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3 years ago

Ill mark brainliest, question is attached

Mathematics

2 answers:

Evgesh-ka [11]3 years ago

6 0

Answer:

-3

Step-by-step explanation:

-9u+9u is 0 so all you are left with is -3

mart [117]3 years ago

3 0

Answer:

-3

Step-by-step explanation:

-9u + 9u - 3

-9u and 9u cancel out, so it simplifies to -3

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P = $14,300

Arte-miy333 [17]

Answer:

B. 4.29

Step-by-step explanation:

4 0

3 years ago

Ricky and Sam each bought a remote control car for $80. A few months later they both sold it. Ricky sold his car for 25% less th

Elden [556K]

Answer:

Step-by-step explanation:

25/100 x 80/1 = 20/1

80 - 20 = 60

30/100 x 80/1 = 24/1

80 - 24 = 56

60-56=4

6 0

3 years ago

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.

8 0

3 years ago

How many ways can first second and third place assigned

Papessa [141]

Answer:

i cannot help you get the answer to the question without more information but i can explain how to solve it, what your question is, is called a "permutation"

Step-by-step explanation:

To calculate permutations, use the equation nPr, where n is the total number of choices and r is the amount of items being selected. To solve this equation, use the equation nPr = n! / (n - r)! , an exclimation point means the factorial, say we need the factorial of 4, we would do this 4x3x2x1 go left to right, and it is not all at once. you multiply 4 times 3 which equals 12, then multiply 12 times 2... so on and so forth

3 0

3 years ago

Determine how many times larger is 6 x 10^4 than 6 x 10^2

goldfiish [28.3K]

Answer:

100

Step-by-step explanation:

6 . 10^4 = 6 . 10000 = 60000

6 . 10^2 = 6 . 100 = 600

60000 ÷ 600 = 100

8 0

2 years ago