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4 years ago

Write 5 numbers that round to 360 when rounded to the nearest ten

Mathematics

1 answer:

andrezito [222]4 years ago

8 0

360, 356, 364.71547154... , 359.99, and 360.1.

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Let X be a continuous random variable with density functionf(x)={1−|x|0 for −1

BlackZzzverrR [31]

Answer:

$f_Y (y) =\frac{d}{dy} = 2 \frac{1}{2 |\sqrt{y}|} -1 = \frac{1}{|\sqrt{y}|} -1,0 \leq Y\leq 1$

Step-by-step explanation:

For this case we want to find the density function for $Y=X^2$

And we have the following density function for the random variable X:

$f(X) = 1- |X|,-1 \leq X \leq 1$

So we can do this replacing $Y=X^2$

$F_Y (Y \leq y) = P(X^2 \leq y)$

If we apply square root on both sides we got:

$P(-\sqrt{y} \leq X \leq \sqrt{y}) = \int_{-\sqrt{y}}^0 1+t dt +\int_{0}^{\sqrt{y}} 1-t dt$

And if we integrate we got this:

$F_Y (y) = [t+ \frac{t^2}{2}] \Big|_{-\sqrt{y}}^0+ [t -\frac{t^2}{2}] \Big|_{0}^{\sqrt{y}}$

And replacing we got:

$F_Y (y) = [0 -(-\sqrt{y} +\frac{y}{2})] + [\sqrt{y} -\frac{y}{2}]$

$F_Y (y) = 2 |\sqrt{y}| -y$

And if we want to find the density function we just need to derivate the pdf like this:

$f_Y (y) =\frac{d}{dy} = 2 \frac{1}{2 |\sqrt{y}|} -1 = \frac{1}{|\sqrt{y}|} -1,0 \leq Y\leq 1$

3 0

4 years ago

What is the equation for the hyperbola shown?

amid [387]

Answer:

D. y²/5² - x²/8² = 1

Step-by-step explanation:

A and B are both incorrectly oriented, and D is the only hyperbola that contains the points (0,5) and (0,-5).

Verification (0,5) and (0,-5) are in the hyperbola:

First replace x and y with corresponding x and y values (We will start with x=0 and y=5)

$\frac{5^{2}}{5^{2}}-\frac{0^{2}}{8^{2}}=1$

Then simplify.

$\frac{25}{25}-\frac{0}{16}=1$

$1-0=1$

$1=1$

If the result is an equation (where both sides are equal to each other) then the original x and y values inputted are valid. The same is true with x and y inputs x=0 and y=-5, or any other point along the hyperbola.

6 0

4 years ago

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Which equation best represents this situation?

Tems11 [23]

The answer is B. 21 + b = 30

5 0

3 years ago

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What value is added to both sides of the equation x^2 − 6x = 10 in order to solve by completing the square? (1 point)

snow_lady [41]

Answer:

Step-by-step explanation:

use half the x coefficent and square it

x^2 − 6x + (-3)^2 = 10 + (-3)^2

x^2 - 6x + 9 = 10 + 9

(x - 3)^2 = 19

7 0

3 years ago

Which equation represents a proportional relationship?

snow_tiger [21]

Answer:

y = 15x

Step-by-step explanation:

A proportional relationship is of the form

y = kx ← k is the constant of proportionality

The only relationship fitting this relationship is

y = 15x with k = 15

6 0

3 years ago

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