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galben [10]
3 years ago
8

Simplify the following expression:

Mathematics
1 answer:
nalin [4]3 years ago
3 0
-2/2 I think probably
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Middle school Math!!!
makvit [3.9K]
Y=1x+2 (you don’t have to write the 1 in front of the x so you can write it as y=x+1)
6 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
Hi i just need help checking if my answer is correct I have been working on a implicit diffrention problem and I need some help.
SSSSS [86.1K]

Answer:

  \dfrac{dy}{dx}=\dfrac{-4x^3y+2y^3+3x^2y^3}{x^4-6xy^2-3x^3y^2}

Step-by-step explanation:

Normally, when I do this, I differentiate each term first with respect to x then with respect to y. In this solution, I differentiated the entire expression with respect to x, then with respect to y. That makes separating the dx and dy coefficients much easier, so the solution almost falls into your lap.

  -x^4 y + 2 x y^3 + x^3 y^3=0 \qquad\text{given}\\\\(-4x^3y+2y^3+3x^2y^3)dx+(-x^4+6xy^2+3x^3y^2)dy=0\\\\(-4x^3y+2y^3+3x^2y^3)dx=(x^4-6xy^2-3x^3y^2)dy\\\\\boxed{\dfrac{dy}{dx}=\dfrac{-4x^3y+2y^3+3x^2y^3}{x^4-6xy^2-3x^3y^2}}

6 0
3 years ago
The diagram shows rays of light from a point source, P, that reach a detector, D, along two paths. One path is a straight line,
tatiyna
This can solved using the cosine law which is:
c² = a² + b² - 2ab cos θ
Using the values given from the problem
6² = b² + b² - 2bb cos 112.62
And solving for b
36 = 2b² - 2b² cos 112.62
b = 3.6

The answer is the 3rd option.
8 0
3 years ago
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The width is 20 feet 
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3 years ago
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