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2 years ago

Find the mean, variance, and standard deviation of the following probability

Mathematics

1 answer:

ElenaW [278]2 years ago

3 0

Answer:

5.1 ; 8.29 ; 2.879

Step-by-step explanation:

Given :

Number of Computer Sold X

Probability P(x)

0.1

0.2

0.3

0.2

Mean = E(X) = Σ(x*p(x)

E(X) = (0*0.1) + (2*0.2) + (5*0.3) + (7*0.2) + (9*0.2)

E(X) = 5.1

Variance, Var(X) = Σx²p(x) − E(X)²

Var(X) = [(0^2*0.1) + (2^2*0.2) + (5^2*0.3) + (7^2*0.2) + (9^2*0.2)] - 5.1^2

Var(X) = 34.3 - 26.01 = 8.29

Standard deviation ; Std(X) = √Var(X) = √8.29 = 2.879

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Let T: Rn → Rn be an invertible linear transformation, and let S and U be functions from Rn into Rn such that

ipn [44]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

$T: 1 \ R^n \to 1 \ R^n$ is invertible lines transformation

$S[T(x)]=x \ and \ V[T(x)]=x \\\\t'x \ \varepsilon\ 1 R^n\\\\$

T is invertiable linear transformation means that is

$T(x) =A x \\\\ where \\\\ A= n \times n \ \ matrix$

and $\ det(A) \neq 0 \ \ that \ is \ \ A^{-1} \ \ exists$

Let

$V \varepsilon\ a\ R^{n} \ consider \ \ u= A^{-1} v \varepsilon 1 R^n\\\\T(u)= A(A^{-1} v)=(A \ A^{-1}) \\\\ v= I_{n \times n} \cdot v = v$

so,

$s[T(u)]=v[T(u)]\\\\s(v)=v(v) \ \ \forall \ \ v \ \ \varepsilon \ \ 1 R^n$

7 0

2 years ago

What percent of 60 is 51

vova2212 [387]

51=?
60 100

Then you cross-multiply the 51 and the 100 and get 5,100.

Next you take the 5,100 and divide it by the 60 which gives you your final answer of 85%.

4 0

2 years ago

PLEASE!!! WILL MARK BRAINLIEST IF YOU'RE RIGHTIn this triangle, what is the value of x? enter your answer, rounded to the neares

OLga [1]

$\bold{ANSWER:}$
x = 61.1

$\bold{SOLUTION:}$

6 0

1 year ago

If <img src="https://tex.z-dn.net/?f=%20x%20%3D%20%5Cfrac%7B2y%7D%7Ba%7D%20" id="TexFormula1" title=" x = \frac{2y}{a} " alt=" x

inessss [21]

$\bf A) \\\\\\ \cfrac{3\left( \frac{2y}{a} \right)}{y}\implies \cfrac{\frac{6y}{a}}{y}\implies \cfrac{6y}{a}\cdot \cfrac{1}{y}\implies \cfrac{6}{a}\quad \otimes \\\\\\ B) \\\\\\ \cfrac{6\left( \frac{2y}{a} \right)}{y}\implies\cfrac{\frac{12y}{a}}{y}\implies \cfrac{12y}{a}\cdot \cfrac{1}{y}\implies \cfrac{12}{a}\quad \otimes \\\\\\ C) \\\\\\ \cfrac{6y}{\left( \frac{2y}{a} \right)}\implies \cfrac{6y}{1}\cdot \cfrac{a}{2y}\implies 3a\quad \otimes$

$\bf D) \\\\\\ \cfrac{12\left( \frac{2y}{a} \right)}{y}\implies \cfrac{\frac{24y}{a}}{y}\implies \cfrac{24y}{a}\cdot \cfrac{1}{y}\implies \cfrac{24}{a}\quad \otimes \\\\\\ E) \\\\\\ \cfrac{12y}{\left( \frac{2y}{a} \right)}\implies \cfrac{12y}{1}\cdot \cfrac{a}{2y}\implies 6a\quad \checkmark$

4 0

3 years ago

Please help! A-level maths.

astra-53 [7]

$x+y=k\implies y=k-x$

Substitute this into the parabolic equation,

$k-x=6-(x-3)^2\implies x^2-7x+(k+3)=0$

We're told the line $x+y=k$ intersects $C$ twice, which means the quadratic above has two distinct real solutions. Its discriminant must then be positive, so we know

$(-7)^2-4(k+3)=49-4(k+3)>0\implies k$

We can tell from the quadratic equation that $C$ has its vertex at the point (3, 6). Also, note that

$-1\le\sin t\le1\implies3\le3+2\sin t\le5\implies3\le x\le5$

and

$-1\le\cos2t\le1\implies2\le4+2\cos2t\le6\implies2\le y\le6$

so the furthest to the right that $C$ extends is the point (5, 2). The line $x+y=k$ passes through this point for $2+5=k\implies k=7$ . For any value of $k$ , the line $x+y=k$ passes through $C$ either only once, or not at all.

So $7\le k$ ; in set notation,

$\{k\mid 7\le k$

4 0

3 years ago