Since the Σ( of all colors )= 100%, OR 1, then:
a) P(GREEN ∪ BLUEU) = P(G) + P(BL) = 8%+6% = 14% or 0.14
Since we have to choose ONE candy and only ONE candy at random, then tey are mutually exclusive: No. Choosing a green and blue M&M is possible
b) P(YELLOW ∪ RED) = P(Y) + P(R) = 18%+18% = 36% or 0.36
SAME ANSWER AS BEFORE: mutually exclusive
c) P(NOT PURPLE), Let's calculate 1st, the probability of having a PURPLE:
P(PURPLE) = 21% or 0.21
And the Probability of NOT having a PURPLE is 1-0.21 = 0.79
30 were adults and 10 were children :) just add it up
Step 1: If there is a common factor, factor out the GCF. Step 2<span>: Identify the number of terms: (i) If polynomial has two terms, convert polynomial into difference of two squares or sum of two cubes or difference of two cubes.</span>
Symmetric property is the correct term
Step-by-step explanation:
a) 7!
If there are no restrictions, answer is 7! as it is the permutation of all animals.
b) 4! x 3!
As cats are 6 and Dogs are 5, thus 1st and last must be cats in order to have alternate arrangements. Therefore the only choices are the order of the cats among themselves and the order of the dogs among themselves. There
are 4! permutations of the cats and 3! permutations of the dogs,
so there are a total of 4! x 3! possible arrangements of the suites.
c) 3! x 5!
There are 3! possible arrangements of the dogs among themselves. Now, if we consider the dogs as one ”object” together, then we can think of arranging the 4 cats together with this 1 additional object. There are 5! such arrangements possible, so there are a total of 3! · 5! possible arrangements of the suites.
d) 2 x 4! x 3!
As required that all the cats must be together and all the dogs must be together, either the cats are all before the dogs or the dogs are all before the cats. There are two possible arrangements thus two times of both possibilities is the answer i.e. 2 x 4! x 3!
