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3 years ago

Why must eye protection be worn at all times in welding shop?

2 answers:

6 0

The fumes are way to strong. Plus you'd have to rush to the hospital if something flew in your eyes or cut your eyes. Its always good to have protection.

kati45 [8]3 years ago

4 0

Because of the ligting of the machine and you can go blind if theres no eye protection

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You are busy baking cookies. You have baked 32 cookies and you want to make sure that each of your eight friend’s cookies look t

Alona [7]

Answer: 128 pieces of candy for all the cookies.

8 0

3 years ago

2) X and Y are jointly continuous with joint pdf

Nady [450]

From what I gather from your latest comments, the PDF is given to be

$f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0\le x,y \le1\\0&\text{otherwise}\end{cases}$

and in particular, f(x, y) = cxy over the unit square [0, 1]², meaning for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. (As opposed to the unbounded domain, x ≤ 0 *and* y ≤ 1.)

(a) Find c such that f is a proper density function. This would require

$\displaystyle\int_0^1\int_0^1 cxy\,\mathrm dx\,\mathrm dy=c\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\frac c{2^2}=1\implies \boxed{c=4}$

(b) Get the marginal density of X by integrating the joint density with respect to y :

$f_X(x)=\displaystyle\int_0^1 4xy\,\mathrm dy=(2xy^2)\bigg|_{y=0}^{y=1}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

$f_Y(y)=\displaystyle\int_0^14xy\,\mathrm dx=\begin{cases}2y&\text{for }0\le y\le1\\0&\text{otherwise}\end{cases}$

(d) The conditional distribution of X given Y can obtained by dividing the joint density by the marginal density of Y (which follows directly from the definition of conditional probability):

$f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

(e) From the definition of expectation:

$E[X]=\displaystyle\int_0^1\int_0^1 x\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\boxed{\frac23}$

$E[Y]=\displaystyle\int_0^1\int_0^1 y\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac23}$

$E[XY]=\displaystyle\int_0^1\int_0^1xy\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac49}$

(f) Note that the density of X | Y in part (d) identical to the marginal density of X found in (b), so yes, X and Y are indeed independent.

The result in (e) agrees with this conclusion, since E[XY] = E[X] E[Y] (but keep in mind that this is a property of independent random variables; equality alone does not imply independence.)

8 0

3 years ago

45% of 120 is what answer please

marishachu [46]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

3.1 Write the following in scientific notation a) 217 000 000 b) 0,000 253

AlekseyPX

Answer:

2.17 x 10^8

0.253 x 10^-4 if number is 0.000253 or 2.53 x 10^2 if number is 253

Step-by-step explanation:

6 0

3 years ago

Ben works for software company in Cleveland. To train new employees, Ben travels to their two other locations, Buffalo and Phila

Andru [333]

Answer:

Explanation:we will solve this question with help of linear programming.

Now, let Ben goes x times in buffalo trip and goes y times in Philadelphia trip. Thus, according to the question, Min Z= 180x+300y

And s.t.r. 2x+3y≤105, where x≥5, y≥15.

since at (0,0) 0≤105(true), so area of 2x+3y≤105 will cover the origin.

But, 0≥5 and 0≥15 are false, so their area will not cover the origin.

Thus, after drawing the graph of these three lines we will get a feasible reason of triangle ABC, where point A≡(5,31.667), B≡(5,15), and C≡(30,15)

Z at A- 180×5+300×31.667=10,400.1

at B-180×5+300×15=1350( which is minimum.)

at C- 180×30+300×15=5,850

Thus, Ben should go 5 times to buffalo trip and 15 times to Philadelphia to minimize his cost which is $1350.

3 0

3 years ago