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Nataly [62]
2 years ago
5

Which is the constant term of the algebraic expression -2x^2+17-15x+7xy?

Mathematics
1 answer:
andrew11 [14]2 years ago
7 0
The given expression is
-2x² + 17 - 15x + 7xy

The algebraic expression may be written as
-2x² + 7xy - 15x + 17

The constant term does not contain either x or y.
Therefore  17 is the constant term.

Answer: 17

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Is the following relation a function? {(0.3, 0.6), (0.4, 0.8), (0.3, 0.7), (0.5, 0.5)}
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Answer:

no

Step-by-step explanation:

To be a function, the x values must correspond to one and only one y value.

In the coordinates given, (0.3, 0.6) and (0.3, 0.7) have the same x- value.

That means that they cannot be a function.

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A point is ____<br> •imaginary <br> •real
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Read 2 more answers
Suppose you roll a fair die. Let X be the value of the roll. What is the Moment Generating Function of X?
Alex_Xolod [135]

Answer:

μ₁`= 1/6

μ₂=  5/36

Step-by-step explanation:

The rolling of a fair die is described by the binomial distribution, as  the

  1. the probability of success remains constant for all trials, p.
  2. the successive trials are all independent
  3. the experiment is repeated a fixed number of times
  4. there are two outcomes success, p, and failure ,q.

The moment generating function of the binomial distribution is derived as below

M₀(t) = E (e^tx)

        = ∑ (e^tx) (nCx)pˣ (q^n-x)

        = ∑ (e^tx) (nCx)(pe^t)ˣ (q^n-x)

        = (q+pe^t)^n

the expansion of the binomial is purely algebraic and needs not to be interpreted in terms of probabilities.

We get the moments by differentiating the M₀(t) once, twice with respect to t and putting t= 0

μ₁`= E (x) = [ d/dt (q+pe^t)^n]  t= 0

            = np

μ₂`=  E (x)² =[ d²/dt² (q+pe^t)^n]  t= 0

              = np +n(n-1)p²

μ₂=μ₂`-μ₁` =npq

in similar way the higher moments are obtained.

μ₁`=1(1/6)= 1/6

μ₂= 1(1/6)5/6

   = 5/36

4 0
2 years ago
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