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dusya [7]
3 years ago
9

Which relationship describes angles 1 and 2?

Mathematics
2 answers:
umka2103 [35]3 years ago
8 0
I belive its B,and D,
B because, the lines are supplementary(because supplementary means it equal 180 degrees in length, which is those two angles.)
D, because they are put in such an angle in which making them adjecant.
Hope I helped:-) Bye bye.
Feliz [49]3 years ago
5 0

1) supplementary angles => sum of angles' measure is 180°

Two lines intersecting creating four angles with the left obtuse angle labeled one and the acute angle to the right labeled two

= > angle 1 +  angle 2 = 180°

Then those two match.

2) complementary angles => sum of the two angles = 90°

Two lines intersecting in a right angle with a square indicating a right angle in quadrant one and a ray splitting quadrant two with a two labeling the left angle and one labeling the angle on the right

=> angle 2 + angle 1 = 90°

Then those two match

3) vertical angles

Two parallel horizontal lines intersected by a vertical line with a square indicating a right angle in quadrant one of the top line and a one labeling the angle in quadrant two and two labeling quadrant four of the top line

Those two matches because the angle label 1 and the angle label 2 are vertical angles as per the definittion.

4) adjacent angles

Two lines intersecting in a V shape with the left angle of one hundred fifty seven degrees and a bottom angle of X degrees

Those two match becasue the angle of 157° and the angle of X° are adjacent.

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Answer:

a) Mean, E(X) = - 0.5

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

or

⇒ =\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

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or

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

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

Thus,

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

or

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

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Variance(X) = E(X²) - E(X)²

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Thus,

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

or

⇒ =\sum e^{tx}p(x)

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

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Thus,

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

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Var(X) = E(X²) - E(X)²

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