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Dovator [93]
3 years ago
6

3-x/4>2 plz help me!!!

Mathematics
1 answer:
Vesnalui [34]3 years ago
7 0
3 -x/4 > 2
⇒ -x/4 > 2-3
⇒ -x/4 > -1
⇒ -x >  -1*4
⇒ -x > -4
⇒ x < 4
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sukhopar [10]

Answer:

The median. 80 because the mean is 80.454545

Step-by-step explanation:

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4 years ago
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Explain how you can use the triangle sum theorem to find the measures of the angles of an equilateral triangle
Rudiy27
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aleksley [76]
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3 years ago
What is the constant of variation for the quadratic variation?
Lorico [155]
Given:
x  =   2     3     4      5     6
y  = 20 ; 45 ; 80; 125;180

The constant variation for the quadratic variation is 5.

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y = 5(2²) = 5(4) = 20
y = 5(3²) = 5(9) = 45
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7 0
4 years ago
What is the probability of these events when we randomly select a permutation of {1,2, ..., n} where n≥4? a) 1 precedes 2 c) 1 i
Paraphin [41]

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.

3 0
3 years ago
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