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3 years ago

What is the midpoint of a segment in the complex plane with endpoints at 6 -2i and -4 + 6i

Mathematics

2 answers:

Katyanochek1 [597]3 years ago

7 0

Answer: 2 + 4 i

Step-by-step explanation:

Hi, to solve this we have to apply the next expression:

(a1 +a2)/ 2 + (b1 +b2 )/2 i=

Where a is the real part, and b is the imaginary part (with i)

For example, for our case:

6 -2i , 6 is the real part (2) and -2 is the imaginary part (b)

Replacing with the values given

(6 -4) /2+ (-2 +6) /2 i = 2 + 4 i

Feel free to ask for more if needed or if you did not understand something.

alexdok [17]3 years ago

5 0

Answer:

Midpoint of a segment in the complex plane with endpoints at 6 -2i and -4 + 6i is:

1+2i

Step-by-step explanation:

The midpoint of a segment in the complex plane with endpoints at 6 -2i and -4 + 6i is:

$\dfrac{6-2i-4+6i}{2} \\\\=\dfrac{2+4i}{2}\\ \\=1+2i$

Hence, midpoint of a segment in the complex plane with endpoints at 6 -2i and -4 + 6i is:

1+2i

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15 times what equals 45

Olenka [21]

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3 years ago

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Exercise 5.2. Suppose that X has moment generating function

soldi70 [24.7K]

Answer:

a) Mean, E(X) = - 0.5

Variance = = 9.25

b) $M_{X}(t)=E(e^{tX})$

⇒ $=\sum e^{tx}p(x)$

⇒ $=\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}$

Step-by-step explanation:

Given:

moment generating function of X as:

MX(t) = $\frac{1}{2} + \frac{1}{3}e^{-4t} + \frac{1}{6} e^{5t}$

a) Now

Mean, E(X) = $M_{X}'(t=0)$

Thus,

$M_{X}'(t)=\frac{1}{3}(-4)e^{-4t}+\frac{1}{6}(5)e^{5t}$

$M_{X}'(t)=\frac{-4}{3}e^{-4t}+\frac{5}{6}e^{5t}$

also,

$E(X^{2})=M_{X}''(t=0)$

Thus,

$M_{X}''(t)=\frac{-4}{3}(-4)e^{-4t}+\frac{5}{6}(5)e^{5t}$

$M_{X}''(t)=\frac{16}{3}e^{-4t}+\frac{25}{6}e^{5t}$

Therefore,

Mean, E(X) = $M_{X}'(t=0)=\frac{-4}{3}e^{-4(0)}+\frac{5}{6}e^{5(0)}$

Mean, E(X) = - 0.5

and

$E(X^{2})=M_{X}''(t=0)=\frac{16}{3}e^{-4(0)}+\frac{25}{6}e^{5(0)}$

$E(X^{2})$ = 9.5

also,

Variance(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

b) Now,

Let f(x) be the PMF of X

Thus,

$M_{X}(t)=E(e^{tX})$

⇒ $=\sum e^{tx}p(x)$

⇒ $=\frac{1}{2}+\frac{1}{3}e^{-4t}+\frac{1}{6}e^{5t}$

Therefore,

at x = 0, P(x) = $\frac{1}{2}$

at x= - 4 ,P(x) = $\frac{1}{3}$

at x = 5, P(x) = $\frac{1}{6}$

Thus,

E(X) = $\sum xP(x)=0(\frac{1}{2})+(-4)(\frac{1}{3})+5(\frac{1}{6})$

E(X) = - 0.5

also, $E(X^{2})=\sum x^{2}P(x)=0^{2}(\frac{1}{2})+(-4)^{2}(\frac{1}{3})+5^{2}(\frac{1}{6})$

$E(X^{2})$ = 9.5

Hence,

Var(X) = E(X²) - E(X)²

⇒ 9.5 - (-0.5)²

= 9.25

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