Solve the equation for x by finding a,b, and c of the quadratic then apply the quadratic formula,
X= 
The weight average of the coordinates is 4
<h3>How to determine the weight average?</h3>
The given parameters are:
Coordinate 2 has a weight of 2
Coordinate 3 has a weight of 2
Coordinate 10 has a weight of 1
The weight average is then calculated as:
Weight average = Sum of (Weight * Coordinate)/Sum of Weights
So, we have:
Weight average = (2 * 2 + 3 * 2 + 10 * 1)/(2 + 2 + 1)
Evaluate the quotient
Weight average = 4
Hence, the weight average of the coordinates is 4
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Answer:
-12
Step-by-step explanation:
<u>Step 1: Substitute</u>
ab -> -4 for a and 3 for b
(-4) * (3)
<em>-12</em>
Answer: -12
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.
