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3 years ago

1

Mathematics

1 answer:

Mama L [17]3 years ago

7 0

Hole? It depends how big it is. Since there is no hole being depicted in this question, the answer is inexcusably 2ft

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What is the solution to to the equation x − 2/5 = −1?

ZanzabumX [31]

Answer:

x = -0.6

Step-by-step Solution:

<u>Isolate the variable x to get the answer.</u>

x − 2/5 = −1
=> x = -1 + 2/5
=> x = -1 + 0.4
=> x = -0.6

Hence, the value of x is -0.6.

6 0

2 years ago

Read 2 more answers

Pls pls pls help <br>Find the perimeter and the area

Illusion [34]

Answer:

Perimeter = 42 units Area = 57.73 square units

Step-by-step explanation:

Perimeter of an Astroid = 6l ---> 6 x 7 = 42 units

Area of an Astroid = (3 x pi x a^2)/8 = 57.73 square units

3 0

3 years ago

What is the equation of the blue line

Sauron [17]

Answer:

y=2x+3

Step-by-step explanation:

Use the equation: y=mx+b

m is the slope (2 in this case)

b is the y intercept (3 in this case)

So the equation of the line would be y=2x+3

5 0

2 years ago

The scatter plot below shows the distances some employees of a company commute to work and the annual

salantis [7]

Answer:

It is a

Step-by-step explanation: I looked it up

6 0

3 years ago

Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z

kompoz [17]

$f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}$

is a proper joint density function if, over its support, $f$ is non-negative and the integral of $f$ is 1. The first condition is easily met as long as $C\ge0$ . To meet the second condition, we require

$\displaystyle\int_0^\infty\int_0^\infty\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=100C=1\implies \boxed{C=0.01}$

b. Find the marginal joint density of $X$ and $Y$ by integrating the joint density with respect to $z$ :

$f_{X,Y}(x,y)=\displaystyle\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dz=0.01e^{-(0.5x+0.2y)}\int_0^\infty e^{-0.1z}\,\mathrm dz$

$\implies f_{X,Y}(x,y)=\begin{cases}0.1e^{-(0.5x+0.2y)}&\text{for }x\ge0,y\ge0\\0&\text{otherwise}\end{cases}$

Then

$\displaystyle P(X\le1.375,Y\le1.5)=\int_0^{1.5}\int_0^{1.375}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy$

$\approx\boxed{0.12886}$

c. This probability can be found by simply integrating the joint density:

$\displaystyle P(X\le1.375,Y\le1.5,Z\le1)=\int_0^1\int_0^{1.5}\int_0^{1.375}f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz$

$\approx\boxed{0.012262}$

7 0

3 years ago