Examples of M&A in the following topics:
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Alternating Series
- If $m$ is odd and $S_m - S_n < a_{m}$ via the following calculation:
- $\begin{aligned} S_m - S_n & = \sum_{k=0}^m(-1)^k\,a_k\,-\,\sum_{k=0}^n\,(-1)^k\,a_k\ \\& = \sum_{k=m+1}^n\,(-1)^k\,a_k \\ & =a_{m+1}-a_{m+2}+a_{m+3}-a_{m+4}+\cdots+a_n\\ & =\displaystyle a_{m+1}-(a_{m+2}-a_{m+3}) -\cdots-a_n \le a_{m+1}\le a_{m} \\& \quad [a_{n} \text{ decreasing}].
- Thus, we have the final inequality $S_m - S_n \le a_{m}$.
- Similarly, it can be shown that, since $a_m$ converges to $0$, $S_m - S_n$ converges to $0$ for $m, n \rightarrow \infty$.
- (The sequence $\{ S_m \}$ is said to form a Cauchy sequence, meaning that elements of the sequence become arbitrarily close to each other as the sequence progresses.)
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Dot Plots
- Judge whether a dot plot would be appropriate for a given data set
- Figure 1 uses a dot plot to display the number of M & M's of each color found in a bag of M & M's.
- Each dot represents a single M & M.
- From the figure, you can see that there were 3 blue M & M's, 19 brown M & M's, etc.
- A dot plot showing the number of M & M's of various colors in a bag of M & M's.
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mRNA Processing
- Eukaryotic pre-mRNA receives a 5' cap and a 3' poly (A) tail before introns are removed and the mRNA is considered ready for translation.
- The additional steps involved in eukaryotic mRNA maturation create a molecule with a much longer half-life than a prokaryotic mRNA.
- An enzyme called poly (A) polymerase (PAP) is part of the same protein complex that cleaves the pre-mRNA and it immediately adds a string of approximately 200 A nucleotides, called the poly (A) tail, to the 3' end of the just-cleaved pre-mRNA.
- All introns in a pre-mRNA must be completely and precisely removed before protein synthesis.
- Poly (A) Polymerase adds a 3' poly (A) tail to the pre-mRNA.
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Lab: Probability Topics
- Next, put the M&M's in a cup.
- The experiment is to pick 2 M&M's, one at a time.
- After you record the pick, put both M&M's back.
- Do this a total of 24 times, also.
- Round to 4 decimal places. a.
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Molarity
- How many moles of potassium chloride (KCl) are in 4.0 L of a 0.65 M solution?
- There are 2.6 moles of KCl in a 0.65 M solution that occupies 4.0 L.
- A scientist has a 5.0 M solution of hydrochloric acid (HCl) and his new experiment requires 150.0 mL of 2.0 M HCl.
- Water was added to 25 mL of a stock solution of 5.0 M HBr until the total volume of the solution was 2.5 L.
- We calculate that we will have a 0.05 M solution, which is consistent with our expectations considering we diluted 25 mL of a concentrated solution to 2500 mL.
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Summary of the Uniform and Exponential Probability
- X = a real number between a and b (in some instances, X can take on the values a and b). a = smallest X ; b = largest X
- Probability density function: f (X) = $\frac{1}{ ba }$ for a≤ X ≤b
- X = a real number, 0 or larger. m = the parameter that controls the rate of decay or decline
- µ = σ = 1/m and m = 1/µ = 1/σ
- Area Between c and d: P(c < X < d) = P(X < d)−P(X < c) =1−e−m·d−(1−e−m·c) = e−m·c−e−m·d
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References
- A Vygotskian interpretation of Reading Recovery.
- Hobsbaum, A., & Peters, S. (1996).
- A., & Beaver, J. (1995).
- Allington & S.A.
- A., & Simic, O. (2002).
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An Example
- Here is a simple piece of Mathematica code that will draw the modes of a rectangular plate.
- Lx = 1.5; Ly = 1; c = 1; d[x_,y_,m_,n_]= Sin[m Pi x/Lx]Sin[n Pi y/Ly]; w[n_,m_] = c Sqrt[(m Pi /Lx)^2 + (n Pi/Ly)^2]; Do[Do[ContourPlot[d[x,y,m,n],{x,0,Lx},{y,0,Ly},AspectRatio->Ly/Lx];,{m,2}];,{n,2}];
- And in Figure 2.3 is a 3D perspective view of one of the modes.
- On the WWW page you'll find a Mathematica notebook that animates this.
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Elastic Collisions in One Dimension
- An elastic collision is a collision between two or more bodies in which kinetic energy is conserved.
- An elastic collision is a collision between two or more bodies in which the total kinetic energy of the bodies before the collision is equal to the total kinetic energy of the bodies after the collision.
- Consider a first particle with mass $m_{1}$ and velocity $v_{1i}$ and a second particle with mass $m_{2}$ and velocity $v_{2i}$.
- $\frac{1}{2}m_1\cdot v_{1i}^2+\frac{1}{2}m_2\cdot v_{2i}^2=\frac{1}{2}m_1\cdot v_{1f}^2+\frac{1}{2}m_2\cdot v_{2f}^2$ (due to conservation of kinetic energy)
- After doing a little bit of algebra on Eq. 5 we find:
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Matrices for two degrees of freedom
- $m_1 {\ddot{x}_1} + k_1 x_1 + k_2 (x_1 - x_2) = 0$
- $\displaystyle{ M = \left[ \begin{array}{cc} m_1 & 0 \\ 0 & m_2 \end{array} \right] }$
- $\displaystyle{ M^{-1} = \left[ \begin{array}{cc} {m_1}^{-1} & 0 \\ 0 & {m_2}^{-1} \end{array} \right]. }$
- $\displaystyle{M^{-1} K = \left[ \begin{array}{cc} \frac{k_1 + k_2}{m_1} & \frac{-k_2}{m_1} \\ \frac{-k_2}{m_2} & \frac{k_2+k_3}{m_2} \end{array} \right]. }$
- Letting $\omega_0 = \sqrt{k/m}$ as usual and defining $\Omega = \sqrt{k_2/m}$ , we have the following beautiful form for the matrix $M^{-1} K$ :