Notes, summaries, assignments, exams, and problems for Mathematics

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Principles of Design: Proximity, Alignment, Repetition, Contrast, and More

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PRINCIPLES OF DESIGN

They affect content and message.

1. Proximity

Proximity basically means space; unfortunately, most people simply try to fill up empty space. Space should be organized, so, information can be easily understood.

2. Alignment

Alignment is like rulers or margins. We need to avoid to use more than one alignment, and to use the centered alignment.

3. Repetition

There are many ways to create repetition: -Bullets, bold fonts, color, line, a design element… Repetition unifies and strengthens. It creates visual interest.

4. Contrast

There are many ways to create contrasts: -Large/small type, warm/cold colors, old/new fonts, and horizontal/vertical… Contrast has 2 purposes: 1. Create interest on a page. 2. Aid organization on a page

5.

... Continue reading "Principles of Design: Proximity, Alignment, Repetition, Contrast, and More" »

Understanding Business Statistics: Concepts and Techniques

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1. The role of business statistics is to convert data into meaningful info.

2. The two primary categories of statistical interference tools are estimation and hypothesis testing.
3. The sampling method by which the population is divided into groups that are each intended to be mini-populations is known as cluster sampling.
4. The set of all objects or individuals of interest or the measurements obtained from all objects or individuals of interest is a population.
5. Descriptive numerical measures that are computed from the entire population are called parameters.
6. A sampling technique that selects the items from the population based on accessibility and ease of selection is called convenience sampling
7. In which of the following sampling methods
... Continue reading "Understanding Business Statistics: Concepts and Techniques" »

Relationship between Interest Rates and Bond Prices

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Which of the following statements about the relationship between interest rates and bond prices is true?

  1. There is an inverse relationship between bond prices and interest rates.
  2. There is a direct relationship between bond prices and interest rates.
  3. The price of short-term bonds fluctuates more than the price of long-term bonds for a given change in interest rates. (Assuming that coupon rate is the same for both)
  4. The price of long-term bonds fluctuates more than the price of short-term bonds for a given change in interest rates. (Assuming that the coupon rate is the same for both)

Answer: I and IV only

Bond Duration

Consider a bond with a face value of $1,000, a coupon rate of 8%, a yield to maturity of 9%, and ten years to maturity. This bond's duration... Continue reading "Relationship between Interest Rates and Bond Prices" »

Vector Operations, Dot and Cross Products, and Lines

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Vector Determination by Length and Angle

V = <||V|| Cosθ, ||V|| Sinθ> ---> ||V||Cosθi + ||V||Sinθj

Example:

a) Find the vector of length 2 that makes an angle of π/4 with the positive x-axis.

b) Find the angle that the vector V = -\sqrt{\ }3 i + j makes with the positive x-axis.

a) <||V||Cosθ , ||V||Sinθ> = <2cos45, 2sin45> ---> <\sqrt{\ }2, \sqrt{\ }2>

b) Normalize... ||V|| = \sqrt{\ }(-3)2 + 12 = \sqrt{\ }4 = 2 -----> V/||V|| = <-\sqrt{\ }3/2 , 1/2> = <cosθ, sinθ> ----> cosθ = -\sqrt{\ }3/2, sinθ = 1/2 ---> θ = 5π/6

Dot Product

If U = <U1, U2> and V = <V1, V2>, then the dot product is UV = U1V1 + U2V2.

Example:

a) U = <3, 5>, V = <-1, 2> -----> UV = (3)(-1) + (5)(2) ---> UV = -3 + 10 --> UV = 7

b) U = &... Continue reading "Vector Operations, Dot and Cross Products, and Lines" »

Valuation Methods and Cash Flow Analysis

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Multiples consist of using the multiples of a company similar to yours to apply to your numbers and get an approximation of what yours is worth. It is not the most accurate in my opinion.

An example would be the EBITDA multiple, you multiply EBITDA by that figure obtained by dividing the value of the company you are comparing yourself to by its EBITDA.

This method is useful because although the estimated value of your company is not precise, with this method you can see the evolution over time.

250. The increase in provisions is not considered an outflow of cash, so it goes to the income statement.

– 150. The following are considered cash outflows

Add the cash paid instantly by the customer recorded in the cash flow statement as cash received... Continue reading "Valuation Methods and Cash Flow Analysis" »

Dfsfdsf

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hormigones no controlados
-hormigon simple 170kg/m3
-hormigon armado h-20 300kg/m3
hormigones controlados
-hormigon armado protegido de la interperie 240kg/m3
-hormigon armado expuesto a la interperie 270kg/m3
proteccion y curado
-primera 24 hrs. es de vital importancia (7dias) curado (4dias) alta resistencia.
aditivos y adiciones
aditivos: se agregan pequeñas cantidades para modificar sus propiedades.
adiciones: materiales solidos inertes que se agregan
clasificacion de morteros
-mortero estuco: interior y exterior
-mortero industrial:·Mortero mezclado en una central de mortero.
Mortero de premezclado en seco
-revestimiento continuo y discontinuo
-junta albañileria
cemento clasificacion
por composicion, grado, (alta resistencia - corriente)
tipos de cemento
cemento
... Continue reading "Dfsfdsf" »

Understanding Cryptographic Functions and RSA Security

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What is a One-Way Function?

A one-way function is a function that is easy to compute in one direction but difficult to compute in the reverse. For example, given an input, a hash is easy to compute. However, given the output, it is extremely difficult (time-consuming) to determine the input.

What is a Trapdoor Function?

A trapdoor function is like a one-way function; however, a trapdoor function can reverse the one-way function if the trapdoor is known. For example, finding the two prime divisors of 6,895,601 is difficult. However, if you know that 1,931 is one of the numbers, it will be easy to divide 6,895,601 by 1,931 to determine the answer. In this case, 1,931 is the trapdoor.

What Makes RSA Difficult to Break?

The large number factorization... Continue reading "Understanding Cryptographic Functions and RSA Security" »

Key Statistical Measures: Location and Variation in Data

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Sample Characteristics: Measures of Location and Variation

Measures of Location

  • Average

    The average is a central value, often referred to as a measure of central tendency.

  • Mean

    • Also known as the arithmetic mean.
    • Found by dividing the sum of all values by the total number of values.
    • Formula: Mean = (Sum of Quantities) / (Number of Quantities)
    • Example: Find the mean of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
      • Sum of values: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55
      • Number of values = 10
      • Mean of values = 55 / 10 = 5.5
  • Median

    The median of a set is the middle number when the values are arranged in order from lowest to highest.

  • Mode

    • The mode is the value that occurs most frequently in a set of values, also known as the modal value.
    • Example: Find the mode of 1, 2, 2, 3,
... Continue reading "Key Statistical Measures: Location and Variation in Data" »

Benefits of Using Games, Songs, and Stories in Language Teaching

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10 reasons to use games in language teaching

1.Games create a context for meaningful communication. 2. This meaningful communication serves as a basis for comprehensible input. 3. Games add interest to what learners find boring. 4. Games can be used with all language skills. 5. Games offer a fun experience. 6. Games encourage participation from all learners. 7. Games are learner-centered activities. 8. Games work outside of class. 9. Games promote cooperative learning. 10. Games fit into multiple intelligence theory.

Using Games in Teaching English to Young Learners: How to Choose a Game

Students may wish to play games purely for fun. Teachers need more convincing reasons. The key to a successful language game is that the rules are clear. The... Continue reading "Benefits of Using Games, Songs, and Stories in Language Teaching" »

Normal Distribution Problems: Z‑Scores, Percentages & Parameters

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Problem 4: Bank Savings Distribution

The average savings in the clients' accounts of a certain bank follow a normal distribution N(50,000, 15,000) (mean = 50,000 €, standard deviation = 15,000 €). The bank distributes its clients into three categories:

  • Category A: Those clients that have less than 20,000 € in their savings account
  • Category B: Those clients that have between 20,000 € and 80,000 € in their savings account
  • Category C: Those clients that have more than 80,000 € in their savings account

Calculate:

4a) What percentage of clients belong to Category A?

Use the standard normal variable Z = (X - μ) / σ.

For X = 20,000 €:

Z = (20,000 - 50,000) / 15,000 = -30,000 / 15,000 = -2.000

The cumulative probability Φ(-2.00) ≈ 0.0228,... Continue reading "Normal Distribution Problems: Z‑Scores, Percentages & Parameters" »