All categories

3 years ago

I need help on the part where it says devonte

Mathematics

1 answer:

disa [49]3 years ago

6 0

X = # of cd's in his collection
x - 24 = 35
x = 35 + 24
x = 59...Devonte has 59 alternative rock cd's

You might be interested in

What I do by step <br><br> Simplify

ch4aika [34]

Multiply everything by 2 then combine like terms

4 0

3 years ago

Which ordered pair is a solution of the equation −1/4x + 6 = y?

il63 [147K]

Answer:

Step-by-step explanation:

6 0

2 years ago

what is the curvature of my B I G Bal ls are they are stroking your face while you su ck my a n us? What is 5 times 5?

salantis [7]

<span>Often the university sites exist primarily for internal use, listing equipment and demos available to physics professors teaching large sections of introductory courses. They usually give lists of required equipment, they don't always give much information about setup, effective presentation and explanation of the physics.</span>

7 0

2 years ago

Why is radian measure used in Geometry?

Vladimir [108]

The infinite series description of trig functions is much neater when the argument is radians. For example, for small angles, sin(x) ≈ x when x is in radians. You could say that radians is the "natural" measurement unit for angles, just as "e" is the "natural" base of logarithms.

If the angle measure were degrees or grads or arcseconds, obnoxious scale factors would show up everywhere.

3 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