Answer:
Step-by-step explanation:
The set {1,2,3,4,5,6} has a total of 6! permutations
a. Of those 6! permutations, 5!=120 begin with 1. So first 120 numbers would contain 1 as the unit digit.
b. The next 120, including the 124th, would begin with '2'
c. Then of the 5! numbers beginning with 2, there are 4!=24 including the 124th number, which have the second digit =1
d. Of these 4! permutations beginning with 21, there are 3!=6 including the 124th permutation which have third digit 3
e. Among these 3! permutations beginning with 213, there are 2 numbers with the fourth digit =4 (121th & 122th), 2 with fourth digit 5 (numbers 123 & 124) and 2 with fourth digit 6 (numbers 125 and 126).
Lastly, of the 2! permutations beginning with 2135, there is one with 5th digit 4 (number 123) and one with 5 digit 6 (number 124).
∴ The 124th number is 213564
Similarly reversing the above procedure we can determine the position of 321546 to be 267th on the list.
Answer:
The relation is not a function
The domain is {1, 2, 3}
The range is {3, 4, 5}
Step-by-step explanation:
A relation of a set of ordered pairs x and y is a function if
- Every x has only one value of y
- x appears once in ordered pairs
<u><em>Examples:</em></u>
- The relation {(1, 2), (-2, 3), (4, 5)} is a function because every x has only one value of y (x = 1 has y = 2, x = -2 has y = 3, x = 4 has y = 5)
- The relation {(1, 2), (-2, 3), (1, 5)} is not a function because one x has two values of y (x = 1 has values of y = 2 and 5)
- The domain is the set of values of x
- The range is the set of values of y
Let us solve the question
∵ The relation = {(1, 3), (2, 3), (3, 4), (2, 5)}
∵ x = 1 has y = 3
∵ x = 2 has y = 3
∵ x = 3 has y = 4
∵ x = 2 has y = 5
→ One x appears twice in the ordered pairs
∵ x = 2 has y = 3 and 5
∴ The relation is not a function because one x has two values of y
∵ The domain is the set of values of x
∴ The domain = {1, 2, 3}
∵ The range is the set of values of y
∴ The range = {3, 4, 5}
Answer:
c,-0.45 ,
e. 1.16
g. 5/3
h. √2
Cannot be probability.
Step-by-step explanation:
Probability of any occurrence is always in the interval of 0 to 1. 0 and 1 inclusive.
0≤P(A)≤1
Any value outside this boundaries cannot be probability.
Therefore, -0.45,1.16,5/3 and √2 cannot be probability.
Subtract the second equation from double the first equation:
2x+6y-2x-(-4y)=0
10y=0
Therefore, y=0. Substituting this into the first equation, we see that:
x+3(0)=3
x=3
Therefore, x=3. The solution to this system is x=3, y=0, or (3, 0).
