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MArishka [77]
3 years ago
15

Solve the system of equations.

Mathematics
1 answer:
-BARSIC- [3]3 years ago
3 0

Step-by-step explanation:

it may be `D´D D D D DD D D D D D

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Please help me with these!
Dvinal [7]

Answer:

Step-by-step explanation:

A1. C = 104°, b = 16, c = 25

Law of Sines: B = arcsin[b·sinC/c} ≅ 38.4°

A = 180-C-B = 37.6°

Law of Sines: a = c·sinA/sinC ≅ 15.7

A2. B = 56°, b = 17, c = 14

Law of Sines: C = arcsin[c·sinB/b] ≅43.1°

A = 180-B-C = 80.9°

Law of Sines: a = b·sinA/sinB ≅ 20.2

B1.  B = 116°, a = 11, c = 15

Law of Cosines: b = √(a² + c² - 2ac·cosB) = 22.2

A = arccos{(b²+c²-a²)/(2bc) ≅26.5°

C = 180-A-B = 37.5°

B2. a=18, b=29, c=30

Law of Cosines: A = arccos{(b²+c²-a²)/(2bc) ≅ 35.5°

Law of Cosines: B = arccos[(a²+c²-b²)/(2ac) = 69.2°

C = 180-A-B = 75.3°

6 0
2 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
The function f(x)=30+0.25x/x yields the average cost in dollars per cup of lemonade made by Lincoln at his lemonade stand when h
trapecia [35]

The answer is C: Lincoln uses 25 cents worth of supplies per cup.

8 0
3 years ago
Read 2 more answers
Answer sooon pls i will give thebrainliest for the correct answer☺
OleMash [197]
The correct answer is 5
3 0
3 years ago
Is 5/8 closer to 0,1/2, or 1 and why
Musya8 [376]
It is closer to 1/2 because 5/8 is close to 4/8 which is 1/2.
3 0
3 years ago
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