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3 years ago

What is 86.92 in pi terms

Mathematics

1 answer:

trasher [3.6K]3 years ago

6 0

Assume that pi is approx. 3.14. Then 86.92 = approx. 27 2/3 times pi, or

86.92 is approx. equal to 83pi/3.

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Quiz

Serjik [45]

Answer:

Step-by-step explanation:

5^2 = 25 5^3 = 125 them two added together equals 150

6^2 = 36 so 36 plus 36 equals 72

150-72=78

8 0

2 years ago

Read 2 more answers

Simplify the expression:<br><br> –3(9 − x) =

steposvetlana [31]

Answer:

Expand $-3(9-x): -27 + 3x$

Step-by-step explanation:

$-3(9-x)$

Apply the distributive law: $a(b-c)=ab-ac$

$a = -3, b=9, c=x$

$=-3*9-(-3)x$

Apply minus - plus rules:

$-(-a)=a$

$= -3*9+3x$

Multiply the numbers: $3 * 9 = 27$

$=-27+3x$

Hope this helps! If so, may I get Brainliest and a Thanks?

Thank you, have a good one! =)

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

Is a dodecahedron a 3 dimensional or a dodecagon a 3 dimensional ?

navik [9.2K]

Dodecahedron is a 3D shape, usually when it ends with hedron it is 3D.

8 0

3 years ago

Read 2 more answers

What are the real zeros of x^3+4x^2-9x-36

Katyanochek1 [597]

Answer:

$x = -4\\x = -3\\x = 3$

Step-by-step explanation:

To find the real zeros you must match the function to zero and factor the expression.

We have a polynomial of degree 3.

We try to group the terms to perform the factorization

$x ^ 3 + 4x ^ 2-9x-36 = 0\\\\x ^ 2(x + 4) - 9(x + 4) = 0$

Now we take (x + 4) as a common factor

$(x + 4)(x ^ 2 -9) = 0$

If we have an expression of the form $(a ^ 2-b ^ 2)$ we know that this expression is equivalent to:

$(a ^ 2-b ^ 2) = (a + b)(a-b)$

In this case

$a = x\\b = 3$

So:

$(x ^ 2 -9) = (x + 3)(x-3)$

Finally:

$x ^ 3 + 4x ^ 2-9x-36 = (x + 4)(x + 3)(x-3) = 0$

The solutions are:

$x = -4\\x = -3\\x = 3$

6 0

3 years ago