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3 years ago

I need help badly with this Algebra question, if someone could explain how to solve this It'd really help.

Mathematics

1 answer:

Alex3 years ago

7 0

Answer:

The answer is A, or the top left function.

Step-by-step explanation:

Use the vertical line test. If you can draw a straight, vertical line through the graph, and the function intersects it at 2 or more points, it is NOT a function. Using this logic, only A is the answer.

Y33T

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Solve for the variables

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Answer:

Step-by-step explanation:

The angle shown is a straight angle, so it is 180°.

x + 124 = 180

124 - 124 = x

180 - 124 = 56

x = 56

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2 years ago

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Which graph has the same end behavior as the graph of f(x) = –3x3 – x2 + 1?

Marta_Voda [28]

The answer to the question is D.

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3 years ago

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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4 years ago

Which is a pair of vertical angles?

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Answer:

< TRS and <VRW

Step-by-step explanation:

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3 years ago

What is the difference between a rational number and anirrational number?

pashok25 [27]

Step-by-step explanation:

I'm sure u could've googled this answer.

Rational numbers refers to a number that can be expressed in a ratio of two integers.

An irrational number is one which can't be written as a ratio of two integers.

Rational numbers are the numbers which are integers and fractions

Irrational numbers are the numbers whose expression as a fraction is not possible

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4 years ago