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3 years ago

Rearrange the equation so xxx is the independent variable. 6x+y=4x+11y6x+y=4x+11y

Mathematics

2 answers:

ANEK [815]3 years ago

5 0

Answer:

$y = \frac{1}{5} \cdot x$

Step-by-step explanation:

The following equation is rearranged by some algebraic handling:

$6\cdot x + y = 4\cdot x + 11\cdot y$

$2\cdot x = 10\cdot y$

$\frac{1}{5}\cdot x = y$

$y = \frac{1}{5} \cdot x$

slava [35]3 years ago

4 0

Answer:

10x=10y so x and y are the same or x=y

Step-by-step explanation:

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3 years ago

The probability that a student uses Smarthinking Online Tutoring on a regular basis is 0.31 . In a group of 21 students, what is

Ivenika [448]

Answer: 0.0241

Step-by-step explanation:

This is solved using the probability distribution formula for random variables where the combination formula for selection is used to determine the probability of these random variables occurring. This formula is denoted by:

P(X=r) = nCr × p^r × q^n-r

Where:

n = number of sampled variable which in this case = 21

r = variable outcome being determined which in this case = 5

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4 0

4 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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3 years ago

Mrs. Yin bought 0.81 of a pound of turkey and 0.69 of a pound of chicken. Did Mrs. Yin buy more turkey or more chicken? Show how

S_A_V [24]

Mrs. Yin bought more turkey than chicken. I know because 0.80 is greater than 0.69

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3 years ago