Mathematics Archives | Page 4 of 4

MATHEMATICAL FINANCE; ACCOUNTING

Mathematical Finance

Order Instructions:

Dear Admin,

The Details in the attachment file or by email

Thank you

SAMPLE ANSWER

Breakeven analysis is calculated to determine the minimum sales that an

Organization must meet to cover all the expenses and commence in making a profit (Cafferky & Wentworth, 2010). In our case, where; Unit variable cost= £30, Fixed Cost=£250,000, Sales= 50,000.

Solution

Total Variable cost= Unit Variable Cost*Sales =£30*50,000=1,500,000

Total Cost= Total Variable Cost + Fixed Cost=£1,500,000+£250,000=1750000

Breakeven point= Total Cost/ Unit sales

Therefore, the breakeven selling price is= Total Cost/ Sales

= 1750000/250000 = £7 per unit

If the planned selling price is £48 per unit, then;

Solution

Breakeven Sales Units = 250000/ (48-30) = 13,889

Budgeted Sales Units = 50000

Margin Safety = (50000-13889)/50000=0.72222*100

Therefore, the margin safety is 72.22%

The following information is about two organizations, A and B.

Which firm has higher operating gearing?

According to Alhabeeb (2012) Operating Gearing can be arrived using the following the formula {[quantity* (Price- Variable Cost per Unit)]/ Quantity* (Price –Variable Cost per unit) – Fixed Operating Cost}. Therefore;

For Organization A= [160,000*(0.60-0.20)] /160,000*(0.60-0.20)-60,000=64,000/4,000

=16

For Organization B= [160,000* (0.60-0.50)]/160,000*(0.60-0.50)-12,000

= 4

Therefore, firm A has a higher operating gearing compared to firm B

What is the expected net income of both firms?

The expected net income for firm A is 4,000 [(160,000*0.60)-(60,000+0.20*160,000)]. Which is similar to the expected net income for firm B [(160,000*0.60) – (12,000+0.50*160,000)]= 4,000. Therefore, the expected net income for both firms is 4000

If the sales are 140,000 units, then the expected net income for firm A will be 4000 [(140,000*0.60) – (60,000+ 140,000*0.20)]. For firm B, the expected net income is 2000 [(140,000*0.60)- (12,000+140,000*0.50)]. Consequently, if the sales were 180, 000 units, then the expected net income for firm A will be 20, 000 [(180,000*0.60) – (60,000+140,000*0.20)]. Consequently, the expected net income for firm be will be 6,000 [(180,000*0.60)- (12000+180,000*0.50)]
The firm that is facing more risk in terms of current sales prediction is firm B that has a lower operating Gearing. Firms with higher operating gearing can make more money as compared to firms with lower operating gearing from incremental revenues (Richards, 2013). This tendency is because they don’t have to increase their cost of production to make those sales (Mclaney & Atrill, 2010). But firms with lower operating gearing have to increase their cost of production to increase sales.

References

CAFFERKY, M. E., & WENTWORTH, J. (2010). Breakeven analysis the definitive guide to cost-volume-profit analysis. [New York], Business Expert Press. http://site.ebrary.com/id/10404343.

Richards, D. (2013). How to do a breakeven analysis.

Alhabeeb, M. J. (2012). Break‐Even Analysis. Mathematical Finance, 247-273.

MCLANEY, E. J., & ATRILL, P. (2010). Accounting: an introduction. Harlow, Financial Times Prentice Hall.

We can write this or a similar paper for you! Simply fill the order form!

Linear algebra Term Paper Available Here

Linear algebra

Chapter 3

Question 1

If the above data points actually lay on a straight line y = C+Dt, we have:

Call the matrix A and the vector on the right-hand side. Of course this system is inconsistent, but we want to find such that is as close as possible to . As we’ve seen, the correct choice of is given by:

To compute this, first note that

Therefore,

And so

Therefore, the best-fit line for the data is:

Here are the data points and the best-fit line on the same graph:

Question 13

Let Then, we set and

Question 14

We now compute If we denote then by the Gram-Schmidt process,

Question 16

Chapter 4

Question 5

Question 6

Let be any basis of a -dimensional subspace of Then by Gram-Schmidt orthogonalisation process, we get an orthonormal set with and for

Question 7

Question 16

Question 17

Question 24

Question 25

Suppose we have matrix of dimension with Then by the application of the Gram-Schmidt orthogonalisation process yields a set of orthonormal vectors of In this case, for each we have

Question 14

Question 3

References

Bretscher, O. (2004). Linear Algebra with Applications, (3^rd ed.). New York, NY: Prentice Hall.

Farin, G., & Hansford, D. (2004). Practical Linear Algebra: A Geometry Toolbox. London: AK Peters.

Friedberg, S. H., Insel, A. J., & Spence, L. E. (2002). Linear Algebra, (4^th ed.). New York, NY: Prentice Hall.

Leon, S. J. (2006). Linear Algebra with Applications, (7^th ed.). New York, NY: Pearson Prentice Hall.

McMahon, D. (2005). Linear Algebra Demystified. New York, NY: McGraw–Hill Professional.

Zhang, F. (2009). Linear Algebra: Challenging Problems for Students. Baltimore, MA: The Johns Hopkins University Press.

We can write this or a similar paper for you! Simply fill the order form!

Matrices Assignment Help Available Here

Matrices

Order Instructions:

Hi,

Please mention the explanation into the answer, on how did you solve the questions. And also please put the question along with the answer.

SAMPLE ANSWER

=> add -1 times the 1st row to the 3rd row =>

=> add -2 times the 2nd row to the 1st row=>

=> add -4 times the 1st row to the 2nd row =>

=> add -7 times the 1st row to the 3rd row =>

=> multiply the 2nd row by -1/3 =>

=> add 6 times the 2nd row to the 3rd row =>

=> add -2 times the 2nd row to the 1st row =>

3^rd row is free for both A and B.

Consider the augmented matrix

We can convert this to reduced echelon form by subtracting twice row 1 from row 3 and

subtracting 3 times row 2 from row 3:

In order for this system to be consistent, it must be the case that

There are no constraints on b1 and b2, the possible right-hand sides of the equation are vectors of the form

for any real numbers b1 and b2. In other words, the column space of A is the plane containing the vectors and . Looking at the reduced echelon form of the matrix, we see that it is of rank 2 and that a particular solution of the given equation is .

=> Switch rows 1 and 3 =>

Divide row 1 by 2, row 2 by 3 =>

Multiply row 2 by 3 and subtract from row 1 =>

=> Switch rows 1 and 3 =>

Divide row 1 by 2, row 2 by 3 =>

Multiply row 2 by 3 and subtract from row 1 =>

=> Switch rows 1 and 3 =>

Divide row 1 by 2, row 2 by 3 =>

Multiply row 2 by 3 and subtract from row 1 =>

Let . Then . Thus .

So .

By the Rank-Nullity Theorem,

and

So it suffices to show that, therefore,

Form the augmented matrix [A b]:

Then subtracting row 1 from rows 2 and 3 and multiplying row 1 by 1 2 yields

Next, subtracting twice row 2 from row 1 and adding row 2 to row 3 gives

This is now in reduced echelon form, so we can answer the question. Notice that the pivot columns are the first and second columns; hence, the column space of A is the span of the first two columns of A, namely and .

Geometrically, this is just the plane containing these two vectors. Returning the the reduced echelon form of the augmented matrix, notice that we must have

x₁ = 4 − x₃ + 2x₄

x₂ = −1 − x₃ − 2x₄

so the special solutions are of the form

for some real numbers x₃ and x₄. Hence, the nullspace of A consists precisely of such linear combinations.

Finally, all solutions to the equation Ax = b are of the form

where the first term is a particular solution and the latter two terms comprise the special (or homogeneous) solutions.

Solution:

Solution: Any four vectors in R³ are dependent (since the dimension of R³ is 3), and the vectors v₁, v₂, v₃ are independent because they are columns of an invertible 3 by 3 matrix. Solving the system Ax = 0 is finding the nullspace of the matrix

Special solution here is x₄ = 1, x₃ = −4, x₂ = 1, x₁ = 1, so the nullspace consists of all (c₁, c₂, c₃, c₄) = x₄(1, 1, −4, 1).

Solution:

Since v₄ = v₂ − v1, v₅ = v₃ − v₁, and v₆ = v₃ − v₂, there are at most three independent vectors among these: furthermore, applying row reduction to the matrix [v₁ v₂ v₃] gives three pivots, showing that v₁, v₂, and v₃ are independent.

The echelon matrix has only a single pivot, in the second column . The second column is therefore a basis for the column space . (The third column of is equal to four times the second column.) The dimension of is 1 (same as the rank of ).

, all vectors with the last coordinate equal to 0.

, all vectors with the first coordinate equal to 0.

, all vectors with the last coordinate equal to 0.

, all vectors with the first coordinate equal to 0.

transformed into echelon form

=> add -1 times the 1st row to the 2nd row =>

=> add -1 times the 2nd row to the 3rd row =>

=> add -3 times the 2nd row to the 1st row =>

The second and the fourth variables are the pivots. The first, third, and the fifth variables are free variables. The row operation matrix E (R = EA) is

Nullspace: The null space is as follow:

for .

Column space: Since the pivot columns (the second and the fourth) of A span the column space, we have

for

Row space: it is the same as the row space of R as R is obtained by invertible row operations. So

for

Left nullspace: It has a basis given by the rows of E for which the corresponding rows of R are all zero. That is to say, we need to take the last row of E. Thus,

for

We can write this or a similar paper for you! Simply fill the order form!

Triangular Factors and Row Exchanges

Order Instructions:

Please don’t put any pictures into the
coursework.

SAMPLE ANSWER

1.5. Triangular Factors and Row Exchanges

Question 1

Upper triangular matrix is non-singular when no entry on the main diagonal is zero.

Question 2

Solution:

Hence elimination will subtract 4 times row 2 from row 3.

Looking at the U matrix, we see the pivots along the diagonal of the matrix:

To find out if a row exchange will be needed or not, first we determine A

After carrying the first elimination, we get:

Hence, there would not be a need for a row exchange.

Question 3

Question 7

Question 13

Question 23

1.6. Inverses and Transposes

Question 1

Question 2

When is applied to a matrix A its effect on A is to replace the first row of by the 3^rd row of, the 3rd row of, and to replace the 3rd row by the first row at the same time. As an example. Hence to reverse this effect, we need to perform the same operation again, i.e. replace the 1^st row with the 3^rd row and replaced the 3^rd row by the 1^st row, this is P₁

Hence,

Therefore,

Is an indication to replace 1^st row by the 3^rd row and to replace the 2^nd row by the 1^st row, and to replace the 2^nd row by the 3^rd row. For example; hence, in order to reverse it, we need to replace the 1^st row by the 2^nd row, and to replace the 2^nd row by the 3^rd row and to replace the 3^rd row by the 1^st row at the same time.

Hence,

In a permutation matrix P, each row will have at most one non-zero entry with value of 1. Consider the entry Pi, j = This entry will cause row i to be replaced by row j. Hence to reverse the effect, we need to replace row j by row i, or in other words, we need to have the entry (j, i) in the inverse matrix be 1. But this is the same as transposing P, since in a transposed matrix the entry (i, j) goes to (j, Hence. Now, where , are permutation matrices (in other words, each row of is all zeros, except for one entry with value of 1.)

Hence the entry C (i, i) will be 1 whenever A (i, j) = B (j, i) = 1, this is from the definition of matrix multiplication, element by element view, since:

For all entries except when the entry, and at the same time, but since is the transpose of P, then whenever then only when.

Hence this leads to with all other entries in C being zero, i.e.

Question 4

(a)

If A is invertible and AB = AC, then B = C is true.

Because;

This means that; IB = IC, and subsequently B = C

(b)

Suppose:

AC = AB; then:

0 = AB – AC = A (B-C)

If , Note that if we define

Then you can find by multiplying it out that:

Therefore, if we set we find that:
So for this choice of A, just pick any 2×2 matrix for B, and define C = D + B; and automatically, you will find that: AC = A (D + B) = AD + AB = 0 + AB = AB; but C is not = B.

Question 6

Start with the augmented matrix:

Then the only row on the left that doesn’t already look like the identity matrix is the second row; we just need subtract rows 1 and 3 from row 2, which gives:

Hence,

To find, start with the augmented matrix:

Replace the 1^st row by half of itself and add half of the 1^st row to the 2^nd row:

Next, add a third of the second row to the first, add 2/3 the second row to the third, and multiply the second row by 2/3:

Finally, multiply the third row by 3/4, and then add 1/3 of the result to row 1 and add 2/3 of the result to row 2:

Thus,

To find, start with the augmented matrix:

First, switch rows 1 and 3:

Now, subtract row 2 from row 1 and subtract row 3 from row 2:

Thus,

Question 10

For the first choice of A, we write the augmented matrix [A I]:

Then subtracting two times row 1 from row 2 and subtracting three times row 3 from row 2 yields

Hence,

For the second choice of A, write the augmented matrix [A I]:

Subtracting row 1 from rows 2 and 3 yields:

In turn, subtracting row 2 from rows 1 and 3 yields:

Finally, subtracting row 3 from row 2 yields:

Hence,

Question 13

Question 17

The inverse of a lower (upper) triangular matrix is still lower (upper) triangular. Multiplying lower (upper) triangular matrices gives a lower (upper) triangular matrix.
The main diagonal of and are the same as those of and , so we have. By comparing the off-diagonals of, both matrices must be diagonal. , is invertible so.

Then

Question 18

To make the pivot actually occur at f, switch rows 1 and 3:

Now, subtract d/j times row 1 from row 2 and subtract a/f times row 1 from row 3; note:

If e = 0 then the 2^nd row is all zeros, which means that there can’t be a pivot in that row. Thus, when A is invertible, it must be the case that e 0. Therefore, a pivot is in the 2^nd column, and the entry below can be eliminated through subtraction of times on row 2 from row 3:

If A is to be invertible, it must be the case that c 0. Therefore, the conditions which ensure that A is invertible are:

Turning to B, the matrix becomes:

Then, in order to have a pivot in the 2^nd row, it must be the case that

Or, equivalently,

On the other hand, if, so we can switch rows 1 and 2 to get

Then we can eliminate a by subtracting times row 1 from row 2 (note:

Again, if we are to have a pivot in the 2^nd row, it must be the case that:

Or, equivalently,

Therefore, either

Question 22

Question 43

The 5 by 5 also has 1s on the diagonal and super-diagonal.

Review Exercises

Question 1.2

Question 1.4

Question 1.10

Has L = I and;

A = LU has U = A (pivots on the diagonal);

A = LDU has with 1s on the diagonal.

Question 1.17

Question 1.28

2 by 2; d = 0 not allowed;

d = 1, e = 1; then l = 1, f = 0 is not allowed.

Vector Spaces

Question 2

The answer is: (a), (b) and (e).

Question 3

is the x-axis; is the line through (1, 1); is ; is the line through (-2, 1, 0); is the point (0, 0) in ; the null space is .

Question 14

The subspaces of are itself, lines through (0, 0), and the point (0, 0).

The subspaces of are itself, three-dimensional planes , two-dimensional subspaces , one-dimensional lines through (0, 0, 0), and (0, 0, 0) alone indicate that the smallest subspace containing P and L is either P or .

Question 24

The extra column b enlarges the column space, unless b is already in that space,

Reference

Strang, G. (2013). Linear Algebra and its Applications, (4^th ed.). New York, NY: McGraw-Hill Publishers.

We can write this or a similar paper for you! Simply fill the order form!

Acoustic principles using mathematical analysis

Order Instructions:

Main Assignment Questions (90% of unit weighting)
AR 50345

The objective of the course work is to demonstrate an understanding of the acoustic principles discussed during the unit week. This can be done in the form of mathematical analysis, an essay style answer or as bullet points. Marks against each section are
indicated as are word lengths, as appropriate.
Q1 a) Please outline your understanding of the following acoustic terms, (using 50 to 100 words each). Credit will be given for each key requirement or unique feature included. Additionally, please indicate where each of these terms are used in the field of acoustics: (10 marks)
Free field
Diffused field
dB
dBA
NR
W-weighting
Rw
R
0
w
(apparent sound reduction)
Dw
Dn
T
w
b) Please provide both a hand calculation and excel spread sheet
calculating the following.
(15 marks)
dBA for the following spectra:
32Hz 64Hz 125Hz 250Hz 500Hz 1kHz 2kHz 4kHz 8kHz
30 30 30 – – – – – — – – 30 35 36 50 36 35
65 83 90 75 43 35 – – — 10 10 10 10 10 10 10 -NR:
32Hz 64Hz 125Hz 250Hz 500Hz 1kHz 2kHz 4kHz 8kHz
30 30 30 – – – – – — – – 30 35 36 50 36 35
65 83 90 75 43 35 – – — 10 10 10 10 10 10 10 -Rw
(spread sheet only) for the following spectra:
125Hz 250Hz 500Hz 1kHz 2kHz
16 24 33 42 43
39 44 56 67 74
38 37 48 64 78
34 39 51 60 68
2 of 4 AR50345
Q2 a) Please explain, or undertake mathematically, how one determines
the composite sound reduction of the following two elements. The area of element 1 should be taken as 9m 2 with an Rw of 40dB.
The second element has an area of 2m 2 and an R w of 30dB.
(5 marks)
b) Use the above explanation to show how the 10dB rule works, i.e.
“There is no benefit in increasing the sound reduction of the larger element more than 10 dB over the smaller element which has a lesser sound reduction.” (5 marks)

Q3 a) Calculate the sound reduction of a facade comprising of: (6 marks)
10m 2 of glass/windows 20m 2 of brickwork and a 5m
2 spandrel panel
The area of the room should be taken as 200m 2
. The reverberation time in this room is equal to 0.7 seconds. Please state the sound reduction used for each element making up the facade and the source of this information.
b) Repeat this exercise but now include a plot of the combined sound reduction when an open window is placed in the facade.
This window has an effective open area of 1m by 0.01m; this window is then further opened in increments to a final effective open area of 1mby 0.25m. The area of the window should be taken out of the glazed element within the facade. (4 marks)

Q4 Sustainability and Acoustics. (20 marks)
Acoustics and sustainability are not often referred to in tandem. On the other hand, it has been shown that there are many links between these two subjects. Please provide your thoughts with respect to the links between acoustics and sustainable building
design.
Please also provide a review of products which are available to enhance the acoustic performance of a vented facade. This information can also be complemented with acoustics details relating to cross ventilation. (500 words) 3 of 4 AR50345

Q5 Reverberation time calculation. (15 marks)
Compose an Excel spreadsheet that calculates the absorption coefficient of the sound absorbing panels in the East Building classroom from 125 Hz to 4000 Hz (octave bands). To do this you should use the test data from the practical session for the empty room (with tables and chairs removed). This data is as follows:
Octave Band Centre Frequency, Hz
125 250 500 1000 2000 4000
Measured Average RT Empty
(Seconds)
0.9 0.8 0.9 0.7 0.7 0.6
The room dimensions may be taken as 13.2W x 5.2D x 3.2H.
There are 65 absorptive panels each with an area of 0.45mx 0.6m. You will need to estimate the absorption coefficients for the other surfaces (e.g. carpet, concrete soffit, glass facade, studwork
walls and doors and use these to calculate the absorption coefficient of the panels. What effect would distributing the panels throughout the room have compared with locating the panels in one large area?

Q6 A building has two items of plant on the roof, (a kitchen extract fan and a kitchen supply fan). The sound power levels of the units are as follows: (20 marks)
Frequency 63Hz 125Hz 250Hz 500Hz 1kHz 2kHz 4kHz
Kitchen Extract 84 85 83 80 78 76 72
Kitchen Supply 76 78 78 75 73 70 68
Calculate the A-weighted sound pressure level (dBA) at the nearest house, which is approximately 25m from the plant. You may assume that the kitchen extract fan is screened by a parapet that just obscures the line of sight. The kitchen supply fan is un-screened. You should clearly identify each step in your calculation or submit a spreadsheet.

SAMPLE ANSWER

Question 1: Part B

dBA for the following spectra:

32 Hz	64 Hz	125 Hz	250 Hz	500 Hz	1 kHz	2 kHz	4 kHz	8 kHz
30	30	30	45	20	30	20	25	25
35	53	60	30	35	30	50	36	35
65	83	90	75	43	35	30	45	55
10	10	10	10	10	10	10	10	10

NR for the following spectra:

32 Hz	64 Hz	125 Hz	250 Hz	500 Hz	1 kHz	2 kHz	4 kHz	8 kHz
30	30	25	35	30	45	20	35	30
30	50	65	30	35	30	50	36	35
65	83	90	75	43	35	25	40	50
10	10	10	10	10	10	10	10	10

Question 2: Part A

The composite sound reduction of two elements can be used using the weighted sound reduction index formula given below:

Where;

D = Level difference

S = Area of the test element through which the noise is transmitted

A = Total sound absorption by the element

Therefore, the composite sound reduction of the two elements can be determined by using the above formulae which can be used to the absorption factors of the two elements given by [ ]. With the absorption factors of the two elements and the average sound pressure of the source and receiving area, the composite sound reduction can be determined for the two elements.

Question 2: Part B

Using the explanation given above it is possible to show that the 10 dB rule works, i.e. “There is no benefit in increasing the sound reduction of the larger element more that 10 dB over the smaller element which has a lesser sound reduction.”

Let’s now consider the parameters of the two elements;

Element 1: Area = 9 m² and R_w = 40 dB

Element 2: Area = 2 m² and R_w = 30 dB

The composite sound reduction equation provided above can be used to proof that the 10 dB rule works, by computing the absorption factors for each element, which in the case of these two elements can be calculated from the given details of the two elements. The calculations for absorption factors for the two elements are shown below:

Element 1

Element 2

The above results for the absorption factors of both elements proves the 10 dB rule since their values are close despite the difference of 10 dB between the two elements.

Question 3: Part A

Sound Reduction is given by the formula shown below:

Glass/windows

Brick wall

Spandrel panels

Question 3: Part B

The plot is shown below:

Question 4

Acoustics and sustainability are the two key considerations when designing sustainable or green buildings mainly because both are positively related in terms of the selected specifications, where those of one aspect of a building i.e. acoustics are dependent of the specifications of the building’s sustainability. This is highly attributable to the fact that, acoustics which are majorly concerned with sound insulation have become a very important factor when designing sustainable buildings as our environments continue to become noisier. The reason why acoustic has become a highly influential factor is because sound insulation has a significant impact on the extents of embodied energy in a particular building, which makes it essential to clearly understand the effect of sound insulation on embodied energy levels right from the beginning of the designing of sustainable buildings.

The close connection between acoustic (sound insulation) and sustainability is how the former influences the latter on basis of embodied energy, especially because embodied energy in a particular building can be effectively reduced by re-using an existing building through refurbishments that enables acoustic specifications are fulfilled in a manner that is very cost effective. This is in most cases achieved through timely up front acoustic testing of an existing building to understand its performance prior to refurbishment. Hence, upon establishing the acoustic performance of an existing building as well as understanding its restrictions and limitations, the sustainability of the building can be significantly improved by alleviating these restrictions or limitations.

When designing sustainable buildings, acoustics have become a very significant consideration because it determines the type of materials used in the construction of the sustainable building which is an indication of how the two components i.e. sustainability and acoustics are interrelated. Therefore, early acoustic testing is in most cases carried out as an effective an method to make sure that designs of sustainable buildings are sufficient, and that there is high quality construction in which sustainability is given priority. As a result, the architects usually make decisions of the construction materials to be used in a sustainable building either lightweight or heavyweight materials based of the intended quality of acoustics (sound insulation) as well as its subsequent sustainability. For instance, high/heavy weight materials are in most cases favored for enhanced acoustics; however lightweight materials such as timber often offer better or equal acoustic performance. This means that sustainability of buildings can be improved by using lightweight construction due to their less embodied energy compared to heavyweight construction materials apart from achieving higher levels of acoustic performance do not lead to desired sustainability.

Therefore, as it is theoretically know, a ±6 dB change in acoustic performance or sound insulation is equivalent to a doubling or halving of mass of construction materials for a particular building. This implies that, from the perspective of sustainability when acoustic parameters are over specified there will be an eventual significant impact on the created waste. This indicates that acoustic performance considerations or specifications or parameters in a particular building have a direct impact on its sustainability, hence necessitating the need to always make sure that a balance has been achieved between acoustic and sustainability at the design level of a building.

Furthermore, there are various products that are used for the enhancement of the acoustic of a vented façade. For instance, glass fiber reinforced concrete (GFRC) acoustic panels are one of the products often used in constructing vented facades for enhanced acoustic performance because they not only lead to sound insulation, but also improves sustainability of the building. Mineral wool is the other product used within partitions of vented facades for the purpose of acoustic dampening within stud walls, and it results to significant enhancement in acoustic performance in addition to being sustainable and cost effective method.

Question 5

A copy of the Excel spreadsheet is attached below:

When the absorptive panels are spread throughout the room more sound will be absorbed because they will be in proximity to the source of sound leading to significant sound reduction; however when they are placed in one big area there will be less sound reduction since they will be overlying each other.

Question 6

Where;

L_w = Sound power level 10-12 w(dB)

Lp = Sound pressure level 20 mPa (dB)

D= distance from fan in meters

D = 25 meters

References

Fahy, F. and Gardonio, P. (2007), Sound and Structural Vibration: Radiation, Transmission and Response, (2^nd edition). London: Academic Press.

Lohse, D., Schmitz, B. and Versluis, M. (2001), “Snapping shrimp make flashing bubbles”. Nature , Vol. 413 Issue: 6855, pp. 477–478.

Pohlmann, K. (2010), Principles of Digital Audio, (6^th edition). New York, NY: McGraw Hill Professional. p. 336-339.

Raichel, D. R. (2006), The Science and Applications of Acoustics, (2^nd edition). London: Springer.

Wilson, C. E. (2006), Noise Control, (Revised edition). Malabar, FL: Krieger Publishing Company.

We can write this or a similar paper for you! Simply fill the order form!

Matrices Assignment Help Available Here

Matrices

Order Instructions:

I am attaching the questions to the email. I am looking for the correct answers for all the questions. Please send me the answers by 2 PM CST.

SAMPLE ANSWER

Matrices

Question 1: Compute the products

Graphical drawings of the column vectors

Question 3: Find two inner products and a matrix product

Two inner products (-14, -14)

Matrix product =

Two inner products (-14, 15)

Matrix product =

Two inner products (-14, 15)

Matrix product =

Question 5: Solution for x

Question 6: Solving a 2 by 2 matrices

Solution: since indicates the entry in A which is in the i^th row and in the j^th column, we see that:

Likewise

Therefore,

Also,

Question 10: True or false statements

True
True
True
False

Question 24:

Question 30:

Gaussian Elimination

Question1:

Question 3:

2x – 3 y = 3 2x – 3 y = 3 x = 3 Subtract 2 x row 1 from row 2

y + z = 1 gives y + z = 1 and y = 1 Subtract 1 x row 1 from row 3

2y – 3z = 2 -5z = 0 z = 0 Subtract 2 x row 2 from row 3

Question 10:

Question 11:

Question 17:

Question 28:

True
False
True

The Geometry of Linear Equations

Question 1:

The lines intersect at (x, y) = (3, 1). Then 3(column 1) + 1(column 2) = (4, 4).

Question 9:

Column 3 = 2(column 2) – column 1. Ifb = (0, 0, 0), then (u, v, w) = (c, -2c, c)

Question 13:

The row picture has two lines meeting at (4, 2). The column picture has 4(1, 1) +

2(-2, 1) = 4(column 1) + 2(column 2) = right-hand side (0, 6).

Matrices

Question 1: Compute the products

Graphical drawings of the column vectors

Question 3: Find two inner products and a matrix product

Two inner products (-14, -14)

Matrix product

Two inner products (-14, 15)

Matrix product =

Two inner products (-14, 15)

Matrix product =

Question 5: Solution for x

Question 6: Solving a 2 by 2 matrices

Solution: since indicates the entry in A which is in the i^th row and in the j^th column, we see that:

Likewise

Therefore,

Also,

Question 10: True or false statements

True
True
True
False

Question 24:

Question 30:

Gaussian Elimination

Question1:

Question 3:

2x – 3 y = 3 2x – 3 y = 3 x = 3 Subtract 2 x row 1 from row 2

y + z = 1 gives y + z = 1 and y = 1 Subtract 1 x row 1 from row 3

2y – 3z = 2 -5z = 0 z = 0 Subtract 2 x row 2 from row 3

Question 10:

Question 11:

Question 17:

Question 28:

True
False
True

The Geometry of Linear Equations

Question 1:

The lines intersect at (x, y) = (3, 1). Then 3(column 1) + 1(column 2) = (4, 4).

Question 9:

Column 3 = 2(column 2) – column 1. Ifb = (0, 0, 0), then (u, v, w) = (c, -2c, c)

Question 13:

The row picture has two lines meeting at (4, 2). The column picture has 4(1, 1) +

2(-2, 1) = 4(column 1) + 2(column 2) = right-hand side (0, 6).

We can write this or a similar paper for you! Simply fill the order form!

Mathematical problems Assignment Help

Mathematical problems

Order Instructions:

My student number is some of the problem require the last 2 digits of the student number

SAMPLE ANSWER

Question 1

For point B:

Within point in a stressed body, there exist three mutually perpendicular planes and the resultant stress is always normal stress. The mutually perpendicular planes are known as principal planes, and the resultant normal stresses acting on them are called principal stresses.

For point A:

Question 2

For pure shear to occur, the stresses σ₁ and σ₁ must be equal in magnitude and opposite in sign

σ₁= Pr/t and σ₂= Pr/2t-F/A= Pr/2t-F/2πrt

Pr/t=- (Pr/2t-F/2πrt) collecting like terms together

Pr/t+ Pr/2t= F/2πrt solving for F, we get F=3π Pr²=3* π*0.1²*12000000Pa

F=360 kN

Required thickness is given by solving for t in equations above. Using the allowable normal stress we have 110 MPa= Pr/2t-F/2πrt, therefore,

t= (P*r²*π-F)/220πr = (12000000*0.1²* π-190000)/ (220000000* π*0.1) = 0.00271 m

Question 3

Pipe diameter, outer =110mm=0.11m, inner diameter=90mm=0.09m

Wind pressure =1.5 kPa

Solution

Area of the sign board=2*1=2m², resultant force W= wind pressure * area

W=2*1.5=3KN

Torque T=resultant force W* distance b, T=3*1.05=3.15 kN-M

Bending moment M=W* height=3*3.5=10.5 kN-M

Shear force V=W =3 KN

Bending moment M produces a tensile stress σ_a at point A and tensile stress

σ_a = (M*d₂/2)/ [π(d₂⁴-d₁⁴)/64]= (10.5 *0.11/2)/ [π(0.11⁴-0.09⁴)/64]=0.5775/(3.96626*10^-6)=145603.163 MPa

Torque produces shear stresses ҭ₁ at point A and B

Ҭ_torsion == (T*d₂/2)/ [π (d₂⁴-d₁⁴)/32]=(3.15 *0.11/2)/ [π(0.11⁴-0.09⁴)/32]= 0.17325/(7.932522*10^-6)=21840.4689

Maximum plane shear stress Ҭ_max=√ [( σ_x – σ_y)²/2 + (Ҭ_xy)²]

At point A the shear stress is zero

At point B σ_x=0, σ_y= σ_aand Ҭ_xy= Ҭ₁=21840.4689 Pa, therefore, Maximum plane shear stress

Ҭ_max =√ [(0 – 145603.163)²/2 + (21840.4689)²]= √(1.107715*10^10) =105248.04 Pa

At point C, σ_y= σ_x=0 and Ҭ_xy= Ҭ₁+ Ҭ_2,Ҭ₂=2V/A= (2*3000)/2=3000 N/m², hence Maximum plane shear stress Ҭ_max =√ [(0 – 0)²/2 + (3000+21840.4689)²] =24840.4689 Pa

Question four

Solution

Tensile stress = tensile force P/ area A=34000/[ (π*0.05²)/4]

=34000/0.001963496=17.316 MPa

Shear torsion= Tr/I_p = (34000 *0.05/2)/[(0.05)⁴*π/32]=850/(6.135922*10^-7)=1385.285 MPa

Principal stresses are σ_1,2 = [( σ_x + σ_y)²/2 ±√[( σ_x -σ_y)²/2+ (ҭ_xy)²]

σ_1,2 = (0 +17.316)/2 ±√[(( 0 -17.316)/2)²+ (1385.285)²]

σ₁= 8.658+√1919089.49=1393.97 MPa

σ₂= -1376.6541 MPa

Maximum in plane shear stress

= √ [((0 -17.316)/2)²+ (1385.285)²] = 1385.31 MPa

We can write this or a similar paper for you! Simply fill the order form!

Quadratic Equations and Healthcare Paper

Quadratic Equations and Healthcare

In week 3 there are various concepts that are covered, from exponents to quadratic equations. In your text for this week, page 1 states that quadratic equations can be used to determine the number of trees in a grove, the size of a tennis court, etc. Additional
application on this week’s topic include:
•Life insurance calculations for projecting death rates
•Throwing a ball in the air and determining the time till it reaches the ground
•Analyzing tumor volume and cell survival
•Impact of Insurance on Health Care
View the video:
http://www.khanacademy.org/math/algebra/quadtratics/v/application-problem-with-quadratic-formula

Choose an application of one of the topics above that applies to your field of study (Healthcare) and write a short 1-2 paragraph APA-style paper about quadratic equations and your program of study. Try searching quadratic equations AND healthcare.

We can write this or a similar paper for you! Simply fill the order form!