## STAT0021｜Introductory Statistical Methods and Computing 统计方法和计算入门 伦敦大学学院

statistics-labTM为您提供澳洲国立大学Australian National University STAT0021 Introductory Statistical Methods and Computing 统计方法英国代写代考辅导服务！

This module aims to provide an introduction to statistical methods and interpretation of data, along with associated computing, and to provide some expertise in applying quantitative methods in the Life and Physical Sciences. The statistical methods covered are useful in the routine analysis of scientific methods, as might be encountered in other modules.

## Introductory Statistical Methods and Computing 统计方法和计算入门作业案例

Suppose an economist is organizing a survey of American minimum wage workers, and is interested in understanding how many workers that earn the minimum wage are teenagers. ${ }^1$ Suppose further that one out of every four minimum wage workers is a teenager. If the economist finds 80 minimum wage workers for his survey, what’s the probability that he interviews exactly 14 teenagers? 35 teenagers? What’s the probability that he gets at least 5 teenagers in his survey?

Solution to 2: In this case, we have $p=0.25$ again and $n=80$. We again use the binomial formula to obtain the solution to the first question:
$$P(K=14)=\left(\begin{array}{c} 80 \ 14 \end{array}\right) \cdot 25^{14}(1-.25)^{80-14}=0.0319 .$$
The second uses the same formula:
$$P(K=35)=\left(\begin{array}{l} 80 \ 35 \end{array}\right) \cdot 25^{35}(1-.25)^{80-35}=0.00011704$$

Suppose you flip a weighted coin (probability of heads is $p$ and probability of tails is $q=1-p$ ) $n$ times.

1. What is the probability that you get a particular ordering of $k$ heads and $n-k$ tails?

Solution to 2: From part 1, the probability of getting any particular ordering of $k$ heads and $n-k$ tails would be $p^k q^{n-k}$, but we have to take into account the number of permutations that exist for getting $k$ heads and $n-k$ tails. We can think about this as the “ALGEBRA” and “CALCULUS” problem, by simply labeling the heads as ” $\mathrm{H}$ ” and tails as “T” and then figure out how many different words we can write. This is pretty simple: $\frac{n !}{k !(n-k) !}$ which just so happens to be the same as the binomial coefficient, $\left(\begin{array}{l}n \ k\end{array}\right)$. So, it turns out that we just have the binomial formula for this problem:
$$P(k \text { heads })=\left(\begin{array}{c} n \ k \end{array}\right) p^k q^{n-k} .$$

Let $X \equiv$ the number of heads in $n$ flips. What is the probability density function of $X$ ?

Solution to 3 : Well, I just have to rewrite the answer to 2 slightly differently to change the notation to refer to the random variable, $X$ and its PDF, $f(x ; n)$ :
$$P(X=x)=f(x ; n)=\left(\begin{array}{c} n \ x \end{array}\right) p^x(1-p)^{n-x} .$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## Econ8009｜International Monetary Economics 国际货币经济学 澳洲国立大学

statistics-labTM为您提供澳洲国立大学Australian National University Econ8009 International Monetary Economics 国际货币经济学澳洲代写代考辅导服务！

This course will provide students with models that can be used to analyse issues in international monetary economics. The course introduces students to so-called global imbalances and the basic stylized facts about current accounts with cross-section and the time-series of selected countries such as the US and China. Issues of current account sustainability will also be addressed. The underlying intertemporal modelling framework will be applied to the current account and be used to understand the macroeconomic effects of external shocks, such as changes in terms of trade and the world interest rate.

## International Monetary Economics 国际货币经济学 作业案例

A delegation of three is to be chosen from the untenured faculty of the MIT Economics Department (numbering ten) to represent the department in an Institute-wide committee. In how many ways
a) can the delegation be chosen?

• Solution to (a): Since there are 10 untenured faculty members, there are “10 choose 3” $\left(\begin{array}{c}10 \ 3\end{array}\right)=\frac{10 !}{3 ! 7 !}=\frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1}=120$ different ways to choose the three faculty members. Further, if the three faculty members are selected to unique positions such as the President, VicePresident, and Secretary of the delegation, then there would be $\frac{10 !}{7 !}=10 \cdot 9 \cdot 8=720$ total ways since each of the 120 different combinations of faculty members can be mixed into the 3 positions in 6 different ways.

Does a monkey have a better chance of rearranging “ACCLLUUS” to spell “CALCULUS,”
or of rearranging “AABEGLR” to spell “ALGEBRA?” (2 points.)

Solution: How many letters is “CALCULUS”? Eight. How many letters is “ALGEBRA”? Seven. How many unique permutations of the letters “ACCLLUUS” are there? We have $\frac{8 !}{2 ! 212 !}=7$ ! different permutations since there are two C’s, L’s, and U’s. How many unique permutations of the letters “AABEGLR” are there? We start out with at most 7!, but we have duplicate A’s, so we know that there must be $7 ! / 2$ ! different permutations, which is less than 7 !. Thus, we know that “ALGEBRA” is twice as likely to come up randomly on a monkey’s Shakespearean typewriter.

can it be chosen, if two people must be chosen from MIT assistant faculty (6 professors) and one person must be chosen from visiting assistant faculty

Solution to (d): This should be easy if you’ve already figured out parts (a)-(c), not because you’ve computed the answer yet, but because the methods are similar. We just break the problem into its two parts: choosing MIT assistant faculty and choosing visiting assistant faculty. This give us $\underbrace{\frac{6 \cdot 5}{2}}{\text {MIT }} \cdot \underbrace{4}{\text {Visiting }}=60$. Does this answer make sense? It is less than if there were just two people who hated each other, as this is equivalent to the problem where four people refuse to go together, but we have to have one of them. So, we would expect the count to be less than what we found in (b).

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## STAT0013｜Discrete Mathematics 离散数学 伦敦大学学院

statistics-labTM为您提供伦敦大学学院 London’s Global University STAT0013 Mathematical Analysis数学分析英国代写代考辅导服务！

STAT0013 is specified as a formal option for third and fourth year undergraduates from the Department of Mathematics. Any such student who has taken one of the standard prerequisites (simultaneous attendance of STAT0005 is also acceptable) should simply register for STAT0013 on Portico and await a decision. Mathematics students who do not satisfy the standard prerequisites must additionally consult a member of staff in the Department of Statistical Science.

## Discrete Mathematics 离散数学作业案例

Let
$$B=\left(\begin{array}{rrr} -7 & 4 & -3 \ 2 & -5 & 3 \ 1 & -3 & -2 \end{array}\right)$$
Calculate det $B$.

Solution: We calculate that
\begin{aligned} \operatorname{det} B= & (-7) \cdot \operatorname{det}\left(\begin{array}{rr} -5 & 3 \ -3 & -2 \end{array}\right) \ & -4 \cdot \operatorname{det}\left(\begin{array}{rr} 2 & 3 \ 1 & -2 \end{array}\right) \ & +(-3) \cdot \operatorname{det}\left(\begin{array}{rr} 2 & -5 \ 1 & -3 \end{array}\right) \ = & (-7) \cdot 19-4 \cdot(-7)+(-3) \cdot(-1) \ = & -102 \ \neq & 0 . \end{aligned}
It is worth summarizing the main rule for inverses for which we have given an indication of the justification:

Rule for inverses: A square matrix has an inverse if and only if the matrix has nonzero determinant.

This still does not tell us how to find the inverse, but we shall get to that momentarily.
There are in fact a number of ways to calculate the inverse of a matrix, and we shall indicate two of them here. The first is the method of Gaussian elimination, which is a powerful technique that can be used for many purposes. The idea is to take the given square matrix
$$A=\left(\begin{array}{ll} a & b \ c & d \end{array}\right)$$
and to augment it by adjoining the identity matrix as in the display below:
$$\left(\begin{array}{ll|ll} a & b & 1 & 0 \ c & d & 0 & 1 \end{array}\right)$$
Now the method of gaussian elimination allows us to perform certain operations on the rows of this augmented matrix, the goal being to reduce the square matrix on the left to the identity matrix. Whatever matrix results on the right will be the inverse.

The multiplicative identity for a group is unique.

Proof: Let $G$ be a group. Let $e$ and $e^{\prime}$ both be elements of $G$ that satisfy Axiom 2 . Then
$$e=e \cdot e^{\prime}=e^{\prime}$$
Thus $e$ and $e^{\prime}$ must be the same group element.

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

statistics-labTM为您提供悉尼大学University of Sydney Math1904 Mathematical Analysis数学分析澳洲代写代考辅导服务！

This unit of study provides an introduction to programming and numerical methods. Topics covered include computer arithmetic and computational errors, systems of linear equations, interpolation and approximation, solution of nonlinear equations, quadrature, initial value problems for ordinary differential equations and boundary value problems, and optimisation.

Let $g:[1,+\infty) \rightarrow[8,+\infty)$ be given by $g(x)=x^2+3 x+4$. Then $g(1)=8$, and $g$ is strictly increasing without bound. So $g$ is one-to-one and onto. We conclude that $g$ has an inverse. One may use the quadratic formula to solve for that inverse, and find that $g^{-1}(x)=(-3+\sqrt{-7+4 x}) / 2$.

The idea of inverse function lends itself particularly well to the notation of ordered pairs. For $f: S \rightarrow T$ is inverse to $g: T \rightarrow S$ (and vice versa) provided that for every ordered pair $(s, t) \in f$ there is an ordered pair $(t, s) \in g$ and conversely.
Not every function has an inverse. For instance, let $f: S \rightarrow T$. Suppose that $f(s)=t$ and also that $f\left(s^{\prime}\right)=t$ with $s \neq s^{\prime}$ (in other words, suppose that $f$ is not one-to-one). If $g: T \rightarrow S$ then $g(f(s))=g(t)=g\left(f\left(s^{\prime}\right)\right)$ so it cannot be that both $g(f(s))=s$ and $g\left(f\left(s^{\prime}\right)\right)=s^{\prime}$. In other words, $f$ cannot have an irver se. We conclude that a function that does have an inverse must be one-to-one.

On the other hand, suppose that $t \in T$ has the property that there is no $s \in S$ with $f(s)=t$ (in other words, suppose that $f$ is not onto). Then, in particular, it could not be that $f(g(t))=t$ for any function $g: T \rightarrow S$. So $f$ could not be invertible. We conclude that a function that does have an inverse must be onto.

For each $m$, the set $\mathcal{Q}(m)$ is well ordered.

Proof: The proof is by induction on $m$.
When $m=1$ there is nothing to prove.
Assume that the assertion has been proved for $m=k$. Now let $U$ be a subset of $\mathcal{Q}\left(k^{\prime}\right)$. There are now three possibilities:

1. If in fact $U \subset \mathcal{Q}(k)$ then $U$ has a least element by the inductive hypothesis.
2. If $U={k}$ then $U$ has but one element and that element, namely $k$, is the least element that we seek.
3. The last possibility is that $U$ contains $k$ and some other natural numbers as well. But then $U \backslash{k} \subset \mathcal{Q}(k)$. Hence $U \backslash{k}$ has a least element $s$ by the inductive hypothesis. Since $s$ is automatically less than $k$, it follows that $s$ is a least element for the entire set $U$.
Now we have all our tools in place and we can prove the full result:

The natural numbers $\mathrm{N}$ are well ordered.

Proof: Let $\emptyset \neq S \subset \mathbb{N}$. Select an element $m \in S$. There are now two possibilities:

1. If $\mathcal{Q}(m) \cap S=\emptyset$ then $m$ is the least element of $S$ that we seek.
2. If $T=\mathcal{Q}(m) \cap S \neq \emptyset$ then notice that if $x \in T$ and $y \in S \backslash T$ then $x<y$. So it suffices for us to find a least element of $T$. But such an element exists by applying the preceding proposition to $U=\mathcal{Q}(m) \cap S$.
The proof is complete.

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## Math3076｜Mathematical Computing数学计算 悉尼大学

statistics-labTM为您提供悉尼大学University of Sydney MATH0048 Mathematical Analysis数学分析澳洲代写代考辅导服务！

This unit of study provides an introduction to programming and numerical methods. Topics covered include computer arithmetic and computational errors, systems of linear equations, interpolation and approximation, solution of nonlinear equations, quadrature, initial value problems for ordinary differential equations and boundary value problems, and optimisation.

## Mathematical Computing数学计算作业案例

(b) (0 points) Are these system specifications consistent? _________Prove it!

Solution. There are several ways to approach this problem. One is to construct a truth table with sixteen lines-one for each way of assigning truth values to the four variables $M, N, K$, and $I$.

We can avoid the cumbersome truthtable if we reason by cases. Case 1: $I$ is true. Then the last formula (9) is false, and the whole specification is false. Case 2: $I$ is false. Now formula (8) can be true only if $\neg M$ is false, that is, only if $M$ is true. Likewise, formula (7) can be true only if $\neg K$ is true, that is, $K$ is false.

Since $K$ is false, formula (6) can be true only if $N$ is false. Thus, we have deduced that in order to be consistent in this case, we must have
\begin{aligned} I & =\text { false } \ M & =\text { true } \ K & =\text { false } \ N & =\text { false } \end{aligned}
But now formula (5) is false, so it is impossible for all the formulas to be true: the system is inconsistent.

Problem 2 (20 points). For each of the following logical formulas with domain of discourse the natural numbers, $\mathbb{N}$, indicate whether it is a possible formulation of
I: the Induction Axiom,
S: the Strong Induction Axiom,
L: the Least Number Principle (also known as Well-ordering), or
$\mathrm{N}$ : None of these.
For example, the ordinary Induction Axiom could be expressed by the following formula, so it gets labelled “I”.
$$(P(0) \wedge[\forall k P(k) \longrightarrow P(k+1)]) \longrightarrow \forall k P(k) \quad \longrightarrow$$

(a) (0 points) $(P(b) \wedge[\forall k \geq b P(k) \longrightarrow P(k+1)]) \longrightarrow \forall k \geq b P(k)$

Solution. I. This is a perfect formulation of the Induction Axiom. $b$ is used for the base case; $P(k) \longrightarrow P(k+1)$ is the inductive case.
(b)$(P(b) \wedge[\forall k \leq b P(k) \longrightarrow P(k+1)]) \longrightarrow \forall k \leq b P(k)$
Solution. N. The two occurences of $k<=b$ should have been $k>=b$
(c) $[\forall b(\forall k<b P(k)) \longrightarrow P(b)] \longrightarrow \forall k P(k)$
Solution. S. Since you are assuming $P(k)$ for all $k<b$, this is strong induction.
(d) $\quad(\exists n P(n)) \longrightarrow \exists n \forall k<n \overline{P(k)}$
Solution. N. This statement is in fact always true; when $n=0, \forall k<n \overline{P(k)}$. It should say $P(n) \wedge \forall k<n ; \overline{P(k)}$

Problem :To encourage collaborative study, the 6.042 staff is considering assigning each student to a study group with two or three other students. Prove that as long as the enrollment is large enough, the class can always be divided into such study groups.

Solution. Proof. The proof is by strong induction. The induction hypothesis is that a class with $n \geq 6$ students can be divided into teams of 3 or 4 . More precisely
$$P(n)::=n \geq 6 \longrightarrow \exists x, y \in \mathbb{N} 3 x+4 y=n .$$
For any $n \geq 0$, we may assume $P(6), \ldots, P(n-1)$ to prove $P(n)$.
Case 1: $(n<6) . P(n)$ holds because the hypothesis $n \geq 6$ is false.
Case 2: $(n=6,7$, or 8$) P(6)$ is true because there could be two teams of $3, P(7)$ is true because there could be a team of 3 and a team of 4 , and $P(8)$ is true because there could be two teams of 4 .

Case 3: $(n \geq 9)$. Of course $n>n-3$ so $P(n-3)$ holds by the strong induction hypothesis. But $n-3 \geq 6$, so $P(n-3)$ implies $3 x^{\prime}+4 y^{\prime}=n-3$ for some $x^{\prime}, y^{\prime} \in \mathbb{N}$, and therefore $3 x+4 y=n$ where $x::=x^{\prime}+1$ and $y::=y^{\prime}$. So $P(n)$ holds, as required.

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## Math0048｜Mathematical Analysis数学分析 伦敦大学学院

statistics-labTM为您提供伦敦大学学院 London’s Global University MATH0048 Mathematical Analysis数学分析英国代写代考辅导服务！

This module is an introduction to mathematical analysis, one of the most important and welldeveloped strands of pure mathematics with many elegant and beautiful theorems, and also
with applications to many areas of mathematics, theoretical statistics, econometrics, and optimisation.
The aim is to introduce students to the ideas of formal definitions and rigorous proofs (one
of the fundamental features of modern mathematics, and something that is not familiar from
A-level), and to develop their powers of logical thinking.
This module is a prerequisite for Complex Analysis, MATH0013 and provides a useful foundation for courses such as Logic, MATH0050.
The module is intended for second or third year students in departments outside Mathematics,
particularly in Economics or Statistics. Students taking this module should be mathematically
able and will normally have demonstrated this by achieving a strong result in a module such as
MATH0047 or having an A* in Further Mathematics A-level.

## Mathematical Analysis数学分析作业案例

Let $A: X \rightarrow Y$ be a linear operator between two Euclidean or two Hermitian spaces and let $A^: Y \rightarrow X$ be its adjoint. Then $\operatorname{Rank} A^=\operatorname{Rank} A$. Moreover,
$$\begin{array}{ll} (\operatorname{Im} A)^{\perp}=\operatorname{ker} A^, & \operatorname{Im} A=\left(\operatorname{ker} A^\right)^{\perp}, \ \left(\operatorname{Im} A^\right)^{\perp}=\operatorname{ker} A, & \operatorname{Im} A^=(\operatorname{ker} A)^{\perp} . \end{array}$$

Proof. Fix two orthonormal bases on $X$ and $Y$, and let $\mathbf{A}$ be the matrix associated to $A$ using these bases. Then, see (3.8), the matrix associated to $A^$ is $\mathbf{A}^T$, hence $$\operatorname{Rank} A^=\operatorname{Rank} \mathbf{A}^T=\operatorname{Rank} \mathbf{A}=\operatorname{Rank} A,$$
and
$$\operatorname{dim}\left(\operatorname{ker} A^\right)^{\perp}=\operatorname{dim} Y-\operatorname{dim} \operatorname{ker} A^=\operatorname{Rank} A^=\operatorname{Rank} A=\operatorname{dim} \operatorname{Im} A .$$ On the other hand, $\operatorname{Im} A \subset\left(\operatorname{ker} A^\right)^{\perp}$ since, if $y=A(x)$ and $A^(v)=0$, then $(y \mid v)=$ $(A(x) \mid v)=\left(x \mid A^(v)\right)=0$. We then conclude that $\left(\operatorname{ker} A^\right)^{\perp}=\operatorname{Im} A$. The other claims easily follow. In fact, they are all equivalent to $\operatorname{Im} A=\left(\operatorname{ker} A^\right)^{\perp}$.

Let $X$ be a finite-dimensional vector space and let $\left(e_1, e_2, \ldots, e_n\right)$ be a $g$-orthogonal basis for a metric $g$ on $X$. Denote by $n_{+}, n_{-}$and $n_0$ the numbers of elements in the basis such that respectively, we have $g\left(e_i, e_i\right)>0, g\left(e_i, e_i\right)<0, g\left(e_i, e_i\right)=0$. Then $n_{+}=i_{+}(g)$, $n_{-}=i_{-}(g)$ and $n_0=i_0(g)$. In particular, $n_{+}, n_{-}, n_0$ do not depend on the chosen g-orthogonal basis,
$$i_{+}(g)+i_{-}(g)=r(g) \quad \text { and } \quad i_{+}(g)+i_{-}(g)+i_0(g)=n .$$

Proof. Suppose that $g\left(e_i, e_i\right)>0$ for $i=1, \ldots, n_{+}$. For each $v=\sum_{i=1}^{n_{+}} v^i e_i$, we have
$$g(v, v)=\sum_{i=1}^{n_{+}}\left|v^i\right|^2 g\left(e_i, e_i\right)>0,$$
hence $\operatorname{dim} \operatorname{Span}\left{e_1, e_2, \ldots, e_{n_{+}}\right} \leq i_{+}(g)$. On the other hand, if $W \subset X$ is a subspace of dimension $i_{+}(g)$ such that $g(v, v)>0 \forall v \in W$, we have
$$W \cap \operatorname{Span}\left{e_{n_{+}+1}, \ldots, e_n\right}={0}$$
since $g(v, v) \leq 0$ for all $v \in \operatorname{Span}\left{e_{n_{+}+1}, \ldots, e_n\right}$. Therefore we also have $i_{+}(g) \leq$ $n-\left(n-n_{+}\right)=n_{+}$.

Similarly, one proves that $n_{-}=i_{-}(g)$. Finally, since $\mathbf{G}:=\left[g\left(e_i, e_j\right)\right]$ is the matrix associated to $g$ in the basis $\left(e_1, e_2, \ldots, e_n\right)$, we have $i_0(g)=\operatorname{dim} \operatorname{rad}(g)=\operatorname{dim} \operatorname{ker} \mathbf{G}$, and, since $\mathbf{G}$ is diagonal, $\operatorname{dim} \operatorname{ker} G=n_0$.

Let $g$ be a metric on a finite-dimensional real vector space $X$. Then $g$ has a $g$-orthogonal basis.

Proof. Let $r$ be the rank of $g, r:=n-\operatorname{dim} \operatorname{rad}(g)$, and let $\left(w_1, w_2, \ldots, w_{n-r}\right)$ be a basis of $\operatorname{rad}(g)$. If $V$ denotes a supplementary subspace of $\operatorname{rad}(g)$, then $V$ is $g$-orthogonal to $\operatorname{rad} g$ and $\operatorname{dim} V=r$. Moreover, for every $v \in V$ there is $z \in X$ such that $g(v, z) \neq 0$. Decomposing $z$ as $z=w+t, w \in V, t \in \operatorname{rad}(g)$, we then have $g(v, w)=g(v, w)+$ $g(v, t)=g(v, z) \neq 0$, i.e., $g$ is nondegenerate on $V$. Since trivially, $\left(w_1, w_2, \ldots, w_{n-r}\right)$ is $g$-orthogonal and $V$ is $g$-orthogonal to $\left(w_1, w_2, \ldots, w_{n-r}\right)$, in order to conclude it suffices to complete the basis $\left(w_1, w_2, \ldots, w_{n-r}\right)$ with a $g$-orthogonal basis of $V$; in other words, it suffices to prove the claim under the further assumption that $g$ be nondegenerate.

We proceed by induction on the dimension of $X$. Let $\left(f_1, f_2, \ldots, f_n\right)$ be a basis of $X$. We claim that there exists $e_1 \in X$ with $g\left(e_1, e_1\right) \neq 0$. In fact, if for some $f_i$ we have $g\left(f_i, f_i\right) \neq 0$, we simply choose $e_1:=f_i$, otherwise, if $g\left(f_i, f_i\right)=0$ for all $i$, for some $k \neq 0$ we must have $g\left(f_1, f_k\right) \neq 0$, since by assumption $\operatorname{rad}(g)={0}$. In this case, we choose $e_1:=f_1+f_k$ as
$$g\left(f_1+f_k, f_1+f_k\right)=g\left(f_1, f_1\right)+2 g\left(f_1, f_k\right)+g\left(f_k, f_k\right)=0+2 g\left(f_1, f_k\right)+0 \neq 0 .$$
Now it is easily seen that the subspace
$$V_1:=\left{v \in X \mid g\left(e_1, v\right)=0\right}$$
supplements Span $\left{e_1\right}$, and we find a basis $\left(v_2, \ldots, v_n\right)$ of $V_1$ such that $g\left(v_j, e_1\right)=0$ for all $j=2, \ldots, n$ by setting
$$v_j:=f_j-\frac{g\left(f_j, e_1\right)}{g\left(e_1, e_1\right)} e_1 .$$
Since $g$ is nondegenerate on $V_1$, by the induction assumption we find a $g$-orthogonal basis $\left(e_2, \ldots, e_n\right)$ of $V_1$, and the vectors $\left(e_1, e_2, \ldots, e_n\right)$ form a $g$-orthogonal basis of $X$.

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## Math0033｜Numerical Methods 数值方法 伦敦大学学院

statistics-labTM为您提供悉尼大学 London’s Global University MATH0048 Mathematical Analysis数学分析英国代写代考辅导服务！

Many phenomena in engineering and the physical and biological sciences can be described
using mathematical models. Frequently the resulting models cannot be solved analytically,
in which case a common approach is to use a numerical method to find an approximate
solution. The aim of this course is to introduce the basic ideas underpinning computational mathematics, study a series of numerical methods to solve different problems, and
carry out a rigorous mathematical analysis of their accuracy and stability

## Numerical Methods 数值方法作业案例

we derived the an equation to describe a mass that is attached to a spring that would break when its elongation reached $0.03 \mathrm{~m}$ during resonant vibration of the springmass system. We need to determine the time $\mathrm{t}_{\mathrm{f}}$ at which the spring breaks from Equation (a):
$$\left(0.05-\frac{t_f}{20}\right) \cos 10 t_f+\frac{1}{200} \sin 10 t_f-0.03=0$$

We will use the Newton-Raphson’s method to solve the unknown quantity $t_f$ in Equation (a) by first assuming a solution on $t_f=0.75$. We made this assumed solution based on a crude approximated value of $\mathrm{t}{\mathrm{f}}=0.7$ in Example 8.9. Again, let us replace the unknown quantity $t_f$ in Equation (a) by conventional unknown symbol $\mathrm{x}$ in the following alternative form: $$\begin{gathered} \left(0.05-\frac{x}{20}\right) \cos 10 x+\frac{1}{200} \sin 10 x-0.03=0 \ f(x)=\left(0.05-\frac{x}{20}\right) \cos 10 x+\frac{1}{200} \sin 10 x-0.03 \ f^{\prime}(x)=-(0.05-0.5 x) \sin 10 x \end{gathered}$$ Thus, the estimated root $x{i+1}$ after the previously estimated root $x_i$ may be computed by using the expression, as will be shown in the next slide.

Solution $\mathrm{x}$ from the equation: $\left(0.05-\frac{x}{20}\right) \cos 10 x+\frac{1}{200} \sin 10 x-0.03=0$

Evaluate the following integral by using the Gaussian quadrature $$I=\int_0^\pi \cos x d x$$

We have the function $y(x)=\cos x$ over the integration limits $x_a=0$ and $x_b=\pi$. The transformation of coordinates makes use of the relationship $x=\frac{\pi}{2} \xi+\frac{\pi}{2}$ from Equation (10.13), from which we get:
$$y(x)=\cos x=F(\xi)=\cos \left(\frac{\pi}{2} \xi+\frac{\pi}{2}\right)=\sin \left(\frac{\pi}{2} \xi\right)$$
Also, from Equation (10.14) with the use of the trigonometric relationships such as:
$$\sin \left(\frac{\pi}{2}+\theta\right)=\cos \theta \text { and } \cos \left(\frac{\pi}{2}+\theta\right)=\sin \theta$$
We may arrive at the following expression for integrating I in Equation (a) using Gaussian quadrature:
$$I=\int_0^\pi \cos x d x=\int_{-1}^1\left[\sin \left(\frac{\pi}{2} \xi\right)\left(\frac{\pi}{2} d \xi\right)\right]=\frac{\pi}{2} \int_{-1}^1 \sin \frac{\pi}{2} \xi d \xi=\frac{\pi}{2} \sum_{i=1}^n H_i \sin \left(\frac{\pi}{2} a_i\right)$$
Let us take, for example, 3 sampling points, i.e., $\mathrm{n}=3$ from Table 10.3 on P.357 with:
$$\begin{array}{lll} \mathrm{a}_1=0 & \mathrm{a}_2=+0.77459 & \mathrm{a}_3=-0.77459 \ \mathrm{H}_1=0.88888 & \mathrm{H}_2=0.55555 & \mathrm{H}_3=0.55555 \end{array}$$
Substituting the above numbers into Equation (b) will lead to the solution:
\begin{aligned} I & =\frac{\pi}{2}\left[0.88888 \sin (0)+0.55555 \sin \left(\frac{\pi}{2} \times 0.77459\right)+0.55555 \sin \left(-\frac{\pi}{2} \times 0.77459\right)\right] \ & =\frac{\pi}{2}[0.55555 \sin (1.2167)-0.55555 \sin (1.2167)]=0 \end{aligned}
25

Evaluate the following integral in Example 10.8 using Gaussian quadrature method

Solution:
$$I=\int_{0.5}^{3.5} y(x) d x=\int_{0.5}^{3.5} x \sqrt{\left(16-x^2\right)^3} d x$$

We notice the function $y(x)$ in the integral in Equation (a) is identical to what we had in previous Examples 10.5 (p.349), 6 (p.350) and 8 (p.355).
We will first derive the expression of the function $F(\xi)$ for the function $y(x)$ in Equation (a) from the integration limits of 0.5 and 3.5 to the limit -1 and +1 , as required in Gaussian quadrature.
The transformation of coordinate systems as illustrated in Figure 10.15 begins with the transformation of variable from $x$ to $\xi$ using the relationship in Equation (10.13), leading to the following relationship between the variables $x$ and $\xi$ as shown below:
$x=\frac{1}{2}\left(x_b-x_a\right) \xi+\frac{1}{2}\left(x_b+x_a\right)=1.5 \xi+2$, from which we obtain: $\mathrm{dx}=1.5 \mathrm{~d} \xi$
The integral in Equation (a) in the $y(x)$ vs. $x$ coordinates can thus be transformed to the $F(\xi)$ vs. $\xi$ coordinates by using Equation (10.14), yielding the following expression:
\begin{aligned} I=\int_{0.5}^{3.5} y(x) d x=\int_{0.5}^{3.5} x \sqrt{\left(16-x^2\right)^3} d x & =\frac{x_b-x_a}{2} \int_{-1}^1 F(\xi) d \xi \ & =\left.\frac{3.5-0.5}{2} \int_{-1}^1 y(x)\right|{x=1.5 \xi+2} d \xi \end{aligned} We may thus evaluate the integral with the expression in Equation (d) as: $$I=1.5 \int{-1}^1(1.5 \xi+2) \sqrt{\left[16-(1.5 \xi+2)^2\right]^3} d \xi=1.5 \int_{-1}^1(1.5 \xi+2) \sqrt{\left(-2.25 \xi^2-6 \xi+12\right)^3} d \xi We have arrived at the following integral:$$
I=1.5 \int^1(1.5 \xi+2) \sqrt{\left[16-(1.5 \xi+2)^2\right]^3} d \xi=1.5 \int^1(1.5 \xi+2) \sqrt{\left(-2.25 \xi^2-6 \xi+12\right)^3} d \xi
$$统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。 ## 金融工程代写 金融工程是使用数学技术来解决金融问题。金融工程使用计算机科学、统计学、经济学和应用数学领域的工具和知识来解决当前的金融问题，以及设计新的和创新的金融产品。 ## 非参数统计代写 非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。 ## 广义线性模型代考 广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。 术语 广义线性模型（GLM）通常是指给定连续和/或分类预测因素的连续响应变量的常规线性回归模型。它包括多元线性回归，以及方差分析和方差分析（仅含固定效应）。 ## 有限元方法代写 有限元方法（FEM）是一种流行的方法，用于数值解决工程和数学建模中出现的微分方程。典型的问题领域包括结构分析、传热、流体流动、质量运输和电磁势等传统领域。 有限元是一种通用的数值方法，用于解决两个或三个空间变量的偏微分方程（即一些边界值问题）。为了解决一个问题，有限元将一个大系统细分为更小、更简单的部分，称为有限元。这是通过在空间维度上的特定空间离散化来实现的，它是通过构建对象的网格来实现的：用于求解的数值域，它有有限数量的点。边界值问题的有限元方法表述最终导致一个代数方程组。该方法在域上对未知函数进行逼近。[1] 然后将模拟这些有限元的简单方程组合成一个更大的方程系统，以模拟整个问题。然后，有限元通过变化微积分使相关的误差函数最小化来逼近一个解决方案。 tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。 ## 随机分析代写 随机微积分是数学的一个分支，对随机过程进行操作。它允许为随机过程的积分定义一个关于随机过程的一致的积分理论。这个领域是由日本数学家伊藤清在第二次世界大战期间创建并开始的。 ## 时间序列分析代写 随机过程，是依赖于参数的一组随机变量的全体，参数通常是时间。 随机变量是随机现象的数量表现，其时间序列是一组按照时间发生先后顺序进行排列的数据点序列。通常一组时间序列的时间间隔为一恒定值（如1秒，5分钟，12小时，7天，1年），因此时间序列可以作为离散时间数据进行分析处理。研究时间序列数据的意义在于现实中，往往需要研究某个事物其随时间发展变化的规律。这就需要通过研究该事物过去发展的历史记录，以得到其自身发展的规律。 ## 回归分析代写 多元回归分析渐进（Multiple Regression Analysis Asymptotics）属于计量经济学领域，主要是一种数学上的统计分析方法，可以分析复杂情况下各影响因素的数学关系，在自然科学、社会和经济学等多个领域内应用广泛。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 ## Econ8009｜International Monetary Economics 国际货币经济学 澳洲国立大学 statistics-labTM为您提供澳洲国立大学 Australian National University Econ8009 Mathematical Analysis数学分析澳洲代写代考辅导服务！ 课程介绍：This course will provide students with models that can be used to analyse issues in international monetary economics. The course introduces students to so-called global imbalances and the basic stylized facts about current accounts with cross-section and the time-series of selected countries such as the US and China. Issues of current account sustainability will also be addressed. ## International Monetary Economics 国际货币经济学作业案例 问题 1. 2) Jo and Maureen (aka Mo) are the only two people who drive cabs from the Leominster Airport to the Leominster Inn. The two cabs have the following identical total cost functions.$$
\mathrm{TC}{\mathrm{Jo}}=\$3.00 \mathrm{xQ}{\mathrm{Jo}}, \quad \mathrm{TC}{\mathrm{Mo}}=\$ 3.00 \mathrm{x} \mathrm{Q}{\mathrm{Mo}} \text { where } \mathrm{QX}_{\mathrm{X}}=\text { number of trips per day made by driver } \mathrm{X} .
$$Demand for trips is given by the demand function:$$
\text { Price per Trip }=\$6.00-.05 \mathrm{xQ} $$a) Write down the average and marginal cost functions for this cab “industry”. Answer: You should be able to write down the answers by inspection but we can derive them mathematically. Define \mathrm{Q}=\mathrm{Q}{\mathrm{Joe}}+\mathrm{Q}{\mathrm{Moe}}. Then \mathrm{TC}{\text {Industry }}=\ 3.00 \mathrm{xQ}$$ \mathrm{AC}{\text {Industry }}=\mathrm{TC}{\text {Industry }} / \mathrm{Q}=\$ 3.00 \quad \mathrm{MC}{\text {Industry }}=\frac{\mathrm{d}\left(\mathrm{TC}_{\text {Industry })}\right.}{\mathrm{dQ}}=\$3.00 $$i.e. both average and marginal costs are the same horizontal line =\ 3.00 问题 2. b) Assume Jo and Mo agree to cooperate to set the price per trip and split any profits they generate. Determine the industry equilibrium price and quantity. Answer: They will run the industry like a monopoly setting marginal revenue = marginal cost. They will then presumably equally split the profit maximizing number of trips. As we know by now, Marginal Revenue for this demand curve is \ 6.00-.1 \mathrm{Q} so:$$ \$ 6.00-.1 \mathrm{Q}=\$3.00 \text { or } \mathrm{Q}=\$ 3.00 / .1=30 \text { trips }(\text { total) } \text { or } 15 \text { trips per driver. }
$$To get the price, we plug 30 trips back into the demand curve and get \mathrm{P}=\ 6.00-.05 \times 30=\ 4.50 Profit per cab =( price – average cost ) \mathrm{x} quantity =\ 1.50 \mathrm{x} 15=\ 22.50 问题 3. c) The Leominster Environmental Protection Agency (LEPA) has determined that each cab trip between the Leominster Airport and the Leominster Inn generate a pollution cost equal to \ .50 per trip. What is the implication of this cost for your solution in (b)? Discus in as much detail as you can LEPA’s policy options for dealing with this cost. Answer: Using the terms we developed in class, the \ 3.00 \mathrm{MC} is really the private marginal cost per trip while the \ .50 is the social cost of the trip. The economically efficient outcome is that the taxi industry set:$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。