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f X Y ( t , u ) = 3 88 ( 2 t + 3 u 2 ) for 0 t 2 , 0 u 1 + t (see Exercise 15 from "Problems on Random Vectors and Joint Distributions").

Z = I [ 0 , 1 ] ( X ) 4 X + I ( 1 , 2 ] ( X ) ( X + Y )

Determine P ( Z 2 )

P ( Z 2 ) = P ( Z Q = Q 1 M 1 Q 2 M 2 ) , where M 1 = { ( t , u ) : 0 t 1 , 0 u 1 + t }
M 2 = { ( t , u ) : 1 < t 2 , 0 u 1 + t }
Q 1 = { ( t , u ) : 0 t 1 / 2 } , Q 2 = { ( t , u ) : u 2 - t } (see figure)
P = 3 88 0 1 / 2 0 1 + t ( 2 t + 3 u 2 ) d u d t + 3 88 1 2 0 2 - t ( 2 t + 3 u 2 ) d u d t = 563 5632
tuappr Enter matrix [a b]of X-range endpoints [0 2] Enter matrix [c d]of Y-range endpoints [0 3] Enter number of X approximation points 200Enter number of Y approximation points 300 Enter expression for joint density (3/88)*(2*t + 3*u.^2).*(u<=1+t) Use array operations on X, Y, PX, PY, t, u, and PG = 4*t.*(t<=1) + (t+u).*(t>1); [Z,PZ]= csort(G,P); PZ2 = (Z<=2)*PZ' PZ2 = 0.1010 % Theoretical = 563/5632 = 0.1000
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Figure 1 is comprised of two separate figures, one for exercise 11, on the left, and one for exercise 12, on the right. The figure for exercise 11 is a cartesian graph of the first quadrant, with the horizontal axis labeled t, and the vertical axis labeled u. The values on the horizontal axis range from 0 to 2, in increments of 1, with an extra marking in between 0 and 1 for the value 0.5. The values on the vertical axis range from 0 to 3 in increments of 1. The figure contains three major shapes, with two smaller shaded shapes located inside a larger polygon. The large polygon has four sides. The first is a vertical portion located on the vertical axis, from (0, 0) to (0, 1). The upper side is a diagonal line, labeled u = 1 + t, that is drawn from (0, 1) to (2, 3). The rightmost side is another vertical line from the end of the second side at (2, 3) to the horizontal axis, at (2, 0). The bottom side of the shape connects (2, 0) horizontally to the origin, completing the shape. Inside this large polygon are two shaded shapes. The first is a right triangle with right-angle vertex located at (1, 0). One side is drawn vertically from (1, 0) to (1, 1), another is drawn horizontally from (1, 0) to (2, 0), and the hypotenuse is connected from (1, 1) to (2, 0). The hypotenuse is labeled u = 2 - t. There is a dashed vertical line connecting the triangles vertex at (1, 1) to the midpoint of the large polygon's diagonal side, at (1, 2). The second small shape is similar to the large polygon. It shares the same side on the vertical axis from (0, 0) to (0,1), and shares a portion of the diagonal line u = 1 + t from (0, 1) to (0.5, 1.5). Its third side is drawn vertically from that point to (0.5, 0) on the horizontal axis. A fourth side completes the polygon from (0.5, 0) to the origin.  The figure for exercise 12 is a cartesian graph of the first quadrant, with the horizontal axis labeled t, and the vertical axis labeled u. The values on the horizontal axis range from 0 to 2, in increments of 1, with an extra marking for the value 0.5. The vertical axis does not contain any markings for values. There are five lines, which, connecting to the horizontal and vertical axes, create four separate enclosed sections. Two of these sections are shaded. The outermost lines are labeled with equations. A horizontal line from (0, 1) to (1, 1) is labeled u = 1. A diagonal line with negative slope connects to the horizontal line and the t-axis from (1, 1) to (2, 0) and is labeled u = 2 - t. These two labeled lines, along with the u and t-axes, create a four-sided polygon with two horizontal sides, one vertical side on the u-axis, and one diagonal side of negative slope labeled u = 2 - t. There are three interior lines. The longest interior line extends from the origin to point (1, 1), and in doing so, divides the larger aforementioned polygon into two triangles. The first smaller interior line is drawn from (0.5, 1) to (0.5, 0.5), a vertical line. The second is drawn from (0.5, 0.5) to approximately (1.5, 0.5) a horizontal line. These two smaller interior lines divide the two aforementioned triangles into two smaller triangles, a large symmetric trapezoid, and a four-sided polygon. The polygon inside and the trapezoid are shaded, and the smaller triangles remain white. Figure 1 is comprised of two separate figures, one for exercise 11, on the left, and one for exercise 12, on the right. The figure for exercise 11 is a cartesian graph of the first quadrant, with the horizontal axis labeled t, and the vertical axis labeled u. The values on the horizontal axis range from 0 to 2, in increments of 1, with an extra marking in between 0 and 1 for the value 0.5. The values on the vertical axis range from 0 to 3 in increments of 1. The figure contains three major shapes, with two smaller shaded shapes located inside a larger polygon. The large polygon has four sides. The first is a vertical portion located on the vertical axis, from (0, 0) to (0, 1). The upper side is a diagonal line, labeled u = 1 + t, that is drawn from (0, 1) to (2, 3). The rightmost side is another vertical line from the end of the second side at (2, 3) to the horizontal axis, at (2, 0). The bottom side of the shape connects (2, 0) horizontally to the origin, completing the shape. Inside this large polygon are two shaded shapes. The first is a right triangle with right-angle vertex located at (1, 0). One side is drawn vertically from (1, 0) to (1, 1), another is drawn horizontally from (1, 0) to (2, 0), and the hypotenuse is connected from (1, 1) to (2, 0). The hypotenuse is labeled u = 2 - t. There is a dashed vertical line connecting the triangles vertex at (1, 1) to the midpoint of the large polygon's diagonal side, at (1, 2). The second small shape is similar to the large polygon. It shares the same side on the vertical axis from (0, 0) to (0,1), and shares a portion of the diagonal line u = 1 + t from (0, 1) to (0.5, 1.5). Its third side is drawn vertically from that point to (0.5, 0) on the horizontal axis. A fourth side completes the polygon from (0.5, 0) to the origin.  The figure for exercise 12 is a cartesian graph of the first quadrant, with the horizontal axis labeled t, and the vertical axis labeled u. The values on the horizontal axis range from 0 to 2, in increments of 1, with an extra marking for the value 0.5. The vertical axis does not contain any markings for values. There are five lines, which, connecting to the horizontal and vertical axes, create four separate enclosed sections. Two of these sections are shaded. The outermost lines are labeled with equations. A horizontal line from (0, 1) to (1, 1) is labeled u = 1. A diagonal line with negative slope connects to the horizontal line and the t-axis from (1, 1) to (2, 0) and is labeled u = 2 - t. These two labeled lines, along with the u and t-axes, create a four-sided polygon with two horizontal sides, one vertical side on the u-axis, and one diagonal side of negative slope labeled u = 2 - t. There are three interior lines. The longest interior line extends from the origin to point (1, 1), and in doing so, divides the larger aforementioned polygon into two triangles. The first smaller interior line is drawn from (0.5, 1) to (0.5, 0.5), a vertical line. The second is drawn from (0.5, 0.5) to approximately (1.5, 0.5) a horizontal line. These two smaller interior lines divide the two aforementioned triangles into two smaller triangles, a large symmetric trapezoid, and a four-sided polygon. The polygon inside and the trapezoid are shaded, and the smaller triangles remain white.

f X Y ( t , u ) = 24 11 t u for 0 t 2 , 0 u min { 1 , 2 - t } (see Exercise 17 from "Problems on Random Vectors and Joint Distributions").

Z = I M ( X , Y ) 1 2 X + I M c ( X , Y ) Y 2 , M = { ( t , u ) : u > t }

Determine P ( Z 1 / 4 ) .

P ( Z 1 / 4 ) = P ( ( X , Y ) M 1 Q 1 M 2 Q 2 ) , M 1 = { ( t , u ) : 0 t u 1 }
M 2 = { ( t , u ) : 0 t 2 , 0 t min ( t , 2 - t ) }
Q 1 = { ( t , u ) : t 1 / 2 } Q 2 = { ( t , u ) : u 1 / 2 } (see figure)
P = 24 11 0 1 / 2 0 1 t u d u d t + 24 11 1 / 2 3 / 2 0 1 / 2 t u d u d t + 24 11 3 / 2 2 0 2 - t t u d u d t = 85 176
tuappr Enter matrix [a b]of X-range endpoints [0 2] Enter matrix [c d]of Y-range endpoints [0 1] Enter number of X approximation points 400Enter number of Y approximation points 200 Enter expression for joint density (24/11)*t.*u.*(u<=min(1,2-t)) Use array operations on X, Y, PX, PY, t, u, and PG = 0.5*t.*(u>t) + u.^2.*(u<t); [Z,PZ]= csort(G,P); pp = (Z<=1/4)*PZ' pp = 0.4844 % Theoretical = 85/176 = 0.4830
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f X Y ( t , u ) = 3 23 ( t + 2 u ) for 0 t 2 , 0 u max { 2 - t , t } (see Exercise 18 from "Problems on Random Vectors and Joint Distributions").

Z = I M ( X , Y ) ( X + Y ) + I M c ( X , Y ) 2 Y , M = { ( t , u ) : max ( t , u ) 1 }

Determine P ( Z 1 ) .

P ( Z 1 ) = P ( ( X , Y ) M 1 Q 1 M 2 Q 2 ) , M 1 = { ( t , u ) : 0 t 1 , 0 u 1 - t }
M 2 = { ( t , u ) : 1 t 2 , 0 u t }
Q 1 = { ( t , u ) : u 1 - t } Q 2 = { ( t , u ) : u 1 / 2 } (see figure)
P = 3 23 0 1 0 1 - t ( t + 2 u ) d u d t + 3 23 1 2 0 1 / 2 ( t + 2 u ) d u d t = 9 46
tuappr Enter matrix [a b]of X-range endpoints [0 2] Enter matrix [c d]of Y-range endpoints [0 2] Enter number of X approximation points 300Enter number of Y approximation points 300 Enter expression for joint density (3/23)*(t + 2*u).*(u<=max(2-t,t)) Use array operations on X, Y, PX, PY, t, u, and PM = max(t,u)<= 1; G = M.*(t + u) + (1 - M)*2.*u;p = total((G<=1).*P) p = 0.1960 % Theoretical = 9/46 = 0.1957
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Figure two is comprised of two cartesian graphs, both displaying only the first coordinate. Both graphs use t as the horizontal axis and u as the vertical axis. Both axes on the graphs range in value from 0 to 2 in increments of 1. The first graph, for exercise 13,  contains five labeled lines, and two unlabeled lines. The lines are labeled with their equations. From (0, 2) to (1, 1) is a line segment labeled u = 2 - t. From (1, 1) to (2, 2) is a line segment labeled u = t. From (0, 1) to (1, 0) is a line segment labeled u = 1 - t. From (1, 0.5) to (2, 0.5) is a line segment labeled u = 0.5. From (2, 0) to (2, 2) is a line labeled t = 2. From (0, 1) to (1, 1) is an unlabeled line segment. From (1, 0) to (1, 1) is the final unlabeled line segment. Two shapes created by the lines on this graph are shaded. The first is a right triangle bound by u = 1 - t and the horizontal and vertical axes. The second is a rectangle bound by u = 0.5, t = 2, the vertical unlabeled line segment from (1, 0) to (1, 1), and the horizontal axis. The second graph, for exercise 14, contains 5 labeled line segments, one unlabeled line segment, and one dashed, unlabeled line segment. From (0, 2) to (1, 2) is a line segment labeled u = 2. From (0, 2) to (1, 1) is a line segment labeled u = 2 - t. From (1, 2) to (2, 1) is a line segment labeled u = 3 - t. From (1, 1) to (2, 1) is a line segment labeled u = 1. From (2, 1) to (2, 0) is a line segment labeled t = 2. The unlabeled solid line segment is drawn from (1, 2) to (1, 1). The unlabeled dashed line segment is drawn from (0, 1) to (1, 1). A large region, bounded by u = 2 - t, u = 1, t = 2, the vertical axis, and the horizontal axis, is shaded grey. Figure two is comprised of two cartesian graphs, both displaying only the first coordinate. Both graphs use t as the horizontal axis and u as the vertical axis. Both axes on the graphs range in value from 0 to 2 in increments of 1. The first graph, for exercise 13,  contains five labeled lines, and two unlabeled lines. The lines are labeled with their equations. From (0, 2) to (1, 1) is a line segment labeled u = 2 - t. From (1, 1) to (2, 2) is a line segment labeled u = t. From (0, 1) to (1, 0) is a line segment labeled u = 1 - t. From (1, 0.5) to (2, 0.5) is a line segment labeled u = 0.5. From (2, 0) to (2, 2) is a line labeled t = 2. From (0, 1) to (1, 1) is an unlabeled line segment. From (1, 0) to (1, 1) is the final unlabeled line segment. Two shapes created by the lines on this graph are shaded. The first is a right triangle bound by u = 1 - t and the horizontal and vertical axes. The second is a rectangle bound by u = 0.5, t = 2, the vertical unlabeled line segment from (1, 0) to (1, 1), and the horizontal axis. The second graph, for exercise 14, contains 5 labeled line segments, one unlabeled line segment, and one dashed, unlabeled line segment. From (0, 2) to (1, 2) is a line segment labeled u = 2. From (0, 2) to (1, 1) is a line segment labeled u = 2 - t. From (1, 2) to (2, 1) is a line segment labeled u = 3 - t. From (1, 1) to (2, 1) is a line segment labeled u = 1. From (2, 1) to (2, 0) is a line segment labeled t = 2. The unlabeled solid line segment is drawn from (1, 2) to (1, 1). The unlabeled dashed line segment is drawn from (0, 1) to (1, 1). A large region, bounded by u = 2 - t, u = 1, t = 2, the vertical axis, and the horizontal axis, is shaded grey.

f X Y ( t , u ) = 12 179 ( 3 t 2 + u ) , for 0 t 2 , 0 u min { 2 , 3 - t } (see Exercise 19 from "Problems on Random Vectors and Joint Distributions").

Z = I M ( X , Y ) ( X + Y ) + I M c ( X , Y ) 2 Y 2 , M = { ( t , u ) : t 1 , u 1 }

Determine P ( Z 2 ) .

P ( Z 2 ) = P ( ( , Y ) M 1 Q 1 ( M 2 M 3 ) Q 2 ) , M 1 = { ( t , u ) : 0 t 1 , 1 u 2 }
M 2 = { ( t , u ) : 0 t 1 , 0 u 1 } M 3 = { ( t , u ) : 1 t 2 , 0 u 3 - t }
Q 1 = { ( t , u ) : u 1 - t } Q 2 = { ( t , u ) : u 1 / 2 } (see figure)
P = 12 179 0 1 0 2 - t ( 3 t 2 + u ) d u d t + 12 179 1 2 0 1 ( 3 t 2 + u ) d u d t = 119 179
tuappr Enter matrix [a b]of X-range endpoints [0 2] Enter matrix [c d]of Y-range endpoints [0 2] Enter number of X approximation points 300Enter number of Y approximation points 300 Enter expression for joint density (12/179)*(3*t.^2 + u).*(u<=min(2,3-t)) Use array operations on X, Y, PX, PY, t, u, and PM = (t<=1)&(u>=1); Z = M.*(t + u) + (1 - M)*2.*u.^2;G = M.*(t + u) + (1 - M)*2.*u.^2; p = total((G<=2).*P) p = 0.6662 % Theoretical = 119/179 = 0.6648
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f X Y ( t , u ) = 12 227 ( 3 t + 2 t u ) , for 0 t 2 , 0 u min { 1 + t , 2 } (see Exercise 20 from "Problems on Random Variables and Joint Distributions")

Z = I M ( X , Y ) X + I M c ( X , Y ) Y X , M = { ( t , u ) : u min ( 1 , 2 - t ) }

Detemine P ( Z 1 ) .

Figure three, for exercise 15, is comprised of one cartesian graph, with only the first quadrant displayed. The horizontal axis is labeled t, and the vertical axis is labeled u. The values on both axes range from 0 to 2 in increments of 1. There are 6 labeled line segments, one unlabeled line segment, and two resulting shaded shapes in this figure. The line segment from (0, 1) to (1, 2) is labeled u = 1 + t. The line segment from (0, 1) to (1, 1) is labeled u = 1. The line segment from (1, 2) to (2, 2) is labeled u = 2. The line segment from (1, 1) to (2, 2) is labeled u = t. The line segment from (1, 1) to (2, 0) is labeled u = 2 - t. The line segment from (2, 0) to (2, 2) is labeled t = 2. The unlabeled line segment is drawn from (1, 0) to (1, 1). A shaded square resulting from the drawn segments is bound by u = 1, the vertical axis, the horizontal axis, and the unlabeled line segment. A shaded triangle is bound by u = t, u = 2 - t, and t = 2. Figure three, for exercise 15, is comprised of one cartesian graph, with only the first quadrant displayed. The horizontal axis is labeled t, and the vertical axis is labeled u. The values on both axes range from 0 to 2 in increments of 1. There are 6 labeled line segments, one unlabeled line segment, and two resulting shaded shapes in this figure. The line segment from (0, 1) to (1, 2) is labeled u = 1 + t. The line segment from (0, 1) to (1, 1) is labeled u = 1. The line segment from (1, 2) to (2, 2) is labeled u = 2. The line segment from (1, 1) to (2, 2) is labeled u = t. The line segment from (1, 1) to (2, 0) is labeled u = 2 - t. The line segment from (2, 0) to (2, 2) is labeled t = 2. The unlabeled line segment is drawn from (1, 0) to (1, 1). A shaded square resulting from the drawn segments is bound by u = 1, the vertical axis, the horizontal axis, and the unlabeled line segment. A shaded triangle is bound by u = t, u = 2 - t, and t = 2.
P ( Z 1 ) = P ( ( X , Y ) M 1 Q 1 M 2 Q 2 ) , M 1 = M , M 2 = M c
Q 1 = { ( t , u ) : 0 t 1 } Q 2 = { ( t , u ) : u t } (see figure)
P = 12 227 0 1 0 1 ( 3 t + 2 t u ) d u d t + 12 227 1 2 2 - t t ( 3 t + 2 t u ) d u d t = 124 227
tuappr Enter matrix [a b]of X-range endpoints [0 2] Enter matrix [c d]of Y-range endpoints [0 2] Enter number of X approximation points 400Enter number of Y approximation points 400 Enter expression for joint density (12/227)*(3*t+2*t.*u).*(u<=min(1+t,2)) Use array operations on X, Y, PX, PY, t, u, and PQ = (u<=1).*(t<=1) + (t>1).*(u>=2-t).*(u<=t); P = total(Q.*P)P = 0.5478 % Theoretical = 124/227 = 0.5463
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The class { X , Y , Z } is independent.

          X = - 2 I A + I B + 3 I C . Minterm probabilities are (in the usual order)

0 . 255 0 . 025 0 . 375 0 . 045 0 . 108 0 . 012 0 . 162 0 . 018

          Y = I D + 3 I E + I F - 3 . The class { D , E , F } is independent with

P ( D ) = 0 . 32 P ( E ) = 0 . 56 P ( F ) = 0 . 40

          Z has distribution

Value -1.3 1.2 2.7 3.4 5.8
Probability 0.12 0.24 0.43 0.13 0.08

Determine P ( X 2 + 3 X Y 2 > 3 Z ) .

% file npr10_16.m Data for [link] cx = [-2 1 3 0];pmx = 0.001*[255 25 375 45 108 12 162 18];cy = [1 3 1 -3];pmy = minprob(0.01*[32 56 40]);Z = [-1.3 1.2 2.7 3.4 5.8];PZ = 0.01*[12 24 43 13 8];disp('Data are in cx, pmx, cy, pmy, Z, PZ') npr10_16 % Call for dataData are in cx, pmx, cy, pmy, Z, PZ [X,PX]= canonicf(cx,pmx); [Y,PY]= canonicf(cy,pmy); icalc3Enter row matrix of X-values X Enter row matrix of Y-values YEnter row matrix of Z-values Z Enter X probabilities PXEnter Y probabilities PY Enter Z probabilities PZUse array operations on matrices X, Y, Z, PX, PY, PZ, t, u, v, and PM = t.^2 + 3*t.*u.^2>3*v; PM = total(M.*P)PM = 0.3587
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The simple random variable X has distribution

X = [ - 3 . 1 - 0 . 5 1 . 2 2 . 4 3 . 7 4 . 9 ] P X = [ 0 . 15 0 . 22 0 . 33 0 . 12 0 . 11 0 . 07 ]
  1. Plot the distribution function F X and the quantile function Q X .
  2. Take a random sample of size n = 10 , 000 . Compare the relative frequency for each value with the probability that value is taken on.
X = [-3.1 -0.5 1.2 2.4 3.7 4.9];PX = 0.01*[15 22 33 12 11 7];ddbn Enter row matrix of VALUES XEnter row matrix of PROBABILITIES PX % Plot not reproduced here dquanplotEnter VALUES for X X Enter PROBABILITIES for X PX % Plot not reproduced hererand('seed',0) % Reset random number generator dsample % for comparison purposesEnter row matrix of VALUES X Enter row matrix of PROBABILITIES PXSample size n 10000 Value Prob Rel freq-3.1000 0.1500 0.1490 -0.5000 0.2200 0.21641.2000 0.3300 0.3340 2.4000 0.1200 0.11843.7000 0.1100 0.1070 4.9000 0.0700 0.0752Sample average ex = 0.8792 Population mean E[X]= 0.859 Sample variance vx = 5.146Population variance Var[X] = 5.112
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Questions & Answers

what is Nano technology ?
Bob Reply
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
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Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
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Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
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Adin
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Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
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it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
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Cied
how did you get the value of 2000N.What calculations are needed to arrive at it
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A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive
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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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