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faltersainse [42]
2 years ago
12

(2a² – b) (за² + 3b)

Mathematics
1 answer:
nataly862011 [7]2 years ago
6 0

Answer:

a^4+3a^2-3b^2

Step-by-step explanation:

\hookrightarrow \: (2 {a}^{2}  - b)(3 {a}^{2}  + 3b) \\  \hookrightarrow \:2 {a}^{2} (3 {a}^{2}  + 3b) - b(3 {a}^{2}  + 3b) \\  \hookrightarrow \:6 {a}^{4}  + 6 {a}^{2} b - 3 {a}^{2} b - 3 {b}^{2} \\   \hookrightarrow \:{a}^{4}  + 3 {a}^{2} b- 3 {b}^{2}

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9/10=a/30. and value of a is 27

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4 years ago
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Suppose that W1, W2, and W3 are independent uniform random variables with the following distributions: Wi ~ Uni(0,10*i). What is
nadya68 [22]

I'll leave the computation via R to you. The W_i are distributed uniformly on the intervals [0,10i], so that

f_{W_i}(w)=\begin{cases}\dfrac1{10i}&\text{for }0\le w\le10i\\\\0&\text{otherwise}\end{cases}

each with mean/expectation

E[W_i]=\displaystyle\int_{-\infty}^\infty wf_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac w{10i}\,\mathrm dw=5i

and variance

\mathrm{Var}[W_i]=E[(W_i-E[W_i])^2]=E[{W_i}^2]-E[W_i]^2

We have

E[{W_i}^2]=\displaystyle\int_{-\infty}^\infty w^2f_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac{w^2}{10i}\,\mathrm dw=\frac{100i^2}3

so that

\mathrm{Var}[W_i]=\dfrac{25i^2}3

Now,

E[W_1+W_2+W_3]=E[W_1]+E[W_2]+E[W_3]=5+10+15=30

and

\mathrm{Var}[W_1+W_2+W_3]=E\left[\big((W_1+W_2+W_3)-E[W_1+W_2+W_3]\big)^2\right]

\mathrm{Var}[W_1+W_2+W_3]=E[(W_1+W_2+W_3)^2]-E[W_1+W_2+W_3]^2

We have

(W_1+W_2+W_3)^2={W_1}^2+{W_2}^2+{W_3}^2+2(W_1W_2+W_1W_3+W_2W_3)

E[(W_1+W_2+W_3)^2]

=E[{W_1}^2]+E[{W_2}^2]+E[{W_3}^2]+2(E[W_1]E[W_2]+E[W_1]E[W_3]+E[W_2]E[W_3])

because W_i and W_j are independent when i\neq j, and so

E[(W_1+W_2+W_3)^2]=\dfrac{100}3+\dfrac{400}3+300+2(50+75+150)=\dfrac{3050}3

giving a variance of

\mathrm{Var}[W_1+W_2+W_3]=\dfrac{3050}3-30^2=\dfrac{350}3

and so the standard deviation is \sqrt{\dfrac{350}3}\approx\boxed{116.67}

# # #

A faster way, assuming you know the variance of a linear combination of independent random variables, is to compute

\mathrm{Var}[W_1+W_2+W_3]

=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]+2(\mathrm{Cov}[W_1,W_2]+\mathrm{Cov}[W_1,W_3]+\mathrm{Cov}[W_2,W_3])

and since the W_i are independent, each covariance is 0. Then

\mathrm{Var}[W_1+W_2+W_3]=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]

\mathrm{Var}[W_1+W_2+W_3]=\dfrac{25}3+\dfrac{100}3+75=\dfrac{350}3

and take the square root to get the standard deviation.

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Answer:

I think the answer is 2n-5

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