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2 years ago

Match each division problem to its quotient

Mathematics

1 answer:

Svetlanka [38]2 years ago

4 0

Answer:

Step-by-step explanation:

-2 one goes with 3

-10 one goes with -4

2.75 -1.25

last one 2/3

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Let X be a positive continuous random variable with density fXpxq. Let Y " lnpXq. (a) Find the density fY pyq of Y in terms of t

salantis [7]

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X density = fXpxq and

Y" =InpXq

Now to find Y density FYpyq interms of the density of X we compare the density of X with Y"

fX = In

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2 years ago

Read 2 more answers

Which of the following is the standard equation of the ellipse with vertices at (1,0) and (27,0) and an eccentricity of 5/13?

Dovator [93]

The equation of the elipse is given by:

$\frac{(x - 14)^2}{169} + \frac{y^2}{144} = 1$

The equation of an elipse of center $(x_0, y_0)$ is given by:

$\frac{(x - x_0)^2}{a^2} + \frac{(y - y_0)^2}{b^2} = 0$

Values a and b are found according to the <u>vertices and the eccentricity</u>.

It has vertices at (1,0) and (27,0), thus:

$x_0 = \frac{27 + 1}{2} = 14$

$y_0 = \frac{0 + 0}{2} = 0$

$a = \frac{27 - 1}{2} = 13$

$a^2 = 169$

It has eccentricity of $\frac{5}{13}$ , thus:

$\frac{5}{13} = \frac{c}{a}$

$\frac{5}{13} = \frac{c}{13}$

$c = 13$

Thus, b is given according to the following equation:

$c^2 = a^2 - b^2$

$b^2 = a^2 - c^2$

$b^2 = 169 - 25$

$b = \sqrt{144}$

$b = 12$

The equation of the elipse is:

$\frac{(x - 14)^2}{169} + \frac{y^2}{144} = 1$

A similar problem is given at brainly.com/question/21405803

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2 years ago

Match each division problem to its quotient​

Match each division problem to its quotient