The trial balances shown below are before and after adjustment for he.pdf
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The trial balances shown below are before and after adjustment for here Company at the end of its fiscal year. Solution DR CR Account Receivable 750 Service Revenue 750 Account Receivable(8380-9130) (to record service revenue) Supplies Expense 1640 Supplies 1640 (Supplies (2900- 1260) Insurance expenses 1490 Prepaid Inurance 1490 (Prepaid inurance (4500-3010) Depreciation 1200 Accumulated depreciation 1200 ACC dep (4600-3400) Salaries and wages 1200 Salaries and wage Payable 1200 Unearned rent revenue 840 Rent revenue 840.
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This starts off with five points equally spaced along the x-axis. Thu.pdf
This starts off with five points equally spaced along the x-axis. Thus they initially lie on a line
with intercept zero and slope zero. Then each point is nudged randomly, as if they were moving
under Brownian motion (like pollen grains in water seen under a microscope). Run the file to see
the five points as they slowly drift about randomly But there is a problem! The animation is
supposed to update the best-fit line but it just stays motionless along the x-axis. Study the code to
see how it works. Then add one or two lines at the indicated spot, so the best-fit line updates
every time the points drift.
Solution
/*********************************************************
* Water molecules colliding with pollen particles,
* causing a random motion.
*********************************************************/
var BACKGROUND = color(230, 240, 250);
frameRate(30);
// Physical variables to play with
var gravity = 0; // Force due to gravity
var friction = 0.9; // Energy lost on collision
var wallTemp = 1; // Speed of atoms in the walls
// Initial particle speed
var initialSpeed = 5;
// How many iterations before updating the display
var framesNotShown = 3;
var waterCount = 140;
var waterSize = 2;
var waterMass = 10;
var pollenCount = 4;
var pollenSize = 25;
var pollenMass = 100;
var dotProduct = function(ax, ay, bx, by) {
return ax * bx + ay * by;
};
// Particles have a position, speed, colour and masss
var Particle = function(x, y, r, c, mass) {
// Position
this.x = x;
this.y = y;
// Radius
this.r = r;
this.mass = mass || this.r * this.r;
// Colour
this.c = c;
// Velocity
this.dx = initialSpeed * (random() - 0.5) / this.r;
this.dy = initialSpeed * (random() - 0.5) / this.r;
};
Particle.prototype.draw = function() {
fill(this.c);
ellipse(this.x, this.y, this.r * 2, this.r * 2);
};
// Move ball based on its velocity
Particle.prototype.move = function() {
this.x += this.dx;
this.y += this.dy;
};
Particle.prototype.collide = function(that) {
var dx = this.x - that.x;
var dy = this.y - that.y;
var dr = this.r + that.r;
var d = dx * dx + dy * dy;
if (d < dr * dr) {
// Particles collide
var collisionDist = sqrt(d + 0.1);
// Find unit vector in direction of collision
var collisionVi = dx / collisionDist;
var collisionVj = dy / collisionDist;
// Find velocity of particle projected on to collision vector
var collisionV1 = dotProduct(this.dx, this.dy, dx, dy) / collisionDist;
var collisionV2 = dotProduct(that.dx, that.dy, dx, dy) / collisionDist;
// Find velocity of particle perpendicular to collision vector
var perpV1 = dotProduct(this.dx, this.dy, -dy, dx) / collisionDist;
var perpV2 = dotProduct(that.dx, that.dy, -dy, dx) / collisionDist;
// Find movement in direction of collision
var sumMass = this.mass + that.mass;
var diffMass = this.mass - that.mass;
var v1p = (diffMass * collisionV1 + 2 * that.mass * collisionV2) / sumMass;
var v2p = (2 * this.mass * collisionV1 - diffMass * collisionV2) / sumMass;
// Update velocities
this.dx = v1p * collisionVi - per.
This it an elaboration of Problem 14. Assume U is a unit vector. In w.pdf
This it an elaboration of Problem 14. Assume U is a unit vector. In what direction u docs the
function f(x.y) = sin (3r - y) increase most rapidly at the point (pi /6, pi/4)? Assume u is a unit
vector. u = . In what direction V (again, a unit vector) docs the same function increase most
rapidly at the point ( 0.pi)? v = .
Solution
3/4.
The system was initially at rest in static equilibrium condition. If .pdf
The system was initially at rest in static equilibrium condition. If p(t) = sin(omega t) is applied
at t = 0, write the response of total displacement u(t).
Solution
Displacement in case of cantilever is given as (PL^3)/3EI
Where P is the external point loading
Thus U(t) should be PL^3/3EI.
The Thiessen polygons method is a common method used to interpolate p.pdf
The Thiessen polygons method is a common method used to interpolate precipitation data over
time. Two common methods to develop a unit hydrograph for ungagged basins are the Snyder
and the SCS. The temperature is the only environmental factor affecting potential
evapotranspiration. Both the phi Index and the Green-Ampt are common methods in engineering
to estimate water infiltration. Early this week the Mason Campus was closed due the effects of
Winter Storm Titan. This storm is a clear example of a convective precipitation event. The Time
of rise (Tr) as defined when using the unit hydrograph method represents the total duration of the
runoff. Base flow in a stream results from flows from the groundwater into the stream.
Hydraulics is the field that studies the occurrence, distribution, movement and properties of
water on earth. Rain gages are instruments used to measure precipitation at a point scale. On the
other hand radars such as the NEXRAD can be used to measure precipitation over a larger spatial
scale. Important parameters for design storms are volume, duration, frequency and intensity
The rational method is designed to calculate runoff for very large watersheds (e.g.: Mississippi
river watershed) given the runoff coefficient, rainfall intensity and watershed areas. River flow
is usually determined by measuring water levels and applying a rating curse to convert water
levels to discharge.
Solution
1. True - theissen polygon method is the most common method used to calculate precipitation
data over time and when there is more than one measurement.in this the area is divided into no.of
polygons eachone around a measurement point,and then take a weighted avg of polygon based
on size of each polygon.
2. True - these both snyder and SCS are the two methods along with commons dimensionless
method.in these three methods snyder is the mostly used method to develop unit hydrograph for
ungagged basins.
3. False - the evapotransparation process is not only effected by temperature but also due to soil
characteristics,physical attributes of vegitation,water availaability etc.
4. False - green and ampt method is used to estimate water infiltration along with some other
methods like richards equation,hoston equation,darchys law etc.and phi-index method is used to
compute excess rainfall hetrograph using total rainfall hetrograph using constant phi.
5.True - this strom is a convective precipitation as these type of rainfall mostly occurs in the
form of snow or freezing liquids
6. False - the time of rise is defined as the time passt a specified point in rising limb.
7. True - the portion of stream flow that is not runoff and results from seepage of water from the
ground into a channel slowly over time.
8. False - hydraulics is a science and technology concerned with the conveyance of liquids
through pipes and channels, especially as a source of mechanical force.
9. True - NEXRAD is a network of 160 high-resolution S-band Doppler weather radars opera.
The two-asset case The expected return for asset A is 6.25 with a s.pdf
The two-asset case The expected return for asset A is 6.25% with a standard deviation of
5.00%, and the expected return for asset B is 4.50% with a standard deviation of 5.00%. Based
on your knowledge of efficient portfolis, fill in the blanks in the following table with the
appropriate answers. What is the efficient proportion of asset A within the portfolio for case II?
0.50 1.00 0.00 0.43
Solution
Answer
Answer 1
Security A
Security B
Expected portfolio return
Proportion
Expected Return
SD
Proportion
Expected Return
SD
A
B
C
D
E
F
(A*B)+(D*E)
1
6.25
5
0
4.5
5
6.25
0.75
6.25
5
0.25
4.5
5
5.81
0.5
6.25
5
0.5
4.5
5
5.38
0.25
6.25
5
0.75
4.5
5
4.94
0
6.25
5
1
4.5
5
4.50
CASE I
Security A
Security B
SD of Portfolio
Proportion
SD
Proportion
SD
Pab
A
M
C
N
O
D
P
F
Q
R
S
T
U
A*A
C*C
M*N
D*D
F*F
P*Q
2*A*C*D*F*S
O+R+T
Squreroot of U
1.00
1.0
5.0
25.0
25.0
0.00
0.0
5.0
25.0
0.0
-0.5
0.0
25.0
5.0
0.75
0.6
5.0
25.0
14.1
0.25
0.1
5.0
25.0
1.6
-0.5
-4.7
10.9
3.3
0.50
0.3
5.0
25.0
6.3
0.50
0.3
5.0
25.0
6.3
-0.5
-6.3
6.3
2.5
0.25
0.1
5.0
25.0
1.6
0.75
0.6
5.0
25.0
14.1
-0.5
-4.7
10.9
3.3
0.00
0.0
5.0
25.0
0.0
1.00
1.0
5.0
25.0
25.0
-0.5
0.0
25.0
5.0
CASE II
Security A
Security B
SD of Portfolio
Proportion
SD
Proportion
SD
Pab
A
M
C
N
O
D
P
F
Q
R
S
T
U
A*A
C*C
M*N
D*D
F*F
P*Q
2*A*C*D*F*S
O+R+T
Squreroot of U
1.00
1.0
5.0
25.0
25.0
0.00
0.0
5.0
25.0
0.0
0.4
0.0
25.0
5.0
0.75
0.6
5.0
25.0
14.1
0.25
0.1
5.0
25.0
1.6
0.4
3.8
19.4
4.4
0.50
0.3
5.0
25.0
6.3
0.50
0.3
5.0
25.0
6.3
0.4
5.0
17.5
4.2
0.25
0.1
5.0
25.0
1.6
0.75
0.6
5.0
25.0
14.1
0.4
3.8
19.4
4.4
0.00
0.0
5.0
25.0
0.0
1.00
1.0
5.0
25.0
25.0
0.4
0.0
25.0
5.0
CASE III
Security A
Security B
SD of Portfolio
Proportion
SD
Proportion
SD
Pab
A
M
C
N
O
D
P
F
Q
R
S
T
U
A*A
C*C
M*N
D*D
F*F
P*Q
2*A*C*D*F*S
O+R+T
Squreroot of U
1.00
1.0
5.0
25.0
25.0
0.00
0.0
5.0
25.0
0.0
0.7
0.0
25.0
5.0
0.75
0.6
5.0
25.0
14.1
0.25
0.1
5.0
25.0
1.6
0.7
6.6
22.2
4.7
0.50
0.3
5.0
25.0
6.3
0.50
0.3
5.0
25.0
6.3
0.7
8.8
21.3
4.6
0.25
0.1
5.0
25.0
1.6
0.75
0.6
5.0
25.0
14.1
0.7
6.6
22.2
4.7
0.00
0.0
5.0
25.0
0.0
1.00
1.0
5.0
25.0
25.0
0.7
0.0
25.0
5.0
Answer 2
What is the efficient proportion of asset A within the portfolio for CASE II?
Efficient portfolio offers the highest expected return for a defined level of risk or the lowest risk
for a given level of expected return.
Answer : efficient proportion of asset A is 0.5 ( Where standard deviation (SD) of portfolio is
lowest i.e. 4.2%)
Security A
Security B
Expected portfolio return
Proportion
Expected Return
SD
Proportion
Expected Return
SD
A
B
C
D
E
F
(A*B)+(D*E)
1
6.25
5
0
4.5
5
6.25
0.75
6.25
5
0.25
4.5
5
5.81
0.5
6.25
5
0.5
4.5
5
5.38
0.25
6.25
5
0.75
4.5
5
4.94
0
6.25
5
1
4.5
5
4.50.
This 4 pts The statement of cash flows OA OB Oc OD dlassifies a firm.pdf
This :4 pts The statement of cash flows OA OB Oc OD dlassifies a firm\'s cash fows as
operating, investing, financing, and other activities shows the financial position of a firm at a
given point of time summarizes all the purchase and sale of fixed assets and raw materials
provides insight into a firm\'s operating, investment, and financing cash flows
Solution
The statement of cash flows
Answer
Classifies cashflows as operating, investing, financing and other activities.
Which only focuses on the actual cash balance of the firm
Q2 the present value of a 25000 perpetuity at a 14 percent discount rate is
PV = D/r
Where:-
D = Dividend or coupon period
r = discount rate
PV = Present Value
PV = 25,000/0.14 = $178,571
Q3. Which off the following is true about annuities?
Answer:-
An annuity is an equal stream of cash flows is paid or recieved at the beginning of each period.
There are41students in an elementary statistics class. On the basis.pdf
There are41students in an elementary statistics class. On the basis of years of experience, the
instructor knows that the time needed to grade a randomly chosen first examination paper is a
random variable with an expected value of5min and a standard deviation of4min. (Round your
answers to four decimal places.)
Solution
a) z=(10-5)/(4/sqrt(41))=8.00
probability that he is through grading before the11:00 P.M.TV news begins = 1
b)probability that he misses part of the report if he waits until grading is done before turning on
the TV = 0.
There does not exist a fish that can live on land. Write an equ.pdf
??There does not exist a fish that can live on land.?? Write an equivalent English statement that
begins with the words ??Every fish??? Let P(x) be the predicate ??x can live on land,?? where
the variable x represents animals. Write both statements symbolically using P(x) and quantifiers,
( There exists). For all
Solution
I\'m adding a predicate N(x) to represent \"x is a animal\".
1. Every fish is unable to live on land.
?x N(x) -> ~P(x)
2.The original statement might be represented as
~?x N(x) & P(x).
Recommended
This starts off with five points equally spaced along the x-axis. Thu.pdf
This starts off with five points equally spaced along the x-axis. Thus they initially lie on a line
with intercept zero and slope zero. Then each point is nudged randomly, as if they were moving
under Brownian motion (like pollen grains in water seen under a microscope). Run the file to see
the five points as they slowly drift about randomly But there is a problem! The animation is
supposed to update the best-fit line but it just stays motionless along the x-axis. Study the code to
see how it works. Then add one or two lines at the indicated spot, so the best-fit line updates
every time the points drift.
Solution
/*********************************************************
* Water molecules colliding with pollen particles,
* causing a random motion.
*********************************************************/
var BACKGROUND = color(230, 240, 250);
frameRate(30);
// Physical variables to play with
var gravity = 0; // Force due to gravity
var friction = 0.9; // Energy lost on collision
var wallTemp = 1; // Speed of atoms in the walls
// Initial particle speed
var initialSpeed = 5;
// How many iterations before updating the display
var framesNotShown = 3;
var waterCount = 140;
var waterSize = 2;
var waterMass = 10;
var pollenCount = 4;
var pollenSize = 25;
var pollenMass = 100;
var dotProduct = function(ax, ay, bx, by) {
return ax * bx + ay * by;
};
// Particles have a position, speed, colour and masss
var Particle = function(x, y, r, c, mass) {
// Position
this.x = x;
this.y = y;
// Radius
this.r = r;
this.mass = mass || this.r * this.r;
// Colour
this.c = c;
// Velocity
this.dx = initialSpeed * (random() - 0.5) / this.r;
this.dy = initialSpeed * (random() - 0.5) / this.r;
};
Particle.prototype.draw = function() {
fill(this.c);
ellipse(this.x, this.y, this.r * 2, this.r * 2);
};
// Move ball based on its velocity
Particle.prototype.move = function() {
this.x += this.dx;
this.y += this.dy;
};
Particle.prototype.collide = function(that) {
var dx = this.x - that.x;
var dy = this.y - that.y;
var dr = this.r + that.r;
var d = dx * dx + dy * dy;
if (d < dr * dr) {
// Particles collide
var collisionDist = sqrt(d + 0.1);
// Find unit vector in direction of collision
var collisionVi = dx / collisionDist;
var collisionVj = dy / collisionDist;
// Find velocity of particle projected on to collision vector
var collisionV1 = dotProduct(this.dx, this.dy, dx, dy) / collisionDist;
var collisionV2 = dotProduct(that.dx, that.dy, dx, dy) / collisionDist;
// Find velocity of particle perpendicular to collision vector
var perpV1 = dotProduct(this.dx, this.dy, -dy, dx) / collisionDist;
var perpV2 = dotProduct(that.dx, that.dy, -dy, dx) / collisionDist;
// Find movement in direction of collision
var sumMass = this.mass + that.mass;
var diffMass = this.mass - that.mass;
var v1p = (diffMass * collisionV1 + 2 * that.mass * collisionV2) / sumMass;
var v2p = (2 * this.mass * collisionV1 - diffMass * collisionV2) / sumMass;
// Update velocities
this.dx = v1p * collisionVi - per.
This it an elaboration of Problem 14. Assume U is a unit vector. In w.pdf
This it an elaboration of Problem 14. Assume U is a unit vector. In what direction u docs the
function f(x.y) = sin (3r - y) increase most rapidly at the point (pi /6, pi/4)? Assume u is a unit
vector. u = . In what direction V (again, a unit vector) docs the same function increase most
rapidly at the point ( 0.pi)? v = .
Solution
3/4.
The system was initially at rest in static equilibrium condition. If .pdf
The system was initially at rest in static equilibrium condition. If p(t) = sin(omega t) is applied
at t = 0, write the response of total displacement u(t).
Solution
Displacement in case of cantilever is given as (PL^3)/3EI
Where P is the external point loading
Thus U(t) should be PL^3/3EI.
The Thiessen polygons method is a common method used to interpolate p.pdf
The Thiessen polygons method is a common method used to interpolate precipitation data over
time. Two common methods to develop a unit hydrograph for ungagged basins are the Snyder
and the SCS. The temperature is the only environmental factor affecting potential
evapotranspiration. Both the phi Index and the Green-Ampt are common methods in engineering
to estimate water infiltration. Early this week the Mason Campus was closed due the effects of
Winter Storm Titan. This storm is a clear example of a convective precipitation event. The Time
of rise (Tr) as defined when using the unit hydrograph method represents the total duration of the
runoff. Base flow in a stream results from flows from the groundwater into the stream.
Hydraulics is the field that studies the occurrence, distribution, movement and properties of
water on earth. Rain gages are instruments used to measure precipitation at a point scale. On the
other hand radars such as the NEXRAD can be used to measure precipitation over a larger spatial
scale. Important parameters for design storms are volume, duration, frequency and intensity
The rational method is designed to calculate runoff for very large watersheds (e.g.: Mississippi
river watershed) given the runoff coefficient, rainfall intensity and watershed areas. River flow
is usually determined by measuring water levels and applying a rating curse to convert water
levels to discharge.
Solution
1. True - theissen polygon method is the most common method used to calculate precipitation
data over time and when there is more than one measurement.in this the area is divided into no.of
polygons eachone around a measurement point,and then take a weighted avg of polygon based
on size of each polygon.
2. True - these both snyder and SCS are the two methods along with commons dimensionless
method.in these three methods snyder is the mostly used method to develop unit hydrograph for
ungagged basins.
3. False - the evapotransparation process is not only effected by temperature but also due to soil
characteristics,physical attributes of vegitation,water availaability etc.
4. False - green and ampt method is used to estimate water infiltration along with some other
methods like richards equation,hoston equation,darchys law etc.and phi-index method is used to
compute excess rainfall hetrograph using total rainfall hetrograph using constant phi.
5.True - this strom is a convective precipitation as these type of rainfall mostly occurs in the
form of snow or freezing liquids
6. False - the time of rise is defined as the time passt a specified point in rising limb.
7. True - the portion of stream flow that is not runoff and results from seepage of water from the
ground into a channel slowly over time.
8. False - hydraulics is a science and technology concerned with the conveyance of liquids
through pipes and channels, especially as a source of mechanical force.
9. True - NEXRAD is a network of 160 high-resolution S-band Doppler weather radars opera.
The two-asset case The expected return for asset A is 6.25 with a s.pdf
The two-asset case The expected return for asset A is 6.25% with a standard deviation of
5.00%, and the expected return for asset B is 4.50% with a standard deviation of 5.00%. Based
on your knowledge of efficient portfolis, fill in the blanks in the following table with the
appropriate answers. What is the efficient proportion of asset A within the portfolio for case II?
0.50 1.00 0.00 0.43
Solution
Answer
Answer 1
Security A
Security B
Expected portfolio return
Proportion
Expected Return
SD
Proportion
Expected Return
SD
A
B
C
D
E
F
(A*B)+(D*E)
1
6.25
5
0
4.5
5
6.25
0.75
6.25
5
0.25
4.5
5
5.81
0.5
6.25
5
0.5
4.5
5
5.38
0.25
6.25
5
0.75
4.5
5
4.94
0
6.25
5
1
4.5
5
4.50
CASE I
Security A
Security B
SD of Portfolio
Proportion
SD
Proportion
SD
Pab
A
M
C
N
O
D
P
F
Q
R
S
T
U
A*A
C*C
M*N
D*D
F*F
P*Q
2*A*C*D*F*S
O+R+T
Squreroot of U
1.00
1.0
5.0
25.0
25.0
0.00
0.0
5.0
25.0
0.0
-0.5
0.0
25.0
5.0
0.75
0.6
5.0
25.0
14.1
0.25
0.1
5.0
25.0
1.6
-0.5
-4.7
10.9
3.3
0.50
0.3
5.0
25.0
6.3
0.50
0.3
5.0
25.0
6.3
-0.5
-6.3
6.3
2.5
0.25
0.1
5.0
25.0
1.6
0.75
0.6
5.0
25.0
14.1
-0.5
-4.7
10.9
3.3
0.00
0.0
5.0
25.0
0.0
1.00
1.0
5.0
25.0
25.0
-0.5
0.0
25.0
5.0
CASE II
Security A
Security B
SD of Portfolio
Proportion
SD
Proportion
SD
Pab
A
M
C
N
O
D
P
F
Q
R
S
T
U
A*A
C*C
M*N
D*D
F*F
P*Q
2*A*C*D*F*S
O+R+T
Squreroot of U
1.00
1.0
5.0
25.0
25.0
0.00
0.0
5.0
25.0
0.0
0.4
0.0
25.0
5.0
0.75
0.6
5.0
25.0
14.1
0.25
0.1
5.0
25.0
1.6
0.4
3.8
19.4
4.4
0.50
0.3
5.0
25.0
6.3
0.50
0.3
5.0
25.0
6.3
0.4
5.0
17.5
4.2
0.25
0.1
5.0
25.0
1.6
0.75
0.6
5.0
25.0
14.1
0.4
3.8
19.4
4.4
0.00
0.0
5.0
25.0
0.0
1.00
1.0
5.0
25.0
25.0
0.4
0.0
25.0
5.0
CASE III
Security A
Security B
SD of Portfolio
Proportion
SD
Proportion
SD
Pab
A
M
C
N
O
D
P
F
Q
R
S
T
U
A*A
C*C
M*N
D*D
F*F
P*Q
2*A*C*D*F*S
O+R+T
Squreroot of U
1.00
1.0
5.0
25.0
25.0
0.00
0.0
5.0
25.0
0.0
0.7
0.0
25.0
5.0
0.75
0.6
5.0
25.0
14.1
0.25
0.1
5.0
25.0
1.6
0.7
6.6
22.2
4.7
0.50
0.3
5.0
25.0
6.3
0.50
0.3
5.0
25.0
6.3
0.7
8.8
21.3
4.6
0.25
0.1
5.0
25.0
1.6
0.75
0.6
5.0
25.0
14.1
0.7
6.6
22.2
4.7
0.00
0.0
5.0
25.0
0.0
1.00
1.0
5.0
25.0
25.0
0.7
0.0
25.0
5.0
Answer 2
What is the efficient proportion of asset A within the portfolio for CASE II?
Efficient portfolio offers the highest expected return for a defined level of risk or the lowest risk
for a given level of expected return.
Answer : efficient proportion of asset A is 0.5 ( Where standard deviation (SD) of portfolio is
lowest i.e. 4.2%)
Security A
Security B
Expected portfolio return
Proportion
Expected Return
SD
Proportion
Expected Return
SD
A
B
C
D
E
F
(A*B)+(D*E)
1
6.25
5
0
4.5
5
6.25
0.75
6.25
5
0.25
4.5
5
5.81
0.5
6.25
5
0.5
4.5
5
5.38
0.25
6.25
5
0.75
4.5
5
4.94
0
6.25
5
1
4.5
5
4.50.
This 4 pts The statement of cash flows OA OB Oc OD dlassifies a firm.pdf
This :4 pts The statement of cash flows OA OB Oc OD dlassifies a firm\'s cash fows as
operating, investing, financing, and other activities shows the financial position of a firm at a
given point of time summarizes all the purchase and sale of fixed assets and raw materials
provides insight into a firm\'s operating, investment, and financing cash flows
Solution
The statement of cash flows
Answer
Classifies cashflows as operating, investing, financing and other activities.
Which only focuses on the actual cash balance of the firm
Q2 the present value of a 25000 perpetuity at a 14 percent discount rate is
PV = D/r
Where:-
D = Dividend or coupon period
r = discount rate
PV = Present Value
PV = 25,000/0.14 = $178,571
Q3. Which off the following is true about annuities?
Answer:-
An annuity is an equal stream of cash flows is paid or recieved at the beginning of each period.
There are41students in an elementary statistics class. On the basis.pdf
There are41students in an elementary statistics class. On the basis of years of experience, the
instructor knows that the time needed to grade a randomly chosen first examination paper is a
random variable with an expected value of5min and a standard deviation of4min. (Round your
answers to four decimal places.)
Solution
a) z=(10-5)/(4/sqrt(41))=8.00
probability that he is through grading before the11:00 P.M.TV news begins = 1
b)probability that he misses part of the report if he waits until grading is done before turning on
the TV = 0.
There does not exist a fish that can live on land. Write an equ.pdf
??There does not exist a fish that can live on land.?? Write an equivalent English statement that
begins with the words ??Every fish??? Let P(x) be the predicate ??x can live on land,?? where
the variable x represents animals. Write both statements symbolically using P(x) and quantifiers,
( There exists). For all
Solution
I\'m adding a predicate N(x) to represent \"x is a animal\".
1. Every fish is unable to live on land.
?x N(x) -> ~P(x)
2.The original statement might be represented as
~?x N(x) & P(x).
There are twelve astrological signs Aries, Taurus, Gemini, Cancer, L.pdf
There are twelve astrological signs: Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, Scorpio,
Sagittarius, Capricorn, Aquarius, and Pisces. For example, your sign is Pisces if your birthday
falls between February 19th and March 20th. Assuming all the astrological signs are equally
likely for birthdays, find the probability that among a set of n people, at least two have the same
astrological sign. Hint: remember the pigeonhole principle. Find the minimum number of people
needed to guarantee a greater than 50% chance that two of them share the same astrological sign.
Solution
a) total no of events(sample space) = 12^n Atleast two having same sign = total - no
one having same sign =(12^n) - (12*11*10*...*(13-n)) if n<=12 = 0 if n>12 by pegion hole
principle P(E) = 1 - ( ( 12*11*10*....*(13-n) ) / 12^n ) ----if n<=12 = 0 -------------------------------
------if n>12 b)P(E) > 1/2 So 12*11*10...*13-n / 12^n < 1/2 Solving we get n = 5.
There are two containers. One contains the letters BUMBLEBEES the oth.pdf
There are two containers. One contains the letters BUMBLEBEES the other the letters
KOMONDOROK. You choose a container by coin toss and then randomly select a letter. Find
the following probabilities. P(0) = P( vowel) =
Solution
0.5*0.4 =0.2 0.5*0.4 + 0.5*0.4 =0.4.
There are several documentaries about the engineering disasters assoc.pdf
There are several documentaries about the engineering disasters associated with Hurricane
Katrina There are equally as many publications on the topic. Review at least three publications
(not blogs or media reports or public opinion-based sources) to address the following questions
What failed Why it failed Possible corrective actions Who was at fault and why Write a 3-5
page review addressing the afore-mentioned topics and include the following as well Describe in
detail what ethical issues may have been partly/wholly responsible for the failure Be sure to
address lack of professionalism or conscientiousness that may have helped lead to this failure
You must cite references appropriately and there must be at least three credible references.
Solution
Engineering Review of the Disaster caused by Hurricane Katrina
Even though 5 engineering review bodies operated separately, all of them reacted to IPET’s
work, the main and finest funded of the study groups. Volumes 2 through 7 of the IPET “draft
final” report divide the results into six broad sections:
1. Geodetic Vertical and Water Level Datums
2. Hurricane Protection System
3. The Storm
4. The Performance of the Levees and Floodwalls
5. Performance of Interior Drainage and Pumping
6. The Consequences; and Engineering and Operational Risk and Reliability Analyses
The report is more than 6,000 pages extensive so far, and the last section on risk and reliability
has not yet appeared. The NAE/NRC team and the ASCE team both prepared their commentary
to conform to the IPET pattern.
In reviewing a study of this kind, it is significant to keep in mind that critiques may address the
quality of the design and construction for facilities in place at the time of the hurricane or the
adequacy of the post-hurricane investigations described in the report. For example, one might
find that the wind and water levels used for the design calculations were not adequate but that the
hind-casting calculations of the hurricane’s effects described in the report are excellent. The
NAE/NRC team and the ASCE team had similar reactions, with some minor differences of
detail.
The detailed study of the geodetic levels revealed considerable room for confusion and error.
Two different benchmark levels were used in creating the levee system. In some cases, the water
levels were expressed against one benchmark and the height of the levees against another.
Furthermore, there seemed to be no consistent effort to monitor subsidence of the levee system
or its components. The focus was on building the levees to the “authorized” elevations without
considering whether the corresponding water elevations were measured from the same base or
whether subsequent settlement and subsidence might make the authorized levels irrelevant.
Ethical Issues:
All of the review teams found that the hurricane protection system was a system in name only.
Planning and system-wide design were confused by political and short-term economic
considerations over.
There are many famous (more properly, infamous) examples of catastrop.pdf
There are many famous (more properly, infamous) examples of catastrophic failures of major
engineering structures and systems. Use the Internet to look up each of the following and briefly
describe whether the failure involved brittle or ductile failure initially (as a cause) and brittle or
ductile failure overall (macroscopically). Tacoma (WA) Narrows Bridge, known as Galloping
Gertie Minnesota I-35W Bridge Space Shuttle Challenger Space Shuttle Columbia RMS
Titanic Aloha Airlines Flight 243
Solution
a. Tacoma bridge failure is considered a ductile failure, because the collapse due to the wind
vibrations lasted for long time so that no human casualities were made. Only one dog was the
casuality.
2. The failure of I-35 bridge in Minnesota was a brittle failure because it occured suddenly and
caused enormous catastrophic consequences.
c. Space shuttle Challenger disaster occured due to the occurence of the brittle failure cracks in
O-rings which allowed the fuel leakage and thereby the explosion. No one in the shuttle
surviced, therefore it is a brittle failure. O-rings are ductile in nature but, at low temperature they
become brittle, so such failures come in category of ductile-to-brittle transition failures.
d. Space Shuttle Columbia failure occured due to the damage to the thermal shielding tiles from
impact with a piece of foam, which fell from the insulation of an external tank during the take
off. Failure occured while returning to the earth, while its speed was very very high, and wing
experienced a high temperature of 1540 degree celsius. Damage to the thermal shield panel was
tested to be a brittle failure, and therefore it also comes in category of brittle failure.
e. RMS titanic was floating at very cold temperature and when near freezing temperature was
met, tests showed that steel became extremely brittle. It was also a ductile-to-brittle transition
failure. Overall ship sunk in 2 hours and about 15 minutes, so overall it could be considered a
ductile failure, but no help was available in that time.
f. It was a gatigue failure, which progressed in 19 years, but the end failure was brittle, therefore
a brittle failure. Overall (maroscopically) failure was a ductile failure, since it happened due to
the corrosion and fatigue of 19 years..
The Warren truss is used to support a staircase. Determine the force .pdf
The Warren truss is used to support a staircase. Determine the force in members CE, ED, and
DF, and state whether the members are in tension or compression. Assume all joints are pinned.
Solution
D. ED = 0.400 kN C, DF = 2.26 kN T, CE = 4.53 kN C.
The Wind Mountain archaeological site is located in southwestern Ne.pdf
The Wind Mountain archaeological site is located in southwestern New Mexico. Wind
Mountain was home to an ancient culture of prehistoric Native Americans called Anasazi. A
random sample of excavations at Wind Mountain gave the following depths (in centimeters)
from present-day surface grade to the location of significant archaeological
artifacts%u2020.8545120807555656065959070756568
Solution
(b) Given a=0.02, |t(0.01, df=n-1=14)|=2.62
So 98% CI is
xbar+/- t*s/vn
--> 74.2+/- 2.62*18.3/sqrt(15)
--> (61.8204, 86.5796).
The waiting time Y until delivery of a new component for an industr.pdf
The waiting time Y until delivery of a new component for an industrial operation is uniformly
distributed over the interval from 1 to 5 days. The cost of this delay is given by U = 2Y2 + 3.
Use the method of transformations to derive the density function of U.
I have found that the distribution function of Y is P(Y <= y) = (y-1)/5. Why is this?
Solution
You know F_Y(y) because Y is uniformly distributed.
F_Y(y)= integral_1^y 1/4 dx= (y-1)/4 when y in [1,5]
= 0 when y<1
= 1 when y>5
1/4 because the length of the interval [1,5] is 4. the density function is f_y(y)=1/4 when y in
[1,5]..
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There are twelve astrological signs Aries, Taurus, Gemini, Cancer, L.pdf
There are twelve astrological signs: Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, Scorpio,
Sagittarius, Capricorn, Aquarius, and Pisces. For example, your sign is Pisces if your birthday
falls between February 19th and March 20th. Assuming all the astrological signs are equally
likely for birthdays, find the probability that among a set of n people, at least two have the same
astrological sign. Hint: remember the pigeonhole principle. Find the minimum number of people
needed to guarantee a greater than 50% chance that two of them share the same astrological sign.
Solution
a) total no of events(sample space) = 12^n Atleast two having same sign = total - no
one having same sign =(12^n) - (12*11*10*...*(13-n)) if n<=12 = 0 if n>12 by pegion hole
principle P(E) = 1 - ( ( 12*11*10*....*(13-n) ) / 12^n ) ----if n<=12 = 0 -------------------------------
------if n>12 b)P(E) > 1/2 So 12*11*10...*13-n / 12^n < 1/2 Solving we get n = 5.
There are two containers. One contains the letters BUMBLEBEES the oth.pdf
There are two containers. One contains the letters BUMBLEBEES the other the letters
KOMONDOROK. You choose a container by coin toss and then randomly select a letter. Find
the following probabilities. P(0) = P( vowel) =
Solution
0.5*0.4 =0.2 0.5*0.4 + 0.5*0.4 =0.4.
There are several documentaries about the engineering disasters assoc.pdf
There are several documentaries about the engineering disasters associated with Hurricane
Katrina There are equally as many publications on the topic. Review at least three publications
(not blogs or media reports or public opinion-based sources) to address the following questions
What failed Why it failed Possible corrective actions Who was at fault and why Write a 3-5
page review addressing the afore-mentioned topics and include the following as well Describe in
detail what ethical issues may have been partly/wholly responsible for the failure Be sure to
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You must cite references appropriately and there must be at least three credible references.
Solution
Engineering Review of the Disaster caused by Hurricane Katrina
Even though 5 engineering review bodies operated separately, all of them reacted to IPET’s
work, the main and finest funded of the study groups. Volumes 2 through 7 of the IPET “draft
final” report divide the results into six broad sections:
1. Geodetic Vertical and Water Level Datums
2. Hurricane Protection System
3. The Storm
4. The Performance of the Levees and Floodwalls
5. Performance of Interior Drainage and Pumping
6. The Consequences; and Engineering and Operational Risk and Reliability Analyses
The report is more than 6,000 pages extensive so far, and the last section on risk and reliability
has not yet appeared. The NAE/NRC team and the ASCE team both prepared their commentary
to conform to the IPET pattern.
In reviewing a study of this kind, it is significant to keep in mind that critiques may address the
quality of the design and construction for facilities in place at the time of the hurricane or the
adequacy of the post-hurricane investigations described in the report. For example, one might
find that the wind and water levels used for the design calculations were not adequate but that the
hind-casting calculations of the hurricane’s effects described in the report are excellent. The
NAE/NRC team and the ASCE team had similar reactions, with some minor differences of
detail.
The detailed study of the geodetic levels revealed considerable room for confusion and error.
Two different benchmark levels were used in creating the levee system. In some cases, the water
levels were expressed against one benchmark and the height of the levees against another.
Furthermore, there seemed to be no consistent effort to monitor subsidence of the levee system
or its components. The focus was on building the levees to the “authorized” elevations without
considering whether the corresponding water elevations were measured from the same base or
whether subsequent settlement and subsidence might make the authorized levels irrelevant.
Ethical Issues:
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Planning and system-wide design were confused by political and short-term economic
considerations over.
There are many famous (more properly, infamous) examples of catastrop.pdf
There are many famous (more properly, infamous) examples of catastrophic failures of major
engineering structures and systems. Use the Internet to look up each of the following and briefly
describe whether the failure involved brittle or ductile failure initially (as a cause) and brittle or
ductile failure overall (macroscopically). Tacoma (WA) Narrows Bridge, known as Galloping
Gertie Minnesota I-35W Bridge Space Shuttle Challenger Space Shuttle Columbia RMS
Titanic Aloha Airlines Flight 243
Solution
a. Tacoma bridge failure is considered a ductile failure, because the collapse due to the wind
vibrations lasted for long time so that no human casualities were made. Only one dog was the
casuality.
2. The failure of I-35 bridge in Minnesota was a brittle failure because it occured suddenly and
caused enormous catastrophic consequences.
c. Space shuttle Challenger disaster occured due to the occurence of the brittle failure cracks in
O-rings which allowed the fuel leakage and thereby the explosion. No one in the shuttle
surviced, therefore it is a brittle failure. O-rings are ductile in nature but, at low temperature they
become brittle, so such failures come in category of ductile-to-brittle transition failures.
d. Space Shuttle Columbia failure occured due to the damage to the thermal shielding tiles from
impact with a piece of foam, which fell from the insulation of an external tank during the take
off. Failure occured while returning to the earth, while its speed was very very high, and wing
experienced a high temperature of 1540 degree celsius. Damage to the thermal shield panel was
tested to be a brittle failure, and therefore it also comes in category of brittle failure.
e. RMS titanic was floating at very cold temperature and when near freezing temperature was
met, tests showed that steel became extremely brittle. It was also a ductile-to-brittle transition
failure. Overall ship sunk in 2 hours and about 15 minutes, so overall it could be considered a
ductile failure, but no help was available in that time.
f. It was a gatigue failure, which progressed in 19 years, but the end failure was brittle, therefore
a brittle failure. Overall (maroscopically) failure was a ductile failure, since it happened due to
the corrosion and fatigue of 19 years..
The Warren truss is used to support a staircase. Determine the force .pdf
The Warren truss is used to support a staircase. Determine the force in members CE, ED, and
DF, and state whether the members are in tension or compression. Assume all joints are pinned.
Solution
D. ED = 0.400 kN C, DF = 2.26 kN T, CE = 4.53 kN C.
The Wind Mountain archaeological site is located in southwestern Ne.pdf
The Wind Mountain archaeological site is located in southwestern New Mexico. Wind
Mountain was home to an ancient culture of prehistoric Native Americans called Anasazi. A
random sample of excavations at Wind Mountain gave the following depths (in centimeters)
from present-day surface grade to the location of significant archaeological
artifacts%u2020.8545120807555656065959070756568
Solution
(b) Given a=0.02, |t(0.01, df=n-1=14)|=2.62
So 98% CI is
xbar+/- t*s/vn
--> 74.2+/- 2.62*18.3/sqrt(15)
--> (61.8204, 86.5796).
The waiting time Y until delivery of a new component for an industr.pdf
The waiting time Y until delivery of a new component for an industrial operation is uniformly
distributed over the interval from 1 to 5 days. The cost of this delay is given by U = 2Y2 + 3.
Use the method of transformations to derive the density function of U.
I have found that the distribution function of Y is P(Y <= y) = (y-1)/5. Why is this?
Solution
You know F_Y(y) because Y is uniformly distributed.
F_Y(y)= integral_1^y 1/4 dx= (y-1)/4 when y in [1,5]
= 0 when y<1
= 1 when y>5
1/4 because the length of the interval [1,5] is 4. the density function is f_y(y)=1/4 when y in
[1,5]..
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Community pharmacy- Social and preventive pharmacy UNIT 5
Community pharmacy- Social and preventive pharmacy UNIT 5
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
Pride Month Slides 2024 David Douglas School District
Pride Month Slides 2024 David Douglas School District
The trial balances shown below are before and after adjustment for he.pdf
- 1. The trial balances shown below are before and after adjustment for here Company at the end of its fiscal year. Solution DR CR Account Receivable 750 Service Revenue 750 Account Receivable(8380-9130) (to record service revenue) Supplies Expense 1640 Supplies 1640 (Supplies (2900- 1260) Insurance expenses 1490 Prepaid Inurance 1490 (Prepaid inurance (4500-3010) Depreciation 1200 Accumulated depreciation 1200 ACC dep (4600-3400) Salaries and wages 1200 Salaries and wage Payable 1200 Unearned rent revenue 840 Rent revenue 840