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AVprozaik [17]
3 years ago
5

Hello Guys... I need help...

Mathematics
2 answers:
sesenic [268]3 years ago
8 0

Answer

divide 15 by 5 and what do u get is the awser

Step-by-step explanation:

ahrayia [7]3 years ago
7 0

Answer:

The arrows should end in the negative direction

Step-by-step explanation:

The reason being is -15 / 5 is -3.

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What is the solution to this equation?<br><br> 8x - 5 (x-3) = 18
Kaylis [27]

Answer:

x=1

Step-by-step explanation:

We distribute the negative 5 to x minus three to get 8x-5x+15=18. Then, we cancel out the 15 by subtracting it from both sides. This leaves us with 8x-5x=3. When we subtract, we get 3x=3, and when we divide both sides by 3, we get x=1. Hope this helps :)

3 0
3 years ago
What percent of 75.5 is 81.54
Lady bird [3.3K]
I got the answer of 6127.731 hope this works out for you.
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3 years ago
74°. 2x° what is the value of x
horrorfan [7]

Answer:

2x-7

Step-by-step explanation:

5 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
2 (0815010)
Yuliya22 [10]

Answer:

10

Step-by-step explanation:

7 0
3 years ago
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