Answer:
a) 1/2
b) 1/n
c) 1/4
Step-by-step explanation:
a) For each permutation, either 1 precedes 2 or 2 precedes 1. For each permutation in which 1 precedes 2, we can swap 1 and 2 to obtain a permutation in which 2 preceds 1. Thus, half of the total permutations will involve in 1 preceding 2, hence, the probability for a permutation having 1 before 2 is 1/2.
c) If 2 is at the start of the permutation, then it is impossible for 1 to be before 2. If that is not the case, then 1 has a probability of 1/n-1 to be exactly in the position before 2. We can divide in 2 cases using the theorem of total probability,
P( 1 immediately preceds 2) = P (1 immediately precedes 2 | 2 is at position 1) * P(2 is at position 1) + P(1 immediately precedes 2 | 2 is not at position 1) * P(2 is not at position 1) = 0 * 1/n + (1/n-1)*(n-1/n) = 1/n.
d) We can divide the total of permutations in 4 different groups with equal cardinality:
- Those in which n precedes 1 and n-1 precedes 2
- those in which n precedes 1 and 2 precedes n-1
- those in which 1 precedes n and n-1 precedes 2
- those in which 1 precedes n and 2 precedes n-1
All this groups have equal cardinality because we can obtain any element from one group from another by making a permutations between 1 and n and/or 2 and n-1.
This means that the total amount of favourable cases (elements of the first group) are a quarter of the total, hence, the probability of the event is 1/4.
Step-by-step explanation:
The root of an equation is when x=0.
We need to rewrite this as 3x²-4x+10=0.
The quadratic formula is
(-b+or-sqrt(b²-4ac))/2a
so 4+or-sqrt(16-120)/2a
since the sqrt can no be negative this answer will have imaginary numbers.
(4+2i sqrt(26))6 and (4-2i sqrt(26))6
simplified is
x=(2+i sqrt(26))/3 and x=(2-i sqrt(26))/3
Answer:
E because he is 9/16 standard deviations away from the mean, which is an improvement from 1/10 standard deviations
Hope this helped and brainliest please
Answer:
20%×X=30
x=150
Step-by-step explanation:
30÷20%=150
20%×150=30
