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alexandr402 [8]
2 years ago
7

1. En una tienda de ropa se pusieron en oferta guantes y chamarras. El primer dia se vendieron 6 pares

Mathematics
2 answers:
Softa [21]2 years ago
6 0
10560 so much money all forget to count
Semenov [28]2 years ago
6 0

1. En una tienda de ropa se pusieron en oferta guantes y chamarras. El primer dia se vendieron 6 pares
de guantes y 7 chamarras, sumando la cantidad de $1850. El segundo dia se venden 8 pares de guantes
y 4 chamarras con un monto de $1050. ¿Cuál es el precio de cada par de guantes y de chamarras
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A garden has 8 white roses and 11 red roses. One rose is plucked from the garden and then one more rose is plucked without repla
liberstina [14]
P(r/w) is the probability of picking a red rose at first picking and a white rose at second picking.
P(w/r) is the probability of picking a white rose at first picking and a red rose at second picking.

P(r/w) = \frac{11}{19}×\frac{8}{18} = \frac{88}{342}
P(w/r) = \frac{8}{19}×\frac{11}{18}=\frac{88}{342}

Notice that the second fraction is out of 18 because the second picking of rose will be out of 18 since the first rose is not replaced.

P(r/w) equals to P(w/r)
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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)

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