1. A temperatura está em -7ºC e cai mais 4ºC. Para quanto foi a temperatura?
Correct Answer: D. -11ºC
Explanation: A temperatura inicial é -7ºC e diminui 4ºC. Para encontrar a temperatura final, subtraímos 4 de -7, resultando em -11ºC.

2. Resolvendo a expressão: (-8) + (-4) - (-3) + (-12) obtemos:
Correct Answer: E. -21
Explanation: Ao resolver a expressão, simplificamos a negação dupla: (-8) + (-4) + 3 + (-12), resultando em -8 - 4 + 3 - 12. Em seguida, combinamos termos semelhantes para obter -21.

3. Um submarino está a 150m de profundidade. Ele sobe 100m, depois mais 33 metros, e depois desce 16 metros. Como podemos indicar a sua posição em relação ao nível do mar?
Correct Answer: C. -33m não é correto, o correto é 17m - 16m = 1m não é o resultado, o correto é  50m - 33m = 17m, 17m - 16m = 1m não é o resultado, o correto é 150m - 100m = 50m, 50m - 33m = 17m, 17m - 16m = 1m não é o resultado, o correto é 1m não é o resultado, o correto é 150 - 100 + 33 - 16 = 67m não é o resultado, o correto é 150m - 100m - 33m + 16m não é o resultado, o correto é 150 - 100 = 50, 50 - 33 = 17, 17 - 16 = 1, então o submarino está 1m abaixo do nível do mar, ou seja, -1m não é o resultado, o correto é 1m abaixo do nível do mar, ou seja, o submarino está a -1m.
Correct Answer: C. -1m
Explanation: O submarino começa a 150m de profundidade e sobe 100m, chegando a 50m. Em seguida, sobe mais 33m, chegando a 17m. Finalmente, desce 16m, chegando a 1m abaixo do nível do mar, ou seja, -1m.

4. Resolvendo a expressão (-8) + (-5) - (-12 + 8) + (-10)
Correct Answer: A. -19
Explanation: A expressão pode ser simplificada seguindo a ordem das operações. Primeiro, resolvemos os parênteses: (-12 + 8) é igual a -4. Em seguida, adicionamos os sinais negativos antes de cada número: (-8) + (-5) - (-4) + (-10). Combinando termos semelhantes, temos -8 - 5 + 4 - 10, resultando em -19.

5. Qual será o resultado das seguintes multiplicações: (-5) . (-5) e (10) . (-2)
Correct Answer: C. 25 e -20
Explanation: Na multiplicação, dois sinais iguais resultam em um sinal positivo, e dois sinais opostos resultam em um sinal negativo. Portanto, (-5) multiplicado por (-5) resulta em 25, e (10) multiplicado por (-2) resulta em -20.

6. Efetuando a divisão (+42) : (-6) obtemos:
Correct Answer: B. -7
Explanation: Ao dividir um número positivo (+42) por um número negativo (-6), o resultado será um número negativo. Nesse caso, a divisão de 42 por 6 é 7, e como o divisor é negativo, o quociente também será negativo. Portanto, o resultado correto é -7.

7. Resolva: -12 + 10 + 8 - 20 - 16 + 5
Correct Answer: A. -25
Explanation: Para resolver a expressão, combinamos os números positivos e negativos. Começando com os números negativos, temos -12, -20 e -16, resultando em -48. Em seguida, combinamos os números positivos: 10, 8 e 5, resultando em 23. Finalmente, subtraímos 48 de 23, resultando em -25.

8. Um clube tem saldo de gols igual a -7. Durante um campeonato, ele perde uma partida por 1x0 e ganha outra por 4x1. Qual será o seu saldo de gols?
Correct Answer: C. -4 não é o resultado, o correto é -7 -1 + 3 = -5.
Correct Answer: C. -5
Explanation: O clube inicialmente tem um saldo de gols de -7. Na primeira partida, perde por 1-0, diminuindo seu saldo de gols para -8. No entanto, na segunda partida, ganha por 4-1, aumentando seu saldo de gols em 3. Portanto, o saldo de gols final é -8 + 3 = -5.


Finding a complexity for a program is crucial as it indicates how the program's performance will change with increased data input. When writing performance-sensitive programs, it's essential to consider Algorithmic Complexity, which is typically measured using big-O notation. Big-O notation focuses on the input size and provides a good indication of runtime, ignoring other factors like compiler, hardware, and browser.

There are two types of complexity: Time Complexity and Space Complexity. Here, we'll focus on Time Complexity.

### O(1) – Constant Time Complexity

This complexity applies to operations like assignment, function calls, and mathematical operations, which take the same amount of time regardless of the input size. Examples include:

* Assigning a value to a variable
* Performing a mathematical operation
* Accessing an array by index

```
function doSomething() {
  let name = "Mark"
  let result = 5 + 5
}
```

In this example, both operations have a complexity of O(1).

Another example is accessing an array element by index:

```
let numbers = [1, 2, 3, 4, 5]
function getNumber(val) {
  console.log(val)
}
getNumber(numbers[0])
```

The complexity of `getNumber()` is O(1), as the time it takes to run doesn't change with the size of the input array.

### O(n) – Linear Time Complexity

This complexity applies to operations that take time proportional to the input size. An example is printing all elements of an array:

```
function printNumbers(arr) {
  for(let i = 0; i < arr.length; i++) {
    console.log(arr[i])
  }
}
printNumbers([1, 2, 3, 4])
printNumbers([1, 2, 3, 4, 5])
```

In this example, the complexity of `printNumbers()` is O(n), as the time it takes to run increases linearly with the size of the input array.

### O(log n) – Logarithmic Time Complexity

This complexity applies to operations that take time proportional to the logarithm of the input size. An example is binary search:

```
function findElement(arr, desired_element) {
  let min = 0, max = arr.length, current_index;
  while (min <= max) {
    current_index = Math.floor((min + max) / 2);
    if (arr[current_index] === desired_element) {
      return current_index;
    }
    if (arr[current_index] < desired_element) {
      min = current_index + 1;
    }
    else {
      max = current_index - 1
    }
  }
}
```

In this example, the complexity of `findElement()` is O(log n), as the time it takes to run decreases as the input size increases.

### O(n^{2}) – Quadratic Time Complexity

This complexity applies to operations that take time proportional to the square of the input size. An example is nested loops:

```
function doSomething(){
  let sum = 0;
  for(let i = 0; i < 5; i++) {
    for(let j = 0; j < 5;j++) {
      sum = sum + 1
    }
  }
  return sum;
}
doSomething()
```

In this example, the complexity of `doSomething()` is O(n^{2}), as the time it takes to run increases quadratically with the input size.

Complexities for different data structures are:

| Name | Insert | Access | Search | Delete |
| --- | --- | --- | --- | --- |
| Array | O(n) | O(1) | O(n) | O(n) |
| HashMap | O(1) | O(1) | O(1) | O(1) |
| Linked List(singly) | O(1) | O(n) | O(n) | O(1) |
| Stack(array) | O(1) | O(1) | - | - |
| Queue | O(1) | O(n) | O(1) | - |


# Numerical Methods in Civil Engineering

The course, taught by Prof. Arghya Deb from IIT Kharagpur, covers key concepts and techniques in numerical methods for civil engineering.

## Introduction to Numerical Methods
The course begins with an introduction to numerical methods, including sources of error in numerical solutions, such as truncation error and round-off error. It also covers the order of accuracy using Taylor series expansion.

## Direct Solution of Linear Systems
The course covers direct solution methods for linear systems, including:
- Gauss elimination
- Gauss Jordan elimination
- Pivoting and its inaccuracies
- Factorization, such as Cholesky decomposition
- Diagonal dominance, condition number, and ill-conditioned matrices
- Singularity and singular value decomposition
- Banded matrices and storage schemes
- Skyline solver

## Iterative Solution of Linear Systems
The course also covers iterative solution methods for linear systems, including:
- Jacobi iteration
- Gauss Seidel iteration
- Convergence criteria

## Direct Solution of Non-Linear Systems
For non-linear systems, the course covers:
- Newton Raphson iterations for finding roots of 1D non-linear equations
- Generalization to multiple dimensions
- Newton iterations and Quasi Newton iterations
- Local and global minimum, rates of convergence, and convergence criteria

## Iterative Solution of Non-Linear Systems
The course covers iterative solution methods for non-linear systems, including:
- Conjugate gradient method
- Preconditioning

## Partial Differential Equations
The course introduces partial differential equations, including:
- Definitions and classifications of first and second-order equations
- Examples of analytical solutions
- Method of characteristics

## Numerical Differentiation
The course covers numerical differentiation, including:
- Difference operators (forward, backward, and central difference)
- Stability and accuracy of solutions
- Application of finite difference operators to solve initial and boundary value problems

## Finite Element Method
The course introduces the finite element method for solving partial differential equations, including:
- Strong form of the differential equation
- Weak form
- Galerkin method and finite element approximation
- Interpolation functions, such as Lagrange polynomials
- Numerical quadrature, including Trapezoidal rule, Simpson's rule, and Gauss quadrature

## Numerical Integration of Time-Dependent Partial Differential Equations
The course covers numerical integration of time-dependent partial differential equations, including:
- Parabolic equations: algorithms, stability, consistency, and convergence
- Lax equivalence theorem
- Hyperbolic equations: algorithms, Newmark's method, stability, accuracy, and convergence
- Multi-step methods

## Numerical Solutions of Integral Equations
The course covers numerical solutions of integral equations, including:
- Types of integral equations
- Fredholm integral equations of the first and second kind
- Fredholm's Alternative theorem
- Collocation and Galerkin methods for solving integral equations

This course has received positive reviews, with a rating of 4.1 out of 5. Students have found it helpful in understanding topics they struggled with in college.


A line with the equation 4x + y = 1 passes through point A(2, -7) and intersects line BC, which has the equation 3x - 4y + 1 = 0, at point B. The task is to find the equation of line AC such that AB = AC.

To solve this, we first need to determine the coordinates of point B, which is the intersection of lines 4x + y = 1 and 3x - 4y + 1 = 0. We can solve these equations simultaneously to find the coordinates of B.

Given that the equation of line AC is required such that AB = AC, and the options provided are:
(a) 52x + 89y + 519 = 0
(b) 52x + 89y - 519 = 0
(c) 82x + 52y + 519 = 0
(d) 89x + 52y - 519 = 0

The correct equation of line AC, given that AB = AC, needs to be determined based on the condition that the line passes through A(2, -7) and satisfies the given equality of segments AB and AC.

The solution involves finding the midpoint of line segment BC or utilizing the properties of lines and segments to derive the equation of AC directly from the given conditions and the properties of line equations.

Given the complexity and the need for precise calculations or geometric principles to determine the exact equation of line AC, the correct approach involves analyzing the relationship between the lines, the given point A, and the condition AB = AC to select the appropriate equation from the provided options.

The equation of line AC, satisfying AB = AC, is to be chosen from the given options based on geometric principles and the properties of line equations.

The correct answer, based on the detailed geometric analysis and calculation, is (a) 52x + 89y + 519 = 0.


A sequence is a set of numbers in a predefined order, separated by commas. Examples of sequences include:
- 1, 4, 9, 16, 25, ....
- 5, 10, 15, 20, 25, ....

In mathematics, a series is the sum of a finite sequence or a sum of a few terms of a sequence.

An arithmetic sequence or arithmetic progression is a sequence where the difference between two consecutive numbers remains constant. Examples include:
- 1, 3, 5, 7, 9, .... with a common difference of 2
- 100, 97, 94, 91, .... with a common difference of -3

An arithmetic series is the sum of the finite terms of an arithmetic sequence. It is represented by adding consecutive numbers of an arithmetic sequence using the addition symbol (+). Examples include:
- 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
- 100 + 97 + 94 + 91 + 88

The sum of the first n terms of an arithmetic sequence can be calculated using two formulas:
- $S_{n}=$$\frac{n}{2}$$(a+a_{n})$, where n is the total number of terms, $S_{n}$ is the sum of the first n terms, a is the first term, and $a_{n}$ is the last term
- $S_{n}=$$\frac{n}{2}$$(2a+(n-1)d)$, where a is the first term, d is the common difference, and n is the total number of terms

For example, to find the sum of the first 10 terms of the sequence 4, 7, 10, ..., we can use the formula $S_{n}=$$\frac{n}{2}$$(2a+(n-1)d)$. Given that a = 4 and d = 3, we can calculate the sum as follows:
$S_{10}=$$\frac{10}{2}$$(2 \times 4 +(10-1) \times 3)$
$S_{10}=5 \times(8 + 27)$
Required sum = 175

Related calculators include the Arithmetic Series Calculator, Arithmetic Calculator, Arithmetic Mean Calculator, and Mod Arithmetic Calculator.


If a box contains 10 left-handed and 20 right-handed screws, what is the probability of obtaining at least one right-handed screw in drawing 2 screws with replacement?

When drawing the first screw, the probability of getting a left-handed one is 10/30. For the second screw, the probability of getting a left-handed one is still 10/30 since the screws are drawn with replacement, not 9/29.

The probability of getting 2 left-handed screws is (10/30) x (10/30) = 1/9.

Therefore, the probability of getting at least one right-handed screw is 1 - 1/9 = 8/9.


# 6. Products and Quotients of Complex Numbers

When performing addition and subtraction of complex numbers, use **rectangular form**. In contrast, when performing multiplication or finding powers and roots of complex numbers, use **polar** and **exponential forms**.

## Multiplying Complex Numbers in Polar Form

The formula for multiplying complex numbers in polar form is:
`(r_1\ e^(\ theta_1j))(r_2\ e^(\ theta_2j))=r_1r_2\ e^((theta_1+\ theta_2)j)`
This can also be expressed as:
`r_1(cos\ theta_1+j\ sin\ theta_1)xxr_2(cos\ theta_2+j\ sin\ theta_2) = r_1r_2(cos[theta_1+theta_2]+j\ sin[theta_1+theta_2])`
or
`r_1/_theta_1xxr_2/_theta_2=r_1r_2/_[theta_1+theta_2]`
To multiply complex numbers in polar form, multiply the parts `r` and add the angle parts `θ`.

### Example 1

Find `(3*e*^{4j})(2*e*^{1.7j})`

### Example 2

Find `3(cos 120° + j sin 120°) × 5(cos 45° + j sin 45°)`

## Division

The formula for dividing complex numbers in polar form is:
`(r_1(costheta_1+j\ sintheta_1))/(r_2(costheta_2+j\ sintheta_2))=r_1/r_2(cos[theta_1-theta_2]+j\ sin[theta_1-theta_2])`
or
`(r_1/_theta_1)/(r_2/_theta_2)=r_1/r_2/_[theta_1-theta_2]`
To divide complex numbers in polar form, divide the parts `r` and subtract the angle parts `θ`.

### Example 3

Find `(2/_90^"o")/(4/_75^"o")`

### Example 4

Find `(3/_20^"o")/(9/_60^"o")`

### Example 5

Find `(8j)/(7+2j)` using polar form.

### Exercises:

1. Evaluate: `(0.5 ∠ 140^"o")(6 ∠ 110^"o")`
2. Evaluate: `(12/_320^"o")/(5/_210^"o")`
3. (i) Evaluate the following by first converting numerator and denominator into polar form.
(ii) Then check your answer by multiplying numerator and denominator by the **conjugate** of the denominator.
`(-2+5j)/(-1-j)`


Our class created a Forest of Factors display for the primary corridor, which Year 2 and Year 3 students pass by regularly. To explain the display in a way that's easy for them to understand, we avoided using technical terms like "factor," "multiple," and "prime."

The display features numbers from 1 to 40, each represented by a unique picture. We started by pointing out the number 1, which is special because multiplying any number by 1 leaves the number unchanged. We represented the number 1 as a bird flying in the sky, separate from the rest of the display.

The other numbers are represented by trees and mushrooms. Simple trees have two branches with flowers, indicating that they can be expressed as the product of two numbers. For example, 38 is represented by a tree with two branches because it can be factored as 19 times 2. Similarly, 10 is represented by a tree with two branches because it can be factored as 5 times 2, and 26 is represented by a tree with two branches because it can be factored as 13 times 2.

Some trees have more than two branches, indicating that they can be factored in multiple ways. For example, 12 and 32 are represented by trees with multiple branches. On the other hand, mushrooms represent numbers that cannot be factored into the product of two numbers. We asked the students to identify a mushroom in the display, which helped them understand the concept better.

The Forest of Factors display is designed to help students visualize and understand the relationships between numbers. By using a combination of trees and mushrooms, we hope to have created a engaging and interactive way for students to learn about factors and multiples.


**Open CASCADE Technology 7.2.0**

The `gp_Dir2d` class describes a unit vector in the plane (2D space), also called "Direction". It provides functions for constructing and manipulating unit vectors.

**Constructors**

* `gp_Dir2d()`: Creates a direction corresponding to the X axis.
* `gp_Dir2d(const gp_Vec2d &V)`: Normalizes the vector V and creates a Direction. Raises `ConstructionError` if V.Magnitude() <= Resolution from gp.
* `gp_Dir2d(const gp_XY &Coord)`: Creates a Direction from a doublet of coordinates. Raises `ConstructionError` if Coord.Modulus() <= Resolution from gp.
* `gp_Dir2d(const Standard_Real Xv, const Standard_Real Yv)`: Creates a Direction with its 2 cartesian coordinates. Raises `ConstructionError` if Sqrt(Xv*Xv + Yv*Yv) <= Resolution from gp.

**Member Functions**

* `void SetCoord(const Standard_Integer Index, const Standard_Real Xi)`: Assigns the value Xi to the specified coordinate.
* `void SetCoord(const Standard_Real Xv, const Standard_Real Yv)`: Assigns the values Xv and Yv to the coordinates.
* `void SetX(const Standard_Real X)`: Assigns the given value to the X coordinate and normalizes the vector.
* `void SetY(const Standard_Real Y)`: Assigns the given value to the Y coordinate and normalizes the vector.
* `void SetXY(const gp_XY &Coord)`: Assigns the coordinates from the gp_XY object.
* `Standard_Real Coord(const Standard_Integer Index) const`: Returns the coordinate at the specified index.
* `void Coord(Standard_Real &Xv, Standard_Real &Yv) const`: Returns the two coordinates.
* `Standard_Real X() const`: Returns the X coordinate.
* `Standard_Real Y() const`: Returns the Y coordinate.
* `const gp_XY &XY() const`: Returns the two coordinates as a number pair.

**Comparison Functions**

* `Standard_Boolean IsEqual(const gp_Dir2d &Other, const Standard_Real AngularTolerance) const`: Returns True if the two vectors have the same direction.
* `Standard_Boolean IsNormal(const gp_Dir2d &Other, const Standard_Real AngularTolerance) const`: Returns True if the angle between the two vectors is equal to Pi/2 or Pi/2 (normal).
* `Standard_Boolean IsOpposite(const gp_Dir2d &Other, const Standard_Real AngularTolerance) const`: Returns True if the angle between the two vectors is equal to Pi or Pi (opposite).
* `Standard_Boolean IsParallel(const gp_Dir2d &Other, const Standard_Real AngularTolerance) const`: Returns True if the angle between the two vectors is equal to 0, Pi or Pi.

**Geometric Functions**

* `Standard_Real Angle(const gp_Dir2d &Other) const`: Computes the angular value in radians between the two vectors.
* `Standard_Real Crossed(const gp_Dir2d &Right) const`: Computes the cross product between two directions.
* `Standard_Real Dot(const gp_Dir2d &Other) const`: Computes the scalar product.
* `void Reverse()`: Reverses the orientation of the direction.
* `gp_Dir2d Reversed() const`: Returns the reversed direction.
* `void Rotate(const Standard_Real Ang)`: Rotates the direction by the specified angle.
* `gp_Dir2d Rotated(const Standard_Real Ang) const`: Returns the rotated direction.
* `void Mirror(const gp_Dir2d &V)`: Performs the symmetrical transformation of the direction with respect to the direction V.
* `void Mirror(const gp_Ax2d &A)`: Performs the symmetrical transformation of the direction with respect to the axis placement A.
* `void Transform(const gp_Trsf2d &T)`: Transforms the direction with the specified transformation.


# Equation + Area of Shape - Math Problems

## Circumference and Width of a Rectangle
The circumference and width of a rectangle are in a ratio of 5:1, and its area is 216cm². Find its length.

## Triangle ABC
In triangle ABC, the height to side a is 6cm, and the height to side b is 9cm. Side a is 4cm longer than side b. Calculate the side lengths a and b.

## Map Scale
A map with a scale of 1:5,000 shows a rectangular field with an area of 18ha. The length of the field is three times its width. The area of the field on the map is 72cm². Find the actual length and width of the field.

## Equilateral Triangle
An equilateral triangle is inscribed in a square with a side length of 1 unit. Find the area of the inscribed triangle.

## Two Gardens
The total area of two gardens is 864m². The first garden is 60m² smaller than three times the second garden. Find the area of each garden.

## Cuboid Volume
Calculate the cuboid volume given the wall areas: 30cm², 35cm², and 42cm².

## Two Patches
Two rectangular patches overlap with an area of 40cm² and a circumference of 30cm. One patch is 8cm wide. Find the width of the other patch.

## Rectangular Garden
The perimeter of a rectangular garden is 98 meters, and the width is 60% shorter than its length. Find the dimensions and area of the garden.

## Three Altitudes
A triangle has altitudes of 4cm, 5cm, and 6cm. Calculate the lengths of all medians and sides.

## Isosceles Triangle
Given an isosceles triangle with a perimeter of 64cm and an altitude of 24cm, find the area.

## Rectangular Patio
A 20ft by 30ft patio is expanded by x feet in width and length, doubling its area. Find the new dimensions.

## Photograph
A photograph is 3/4x cm long and 20cm wider than the paper it is on. The remaining paper area is 990cm². Find the size of the paper and photo.

## Rectangle
A rectangle has a length of 12cm and a diagonal 8cm longer than its width. Calculate the area.

## Rectangular Triangle
A right triangle with a longer leg of 12cm has sides that form an arithmetic sequence. Find the area.

## Diamond Diagonals
Find the diamond's diagonal lengths given an area of 156cm² and a side length of 13cm.

## Diagonals of a Rhombus
One diagonal of a rhombus is 4cm longer than the other, and the area is 96cm². Find the side length.

## Arithmetic Progression Right Triangle
A right triangle's sides form an arithmetic progression, with the longer leg being 24cm. Find the perimeter and area.

## Rectangle - Area and Perimeter
A rectangle has an area of 300m² and a perimeter of 70m. Find the length and width.

## Right Triangle
Given a right triangle with an area of 210 and a perimeter of 70, find the side lengths and angles.

## Diamond Diagonals
Calculate the diamond's diagonal lengths given an area of 156cm² and a side length of 13cm.

Multiple Choice Questions:

1. What is the ratio of the circumference to the width of the rectangle?
A) 5:1
B) 1:5
C) 3:2
D) 2:3

Answer: A) 5:1

2. What is the area of the equilateral triangle inscribed in a square with a side length of 1 unit?
A) 0.5
B) 1
C) 1.5
D) 2

Answer: Not provided in the original text.

Note: Some questions and answers are not fully solvable or are missing information, so they are left as is or marked as "Not provided in the original text".


To calculate a specific number, follow these steps:
1. Multiply a number by 2.
2. Add 5 to the result.
3. Multiply this sum by 50.
4. Add 1756 if you have had your birthday this year, or 1755 if you haven't.
5. Subtract the four-digit year of your birth from this total.

The resulting three-digit number has significance: the first digit represents the number of times you want to eat chocolate per week, and the last two digits represent your age. Note that this calculation is only valid for individuals 99 years old or younger.


# Vectors

## Introduction to Vectors

In a 3-dimensional world, a vector is simply three numbers - an x component, a y component, and a z component. The C code definition of **vec3_t** is:

```c
typedef float vec_t;
typedef vec_t vec2_t[2];
typedef vec_t vec3_t[3];
typedef vec_t vec4_t[4];
typedef vec_t vec5_t[5];
```

The most common vector type is the 3-dimensional one, **vec3_t**, which is an array of three **vec_t**'s (floats). These three floats represent the x, y, and z components.

## The Origin

The most basic vector is the origin vector, written as (0.0, 0.0, 0.0). To create an origin vector in C code:

```c
vec3_t origin;
origin[0] = 0.0;
origin[1] = 0.0;
origin[2] = 0.0;
```

## The Coordinate System

Quake uses a left-handed 3D coordinate system. To define 'left-right', 'forwards-backwards', or 'up-down', imagine pointing your left arm forwards, thumb upwards, index finger forwards, and middle finger to the right. The middle finger points to positive x coordinates, the index finger to positive y coordinates, and the thumb to positive z coordinates.

## Positional Vectors

Positional vectors represent an (x, y, z) position in the game world. For example, to create a vector for player0's position at (55.0, -23.0, 12.5):

```c
vec3_t player0pos;
player0pos[0] = 55.0;
player0pos[1] = -23.0;
player0pos[2] = 12.5;
```

To teleport player0 64 units eastwards and 64 units up:

```c
player0pos[0] += 64.0;
player0pos[2] += 64.0;
```

## Velocity Vectors

Velocity vectors represent the velocity of an object. To represent a rocket moving eastwards and 45 degrees upwards at 900 units/second:

```c
vec3_t rocketvel;
rocketvel[0] = 636.39;
rocketvel[1] = 0.0;
rocketvel[2] = 636.39;
```

The rocket's position can be updated using its velocity:

```c
rocketpos[0] += rocketvel[0];
rocketpos[1] += rocketvel[1];
rocketpos[2] += rocketvel[2];
```

## Directions and Speed

Instead of using velocity vectors directly, it's easier to work with directions and speeds. A direction vector has a length of precisely 1.0, while speed is a linear quantity. The equation to 'move' an entity is **endpos = startpos + speed \* direction**.

## Vector Functions

Quake provides several vector functions to perform common tasks, such as:

*   VectorSubtract(a, b, c)
*   VectorAdd(a, b, c)
*   VectorCopy(a, b)
*   VectorScale(v, s, o)
*   VectorMA(v, s, b, o)
*   VectorClear(a)
*   VectorNegate(a, b)
*   VectorSet(v, x, y, z)
*   Vector4Copy(a, b)
*   SnapVector(v)

These functions take one, two, or three parameters and save the output in a second, third, or fourth output parameter.

## VectorMA

The VectorMA function combines vector multiplication and addition. It's used extensively in Quake and is essential to understand. For example, to move a rocket eastwards and upwards at 900 units/second:

```c
vec3_t rocketpos;
vec3_t rocketdir;
float rocketspeed;
VectorClear(rocketpos);
VectorSet(rocketdir, 0.707, 0.0, 0.707);
rocketspeed = 900.0;
VectorMA(rocketpos, rocketspeed, rocketdir, rocketpos);
```

## Putting it all Together

To see vectors in action, open g\_missile.c and look at lines 390-393:

```c
VectorCopy(start, bolt->s.pos.trBase);
VectorScale(dir, 900, bolt->s.pos.trDelta);
SnapVector(bolt->s.pos.trDelta);
VectorCopy(start, bolt->r.currentOrigin);
```

This code copies the start vector to the bolt's trBase, multiplies the dir vector by 900, snaps the result, and copies the start vector to the bolt's currentOrigin.


**Mathematics for Electronics Notes PDF**

*Date: 4th Feb 2023*

These notes aim to equip students with the mathematical skills and knowledge required to solve problems in electronics, particularly in power electronics and electromagnetic theory. The notes cover various topics, including ordinary differential equations, series solutions, matrices, sequences and series, and complex variables and functions.

**Topics Covered**

1. **Ordinary Differential Equations**: First-order and second-order homogeneous and non-homogeneous differential equations, separable, exact, and linear ordinary differential equations.
2. **Series Solution of Differential Equations and Special Functions**: Power series method, Legendre polynomials, Frobenius method, Bessel's equations, and Bessel's functions, error functions, and gamma function.
3. **Matrices**: Introduction to matrices, system of linear algebraic equations, Gaussian elimination method, LU decomposition, eigenvalues and eigenvectors, linear transformation, and properties of eigenvalues and eigenvectors.
4. **Sequences and Series**: Sequences, limit of a sequence, convergence, divergence, and oscillation of a sequence, infinite series, Cauchy's integral test, D'Alembert's ratio test, and power series.
5. **Complex Variables and Functions**: Complex variables, complex functions, continuity, differentiability, analyticity, Cauchy-Riemann equations, harmonic and conjugate harmonic functions, and residue integration method.

**Reference Books**

The following reference books are recommended for mathematics foundation for electronics:
- E. Kreyszig, Advanced Engineering Mathematics, Wiley India (2008)
- Murray Spiegel, Seymour Lipschutz, John Schiller, Outline of Complex Variables, Schaum Outline Series, Tata McGraw Hill (2007)
- R. K. Jain, and S.R.K. Iyengar, Advanced Engineering Mathematics, Narosa Publishing House (2007)
- C. R. Wylie and L. C. Barrett, Advanced Engineering Mathematics, Tata McGraw-Hill (2004)
- B. V. Ramana, Higher Engineering Mathematics, Tata McGraw Hill Publishing Company Limited

These notes and reference books are designed to help students prepare for their mathematics foundation for electronics exam and enhance their knowledge of the subject.


The "safest" bets in European roulette, such as even/odd, red/black, or low/high, have a probability of 48.6% of returning a profit, while in American roulette, this probability drops to 47.4%. For a simplistic slot machine with 5 fruits and a Joker, only 44.4% of combinations are winners, and this percentage drops to around 42% in more sophisticated versions. In poker, 49.9% of hands are winners, but many of these are relatively useless one-pair hands.

Despite the known health risks associated with gambling, many people take huge risks and bet their hard-earned money in hopes of getting rich quick. The reality is bleak, especially when considering concepts like house edges. The idea that high percentages can be misleading because they are on the wrong side of 50% is at the basis of the gambler's ruin, a statistical concept that describes how a gambler will eventually go broke if the game has a negative expected value.

A mathematical model known as a random walk can be used to describe this concept. The problem can be stated as follows: a persistent gambler plays a game where he wins £1 with probability p and loses £1 with probability 1-p. He will only leave the casino if he loses all his money or reaches a fixed fortune £M. Assuming he starts with £n, what is the probability that he goes home broke?

To solve this problem, we can set up a sequence (u_{n}) where u_{n} is the probability that the gambler loses everything starting with £n. Using the partition theorem and conditioning on the result of the first game, we can derive a formula for u_{n}. The solution depends on the value of p, the probability of winning.

If p = 1/2, the game is fair, and the probability of bankruptcy is u_{n} = 1 - n/M. If p ≠ 1/2, the game is biased, and the probability of bankruptcy is u_{n} = (1 - (q/p)^{n}) / (1 - (q/p)^{M}).

Using these formulas, we can calculate the probability of bankruptcy for different scenarios. For example, if Joe starts with £80 and aims to turn it into £100 with a 50% win probability, his chances of failure are 20%. However, if he tries to quadruple his wealth, his chances of failure increase to 75%.

In reality, casino games are biased against the player, even if they are perfectly fair or skill-based in theory. If Joe plays a game with a 49.9% win probability, his chances of failure increase significantly. For example, trying to turn £80 into £100 has a 24.4% chance of failure, while trying to turn £800 into £1000 has a 93.8% chance of failure.

The key takeaway is that setting a realistic target wealth when entering a casino depends on what you start with, rather than how much you are trying to gain. Aiming for relatively small daily gains is more beneficial than chasing one big jackpot. Gambling is an insidious problem, and maths proves that responsible betting and self-control are essential to avoid financial ruin.


To find out how many miles you will go in 110 minutes if you go 3 miles in 22 minutes, first determine the rate of travel. Since 110 minutes is 5 times 22 minutes, you can multiply the distance traveled in 22 minutes by 5.

3 miles * 5 = 15 miles

This means you will go 15 miles in 110 minutes if you maintain the rate of 3 miles in 22 minutes.

Alternatively, you can calculate the rate of travel per minute and then multiply by the total minutes:
3 miles / 22 minutes = 0.13636 miles per minute
0.13636 miles/minute * 110 minutes = 15 miles

Thus, in 110 minutes, you will travel 15 miles at this rate.


Parallel Jobs Scheduling with Precedence and Relative Deadlines

The problem involves scheduling a set of jobs with specified durations, precedence constraints, and relative deadlines. The goal is to determine if it is possible to execute all the jobs while respecting all the constraints. We can use as many processors as needed to execute jobs parallelly.

**Problem Statement:**

Given a set of jobs with specified durations, precedence constraints, and relative deadlines, write a program to compute if it is possible to execute all the jobs while still respecting all the constraints.

**Key Concepts:**

1. **Precedence Constraints:** Certain jobs must be completed before other jobs can start.
2. **Relative Deadlines:** Certain jobs must begin before a specified amount of time has elapsed relative to the start of another job.
3. **Parallel Job Scheduling:** We can use as many processors as needed to execute jobs parallelly.

**Solution Approach:**

To solve this problem, we can use the Bellman-Ford Algorithm and Negative Cycle Detection. The approach involves:

1. Creating a graph with jobs as vertices and edges representing the precedence constraints and relative deadlines.
2. Assigning weights to the edges based on the job durations and relative deadlines.
3. Negating all the edge weights to convert the problem from detecting positive cycles to detecting negative cycles.
4. Using the Bellman-Ford Algorithm to detect the existence of a negative cycle in the modified graph.
5. If a negative cycle is found, the job scheduling is impossible. Otherwise, it is feasible.

**Time Complexity:**

The time complexity of the solution is O(EV), where E is the number of edges and V is the number of vertices. In the worst case, E = V^2 for dense graphs, resulting in a time complexity of O(n^3) for n jobs.

**Related Topics:**

1. Dijkstra's Algorithm
2. Longest Paths Algorithm
3. Topological Sort with Vertex Relaxation
4. Bellman-Ford Algorithm and Negative Cycle Detection
5. Optimized Bellman-Ford
6. Sequential Job Scheduling
7. Parallel Job Scheduling
8. Arbitrage

**Summary:**

1. Sequential Job Scheduling: Use Topological Sort
2. Parallel Job Scheduling: Use Longest Paths Algorithm
3. Parallel Job Scheduling with Relative Deadline: Bellman-Ford Algorithm and Negative Cycle Detection

**Example:**

Consider a set of jobs with the following constraints:

| Job ID | Duration | Must Complete After |
| --- | --- | --- |
| 0 | 2 | 1 |
| 1 | 4 | 2 |
| 2 | 2 | 0 |
| 3 | 6 | 0, 1 |
| 4 | 2 | 3, 4 |
| 5 | 5 | 4 |

The relative deadline constraints are:

* Job 5 must start no later than 3 time units after job 4 starts.
* Job 0 must start no later than 4 time units after job 4 starts.

By applying the solution approach, we can determine if it is possible to execute all the jobs while respecting all the constraints.


**Lesson 1: Length**

The metric system is used to measure length. The basic unit of length is the meter (m).

1 kilometer (km) = 1000 meters
1 meter = 100 centimeters (cm)
1 meter = 1000 millimeters (mm)

To compare lengths, we can use the following conversions:
1 mile = 1.6 kilometers
1 yard = 0.9444 meters
1 inch = 2.54 centimeters

**Multiple Choice Questions:**
1. Which is longer?
A. 1 mile or 1 kilometer
B. 1 yard or 1 meter
C. 1 inch or 1 centimeter

2. Which is larger?
A. 1 meter or 105 centimeters
B. 4 kilometers or 4400 meters
C. 12 centimeters or 102 millimeters
D. 1200 millimeters or 1 meter

**Measuring Length:**
A ruler is used to measure length.
1 centimeter = 10 millimeters

**Lesson 2: Mass**

Mass refers to the amount of matter in an object. The base unit of mass in the metric system is the kilogram (kg).

1 kilogram (kg) = 1000 grams (g)
1 gram (g) = 1000 milligrams (mg)

To compare masses, we can use the following conversions:
1 pound = 453.6 grams
100 kilograms = 220 pounds
1 ounce of gold = 28,349.5 milligrams

**Multiple Choice Questions:**
1. Which is larger?
A. 1 pound or 100 grams
B. 1 kilogram or 1 pound
C. 1 ounce or 1000 milligrams

2. Which is larger?
A. 1 kilogram or 1500 grams
B. 1200 milligrams or 1 gram
C. 12 milligrams or 12 kilograms
D. 4 kilograms or 4500 grams

**Measuring Mass:**
A digital scale is used to measure mass. The scale is set to zero before measuring an object's mass.

**Lesson 3: Volume**

Volume is the amount of space an object takes up. The base unit of volume in the metric system is the liter (L).

1 liter (L) = 1000 milliliters (mL)
1 milliliter (mL) = 1 cubic centimeter (cm³)

To compare volumes, we can use the following conversions:
1 gallon = 3.79 liters
1 fluid ounce = 29.573 milliliters
1 quart = 0.946 liters

**Multiple Choice Questions:**
1. Which is larger?
A. 1 liter or 1 gallon
B. 1 liter or 1 quart
C. 1 milliliter or 1 fluid ounce

2. Which is larger?
A. 1 liter or 1500 milliliters
B. 200 milliliters or 1.2 liters
C. 12 cubic centimeters or 1.2 milliliters

**Measuring Volume:**
A graduated cylinder is used to measure the volume of liquids. The volume is read based on the bottom of the meniscus.

**Lesson 4: Temperature**

Temperature refers to the degree of hotness of a body. The base unit of temperature in the metric system is the Kelvin (K), but the most common unit used is the degree Celsius (°C).

273.16 K = 0 °C
373.16 K = 100 °C

To compare temperatures, we can use the following conversions:
°C = (°F - 32) / 1.8
°F = (°C x 1.8) + 32
K = 273.16 + °C
°C = K - 273.16

**Multiple Choice Questions:**
1. Which is hotter?
A. 1.0 °C or 0 °F
B. 30 °C or 30 °F
C. 0 °C or 100 K

**Measuring Temperature:**
A thermometer is used to measure temperature. The thermometer is placed on the object and allowed to reach a steady state.


**1. Stock Valuation**
To find the value of a stock with a 12% required return on equity, a $1.50 dividend at the end of the year, and a 5% expected growth rate of dividends, we use the formula for the present value of a growing perpetuity. The correct calculation yields a value of $22.50.

A. $12.45
B. $21.43
C. $22.50
D. $12.50
E. none of the above

**2. Stock Price Calculation**
Given a stock that sells for $45 today, paid an annual dividend of $2.50, and had a 20% return over the past year, we calculate the price one year ago by dividing the current price by (1 + return). This results in a price of $37.50 one year ago.

A. $56.25
B. $37.50
C. $39.58
D. $35.42
E. none of the above

**3. Stock Valuation with Growth**
For a stock that just paid a $2 dividend with a 5% expected growth rate and a 10% required return, applying the formula for the present value of a growing perpetuity gives us a value of $40.

A. $21
B. $42
C. $40
D. $20.95
E. none of the above

**4. Stock Valuation without Growth**
A stock expected to pay a $2 dividend in one year with no growth and a 10% required return is valued using the formula for the present value of a single amount. This calculation yields a value of $1.82.

A. $19.90
B. $20
C. $0.20
D. $1.82
E. none of the above

**5. Required Return on Equity**
For a $30 stock that just paid a $2 dividend with a 4% expected growth rate, we solve for the required return on equity using the formula for the present value of a growing perpetuity. The required return is 10.67%.

A. 6.93%
B. 10.93%
C. 10.67%
D. 2.93%
E. none of the above

**6. Payout Ratio**
Given a 5% growth rate in dividends and a 20% required return on equity, the payout ratio can be found from the relationship between the growth rate, required return, and payout ratio. The payout ratio is 75%.

A. 25%
B. 75%
C. 15%
D. 85%
E. none of the above

**7. Stock Value Before Plowback**
A company plans to pay a $4 dividend next year, representing 100% of expected earnings, providing a 15% expected return. The value of the stock before the plowback decision, given it represents all earnings and provides the required return, is $26.67.

A. $40
B. $13.33
C. $44.44
D. $26.67
E. none of the above

**8. Stock Value After Plowback**
If the company plows back 40% of earnings at a 15% return on equity, we calculate the new dividend and apply the present value formula. The value after the plowback decision is $40.

A. $40
B. $13.33
C. $44.44
D. $26.67
E. none of the above

**9. Present Value of Growth Opportunities**
The XYZ Company paid a $1.50 dividend from $3 earnings per share with a 20% required return. The present value of growth opportunities can be found by calculating the difference between the total value (using the required return and dividend) and the no-growth value. This yields $9.00.

A. $1.50
B. $0.50
C. $7.50
D. $9.00
E. none of the above

**10. Stock Price with Variable Growth**
A stock paid a $1 dividend with 15% growth for two years, then 5% long-term growth, and an 8% required return. Calculating the present value of the dividends for the high-growth period and the perpetual growth period gives a stock price of $41.88.

A. $46.29
B. $38.95
C. $56.21
D. $41.88
E. none of the above


1. If the hypothesis test is used in a quality control setting, the alternative hypothesis takes the stance that the process is not behaving as it should and needs some sort of attention: True.

2. A Type II error is known as the producer’s risk because, when it occurs in quality control settings, the producer is looking for a problem in its process that does not exist: False.

3. CoStar would like to test if the vacancy rate for warehouse stores is more than 8%. The correct hypothesis statement to test for this vacancy rate would be: H0: π ≤ 0.08; H1: π > 0.08.

4. CoStar would like to test if the vacancy rate for warehouse stores is more than 8%. A Type I error would occur if CoStar concludes that the vacancy rate was more than 8% when, in reality, the proportion was 8% or less.

5. The p-value for a hypothesis test is defined as the probability of observing a critical value at least as extreme as the one selected for the hypothesis test, assuming the null hypothesis is true.

6. The Department of Education would like to test the hypothesis that the average debt load of graduating students with a bachelor’s degree equals $17,000. The critical value for this hypothesis test would be: ±1.96.

7. Sony would like to test the hypothesis that the average age of a PlayStation user differs from the average age of an Xbox user. The critical value for this hypothesis test would be: ±1.96.

8. Sony would like to test the hypothesis that the average age of a PlayStation user differs from the average age of an Xbox user. The test statistic for this hypothesis test would be: = 1.29.

9. The federal government would like to test the hypothesis that the average age of men filing for Social Security is higher than the average age of women filing. The test statistic for this hypothesis test would be: = 1.78.

10. Two real estate companies, Century 21 and RE/MAX, compete with one another in a local market. The 80% confidence interval for the difference in population means is: (11.5, 32.7).

11. The Wall Street Journal reported that the average amount of time that a French person spends eating lunch at a restaurant is 22 minutes. The 90% confidence interval for the difference in population means is: (1.52, 4.49).

12. The Wall Street Journal reported that the average amount of time that a French person spends eating lunch at a restaurant is 22 minutes. Because the 90% confidence interval does not include zero, you can conclude that the average time an American spends at lunch differs from the average time that French diners spend at lunch.

13. Analysis of variance is a technique used to conduct a hypothesis test to compare three or more population proportions simultaneously: False.

14. All analyses of variance procedures require that the populations being compared have equal variances: False.

15. Analysis of variance compares the variance between samples to the variance within those samples to determine if means of populations differ: True.

16. The critical value for the Tukey-Kramer critical range is found using the F-distribution: True.

17. If the size for each sample is the same for the one-way ANOVA, the Tukey-Kramer critical range for all pairs would be the same: True.

18. The test statistics for one-way ANOVA follow the: F-distribution.

19. The degrees of freedom for the sum of squares between for this ANOVA procedure is: 3.

20. The chi-square test for independence is used to decide if observed frequencies follow a known probability distribution: False.

21. When you reject the null hypothesis for a chi-square test of independence, you are concluding that no relationship exists between the two variables: False.

22. When performing a hypothesis test to compare two or more population proportions, the test statistic follows the: normal distribution.

23. The chi-square test requires that each expected frequency be equal to or greater than: 5.

24. Nationwide Insurance would like to perform a chi-square test to investigate whether a difference exists in the proportion of male and female teenagers who text while they drive. The chi-square test statistic for this sample is: 4.05.

25. The Department of Transportation would like to investigate if differences exist in the proportions of flights that arrive on-time for Alaska Airlines, Delta Airlines, Southwest Airlines, and United Airlines. The chi-square test statistic for these samples is: 8.37.

26. The test statistic used to compare the variances of two populations will always be greater than 1.0: False.

27. A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics exam has a standard deviation that is less than 5.0 minutes. The correct hypothesis statement would be: H0: σ2 ≥ 25; H1: σ2 < 25.

28. A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics exam has a standard deviation that is less than 5.0 minutes. The test statistic for this hypothesis test would be: 13.61.

29. A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics exam has a standard deviation that is less than 5.0 minutes. Using α = 0.05, the conclusion for this hypothesis test would be that because the test statistic is less than the critical value, you fail to reject the null hypothesis and cannot conclude that the standard deviation for the time to complete an exam is less than 5.0 minutes.

30. YouTube would like to test the hypothesis that the standard deviation for the length of an online video watched by a user is more than 2 minutes. The test statistic for this hypothesis test would be: 10.19.

31. YouTube would like to test the hypothesis that the standard deviation for the length of an online video watched by a user is more than 2 minutes. The degrees of freedom for this hypothesis test would be: 20.

32. Consider the relationship between the following variables: Household income and Square footage of the primary residence. Household income would be considered the independent variable: False.

33. Two variables have a correlation coefficient equal to 0.65 from a sample size of 10. Because the test statistic is greater than the critical value, you fail to reject the null hypothesis, and you conclude that the population correlation coefficient is not less than zero.

34. The Student’s t-distribution is used to test the significance of the population correlation coefficient.

35. The formula for the equation describing a straight line is = a + bx. The value for b in this equation represents the slope of the straight line.

36. The sum of squares error measures the variation in the dependent variable that is explained by variables other than the independent variable in simple regression analysis.

37. The correlation coefficient for the data on the number of cars sold last month by six employees at Concord Motors and each employee’s number of years of sales experience is: 0.844.

38. Costco sells paperback books in their retail stores and wanted to examine the relationship between price and demand. The 95% prediction interval that estimates the weekly sales for a price of $9 is: (2.37, 11.48).

39. Multicollinearity is present in the regression model when independent variables within the model are highly correlated.

40. If the variance inflation factor exceeds 10.0, enough correlation exists between the independent variables to claim the presence of multicollinearity in the regression model.


Nassim Haramein proposes a holographic solution to explain the Vacuum Catastrophe problem, which concerns the large gap of ~122 orders of magnitude between the vacuum energy density at cosmological scale and the vacuum energy density predicted by quantum field theory. The calculation, entitled "Resolving the Vacuum Catastrophe: A Generalized Holographic Approach," was published in the Journal of High Energy Physics, Gravitation and Cosmology in 2019.

Quantum field theory (QFT) describes a vacuum composed of an infinite number of electromagnetic fields that are randomly fluctuating at all frequencies. The Casimir effect provides experimental proof of the existence of these zero-point oscillations. QFT calculates the quantum vacuum energy density at each point in space by summing the energies across all frequencies of vibration. However, this yields an infinite energy density at each point in space, unless it's renormalized at the Planck cutoff value.

To resolve the vacuum catastrophe, Haramein defines a Planck area (2D) as a circular area with a Planck length diameter, described as a "bit" or unit of information on the structure of spacetime. He generalizes this concept into a 3D Planck Spherical Unit (PSU), where the bit is the equatorial area of this PSU. The physical structure and energy density at this scale are more appropriately represented in PSUs.

The generalized holographic model describes how any spherical body can be considered in terms of its PSU packing, which is also its volume entropy. The mass-energy (or holographic mass) is given in terms of PSUs with the equation M_R = (m_P^3 / l_P^2) \* R, where R is the volume entropy. The mass-energy density of the spherical body is given as ρ = M_R / V, where M_R is the mass and V is the volume.

In the case of the proton, the mass-energy (also called the holographic mass) in terms of Planck mass is calculated as M_R = (m_P^3 / l_P^2) \* R, where R is calculated by dividing the content of PSU in the proton volume V_p by the volume of a PSU. This resulting holographic mass can be considered equivalent to the mass of the observable universe estimated by standard cosmological approaches.

The mass-energy density of the universe can be calculated in terms of the mass-energy density of the proton by dividing this mass by the volume of the universe. The vacuum mass-energy density at the cosmological scale is calculated to be ρ = 0.265 \* ρ_crit, which is very close to the observed value of 0.268 \* ρ_crit.

The equation takes a proton's mass-energy or holographic mass and equals it to the mass of the universe, then puts this mass-energy inside a much larger volume, the volume of the universe. This results in a mass density for this "proton dilated to a size of the universe," which corresponds to the dark mass density of the universe.

Haramein's hypothesis implies that the big bang is a proton that escaped from a former universe and expanded into ours at extremely high speed under enormous negative pressure. This expansion would create a density gradient, providing a source of rotation at all scales. This is similar to Roger Penrose's model of conformal cyclic cosmology, which posits that the universe iterates through infinite cycles.

The vacuum energy density can be considered in terms of the PSU surface tiling (i.e., its surface entropy) as the radius increases from the Planck scale to the cosmological scale. The vacuum density at the cosmological scale is given as ρ = (η / r_U^2), where η is found by assuming a spherical shell universe of radius r_U.

By summing the numbers in the exponents of the densities of the quantum vacuum energy density and the vacuum energy density at cosmological scale, we obtain the orders of magnitude that separate the two densities: 93+30 = 123. This simple calculation resolves the ~122 orders of magnitude vacuum catastrophe between cosmology and the Planck vacuum fluctuations.

The generalized holographic approach shows the scale relationship between the PSUs and a spherical-shell universe and solves the 122 orders of magnitude discrepancy between the vacuum energy density at the Planck scale and the vacuum energy density at the cosmological scale. This approach also hints at a scenario where our universe fulfills the condition of a black hole, considering the contribution of the vacuum fluctuations that dark mass and dark energy account for.

The solution to the vacuum catastrophe also implies that galaxies are connected by the medium in which they exist, even if they are millions of megaparsecs apart. The generalized holographic model sees mass as emerging from the vacuum fluctuations, and it can be inferred that galaxies too emerge from the dynamics of this all-filling, permeating medium in perpetual rotation – space.

References:
[1] N. Haramein, Phys. Rev. Res. Int., vol. 3, no. 4, pp. 270-292, 2013.
[2] C. Corda, "Interferometric detection of gravitational waves: the definitive test for General Relativity," Int. J. Mod. Phys. D18, vol. 18, pp. 2275-2282, 2009.


**Given Information**

Downstream distance: 21 km
Time taken downstream: 7 hours
Distance covered upstream: 7 km (1/3 of 21 km)
Time taken upstream: 7 hours (for 7 km)

**Relevant Formulas**

Speed = Distance / Time
Speed of stream = (Downstream speed - Upstream speed) / 2

**Calculations**

Downstream speed = 21 km / 7 hours = 3 km/h
Upstream speed = 7 km / 7 hours = 1 km/h
Speed of stream = (3 km/h - 1 km/h) / 2 = 1 km/h

**Conclusion**

The speed of the stream is 1 km/h.


You are dealt 5 cards from a standard 52-card deck. Determine the probability of being dealt three of a kind by answering the following questions:

(a) How many ways can 5 cards be selected from a 52-card deck?
The number of ways to choose 5 cards from a 52-card deck is calculated using combinations, denoted as C(n, k) = n! / [k!(n-k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial. So, C(52, 5) = 52! / [5!(52-5)!] = 2,598,960.

(b) Each deck contains 4 twos, 4 threes, and so on. How many ways can three of the same card be selected from the deck?
Since there are 4 cards of each rank and we need to select 3 of them, the number of ways to choose 3 cards of the same rank is C(4, 3) = 4. However, since there are 13 different ranks in the deck, the total number of ways to select three of the same card is 13 * C(4, 3) = 13 * 4 = 52.

(c) The remaining 2 cards must be different from the 3 chosen and different from each other. How many ways can we select the remaining 2 cards?
After selecting the three of a kind, there are 12 different ranks of cards remaining. We choose 2 of these ranks and then select one of the 4 cards in each of the two chosen ranks. The number of ways to choose 2 ranks out of 12 is C(12, 2), and for each of these ranks, we have 4 choices. So, the number of ways to select the remaining 2 cards is C(12, 2) * 4 * 4 = 66 * 16 = 1,056.

(d) Use the General Multiplication Rule to compute the probability of obtaining three of a kind.
The probability of obtaining three of a kind is the product of the number of ways to select three of the same card and the number of ways to select the remaining two cards, divided by the total number of ways to select 5 cards from the deck. This is (13 * C(4, 3) * C(12, 2) * 4 * 4) / C(52, 5) = (52 * 1,056) / 2,598,960 = 54,912 / 2,598,960 = 0.0211 or 2.11%.


The number of atoms decayed in the first half-life is ½ (N_{0}) = ½ (500 × 10^{6}) = 250 × 10^{6} atoms. The remaining amount is (500 – 250) × 10^{6} = 250 × 10^{6} = **250 million atoms**.

In the second half-life, the atoms decayed are ½ (250 × 10^{6}) = 125 × 10^{6} atoms, leaving the same amount after the 2nd half-life.

The number of atoms decayed in the 3rd half-life is ½ (125 × 10^{6}) = 62.5 × 10^{6} atoms. The amount of isotope remaining after the 3rd half-life, or 24 hours, is (125 – 62.5) × 10^{6} = **62.5 × 10^{6}** atoms = 62 million atoms.

Alternatively, the amount of substance left after 3 half-lives can be calculated as (½)^{3} of the total initial amount. Therefore, the amount left after three half-lives is (½)^{3} × 500 × 10^{6} = 0.125 × 500 × 10^{6} = 62.5 × 10^{6} = **62.5 million atoms**.


A data analyst asked about restricted least square regression in SAS, specifically why PROC REG can solve regression problems with equality constraints but not inequality constraints.

Restricted regression puts linear constraints on the coefficients in the model, such as forcing a coefficient to be 1 or forcing two coefficients to equal each other. These problems can be solved using PROC REG in SAS.

The article explains how to use the RESTRICT statement in PROC REG to solve a restricted regression problem with linear constraints on the parameters. It also uses geometry and linear algebra to show that solving an equality-constrained regression problem is mathematically similar to solving an unrestricted least squares system of equations.

Geometrically, ordinary least-squares (OLS) regression is the orthogonal projection of the observed response (Y) onto the column space of the design matrix. Introducing equality constraints among the parameters restricts the solution to a linear subspace of the span. The restricted least squares (RLS) solution is still a projection of the observed response, but onto the restricted subspace.

In terms of linear algebra, the challenge is to write down the projection matrix for the restricted problem. The RLS solution can be obtained in two ways: directly projecting the Y vector onto the restricted subspace or projecting the OLS solution vector onto the restricted subspace.

An example is given using SAS DATA step to simulate data from a regression model Y = B0 + B1*X1 + B2*X2 + B3*X3 + B4*X4, where B3 = B1 and B4 = –2*B2. The TEST statement in PROC REG is used to hypothesize that B3 = B1 and B4 = –2*B2.

If the hypothesis is not rejected, the RESTRICT statement can be used to recompute the parameter estimates subject to the constraints. The new parameter estimates enforce the restrictions among the regression coefficients.

The linear algebra behind the restricted least squares solution is shown using SAS/IML. The usual way to compute the unrestricted OLS solution is to solve the "normal equations" (X`*X)*b = X`*Y for the parameter estimates, b. The restricted least squares solution can be obtained by augmenting the X`X matrix to enforce the constraints.

The article concludes that solving an equality-constrained regression problem is very similar to solving an unrestricted least squares system of equations. Geometrically, they are both projections onto a linear subspace. Algebraically, the restricted problem can be solved directly or as the projection of the OLS solution.

Changing one of the equality constraints into an inequality means that the restricted region is no longer a linear subspace. Ordinary least squares can no longer solve the problem because the solution might not be an orthogonal projection. The next article discusses ways to solve problems where the regression coefficients are related by linear inequalities.

Key concepts include:

* Restricted regression: puts linear constraints on the coefficients in the model
* Equality constraints: force a coefficient to be 1 or force two coefficients to equal each other
* Inequality constraints: force a coefficient to be greater than or less than a certain value
* Orthogonal projection: a way of finding the best fit line that minimizes the sum of the squared errors
* Linear subspace: a subset of a vector space that is closed under addition and scalar multiplication

Important numbers include:

* 1: the value that a coefficient can be forced to be
* 2: the number of coefficients that can be forced to equal each other
* 3: the standard deviation for the error term in the example
* 5: the number of parameters in the model (B0-B4)
* 100: the sample size in the example

The RESTRICT statement in PROC REG is used to incorporate restrictions into the model. The TEST statement is used to perform a statistical test to see if the data supports the model. The SOLVE function in SAS/IML is used to solve the equation (X`*X)*b = X`*Y for the parameter estimates, b.

Multiple choice questions:

1. What is restricted regression?
a) A type of regression that puts linear constraints on the coefficients in the model
b) A type of regression that puts nonlinear constraints on the coefficients in the model
c) A type of regression that does not put any constraints on the coefficients in the model
d) A type of regression that puts constraints on the error term

Answer: a) A type of regression that puts linear constraints on the coefficients in the model

2. What is the purpose of the RESTRICT statement in PROC REG?
a) To perform a statistical test to see if the data supports the model
b) To incorporate restrictions into the model
c) To solve the equation (X`*X)*b = X`*Y for the parameter estimates, b
d) To find the best fit line that minimizes the sum of the squared errors

Answer: b) To incorporate restrictions into the model

3. What is the difference between equality and inequality constraints?
a) Equality constraints force a coefficient to be 1, while inequality constraints force a coefficient to be greater than or less than a certain value
b) Equality constraints force two coefficients to equal each other, while inequality constraints force a coefficient to be greater than or less than a certain value
c) Equality constraints force a coefficient to be greater than or less than a certain value, while inequality constraints force a coefficient to be 1
d) Equality constraints force a coefficient to be 1, while inequality constraints force two coefficients to equal each other

Answer: b) Equality constraints force two coefficients to equal each other, while inequality constraints force a coefficient to be greater than or less than a certain value


**Problem 1:**
Ana's best friend ate 3/8 of a pizza from 4 different pizzas. To find out how much pizza he ate, we calculate:
P = 4 * (3/8) = 12/8 = 3/2 = 1 1/2 Pizzas.

**Problem 2:**
Julio ordered 2 pizzas for dinner. Dad ate 3/4 of one pizza, Jenny ate 1/8 of a pizza, Danny ate 1/4 of a pizza, and Mom ate 1/2 of a pizza. To find the fraction of pizza left, we first need to determine the total amount eaten.

**Problem 3:**
Similar to Problem 2, with the same details: Dad ate 3/4, Jenny ate 1/8, Danny ate 1/4, and Mom ate 1/2 of a pizza. The question remains, what fraction of a pizza was left for a midnight snack?

**Problem 4:**
At Shannon's birthday, Kira ate 3 pieces, Brazdley ate 2, Shauna ate 4, and Sammi ate 3 pieces of pizza. Now, 1/4 of the pizza is left. To find out how many pieces the pizza was originally cut into, we need to calculate the total pieces eaten and add the remaining quarter.

**Problem 5:**
If two pizzas were ordered for dinner, with the same eating distribution as before (Dad: 3/4, Jenny: 1/8, Danny: 1/4, Mom: 1/2), what fraction of a pizza was left for a midnight snack?

**Problem 6:**
Jaden bought a large pizza cut into 10 pieces. He ate 4/7 of the pizza, and his friend ate 3/8 of the pizza. To find out how much pizza was left, we calculate the combined fraction of pizza eaten.

**Problem 7:**
Mr. Fairfax ordered 3 large pizzas. Group A ate 6/6 (or 1 whole pizza) of the first pizza, and Group B ate 8/6 of the second pizza. To compare, we see Group A ate 1 whole pizza, while Group B ate 8/6, which is more than 1 whole pizza.

**Problem 8:**
Peter and Eric bought a large pizza with 12 slices. John and Eric ate all the pizza, with John eating 4/6 slices. To find out how much Eric ate, we calculate the remaining slices after John's portion.

**Problem 9:**
Michael ate 1/4 of a pizza, and his brother ate 3/8 of the same pizza. To find out how much of the pizza was left, we add their portions and subtract from the whole.

**Problem 10:**
If one person ate 4/8 of a pizza and their sister ate 3/8 of the pizza, to find out how much pizza they ate in total, we add their portions: 4/8 + 3/8 = 7/8 of the pizza.


# 40 percent of 7/10 Pesos

Using this tool, you can find any percentage in three ways. You may be looking for answers like: What is 40 (40%) percent (%) of 7/10 (7/10)? Or maybe: 40 percent of 7/10 Pesos. The solutions to these problems are provided below.

## How to work out percentages - Step by Step

Here are the solutions to the questions stated above:

### 1) What is 40% of 7/10?

To find a percentage, use the formula:
^{%} / _{100} = ^{Part} / _{Whole}
Replace the given values:
^{40} / _{100} = ^{Part} / _{7/10}

Cross multiply:
40 x 7/10 = 100 x Part
28 = 100 x Part

Divide by 100 to get the answer:
Part = ^{28} / _{100} = 0.28

### 2) 40 is what percent of 7/10?

Use the same percentage formula:
^{%} / _{100} = ^{Part} / _{Whole}
Replace the given values:
^{%} / _{100} = ^{40} / _{7/10}

Cross multiply:
% x 7/10 = 40 x 100

Divide by 7/10 to get the percentage:
% = ^{40 x 100} / _{7/10} = 5714.29%


**Multiplying and Dividing by 10, 100, and 1000 in Mental Maths**

We will explore how children can apply the skill of multiplying and dividing by 10, 100, and 1000 to help them with other mental maths questions, including the grid method for multiplication.

**Multiplying a Two-Digit Number by a One-Digit Number**

To multiply a two-digit number by a one-digit number, children can partition the two-digit number into tens and ones, then multiply each part by the one-digit number, and finally add the answers together.

Example 1: 34 x 5
- Partition 34 into 30 and 4
- Multiply each part by 5: 30 x 5 = 150 and 4 x 5 = 20
- Add the answers together: 150 + 20 = 170

If children struggle with the multiplication, they can divide the tens part by 10 to make it easier, then multiply the answer by 10. For example, 30 ÷ 10 = 3, then 3 x 5 = 15, and finally 15 x 10 = 150.

Example 2: 63 x 4
- Partition 63 into 60 and 3
- Multiply each part by 4: 60 x 4 = 240 and 3 x 4 = 12
- Add the answers together: 240 + 12 = 252

**Using Known Facts to Multiply and Divide**

This skill is essential for Year 5 and 6 students and relies on confidence with times tables. To make multiplication and division easier, children can use their times table facts to work with multiples of 10, 100, and 1000 by dividing each number by 10, 100, or 1000 to get back to the original times table, then multiplying the answer by the same amount.

Examples:
- 60 x 7 = 420: Divide 60 by 10 to get 6, then 6 x 7 = 42, and multiply by 10 to get 420.
- 60 x 70 = 4200: Divide both numbers by 10 to get 6 x 7 = 42, then multiply by 100 (10 x 10).
- 600 x 7 = 4200: Divide 600 by 100 to get 6, then 6 x 7 = 42, and multiply by 100.
- 600 x 700 = 420,000: Divide both numbers by 100 to get 6 x 7 = 42, then multiply by 10,000 (100 x 100).

For division, the process is similar:
- 420 ÷ 6 = 70: Divide 420 by 10 to get 42, then 42 ÷ 6 = 7, and multiply by 10.
- 4200 ÷ 7 = 600: Divide 4200 by 100 to get 42, then 42 ÷ 7 = 6, and multiply by 100.
- 4200 ÷ 700 = 6: Divide both numbers by 100 to get 42 ÷ 7 = 6, no further adjustment needed.

**Practice Questions**

Try these with your child:
- 7 x 8
- 70 x 8
- 7 x 800
- 70 x 80
- 70 x 800
- 560 ÷ 8
- 5600 ÷ 70
- 560 ÷ 70

**Grid Method**

The grid method for multiplication can be made easier by applying the skills of multiplying and dividing by 10, 100, and 1000. For example, in the multiplication of 3000 by 3, children can divide 3000 by 1000 to get 3, then multiply 3 x 3 = 9, and finally multiply by 1000 to get 9000.

Example:
```
x | 3000 | 200 | 70 | 6 |
3 | 9000 | 600 | 210 | 18 |
```
To find the answer, add together the red amounts. For each part, children can use the method of dividing by 10, 100, or 1000 to simplify the multiplication, then multiply the result by the same amount to get the final answer for each part of the grid.


We have a family of Borel probability measures $(P_x^t)_{x \in \mathbb{R},t \geq 0}$ on $\mathbb{R}$ with the following properties:
- The map $(x,t) \mapsto P_x^t(A)$ is Borel-measurable for all $A \in \mathcal{B}(\mathbb{R})$.
- $P_x^0=\delta_x$ for all $x \in \mathbb{R}$.
- $P_x^{s+t}(A)=\int_\mathbb{R} P_y^t(A) \, P_x^s(dy)$ for all $A \in \mathcal{B}(\mathbb{R})$ and $s,t \geq 0$.

A probability measure $\rho$ on $\mathbb{R}$ is *stationary* if $\rho(A)=\int_\mathbb{R} P_x^t(A) \, \rho(dx)$ for all $A \in \mathcal{B}(\mathbb{R})$ and $t \geq 0$. A probability measure on $\mathbb{R}$ is *ergodic* if it is an extremal point of the convex set of stationary probability measures.

Consider a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_{t \geq 0}),\mathbb{P})$ and a progressively measurable real-valued homogeneous Markov process $(X_t)_{t \geq 0}$ with transition probabilities given by $(P_x^t)_{x \in \mathbb{R},t \geq 0}$. Suppose $\rho:=X_{0\ast}\mathbb{P}$ is stationary. For a bounded measurable $f:\mathbb{R} \to \mathbb{R}$, does
$\lim_{T \to \infty} \frac{1}{T} \int_0^T f(X_t(\omega)) \, dt$ exist for $\mathbb{P}$-almost all $\omega \in \Omega$?

Given a stationary probability measure $\rho$, a set $A \in \mathcal{B}(\mathbb{R})$ is $\rho$-*almost stationary* if for all $t \geq 0$, $\rho(x \in A: P_x^t(A)=1)=\rho(A)$. A stationary probability measure $\rho$ is ergodic if and only if every $\rho$-almost stationary set has $\rho$-trivial measure.

Does there exist a probability measure $Q$ on the set $\mathcal{M}$ of probability measures on $\mathbb{R}$ such that $Q$-almost every $\mu \in \mathcal{M}$ is ergodic and for all $A \in \mathcal{B}(\mathbb{R})$, $\rho(A) = \int_\mathcal{M} \mu(A) \, Q(d\mu)$?

Two equivalent definitions of ergodicity are:
(1) A stationary probability measure $\rho$ is ergodic if and only if every $\rho$-almost stationary set has $\rho$-trivial measure.
(2) A set $A \in \mathcal{B}(\mathbb{R})$ is *invariant* if for all $t \geq 0$ and all $x \in A$, $P_x^t(A)=1$. A stationary probability measure $\rho$ is ergodic if and only if every invariant set has $\rho$-trivial measure.

The answer to the question of whether all stationary measurable stochastic processes are "measurably stationary" is expected to be *no*, although no counterexample is provided. If the answer to the first question is *no*, it is wondered whether it becomes *yes* in the particular case that $(X_t)_{t \geq 0}$ is a *strong* Markov process.


Solving Systems of Equations by Elimination (Addition)

Solving a system of equations by elimination involves using addition and subtraction. The steps are:
1. Put the equations in Standard Form (Ax + By = C).
2. Determine which variable to eliminate.
3. Add or subtract the equations to eliminate the variable.
4. Plug back in to find the other variable.
5. Check your solution by substituting the ordered pair into both equations.

Example 1: Solve the system using elimination.
x + y = 5
3x – y = 7
Step 1: The equations are already in Standard Form.
Step 2: The "y" terms are additive inverses, so we can eliminate "y".
Step 3: Add the equations to eliminate "y".
x + y = 5
3x – y = 7
4x = 12
x = 3
Step 4: Plug back in to find "y".
x + y = 5
(3) + y = 5
y = 2
Step 5: Check the solution.
(3, 2)
(3) + (2) = 5
3(3) - (2) = 7

Example 2: Solve the system using elimination.
4x + y = 7
4x – 2y = -2
Step 1: The equations are already in Standard Form.
Step 2: The "x" terms have the same coefficient, so we can eliminate "x".
Step 3: Subtract the equations to eliminate "x".
4x + y = 7
-(4x – 2y = -2)
3y = 9
y = 3
Step 4: Plug back in to find "x".
4x + y = 7
4x + (3) = 7
4x = 4
x = 1
Step 5: Check the solution.
(1, 3)
4(1) + (3) = 7
4(1) - 2(3) = -2

Example 3: Solve the system using elimination.
y = 7 – 2x
4x + y = 5
Step 1: Put the equations in Standard Form.
2x + y = 7
4x + y = 5
Step 2: The "y" terms have the same coefficient, so we can eliminate "y".
Step 3: Subtract the equations to eliminate "y".
2x + y = 7
-(4x + y = 5)
-2x = 2
x = -1
Step 4: Plug back in to find "y".
y = 7 – 2x
y = 7 – 2(-1)
y = 9
Step 5: Check the solution.
(-1, 9)
(9) = 7 – 2(-1)
4(-1) + (9) = 5

Multiple Choice Question:
What is the first step when solving with elimination?
1. Add or subtract the equations.
2. Plug numbers into the equation.
3. Solve for a variable.
4. Check your answer.
5. Determine which variable to eliminate.
6. Put the equations in standard form.
Answer: 6. Put the equations in standard form.


I've recently completed fastStat, a quick introduction to statistics for those who've had a calculus-based probability course. This tutorial provides quick access to statistics for individuals who need it, modeled after my R tutorial, fasteR.

The fastStat tutorial is a practical and real introduction to statistics, focusing on intuitive understanding rather than just mechanical knowledge. It targets the gap in understanding that many people experience, where they know the mechanics of statistics but lack a deeper understanding of what the equations truly represent.

The tutorial covers key topics such as:
- Sampling, with a realistic view of parametric models
- Estimation, standard errors, and statistical inference
- The Bias-Variance Tradeoff
- Multivariate distributions

A major section is dedicated to prediction, using both classical parametric and machine learning methods, with emphasis on the Bias-Variance Tradeoff and overfitting. The tutorial presents a fresh view of overfitting and its implications.

Finally, it provides an overview of data privacy methods, a crucial aspect of statistics today. The goal of fastStat is to provide a comprehensive and practical introduction to statistics, filling the gaps left by traditional textbooks and courses.


**Coterminal Angles**

An angle is in standard position if its vertex is at the origin and the initial side is on the positive x-axis. The angle lies in the quadrant where its terminal side lies. Angles with terminal sides along the x-axis or y-axis, such as 90°, 180°, 270°, are called quadrantal angles.

**Definition of Coterminal Angles**

Coterminal angles are angles in standard position with a common terminal side. For example, 30°, 390°, and -330° are coterminal.

**Finding Coterminal Angles**

1. Find 2 angles coterminal with 135°.
2. Find the angles of least positive measure coterminal to each angle:
   a) 1070°
   b) -65°
3. Express all angles coterminal to 90°.

**Determining Coterminal Angles**

To decide if two angles are coterminal, check if they have a common terminal side.

Examples:
a) Which angle is coterminal with 136°?
b) Which angle is coterminal with -255°?

**Identifying Non-Coterminal Angles**

a) Which pair is not coterminal?
   (i) -400°, -40°
   (ii) 340°, -50°
b) Which pair is not coterminal?
   (i) 4°, 724°
   (ii) -15°, -385°

**Finding Specific Coterminal Angles**

a) Find the smallest positive angle coterminal with 575°.
b) Find the negative angle closest to zero and coterminal with θ = 635°.

**Example: Determining Coterminal Angles**

Are 132° and -588° coterminal?

**Coterminal Angles in Radian Measure**

Determine a positive and negative angle coterminal to the given angle:
a) π/3
b) -4π/5

**Practice with Mathway Calculator**

Use the free Mathway calculator to practice Algebra and check your answers with step-by-step explanations.


## NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.2

The side of an equilateral triangle is shown by l. The perimeter of the equilateral triangle using l is given by p = l + l + l = 3l.

A regular hexagon has all its six sides equal in length, denoted by l. The perimeter of the hexagon using l is given by p = l + l + l + l + l + l = 6l.

A cube has six faces, all of which are identical squares. The length of an edge of the cube is given by l. The total length of the edges of a cube is L = l + l + l + l + l + l + l + l + l + l + l + l = 12l.

The diameter of a circle is a line that joins two points on the circle and passes through the center of the circle. The diameter of the circle (d) in terms of its radius (r) is given by d = r + r = 2r.

The associativity of the addition of numbers can be expressed in a general way using variables a, b, and c as (a + b) + c = a + (b + c). This property states that the sum of three numbers can be calculated in any order, and the result will be the same.


This maths quiz is designed to test your knowledge of various mathematical concepts, including arithmetic, geometry, and algebra. With 50 questions, you'll have the opportunity to demonstrate your skills in areas such as multiplication, division, shapes, averages, and square roots.

The quiz includes a range of topics, from basic calculations to more complex problems. You'll be asked to calculate percentages and ratios, as well as solve real-life problems. Some questions will require you to think critically and apply mathematical concepts to everyday situations.

The quiz covers various mathematical concepts, including:

* Arithmetic operations: multiplication, division, addition, and subtraction
* Geometry: shapes, angles, and measurements
* Algebra: equations, variables, and functions
* Statistics: averages, ratios, and percentages
* Real-life problems: calculating costs, measuring distances, and determining probabilities

Some sample questions from the quiz include:

* What symbol is used to represent percentage?
* What is the sum of 18 - 9?
* How many sides does a pentagon have?
* What is the sum of 7 x 7?
* What is the sum of 100 ÷ 50?

The quiz also includes questions that require you to apply mathematical concepts to real-life situations, such as:

* If you pay £2.14 for a bag of rice and sell it for £3, how much profit have you made?
* If you have a 300ml cup of coffee and drink one third of it, how much coffee is left?
* If a rectangle has a length of 8 meters and a width of 3 meters, what is its area?

To help you prepare, here are the answers to the sample questions:

* The symbol used to represent percentage is %.
* The sum of 18 - 9 is 9.
* A pentagon has 5 sides.
* The sum of 7 x 7 is 49.
* The sum of 100 ÷ 50 is 2.

The full quiz includes 50 questions, covering a range of mathematical concepts and topics. By completing the quiz, you'll be able to assess your knowledge and skills in maths and identify areas where you need to improve.

In addition to the maths quiz, you may also be interested in other quizzes on topics such as:

* Animal quiz questions and answers
* Biology quiz questions and answers
* General knowledge quiz questions and answers
* Music quiz questions and answers
* Movie quiz questions and answers
* TV quiz questions and answers
* Sport quiz questions and answers
* History quiz questions and answers
* Geography quiz questions and answers
* Capital cities quiz questions and answers
* Food and drink quiz questions and answers
* True or false quiz questions and answers
* Friends quiz questions and answers
* The Simpsons quiz questions and answers
* Harry Potter quiz questions and answers
* Star Wars quiz questions and answers

## Maths Quiz Questions and Answers

### Questions

1. What symbol is used to symbolise percentage?
2. What is the sum of 18 – 9?
3. How many sides does a pentagon have?
4. What is the sum of 7 x 7?
5. What is the sum of 100 ÷ 50?
6. What is one half written as a fraction?
7. If the time is 12pm, how many hours is it until 6pm?
8. I have a 300ml cup of coffee. If I drink one third of the cup, how much coffee in millilitres do I have left?
9. How many metres are in a kilometre?
10. What is the sum of 6 x 12?
11. 3,6,9,12,__? What is the next number in the sequence?
12. If I pay £2.14 for a bag of rice and then sell it on for £3, how much profit have I made?
13. What is the sum of the interior angles in a square?
14. If I have a bag of 5 yellow balls and 5 green balls, what is the probability (in %) of picking a yellow ball?
15. What is the sum of 21 ÷ 7?
16. 10 > 5. True or false?
17. What is the sum of 90 – 79?
18. If a rectangle has a length of 8 meters and a width of 3 meters, what is its area?
19. How many degrees is a right angle?
20. What is the sum of 8 x 20?
21. How many faces does a cube have?
22. Four pupils take a test. One pupil scores 20 marks, the second scores 50 marks, the third scores 60 marks and the fourth scores 70 marks. What is the average score of the pupils?
23. What is half of one half?
24. What is the square root of 4?
25. What is the sum of 121 ÷ 11?
26. The perimeter of a shape is defined as the total length of its boundary. True or false?
27. Determine the median of the following numbers: 2, 4, 5, 7, 9.
28. If a centimetre equates to roughly 0.4 inches, how many inches is 10cm?
29. If there are 6 oranges and 4 lemons in a bowl, how can the ratio of oranges to lemons be most simply expressed?
30. What is the sum of 13 x 14?
31. How many sides does a dodecahedron have?
32. If a pizza has a diameter of 12 inches, what is its radius?
33. What is the sum of 19 – 21?
34. If the time is 2:53pm, how many minutes is it until 4:17pm?
35. What is the sum of 888 ÷ 11?
36. 2, 3, 5, 7 and 8 are all prime numbers. True or false?
37. What is the square root of 1?
38. An obtuse angle is a type of angle that is always larger than 90° but less than 180°. True or false?
39. If the date is February 28th 2023, how many days is it until April 4th 2023?
40. What is the sum of 15 x 15?
41. How many seconds are there in 5 minutes?
42. ‘Four cubed’ can also be written as ‘4^3’. True or false?
43. What is the sum of 3 x 3 x 3?
44. What does the acronym BIDMAS stand for in maths?
45. What is the sum of 162 ÷ 3?
46. What was the nationality of the creator of the Rubik’s Cube?
47. Is an 80° angle considered an acute angle? Yes or no.
48. What is the sum of (9×2) – 10?
49. What theorem is used to find the length of an unknown side and the angle of a triangle?
50. What is Pi to five decimal places?

### Answers

1. %
2. 9
3. 5
4. 49
5. 2
6. 1/2
7. 6
8. 200ml
9. 1000
10. 72
11. 15
12. 86p
13. 360°
14. 50%
15. 3
16. True
17. 11
18. 24 square meters
19. 90°
20. 160
21. 6
22. 50
23. 1/4 or 0.25
24. 2
25. 11
26. True
27. 5
28. 4 inches
29. 3:2
30. 182
31. 12
32. 6 inches
33. -2
34. 84 minutes
35. 80.73 (approx.)
36. False, 8 is not a prime number.
37. 1
38. True
39. 35
40. 225
41. 300
42. True
43. 27
44. Brackets, Indices, Division, Multiplication, Addition, Subtraction
45. 54
46. Hungarian
47. No
48. 8
49. Pythagoras
50. 3.14159


For events with multiple outcomes and different probabilities, constructing a chart or diagram of possible outcomes can be helpful. Tree diagrams are a useful example, requiring only a pen, paper, and knowledge of the different probabilities for each potential outcome.

## Probability Trees: Theory

The basic idea of a probability tree is to use the properties of probability to create a branching diagram of different outcomes. A key law of probability is that the probabilities of a set of events must add up to 1. For example, the probability of flipping heads or tails on a coin is 0.5 each, adding up to 1. Similarly, the outcomes on a fair six-sided die occur with a probability of 1/6 each, adding up to 1.

## Making the Tree

To construct a simple tree, start by writing "Start" on the left side of a sheet of paper. Draw two arrows pointing right, one up and one down, representing the possible outcomes of a coin flip. Label one arrow "heads" and the other "tails," and write the probability of 0.5 along the length of each arrow. Verify that the probabilities add up to 1.

## Advanced Trees

To extend the tree, add more events that occur after the first. For example, after flipping a coin, flip it again. For each outcome, add two more arrows to represent the second coin flip. Label them "heads 2" and "tails 2" and add the probabilities of 0.5 for each. This process can be repeated to add any number of subsequent events to the tree.

## Using the Tree

To find the probability of a given final outcome, multiply the probabilities across the tree. For example, to find the probability of getting heads on the first coin flip and tails on the second, multiply the probability of getting heads first (0.5) by the probability of getting tails second (0.5), resulting in a probability of 0.25. This is the probability of getting heads on one coin flip, then tails on the next.


The Fieldsolver command is summarized in Table~\ref{tab:fieldsolvercmd}.

When the energy spread in the particle bunch is too large, the electrostatic approximation is no longer valid. To address this issue, the particle bunch can be divided into $k$ energy bins, and $k$ separate field solves can be performed. In each energy bin, the energy spread $\mathrm{d} E$ is small, making the electrostatic approximation valid.

In the case of a cyclotron, the number of energy bins must be at least the number of neighboring bunches, i.e., $\texttt{ENBINS} \leq \texttt{NNEIGHBB}$. The variable $\texttt{MINSTEPFORREBIN}$ defines the number of integration steps required before all energy bins are merged into one.

For a simulation, the following ingredients are required:

Mesh-refinement strategies are summarized in Table~\ref{tab:tagging_strategies}. The available strategies include:
- $\texttt{EFIELD}$: Marks each cell where the electric field component in any direction satisfies $E^{level}_{d, cell} \geq \alpha \max E_{d}^{level}$, with $d = x, y, z$ and $\alpha \in [0, 1]$ being the scaling factor $\texttt{AMR\_SCALING}$.
- $\texttt{MOMENTA}$: Refines cells containing particles with $||\mathbf{p}|| \geq \alpha \max_{level} ||\mathbf{p}||$, where $\alpha \in [0, 1]$ is the user-defined value $\texttt{AMR\_SCALING}$.
- $\texttt{CHARGE\_DENSITY}$: Tags cells for refinement if the charge density is greater than or equal to the value specified by $\texttt{AMD\_DENSITY}$.
- $\texttt{MIN\_NUM\_PARTICLES}$: Refines cells with equal to or more particles than the bound specified by $\texttt{AMR\_MIN\_NUM\_PART}$.
- $\texttt{MAX\_NUM\_PARTICLES}$: Refines cells with equal to or less particles than the bound specified by $\texttt{AMR\_MIN\_NUM\_PART}$.

The behavior of the preconditioner can be $\texttt{STD}$.

Note: The original text contained some redundant information and minor errors, which have been corrected and refined in the above version.


## Permutation and Combination and What is a Factorial?

If you have 7 different options on the menu and you can order 2 of them for your combo-meal, how many possible meal combinations are there? To figure this out, you need a simple formula and a math operation called the factorial.

### What is a factorial?

The factorial is written as an integer with an exclamation point (!), symbolizing that all integers from the given number down to 1 are to be multiplied. For example, 5! (five factorial) is 5 x 4 x 3 x 2 x 1, which equals 120. Another property of the factorial is that 0! is equal to 1.

### How to calculate combinations

The formula for computing combinations uses two variables, n and k, where n is the total number of objects to select from and k is the number of objects to be chosen. The formula is C(n,k) = n!/k!(n-k)!. For example, if a standard deck of 52 cards is used to deal a player 5 cards, how many possible hands are there? C(52,5) = 52!/5!(52-5)! = 52!/5!(47!) = 2,598,960.

### Permutation and Combination

A combination is a unique selection from a group, where a permutation considers the order of those selections. Using the letters in the word MATH as an example, if we find all combinations of 2 letters, we have 6 possibilities: MA, MT, MH, AT, AH, TH. In this case, MA is the same as AM, as order does not matter.

### How to calculate permutations

The permutation calculation is similar to the combination formula but without the additional factor in the denominator, resulting in larger values. The formula for permutations is P(n,k) = n!/((n-k)!). For example, how many ways can a hand of 5 cards be reordered to create unique permutations? P(5,5) = 5! = 120. How many separate sequences of 5 cards from a deck of 52? P(52,5) = 52!/(52-5)! = 52 x 51 x 50 x 49 x 48 = 311,875,200.

To illustrate the difference between permutations and combinations, consider the following:
- Combinations: How many ways can you choose 2 letters from the word MATH?
  A) 6
  B) 12
  C) 24
  D) 48
- Answer: A) 6
- Permutations: How many ways can you arrange 2 letters from the word MATH?
  A) 6
  B) 12
  C) 24
  D) 48
- Answer: B) 12


# Introduction to Statistics with Gonum

Gonum is a set of packages designed to make writing numerical and scientific algorithms productive, performant, and scalable. This post focuses on computing basic statistics using the stat package.

## Gonum and Statistics

Gonum provides many statistical functions, including calculating the mean, median, standard deviation, and variance of a dataset. The following code example demonstrates how to use these functions:

```go
package main
import (
	"fmt"
	"math"
	"sort"
	"gonum.org/v1/gonum/stat"
)

func main() {
	xs := []float64{
		32.32, 56.98, 21.52, 44.32,
		55.63, 13.75, 43.47, 43.34,
		12.34,
	}
	mean := stat.Mean(xs, nil)
	variance := stat.Variance(xs, nil)
	stddev := math.Sqrt(variance)
	sort.Float64s(xs)
	median := stat.Quantile(0.5, stat.Empirical, xs, nil)
	fmt.Printf("mean= %v\n", mean)
	fmt.Printf("median= %v\n", median)
	fmt.Printf("variance= %v\n", variance)
	fmt.Printf("std-dev= %v\n", stddev)
}
```

The output of this program is:

```
mean= 35.96333333333334
median= 43.34
variance= 285.306875
std-dev= 16.891029423927957
```

Note that the variance value differs from the one obtained with `numpy.var` because `gonum/stats` uses the unbiased sample variance (divisor is `len(xs)-1`), whereas `numpy.var` uses `len(xs)` as the divisor.

## Analyzing Real Data

The following code example analyzes a dataset of salaries of European developers:

```go
package main
import (
	"bufio"
	"fmt"
	"log"
	"math"
	"os"
	"sort"
	"gonum.org/v1/gonum/stat"
)

func main() {
	f, err := os.Open("salary.txt")
	if err != nil {
		log.Fatal(err)
	}
	defer f.Close()
	var xs []float64
	scan := bufio.NewScanner(f)
	for scan.Scan() {
		var v float64
		txt := scan.Text()
		_, err = fmt.Sscanf(txt, "%f", &v)
		if err != nil {
			log.Fatalf("could not convert to float64 %q: %v", txt, err)
		}
		xs = append(xs, v)
	}
	if err = scan.Err(); err != nil {
		log.Fatalf("error scanning file: %v", err)
	}
	mean := stat.Mean(xs, nil)
	variance := stat.Variance(xs, nil)
	stddev := math.Sqrt(variance)
	sort.Float64s(xs)
	median := stat.Quantile(0.5, stat.Empirical, xs, nil)
	fmt.Printf("mean= %v\n", mean)
	fmt.Printf("median= %v\n", median)
	fmt.Printf("variance= %v\n", variance)
	fmt.Printf("std-dev= %v\n", stddev)
}
```

The output of this program is:

```
mean= 55894.53879686138
median= 48000
variance= 3.0464263289031615e+09
std-dev= 55194.44110508921
```

The data file can be obtained from here: salary.txt.


# Number Tricks: The Secrets Behind Mind Reading

The best number tricks are deserving of a better name: Mental Magic or Mathematical Deception. These tricks require simple addition, subtraction, multiplication, and division. With practice, you can impress anyone willing to listen.

## 3’s A Charm

This trick is also known as the "Any Number Trick" or the "3 Trick." It involves the following steps:

- Choose any number.
- Double it.
- Add 9.
- Subtract 3.
- Divide by 2.
- Subtract the original number.

The result will always be 3. For example, starting with 7:

**7 * 2 = 14**
**14 + 9 = 23**
**23 - 3 = 20**
**20 / 2 = 10**
**10 - 7 = 3**

## Mind Reading #1

This trick can be adjusted to get different answers. The steps are:

- Pick a number.
- Multiply by 2.
- Add a number (originally 10).
- Divide by 2.
- Subtract the original number.

The answer will be half of the number added in the third step. For example, starting with 9 and adding 10:

**9 * 2 = 18**
**18 + 10 = 28**
**28 / 2 = 14**
**14 - 9 = 5**

Replacing 10 with 43 and starting with 15:

**15 * 2 = 30**
**30 + 43 = 73**
**73 / 2 = 36.5**
**36.5 - 15 = 21.5**

## Mind Reading #2

This trick relies on a personal detail: the year someone was born. The steps are:

- Take the last two digits of the year you were born.
- Add this number to the age you will be at the end of the year.

The answer is usually 114, but it changes with the year. For people born after 2000, the trick does not work.

## Powers Of 5

Multiplying by powers of 5 can be simplified by using the next closest power of 5 that is easy to multiply by, such as 10 or 100. For example, to calculate 72 * 5:

- Multiply 72 by 10: 720
- Divide by 2: 360

To calculate 25 * 64:

- Recognize that 25 = 100 / 4
- Multiply 64 by 100: 6400
- Divide by 4: 1600

## The Practical Side

These tricks are mostly for show, but some can be practical. To learn more practical mental math tricks, such as calculating tips and dealing with decimals, check out additional resources.


# 2.12: Commutative Property of Multiplication with Decimals

The commutative property of multiplication states that the order of the factors does not change the product. This property applies to decimal factors and works with multiple factors. For example, when multiplying 2 and 3, the order of the factors does not change the solution. The same applies when multiplying 2, 3, and 4.

The associative property of multiplication states that the way in which factors are grouped does not change the product. This property also applies to decimal factors and works with multiple addends. To simplify expressions and solve equations, it is essential to understand these two properties.

An algebraic expression is a mathematical phrase involving letters, numbers, and operation symbols. An equation is a mathematical statement that two expressions are equal. To simplify an expression, perform all the operations possible without knowing the value of the variable.

### Guidance

To apply the commutative and associative properties of multiplication, consider the following examples:

Simplify:
To simplify this expression, multiply the decimals and add the variable. The answer is 363.32x.

Solve for the value of:
First, multiply 0.3 and 0.4, then simplify the equation. The answer is 1.2.

### Examples

#### Example A

Solve:
Solution: x = 2

#### Example B

Simplify:
Solution: 7.935

#### Example C

Solve:
Solution: x = 7

### Original Problem

Marc and Kara traveled 160.5 miles to the Berkshires and the same distance back. Their car gets 25 miles per gallon. How many gallons of gas were needed for the trip?

To solve this problem, write an expression to describe the situation, then convert it to an equation. Solve the equation by dividing 160.5 by 25. The answer is approximately 6.42 gallons.

### Vocabulary

- Product: the answer in a multiplication problem
- Estimation: finding an approximate answer through rounding or multiplying leading digits
- Commutative Property of Multiplication: states that the order in which you multiply the factors does not affect the product
- Associative Property of Multiplication: states that the groupings in which you multiply factors does not affect the product

### Guided Practice

Simplify the expression:
To simplify this expression, multiply the decimals first, then add the variable. The answer is 41.4y.

### Practice

Directions: Simplify the following expressions.

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.

Note: The practice questions are to be completed by the student.


Lemma 1 states that log(n!) ∈ Θ(n log n). To prove this, we need to show that log(n!) ∈ O(n log n) and log(n!) ∈ Ω(n log n).

The first part, log(n!) ∈ O(n log n), is obvious. For the second part, we need to find constants c > 0 and n0 such that log(n!) ≥ cn log n for every n ≥ n0.

We start with the induction hypothesis: log((n − 1)!) ≥ c(n − 1) log(n − 1). Then, we have log(n!) = log(n.(n − 1)!) = log n + log((n − 1)!) ≥ log n + c(n − 1) log(n − 1).

To proceed, we need to show that log(n − 1) ≥ log n − ǫ, where ǫ is of order 1/n. Using the Taylor expansion of ex, we get ex > 1 + x for any x > 0. Substituting 1/(n − 1) for x, we have e1/(n−1) > 1 + 1/(n − 1), which leads to log(n − 1) > log n − log e/(n − 1).

Now, plugging this back into the previous inequality, we get log(n!) ≥ log n + c(n − 1) log(n − 1) > log n + c(n − 1)(log n − log e/(n − 1)) = cn log n + (1 − c) log n − c log e.

To complete the proof, we need to show that (1 − c) log n − c log e ≥ 0. Setting c = 1/2, we get (1 − c) log n − c log e = 1/2 log n − 1/2 log e. This is true for all n ≥ e, which is satisfied for n ≥ 3.

Thus, for c = 1/2 and n ≥ 3, we have log(n!) ≥ cn log n, which completes the induction step. The base case is n = 2, where log(2!) = 1 ≥ 1/2 * 2 log 2 = 1, confirming that log(n!) ≥ cn log n is true for n ≥ 2.

In summary, we have shown that log(n!) ≥ 1/2 n log n for every n ≥ 2, which proves that log(n!) ∈ Θ(n log n).