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3 years ago

What are the solutions to the linear system? X= y = z=

Mathematics

1 answer:

den301095 [7]3 years ago

7 0

Answer:

In a graphic image, the x and y denote width and height; the z denotes depth.x/y/z = (x/y)/z. If you only have division, you should work left to right.

Step-by-step explanation:

The solution to a system of linear equations is the point at which the lines representing the linear equations intersect.

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What is an equation of the line that passes through the points (0, 3) and (5,−3)?

g100num [7]

Answer:

y = -6/5x + 3

Step-by-step explanation:

y = mx + c (or b, depending on where you're from)

Slope = m = -6/5

y = -6/5x + c

You can sub (0, 3) and (5, -3) into y = -6/5x + c and they both have the same answer where c = 3

Therefore, the equation is y = -6/5x + 3

Hop this helps

5 0

3 years ago

Let X and Y be discrete random variables. Let E[X] and var[X] be the expected value and variance, respectively, of a random vari

Ulleksa [173]

Answer:

(a) $E[X+Y]=E[X]+E[Y]$

(b) $Var(X+Y)=Var(X)+Var(Y)$

Step-by-step explanation:

Let X and Y be discrete random variables and E(X) and Var(X) are the Expected Values and Variance of X respectively.

(a)We want to show that E[X + Y ] = E[X] + E[Y ].

When we have two random variables instead of one, we consider their joint distribution function.

For a function f(X,Y) of discrete variables X and Y, we can define

$E[f(X,Y)]=\sum_{x,y}f(x,y)\cdot P(X=x, Y=y).$

Since f(X,Y)=X+Y

$E[X+Y]=\sum_{x,y}(x+y)P(X=x,Y=y)\\=\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y).$

Let us look at the first of these sums.

$\sum_{x,y}xP(X=x,Y=y)\\=\sum_{x}x\sum_{y}P(X=x,Y=y)\\\text{Taking Marginal distribution of x}\\=\sum_{x}xP(X=x)=E[X].$

Similarly,

$\sum_{x,y}yP(X=x,Y=y)\\=\sum_{y}y\sum_{x}P(X=x,Y=y)\\\text{Taking Marginal distribution of y}\\=\sum_{y}yP(Y=y)=E[Y].$

Combining these two gives the formula:

$\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y) =E(X)+E(Y)$

Therefore:

$E[X+Y]=E[X]+E[Y] \text{ as required.}$

(b)We want to show that if X and Y are independent random variables, then:

$Var(X+Y)=Var(X)+Var(Y)$

By definition of Variance, we have that:

$Var(X+Y)=E(X+Y-E[X+Y]^2)$

$=E[(X-\mu_X +Y- \mu_Y)^2]\\=E[(X-\mu_X)^2 +(Y- \mu_Y)^2+2(X-\mu_X)(Y- \mu_Y)]\\$Since we have shown that expectation is linear$\\=E(X-\mu_X)^2 +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2 +E[Y- E(Y)]^2+2Cov (X,Y)$

Since X and Y are independent, Cov(X,Y)=0

$=Var(X)+Var(Y)$

Therefore as required:

$Var(X+Y)=Var(X)+Var(Y)$

7 0

3 years ago

Linda's rental car cost $198

Vsevolod [243]

I got $258.44 but it may be different if the six percent tax was off the entire total which included the insurance, I only took the 6% tax for the cost of the rental car

8 0

3 years ago

Read 2 more answers

Q. A mechanic has a radiator hose that

Dmitriy789 [7]

B 6 whole peices

explanation

3 0

3 years ago

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating

exis [7]

Answer:

Attached please find response.

Step-by-step explanation:

We wish to find the area between the curves 2x+y2=8 and y=x.

Substituting y for x in the equation 2x+y2=8 yields

2y+y2y2+2y−8(y+4)(y−2)=8=0=0

so the line y=x intersects the parabola 2x+y2=8 at the points (−4,−4) and (2,2). Solving the equation 2x+y2=8 for x yields

x=4−12y2

From sketching the graphs of the parabola and the line, we see that the x-values on the parabola are at least those on the line when −4≤y≤2.

6 0

3 years ago