Cboxed%7B0.9%7D)
b. For the same reason,
![P(A\cap B\cap C)=\boxed0](https://tex.z-dn.net/?f=P%28A%5Ccap%20B%5Ccap%20C%29%3D%5Cboxed0)
Another way to see this is to use the result from part (a), and the inclusion/exclusion principle:
![P(A\cap B\cap C)=P(A)+P(B)+P(C)-P(A\cup B\cup C)](https://tex.z-dn.net/?f=P%28A%5Ccap%20B%5Ccap%20C%29%3DP%28A%29%2BP%28B%29%2BP%28C%29-P%28A%5Ccup%20B%5Ccup%20C%29)
and the right side reduces to 0.
c. Same as (b),
![P(A\cap B)=\boxed0](https://tex.z-dn.net/?f=P%28A%5Ccap%20B%29%3D%5Cboxed0)
d. Same as (b),
![P((A\cup B)\cap C)=\boxed0](https://tex.z-dn.net/?f=P%28%28A%5Ccup%20B%29%5Ccap%20C%29%3D%5Cboxed0)
We can also use the distributivity rule for unions and intersections to write
![P((A\cup B)\cap C)=P((A\cap C)\cup(B\cap C))=P(A\cap C)+P(B\cap C)=0](https://tex.z-dn.net/?f=P%28%28A%5Ccup%20B%29%5Ccap%20C%29%3DP%28%28A%5Ccap%20C%29%5Ccup%28B%5Ccap%20C%29%29%3DP%28A%5Ccap%20C%29%2BP%28B%5Ccap%20C%29%3D0)
e. If the
is the complement of
, then by DeMorgan's law,
![P(A'\cap B'\cap C')=P(A\cup B\cup C)'=1-P(A\cup B\cup C)=\boxed{0.1}](https://tex.z-dn.net/?f=P%28A%27%5Ccap%20B%27%5Ccap%20C%27%29%3DP%28A%5Ccup%20B%5Ccup%20C%29%27%3D1-P%28A%5Ccup%20B%5Ccup%20C%29%3D%5Cboxed%7B0.1%7D)
In order to answer this one, it's really really really really helpful if you know what the law of cosines says. In fact it's absolutely necessary.
The law of cosines says if you know two sides of a triangle and the angle between them then you can use that information to find the length of the third side.
In the picture you know the lengths of two sides and you know the angle between them. So you can use the law of cosines to find the length of the third sidethat. That's side AC.
Step-by-step explanation:
Remember when expanding radicals,
![\sqrt{ab} = \sqrt{a} \times \sqrt{b}](https://tex.z-dn.net/?f=%20%5Csqrt%7Bab%7D%20%20%3D%20%20%5Csqrt%7Ba%7D%20%20%5Ctimes%20%20%5Csqrt%7Bb%7D%20)
When expanding radicals into two radicals, we don't let our radicand have two negative answers.
![i = \sqrt{ - 1}](https://tex.z-dn.net/?f=i%20%20%3D%20%20%5Csqrt%7B%20-%201%7D%20)
![{i}^{2} = - 1](https://tex.z-dn.net/?f=%20%7Bi%7D%5E%7B2%7D%20%20%3D%20%20-%201)
We don't do this
![(\sqrt{ - 1} ) {}^{2} = \sqrt{ - 1 \times - 1}](https://tex.z-dn.net/?f=%20%28%5Csqrt%7B%20-%201%7D%20%29%20%7B%7D%5E%7B2%7D%20%20%3D%20%20%5Csqrt%7B%20-%201%20%5Ctimes%20%20-%201%7D%20)
![- 1 = \sqrt{1}](https://tex.z-dn.net/?f=%20-%201%20%3D%20%20%5Csqrt%7B1%7D%20)
![- 1 = 1](https://tex.z-dn.net/?f=%20-%201%20%3D%201)
The number with the same value as 32 tens is 320.
Answer:
9$ per lawn
Step-by-step explanation: