This is a great question!
To determine the probability with which two sweets are not the same, you would have to subtract the probability with which two sweets are the same from 1. That would only be possible if she chose 2 liquorice sweets, 5 mint sweets and 3 humburgs -

As you can see, the first time you were to choose a Liquorice, there would be 12 out of the 20 sweets present. After taking that out however, there would be respectively 11 Liquorice out of 19 remaining. Apply the same concept to each of the other sweets -

____
Calculate the probability of drawing 2 of each, add them together and subtract from one to determine the probability that two sweets will not be the same type of sweet!

<u><em>Thus, the probability should be 111 / 190</em></u>
Answer:
A, C, E
Step-by-step explanation:
As long as all the 3 letters(D, K, and M) are included, it should be a valid name. These are the 3 points in the triangle afterall.
Answer:
Step-by-step explanation:
a).
= 
= 
=
[Since, i =
]
b).
= 
= 
= 5 ± 2i [Since, i =
]
c).
= 
= 
= 
=
[Since, i =
]
The answer is III only, or D.
We can start to solve this by knowing what the HL theorem means. The HL theorem, like its name implies, shows says that if a hypotenuse and leg of a triangle are congruent to the hypotenuse and leg of a different triangle, then the triangles are congruent. The only triangle that we see a hypotenuse congruent in is in figure III. In figure II, those congruent sides are both legs while in figure I we just see 2 congruent angles. Now in figure III, we can also see that two legs are congruent because of the reflexive property. That means that the answer is III, or D.
5≤x≤9 it can also be reversed as 9≥x≥5 which is the same thing