Answer:
The answer for Factorization is (x+6)(x-2) = 0 or Solving for x is 2 and -6.
Step-by-step explanation:
x² - 12 = -4x
Factorisation :
x² + 4x - 12 = 0
x² - 2x + 6x - 12 = 0
x(x-2)x + 6(x-2) = 0
(x+6)(x-2) = 0
Solve for x :
(x+6)(x-2) = 0
x + 6 = 0
x = -6
x - 2 = 0
x = 2
Answer:








Step-by-step explanation:
Given

Convert the probability to decimal

Solving (a): P(ABC)
This is calculated as:

This gives:


Solving (b): 
This is calculated as:

In probability:

So, we have:




Solving (c): 
This is calculated as:

![P(AB^cC) = P(A) * [1 - P(B)] * P(C)](https://tex.z-dn.net/?f=P%28AB%5EcC%29%20%3D%20P%28A%29%20%2A%20%5B1%20-%20P%28B%29%5D%20%2A%20P%28C%29)



Solving (d): 
This is calculated as:

![P(A^cBC) = [1-P(A)] *P(B) * P(C)](https://tex.z-dn.net/?f=P%28A%5EcBC%29%20%3D%20%5B1-P%28A%29%5D%20%2AP%28B%29%20%2A%20P%28C%29)



Solving (e): 
This is calculated as:

![P(AB^cC^c) = P(A) * [1-P(B)] * [1-P(C)]](https://tex.z-dn.net/?f=P%28AB%5EcC%5Ec%29%20%20%3D%20P%28A%29%20%2A%20%5B1-P%28B%29%5D%20%2A%20%5B1-P%28C%29%5D)
![P(AB^cC^c) = 0.48 * [1-0.48] * [1-0.48]](https://tex.z-dn.net/?f=P%28AB%5EcC%5Ec%29%20%20%3D%200.48%20%2A%20%5B1-0.48%5D%20%2A%20%5B1-0.48%5D)


Solving (f): 
This is calculated as:

![P(A^cBC^c) = [1-P(A)] * P(B) * [1-P(C)]](https://tex.z-dn.net/?f=P%28A%5EcBC%5Ec%29%20%20%20%3D%20%5B1-P%28A%29%5D%20%2A%20P%28B%29%20%2A%20%5B1-P%28C%29%5D)
![P(A^cBC^c) = [1-0.48] * 0.48 * [1-0.48]](https://tex.z-dn.net/?f=P%28A%5EcBC%5Ec%29%20%20%20%3D%20%5B1-0.48%5D%20%2A%200.48%20%2A%20%5B1-0.48%5D)


Solving (g): 
This is calculated as:

![P(A^cB^cC) = [1-P(A)] * [1-P(B)] * P(C)](https://tex.z-dn.net/?f=P%28A%5EcB%5EcC%29%20%20%3D%20%5B1-P%28A%29%5D%20%2A%20%5B1-P%28B%29%5D%20%2A%20P%28C%29)
![P(A^cB^cC) = [1-0.48] * [1-0.48] * 0.48](https://tex.z-dn.net/?f=P%28A%5EcB%5EcC%29%20%20%3D%20%5B1-0.48%5D%20%2A%20%5B1-0.48%5D%20%2A%200.48)


Solving (h): 
This is calculated as:

![P(A^cB^cC^c) = [1-P(A)] * [1-P(B)] * [1-P(C)]](https://tex.z-dn.net/?f=P%28A%5EcB%5EcC%5Ec%29%20%20%3D%20%5B1-P%28A%29%5D%20%2A%20%5B1-P%28B%29%5D%20%2A%20%5B1-P%28C%29%5D)
![P(A^cB^cC^c) = [1-0.48] * [1-0.48] * [1-0.48]](https://tex.z-dn.net/?f=P%28A%5EcB%5EcC%5Ec%29%20%20%3D%20%5B1-0.48%5D%20%2A%20%5B1-0.48%5D%20%2A%20%5B1-0.48%5D)


From the information, the mean temperature for January is most likely lower than the mean temperature for March.
There are a couple of clues to this assumption. First, the max temp is lower for January. Second, the interquartile range is lower in March. With both of these factors, it is safe to assume the mean is lower.
Answer:
1.92636657×10^-3
Step-by-step explanation:
This question looks quite awful coz it might be (16+81)^3/4 rather than being (16/81)^3/4....But, if it is (16/81)^3/4 the answer is 1.92636657×10^-3
hope you find it helpful....Thanks very much...