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timofeeve [1]
3 years ago
6

How do I do this? I’m not sure can anyone help?

Mathematics
2 answers:
eduard3 years ago
8 0

value of x= 7√3

Hope it helps you...

kiruha [24]3 years ago
6 0
Maybe same answer is correct
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Z - 5 = 33
evablogger [386]

Answer:

C) 5 less than z is 33

Step-by-step explanation:

z-5=33

z-5+5=33+5

z=38

38-33=5

7 0
3 years ago
2,-10,50,-250, Geometric sequence
DanielleElmas [232]

Answer:

Step-by-step explanation:

t2/t1=r

-10/2=-5

2*-5=-10

5 0
3 years ago
Read 2 more answers
2x+ 20 + 3x +40 =<br> 60 + 5X
umka21 [38]
First answer: 160
Second answer: 300
6 0
3 years ago
The price of a ring was decreased by 25% to £390. what was the price before the decrease?
sergiy2304 [10]

Answer:

The original price was 520

Step-by-step explanation:

To find this, we first need to note that we paid 75% of the price. This is because we took 25% off from the original. Now we take the price we paid and divide it by the percentage of it which we paid. This will give us the original price.

390/75% = Total

390/.75 = Total

520 = Total

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
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