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3 years ago

Identify the correct weight to the nearest 1/8 pound, then reduce the fraction if possible.

Mathematics

1 answer:

MariettaO [177]3 years ago

6 0

2 pounds that’s the nearest to 1/8

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Solve −8−x=−3(2x−4)+3x . Check your solution.

Nastasia [14]

X=10
-8-x=-3(2x-4)+3x
-8-x=-6x+12+3x
-8-x=-3x+12
-x+3x=12+8
2x=20
x=10
You can check your solution by inputting 10 in all the x’s and solving it:)

4 0

3 years ago

Find the x and y intercept of the equation: 8x+4y=24

myrzilka [38]

8x=24 which is 3 so x = 3

4y=24 which is 6 so y = 6
(3,0) for x and (0,6) for y

5 0

4 years ago

Determine the length of AC. 32 units 35.2 units 38.5 units 10.3 units

Alex73 [517]

Answer:

38.5

Step-by-step explanation:

b^2= 29^2+20^2-2×20×29× Cos 102

b^2=841+400-1160 (-0.2079)

=1241- 241.164

b^2=1482.164

b= square root of 1482.164

AC = 38.5

5 0

3 years ago

If −6 + 10 = 17, then what is 3 − 5?

Novay_Z [31]

-6 + 10 = 4
but you say the answer is 17?

okay, so let's assume you add 13 to the final answer

3 - 5 = -2
-2 + 13 = 11

This question doesn't make sense

6 0

4 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