To calculate a test grade, you need to find the percentage of questions you got right. This requires knowing the total number of questions on the test and how many you answered correctly. You can then use a simple equation to find your grade and convert it to a letter grade.

**Method 1: Calculating Your Grade with a Simple Equation**

1. Count your correct answers and write this number down as the top number of a fraction. For example, if you had 21 correct answers, write down 21/. If some questions are worth more points than others, use the total number of points you earned as your top number.
2. Write the total number of questions or points on the bottom of the fraction. For example, if the test had 26 questions, your fraction would be 21/26.
3. Use a calculator to divide the top number by the bottom number. For example, take 21/26 and plug it into the calculator as 21 ÷ 26. You should get the answer 0.8077.
4. Multiply your answer by 100 to get your percentage. For example, 0.8077 x 100 = 80.77. This means your test grade is 80.77%.

**Method 2: Converting a Percentage to a Letter Grade**

1. Check your course materials for a grade range. If you cannot find a range, ask your professor or teacher.
2. Know the typical grading range in the United States:
	* A: 90% to 100%
	* B: 80% to 89%
	* C: 70% to 79%
	* D: 60% to 69%
	* F: 59% and below
3. Learn the British marking system:
	* 70% to 100%: Distinction
	* 60% to 69%: Merit
	* 50% to 59%: Pass
	* 49% or below: Fail (varies by school)

**Community Q&A**

* What would the percentage of 65 questions out of 120 correct be?
Answer: 54% (65/120 = 0.5417, converted to a percentage)
* How many wrong answers can I get on a 46 question test and still pass?
Answer: Assuming a 70% pass rate, you would need to get 32.2 questions correct (0.7 x 46). Since you can't get a fraction of a question correct, you would need to get 33 questions correct. This means you can get 13 questions wrong (46 - 33) and still pass.
* What is my score if a test has 110 questions and I got 45 of them wrong?
Answer: If you got 45 questions wrong, you got 65 questions correct (110 - 45). Your score would be 65/110 = 0.591, converted to a percentage is 59.1%.
* What would your score be with 8 questions wrong on a 25 question test?
Answer: If you got 8 questions wrong, you got 17 questions correct (25 - 8). Your score would be 17/25 = 0.68, converted to a percentage is 68%.
* What is my grade on a 15 question test that I missed 3 on?
Answer: If you missed 3 questions, you got 12 questions correct (15 - 3). Your score would be 12/15 = 0.8, converted to a percentage is 80%.
* Is there a website to grade problems?
Answer: Not really, a calculator is your best bet for checking math problems.
* If I get a 48 on the first marking period what would I need to get on the second to be eligible to play sports?
Answer: Assuming you need a C (70%) to play sports, and you got a 48 on the first marking period, you would need to get a 92% on the second marking period to get a total of 70% (0.7 x 2 = 1.4, 1.4 - 0.48 = 0.92).
* What would you score be when I answered 26 out of 27 questions correctly?
Answer: Your score would be 26/27 = 0.963, converted to a percentage is 96.3%.
* If you received an 88% on a test with 100 questions, how many were answered incorrectly?
Answer: If you got an 88% on a test with 100 questions, you got 88 questions correct. This means you got 12 questions incorrect (100 - 88).
* What would my percentage score be if I got 7 problems wrong out of 44?
Answer: If you got 7 questions wrong, you got 37 questions correct (44 - 7). Your score would be 37/44 = 0.841, converted to a percentage is 84.1%.
* What would my final grade in math be if I had a 44 and got 200 extra points?
Answer: This question cannot be answered without more information.
* What would someone need on a final exam to pass a class if he has an overall math grade of 63?
Answer: This question cannot be answered without more information.
* What if my exam has a total score of 80 and my correct answers are 65? What is my percentage?
Answer: Your score would be 65/80 = 0.8125, converted to a percentage is 81.25%.
* If Sally made one basket out of seven when she was playing basketball, what is her shooting average?
Answer: Her shooting average would be 1/7 = 0.143, converted to a percentage is 14.3%.

**Tips**

* Some calculators have a function to help work out percentages.
* You can also use online grade calculators.

**Warnings**

* It's easy to make simple addition mistakes when adding test questions. Be sure to double-check your work.


**Scientific notation** is used to express very large or very small numbers. A number in scientific notation is written in the form *a* × 10^{n}, where 1 < *a* < 10, and *n* is an integer.

To convert a number to scientific notation, move the decimal point to the left or right until the number is between 1 and 10. For example, 1,200,000 becomes 1.2 by moving the decimal point 6 places to the left. Since the decimal point was moved 6 places, multiply by 10^{6} to maintain the value: 1,200,000 = 1.2 × 10^{6}.

Conversely, to convert from scientific notation to a standard number, move the decimal point to the right if the exponent is positive, or to the left if the exponent is negative. For instance, 1.2 × 10^{-4} becomes 0.00012 by moving the decimal point 4 places to the left.


## Converting Octal to Binary

To convert octal to binary, we first need to understand the basics of both number systems.

### Octal Number System

- The octal number system has a base of 8.
- It consists of 8 digits: 0, 1, 2, 3, 4, 5, 6, 7.
- Example: (156)₈, where "8" represents the base and "156" represents the octal number.

### Binary Number System

- The binary number system has a base of 2.
- It consists of 2 digits: 0 and 1.
- Example: (10111)₂, where "2" represents the base and "10111" represents the binary number.

### Conversion Method

To convert an octal number to binary, we can use a 3-bit representation for each octal digit. The conversion is as follows:

- Digit 0: 000
- Digit 1: 001
- Digit 2: 010
- Digit 3: 011
- Digit 4: 100
- Digit 5: 101
- Digit 6: 110
- Digit 7: 111

### Example Conversion

Let's convert the octal number (55)₈ to binary:
- For the first 5, the binary representation is 101.
- For the second 5, the binary representation is 101.
- The combined binary representation is 101101, which can also be written as (101101)₂.

## Java Programs for Octal to Binary Conversion

There are three methods to convert octal to binary in Java:

### Method 1: Using Switch Case

This method involves taking an octal value as input, using a switch case to convert each digit to its binary equivalent, and then combining the results.

```java
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter a octal Value : ");
        String input1 = sc.next();
        int a = 0;
        String output = "";

        while (a < input1.length()) {
            char c = input1.charAt((int) a);
            switch (c) {
                case '0':
                    output += "000";
                    break;
                case '1':
                    output += "001";
                    break;
                case '2':
                    output += "010";
                    break;
                case '3':
                    output += "011";
                    break;
                case '4':
                    output += "100";
                    break;
                case '5':
                    output += "101";
                    break;
                case '6':
                    output += "110";
                    break;
                case '7':
                    output += "111";
                    break;
                default:
                    System.out.println("\nInvalid Octal Digit " + input1.charAt((int) a));
                    break;
            }
            a++;
        }
        System.out.println("Converted binary is :" + output);
    }
}
```

Output:
```
Enter a octal Value : 55
Converted binary is :101101
```

### Method 2: Using Mathematical Strategy

This method involves converting the octal number to decimal and then converting the decimal number to binary.

```java
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter a octal Value : ");
        int input1 = sc.nextInt();
        int i = 0;
        int decimal = 0;
        int output = 0;

        while (input1 != 0) {
            decimal += (input1 % 10) * (int) Math.pow(8, i);
            ++i;
            input1 /= 10;
        }
        i = 1;

        while (decimal != 0) {
            output += (decimal % 2) * i;
            decimal /= 2;
            i *= 10;
        }
        System.out.println("Converted binary is :" + output);
    }
}
```

Output:
```
Enter a octal Value : 55
Converted binary is :101101
```

### Method 3: Using Built-in Functions

This method involves using the `Integer.parseInt()` function to convert the octal number to decimal and then using the `Integer.toBinaryString()` function to convert the decimal number to binary.

```java
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter a octal Value : ");
        String input1 = sc.next();
        int octal = Integer.parseInt(input1, 8);
        String output = Integer.toBinaryString(octal);
        System.out.println("Converted binary is :" + output);
    }
}
```

Output:
```
Enter a octal Value : 55
Converted binary is :101101
```


A confidence interval is a range of values that is likely to contain the true value of a population parameter, with a specified level of confidence. It consists of a point estimate plus or minus a margin of error. The components of a confidence interval include the confidence level, the margin of error, and the sample size. The confidence level represents the probability that the true value of the parameter falls within the interval. The margin of error represents the maximum amount that the point estimate is likely to differ from the true value. The sample size affects the width of the confidence interval, with larger sample sizes resulting in narrower intervals.

Confidence intervals are used to estimate the range of plausible values for a parameter and to make inferences about the population from which the sample was drawn. They provide a range of values within which the true population parameter is likely to lie, allowing researchers to draw meaningful conclusions from sample data. Confidence intervals are essential in statistical analysis, as they help researchers estimate population parameters, test hypotheses, and make informed decisions.

The confidence level is the probability that the true population parameter falls within the confidence interval. Common confidence levels include 90%, 95%, and 99%. The margin of error is the maximum difference between the sample estimate and the true population parameter. The sample size plays a critical role in determining the accuracy and uncertainty of a confidence interval. Larger sample sizes lead to narrower confidence intervals, reducing the potential error.

The standard error measures the typical deviation between sample estimates and the true population parameter. Z-scores are standardized scores that tell us how many standard deviations a particular sample statistic lies from the population mean. Degrees of freedom measure the amount of information available in a sample, influencing the distribution of Z-scores. T-scores are used when sample size is small, accounting for the increased variability in smaller samples.

The formula for constructing a confidence interval is: Sample statistic ± (Z-score or T-score) × Standard error. The Z-score or T-score is determined by the desired confidence level, which typically ranges from 90% to 99%. By defining this level, we establish the precision of the interval and the level of risk we’re willing to accept when making inferences.

Confidence intervals have profound implications in decision-making. Hypothesis testing relies on confidence intervals to determine if a sample supports or refutes a claim about the population. Parameter estimation allows us to estimate the true value of a population parameter within a specified range. Confidence intervals are indispensable tools in statistical analysis, providing valuable insights into the true nature of population parameters.

When interpreting confidence intervals, the confidence level is crucial. This percentage represents the likelihood that the true population parameter falls within the interval. The interval width provides valuable insights into the precision of the estimate. A narrower interval indicates a higher precision, meaning that the population parameter is likely to be closer to the interval’s midpoint.

Confidence intervals play pivotal roles in both hypothesis testing and parameter estimation. In hypothesis testing, they help in making statistically sound decisions. For example, if a confidence interval for a population mean does not include a hypothesized value, it suggests the hypothesis is incorrect. In parameter estimation, confidence intervals provide a range of plausible values for the unknown parameter, guiding decision-makers by providing a more nuanced understanding of the underlying population.

Confidence intervals are cornerstones of statistical analysis, providing valuable information about population parameters and serving as foundational tools for data-driven decision-making. They help researchers estimate values, draw conclusions, and support informed choices. By understanding and interpreting confidence intervals, we can gain greater insights into the patterns and trends in data, make more accurate predictions, and effectively inform decisions that shape various aspects of life.


## Step 1: Analyze the given text to extract key facts and concepts.
The text includes various mathematical and scientific problems, such as solving for a variable in an equation, calculating the expected value of a game, determining the trajectory of particles, and finding the antiderivative of a function.

## Step 2: Identify and remove any irrelevant or noisy information from the text.
The text contains multiple unrelated problems and questions, so we need to focus on the specific tasks and remove any unnecessary information.

## Step 3: Reorganize the content into a logically structured and concise version.
To refine the text, we will separate the problems into individual sections and provide a clear and step-by-step solution for each problem.

## Step 4: Maintain the original length of the text as much as possible.
We will ensure that the refined text is concise but still provides all the necessary information and solutions to the problems.

## Step 5: Refine multiple-choice questions and answers if any.
There are no multiple-choice questions in the provided text, so this step is not applicable.

The final answer is: $\boxed{1}$


A unique method of growing watermelons involves placing a plastic box around the fruit as it grows, resulting in a cube-shaped watermelon. Although these cubic fruits can be expensive, with a single cube watermelon selling for over $75, it is possible to grow them at home.

The cube watermelons have distinct geometric properties. For instance, they have a box-like shape, which can be observed in various objects, such as shoeboxes or building blocks. A key characteristic of a cube is that it has 6 flat sides, or faces. Additionally, a cube has 8 pointy parts where 3 faces meet, known as corners.

Growing cube watermelons can be space-efficient, as they can be packed more tightly than round watermelons. In fact, it is possible to fit 4 times as many cube watermelons as round watermelons in the same garden space. For example, if a garden can accommodate 5 round watermelons, it could fit 20 cube watermelons instead. When arranging cube watermelons in a rectangular pattern, such as 3 rows of 4, the watermelons in the middle row are not on the edge. In this specific arrangement, there would be 2 watermelons in the middle, not on the edge.


# Find a Midpoint on a Line Segment

To find the coordinates of a line segment's midpoint, use the midpoint formula. This formula calculates the average of the x-coordinates and the average of the y-coordinates of the segment's endpoints, giving the location of the point exactly in the middle of the segment.

The midpoint formula for a segment with endpoints at (x1, y1) and (x2, y2) is:

Note that it doesn't matter which point is (x1, y1) and which is (x2, y2).

The midpoint formula can be applied to problems involving coordinate geometry, such as proving that the diagonals of a rectangle bisect each other. By using the midpoint formula for each diagonal, you can show that each diagonal goes through the midpoint of the other, demonstrating that they bisect each other. For example, given a rectangle with diagonals, the midpoint formula can be used to calculate the midpoint of each diagonal, confirming that they intersect at the same point, such as (6.5, 3).


Students learn to form and simplify algebraic expressions involving multiplication, addition, division, and subtraction in Year 7 Term 2. This unit is followed by Expressions, Equations, and Formulae. Key topics include:

* Using simple formulae
* Generating and describing linear number sequences
* Expressing missing number problems algebraically
* Finding pairs of numbers that satisfy an equation with two unknowns
* Enumerating possibilities of combinations of two variables

Algebraic expressions can be simplified by collecting like terms, which contain the same letter symbol and equal powers. For example, 2 + 2 + 2 can be written as 3 × 2, and a + a + a can be written as 3a. In algebraic notation, × and ÷ symbols are not included, so 3 × a is written as 3a and 3 ÷ a is written as 3/a.

Key concepts include:

* An algebraic expression is a collection of mathematical terms with no stated value
* Multiplicative relationships, such as xy = yx, can be used to simplify expressions
* The order of operations is used when substituting known values into expressions

To develop fluency, students will:

* Use algebra to generalize the structure of arithmetic and formulate mathematical relationships
* Substitute values in expressions, rearrange and simplify expressions, and solve equations

To reason mathematically, students will:

* Identify variables and express relations between variables algebraically and graphically

To solve problems, students will:

* Develop their mathematical knowledge through solving problems and evaluating outcomes, including multi-step problems
* Select appropriate concepts, methods, and techniques to apply to unfamiliar and non-routine problems

Algebraic notation includes:

* ab in place of a × b
* 3y in place of y + y + y and 3 × y
* a^2 in place of a × a, a^3 in place of a × a × a, and a^2b in place of a × a × b
* a/b in place of a ÷ b
* Coefficients written as fractions rather than decimals

Students will simplify and manipulate algebraic expressions to maintain equivalence by:

* Collecting like terms
* Taking out common factors

This unit connects to other topics, such as plotting and interpreting histograms with unequal class widths, and rearranging formulae to solve simultaneous equations related to density, volume, and surface area.


## Brick Puzzle

The Brick puzzle is presented as follows: pack 53 bricks with dimensions 1-by-1-by-4 into a 6-by-6-by-6 box. To approach this problem, we first perform a sanity check. The 6-by-6-by-6 box contains 216 unit cube cells, and 53 bricks would occupy 212 cells, indicating no volume contradiction.

To further analyze the problem, we divide the box into 27 smaller 2-by-2-by-2 boxes and color them in a checkerboard pattern. This results in 13 boxes of one color (white) and 14 boxes of another color (black). Regardless of how a brick is placed inside the original box, it must cover 2 white cells and 2 black cells. However, there are only 104 white cells available, which is sufficient for only 52 bricks.


**Simplification Technique**

Example-1:
- 7.1
- 3.5
- 0.7
- 2.4
- None of these
Solution: The answer is (e) None of these.

Example-2:
- None of these
Solution: The answer is (a).

Example-3:
(9)^3 × (6) ÷ (3)^3 = ?
- 1012
- 984
- 972
- None of these
Solution: (9)^3 × (6) ÷ (3)^3 = ? The answer is (e) None of these.

Example-4:
To find the solution, first look at the last digit, which is 7.


**Law of Cosines**
The law of cosines is used to solve triangles when the measure of two sides and the included angle is known, or when the measure of three sides of a triangle is known. The law of cosines formulas are:

a² = b² + c² - 2bc cos A
b² = a² + c² - 2ac cos B
c² = a² + b² - 2ab cos C

**Example 1: Solving a Triangle with Two Sides and the Included Angle**
Given: b = 14.7, c = 9.3, and A = 46.3°
We need to find the length of side a and the measures of angles B and C.

Using the law of cosines:
a² = b² + c² - 2bc cos A
= (14.7)² + (9.3)² - 2(14.7)(9.3) cos 46.3°
= 216.09 + 86.49 - 273.42(0.6908824)
= 216.09 + 86.49 - 188.90107
= 113.67893
a ≈ 10.7

Now, we can use the law of cosines to find angle B:
b² = a² + c² - 2ac cos B
(14.7)² = (10.7)² + (9.3)² - 2(10.7)(9.3) cos B
216.09 = 114.49 + 86.49 - 199.02 cos B
15.11 = -199.02 cos B
cos B ≈ -0.075922
B ≈ 94.3°

Finally, we can find angle C:
C = 180° - 94.3° - 46.3° = 39.4°

The solution is:
a = 10.7, A = 46.3°, b = 14.7, B = 94.3°, c = 9.3, C = 39.4°

**Example 2: Solving a Triangle with Three Sides**
Given: a = 23.31, b = 27.26, and c = 29.17
We need to find the measures of angles A, B, and C.

Using the law of cosines:
a² = b² + c² - 2bc cos A
(23.31)² = (27.26)² + (29.17)² - 2(27.26)(29.17) cos A
543.36 = 743.11 + 850.89 - 1590.35 cos A
1050.64 = 1590.35 cos A
cos A ≈ 0.660634
A ≈ 48.65°

Now, we can use the law of cosines to find angle B:
b² = a² + c² - 2ac cos B
(27.26)² = (23.31)² + (29.17)² - 2(23.31)(29.17) cos B
743.11 = 543.36 + 850.89 - 1359.91 cos B
651.14 = 1359.91 cos B
cos B ≈ 0.478811
B ≈ 61.39°

Finally, we can find angle C:
C = 180° - 61.39° - 48.65° = 69.96°

The solution is:
a = 23.31, A = 48.65°, b = 27.26, B = 61.39°, c = 29.17, C = 69.96°


### Video Transcript

To find the volume of the solid whose base is bounded by the lines π₯ + 5π¦ = 5, π₯ = 0, and π¦ = 0, with cross sections taken perpendicular to the π₯-axis being semicircles, we use the slicing method. This method involves four steps: determining the base region, finding the shape and orientation of the cross sections, calculating the area of the cross sections, and integrating the area over an appropriate interval.

The base of the solid is a right-angled triangle with height 1 and width 5, bounded by the lines π₯ = 0, π¦ = 0, and π₯ + 5π¦ = 5. The equation π₯ + 5π¦ = 5 can be rearranged to π¦ = -π₯/5 + 1, which is a straight line with slope -1/5 and π¦-intercept 1.

The cross sections are semicircles with diameters equal to the height of the perpendicular line. The diameter of a cross section at π₯ is the length of the vertical line, which is the π¦-coordinate. This π¦-coordinate is -π₯/5 + 1. Thus, the diameter of the cross section is -π₯/5 + 1, and the radius is (-π₯/5 + 1)/2.

The area of a semicircle is (1/2)πr². Substituting the radius, we get the area of the cross section as (1/2)π((-π₯/5 + 1)/2)² = (π/8)(π₯²/25 - 2π₯/5 + 1).

To find the volume, we integrate the area of the cross sections over the interval 0 ≤ π₯ ≤ 5. The volume is ∫[0,5] (π/8)(π₯²/25 - 2π₯/5 + 1) dπ₯.

Evaluating the integral, we get (π/8)[(π₯³/75 - π₯²/5 + π₯)] from 0 to 5. At π₯ = 5, this is (π/8)[(5³/75 - 5²/5 + 5)]. Simplifying, we get (π/8)(125/75 - 25/5 + 5) = (π/8)(5/3) = 5π/24.

Therefore, the volume of the solid is 5π/24.


CHAPTER 1 Exercises

E1.1 Charge is calculated as Current × Time, which equals (2 A) × (10 s) = 20 C.

E1.2 The given equation is used to calculate charge: 2cos(200) + 200cos(200) - 0.01sin(20t) + 0.01cos(20t) = dq/dt = i.

E1.3 In element C, positive charge moves downward due to the positive value of 2. In element E, positive charge moves upward because 3 has a negative value.

E1.4 Energy is calculated as Charge × Voltage, which equals (2 C) × (20 V) = 40 J. Since v_ab is positive, the positive terminal is 'a' and the negative terminal is 'b'. Thus, charge moves from the negative terminal to the positive terminal, and energy is removed from the circuit element.

E1.5 Current enters terminal 'a' since it is positive at that terminal. This indicates a passive reference configuration.

E1.6 (a) The energy calculation yields p = 20 × (2/3) × 20 = 266.67 J and w = ∫p dt = ∫20 × (2/3) × 20 dt = 666.67 J.
(b) With opposite references to the passive sign convention: p = -20 × (2/3) × 20 = -266.67 J and w = ∫-20 × (2/3) × 20 dt = -1000 J.

E1.7 (a) The sum of currents leaving equals the sum of currents entering: i_a = 1 + 3 = 4 A.
(b) 2 = 1 + 3 + b, so b = 2 - 1 - 3 = -2 A.
(c) 0 = 1 + c + 4 + 3, so c = -1 - 4 - 3 = -8 A.

E1.8 Elements A and B are in series, while elements E, F, and G are also in series.

E1.9 Going clockwise around the loop with elements C, D, and E: 3 + 5 + v = 0, so v = -8 V. Then, going clockwise around the loop with elements D, E, and F: 10 + e = 0, so e = -2 V.

E1.10 Elements C and D are in parallel, while elements E and F are in series.


The final part of the 6th-grade Introduction to Algebra unit covers inequalities. The instruction begins with Desmos activities, starting with Polygraph, which is revisited after students have gained more experience with inequalities. A review of inequality symbols is conducted before proceeding to a Desmos activity that introduces graphing. This activity is typically the first time students encounter the greater than or equal to and less than or equal to symbols.

The next day, a "Which One Doesn’t Belong" activity is used for review. Following this, students practice graphing inequalities based on given situations using a Desmos activity. Another review method employed is Mathketball, a game where students arrange their desks in a circle, attempt to solve a problem on the board, and if correct, try to shoot a paper into a garbage can placed in the middle of the room. This engaging game is reserved for concepts where all students can complete the problems in a similar amount of time.

To further reinforce learning, the instructor seeks more visual aids, such as pictures, for which students can write inequalities. An example of student engagement was observed when, upon inquiring about the number of problems on an upcoming quiz, a student suggested writing an inequality for the statement "8 problems tops." This creative approach demonstrates the students' grasp and interest in the subject matter.


### Using GLM Repeated Measures Test in Research

The GLM Repeated Measures procedure analyzes variance when the same measurement is made several times on each subject or case. It tests null hypotheses about the effects of between-subjects and within-subjects factors, including interactions and constant covariates. This general linear model procedure is essentially an ANOVA with repeated measures, used to determine differences between continuous variables measured multiple times in the same group of participants.

For example, to determine the effect of exercise on weight loss in a group of 20 participants, we measure their weight before exercise, after 3 months, and after 6 months, comparing mean values using ANOVA with repeated measures.

### Assumptions of GLM Repeated Measures:

- Continuous dependent variable
- Categorical independent variable with two or more related groups
- Normal data distribution
- No significant outliers
- Equal variances of the differences between all combinations of related groups (sphericity)

### An Example in SPSS: The GLM Repeated Measures

We collected data on Math test scores from 51 students before training, after one month, and after two months to examine the effect of training on scores. With a continuous dependent variable (Math test score) and a categorical independent variable (three groups: before, after 1 month, after 2 months), we use ANOVA with repeated measures.

### Methods and Statistics

Type I, II, III, and IV sums of squares evaluate different hypotheses, with Type III as the default. Post hoc range tests and multiple comparisons include least significant difference, Bonferroni, Sidak, and others. Descriptive statistics provided include observed means, standard deviations, counts, Levene test for homogeneity of variance, Box’s M, and Mauchly’s test of sphericity.

This tutorial guides how to run the GLM Repeated Measures test in SPSS and interpret the results, focusing on testing the effect of an intervention on a dependent variable over time.


In math class, students will apply statistics concepts, such as mean, median, box plots, and distribution, to analyze wealth distribution. The lesson, "Wealth of Nations" from Mathalicious, guides students to think mathematically and draw their own conclusions. The goal is to examine the actual distribution of wealth in the US, compare it to people's perceptions, and discuss whether the distribution is a problem.

The lesson starts with students describing what they think the wealth distribution in the US should be. Each class created a consensus description, represented graphically, with sections proportional to the share of wealth each group has. The "bottom 20%" and "top 20%" refer to 20% of the population, with the same number of people, but differing in personal wealth.

The following are the descriptions from each period on Day 1:
Period 1:
Period 3:
Period 4:
Period 5:
Period 6:

The actual distribution of wealth will be revealed, allowing students to compare their initial thoughts with reality. The "bottom 20%" and "top 20%" have the same number of people, but the people in the bottom have the least personal wealth and the people in the top have the most. The percentage of total wealth each group owns is described in the table.


Decimals

When subtracting numbers with decimals, the process is the same as with whole numbers, with additional columns after the decimal point:

Tens | Units (ones) | . | Tenths | Hundredths |

Whole numbers can be thought of as having zeros after the decimal point, e.g., $87-23$ can be rewritten as $87.00-23.00$ to two decimal places. The process for solving decimals is the same as for whole numbers.

Subtracting decimals involves the same steps as subtracting whole numbers, with the added consideration of values in the columns after the decimal point. When using a place value system and needing to subtract a larger number from a smaller one, it is necessary to exchange a value from the previous place, such as exchanging 1 tenth for 10 hundredths.

Key considerations when subtracting decimals include:

- Starting calculations from the smallest place value column
- Exchanging values from previous places when necessary

Evaluate the following expressions:

- $0.73-0.51$ (write answer as a decimal)
- $6.61-4.42$ (write answer in decimal form)
- $2.75-1.75$

Understanding addition and subtraction of fractions, decimals, and integers is crucial for working with decimals.


# Flow Rate Units Conversion

The conversion of 1 cubic inch per day (cu in/d) to acre feet per minute (ac-ft/min) is as follows:

1 cu in/d = 0.0000000000092 ac-ft/min

This conversion can be reversed to find the equivalent value in cubic inches per day for a given value in acre feet per minute.

## Conversion Chart

The following chart provides the conversion of cubic inches per day to acre feet per minute for values ranging from 1 to 15 cu in/d:

1 cu in/d = 0.0000000000092 ac-ft/min
2 cu in/d = 0.000000000018 ac-ft/min
3 cu in/d = 0.000000000028 ac-ft/min
4 cu in/d = 0.000000000037 ac-ft/min
5 cu in/d = 0.000000000046 ac-ft/min
6 cu in/d = 0.000000000055 ac-ft/min
7 cu in/d = 0.000000000065 ac-ft/min
8 cu in/d = 0.000000000074 ac-ft/min
9 cu in/d = 0.000000000083 ac-ft/min
10 cu in/d = 0.000000000092 ac-ft/min
11 cu in/d = 0.00000000010 ac-ft/min
12 cu in/d = 0.00000000011 ac-ft/min
13 cu in/d = 0.00000000012 ac-ft/min
14 cu in/d = 0.00000000013 ac-ft/min
15 cu in/d = 0.00000000014 ac-ft/min

## Question and Answer

What is the equivalent value in acre feet per minute for 15 cubic inches per day?
Answer: 15 cu in/d = 0.00000000014 ac-ft/min

## Abbreviations

* cu in/d: cubic inch per day
* ac-ft/min: acre foot per minute

## Applications

This flow rate converter can be used for various applications, including:
1. Practicing conversions between cubic inches per day and acre feet per minute.
2. Finding conversion factors between unit pairs.
3. Working with flow rate values and properties.


## Modular Multiplicative Inverse

The modular multiplicative inverse of an integer a modulo m is an integer x such that ax ≡ 1 (mod m). This is equivalent to finding an integer x such that ax = 1 + my, where x and y are integers.

The multiplicative inverse of a modulo m exists if and only if a and m are coprime (i.e., if gcd(a, m) = 1).

### Calculating Modular Multiplicative Inverse

There are several ways to calculate the modular multiplicative inverse:

#### 1. Brute Force

We can calculate the inverse using a brute force approach where we multiply a with all possible values x and find a x such that ax ≡ 1 (mod m).

```cpp
int modInverse(int a, int m) {
    a %= m;
    for(int x = 1; x < m; x++) {
        if((a*x) % m == 1) return x;
    }
}
```

The time complexity of the above code is O(m).

#### 2. Using Extended Euclidean Algorithm

We have to find a number x such that ax = 1 (mod m). This can be written as ax + my = 1, where x and y are integers.

The Extended Euclidean Algorithm is an extension to the Euclidean algorithm. Besides finding the greatest common divisor of integers a and b, as the Euclidean algorithm does, it also finds integers x and y (one of which is typically negative) that satisfy Bézout's identity ax + by = gcd(a,b).

We will look at two ways to find the result of Extended Euclidean Algorithm:

##### Iterative Method

This method computes expressions of the form ri = axi + byi for the remainder in each step i of the Euclidean algorithm.

```cpp
pair<int, pair<int, int> > extendedEuclid(int a, int b) {
    int x = 1, y = 0;
    int xLast = 0, yLast = 1;
    int q, r, m, n;
    while(a != 0) {
        q = b / a;
        r = b % a;
        m = xLast - q * x;
        n = yLast - q * y;
        xLast = x, yLast = y;
        x = m, y = n;
        b = a, a = r;
    }
    return make_pair(b, make_pair(xLast, yLast));
}

int modInverse(int a, int m) {
    return (extendedEuclid(a,m).second.first + m) % m;
}
```

##### Recursive Method

This method attempts to solve the original equation directly, by reducing the dividend and divisor gradually, from the first line to the last line, which can then be substituted with trivial value and work backward to obtain the solution.

```cpp
pair<int, pair<int, int> > extendedEuclid(int a, int b) {
    if(a == 0) return make_pair(b, make_pair(0, 1));
    pair<int, pair<int, int> > p;
    p = extendedEuclid(b % a, a);
    return make_pair(p.first, make_pair(p.second.second - p.second.first*(b/a), p.second.first));
}

int modInverse(int a, int m) {
    return (extendedEuclid(a,m).second.first + m) % m;
}
```

The time complexity of the above codes is O(log(m)^2).

#### 3. Using Fermat’s Little Theorem

Fermat’s little theorem states that if m is a prime and a is an integer co-prime to m, then a^(m-1) ≡ 1 (mod m).

```cpp
int pow(int a, int b, int MOD) {
    int x = 1, y = a;
    while(b > 0) {
        if(b%2 == 1) {
            x=(x*y);
            if(x>MOD) x%=MOD;
        }
        y = (y*y);
        if(y>MOD) y%=MOD;
        b /= 2;
    }
    return x;
}

int modInverse(int a, int m) {
    return pow(a,m-2,m);
}
```

The time complexity of the above codes is O(log(m)).

#### 4. Using Euler’s Theorem

Euler’s theorem is a generalization of Fermat’s Little theorem. According to Euler’s theorem, if a is coprime to m, that is, gcd(a, m) = 1, then a^φ(m) ≡ 1 (mod m), where φ(m) is Euler Totient Function.

### Calculating Modular Multiplicative Inverse for all numbers from 1 to n

We can use sieve to find a factor of composite numbers less than n. So for composite numbers inverse(i) = (inverse(i/factor(i)) * inverse(factor(i))) % m, and we can use either Extended Euclidean Algorithm or Fermat’s Theorem to find inverse for prime numbers.

```cpp
vector<int> inverseArray(int n, int m) {
    vector<int> modInverse(n + 1,0);
    modInverse[1] = 1;
    for(int i = 2; i <= n; i++) {
        modInverse[i] = (-(m/i) * modInverse[m % i]) % m + m;
    }
    return modInverse;
}
```

The time complexity of the above code is O(n).


He sold 4 cycles for Rs. 8400, meaning each cycle costs Rs. 2100. He then sold 3 cycles for Rs. 1900 each. To gain an average profit of Rs. 320 per cycle, the total profit for 10 cycles would be 10 x 320 = Rs. 3200. He already made a profit of Rs. 1850 from selling 7 cycles. The remaining profit needed is Rs. 3200 - Rs. 1850 = Rs. 1350. To achieve this from the remaining 3 cycles, he needs to make a profit of Rs. 1350 / 3 = Rs. 450 per cycle. Therefore, he must sell the remaining 3 cycles at Rs. 2200 each to meet his target.


To increase the average by one, from 14 to 15, he scored 12 more runs than the existing average, which is 26 - 14 = 12. Therefore, to raise the average by five, from 14 to 19, he should score 12 x 5 = 60 more runs than the existing average. Thus, he should score 14 + 60 = 74. The correct answer is 74.


Suppose $f:A→B$ is a given function. Then, there exists a function $g:f(A)→A$ such that $g ∘ f = I_A$ and $f ∘ g = I_{f(A)}$ if and only if $f$ is a one-to-one function and $g = f^{-1}$.

Proof ($→$):
Assume $f$ is one-to-one and $g = f^{-1}$. For all $a ∈ A$, if $f(a) = b$, then $g(b) = a$. Thus, $g(f(a)) = g(b) = a$, which implies $g ∘ f = I_A$. Similarly, for any $b ∈ f(A)$, $f(g(b)) = f(a) = b$, so $f ∘ g = I_{f(A)}$.

To prove ($←$), we need to show that if $g ∘ f = I_A$ and $f ∘ g = I_{f(A)}$, then $f$ is one-to-one and $g = f^{-1}$.


### Affine Subspaces of a Linear Space

We have dealt with linear spaces, where the zero vector is always included. However, it is also important to study spaces translated from zero by a given vector, known as affine subspaces. These subspaces are no longer linear. The term hyperplane describes affine subspaces of dimension one less than the whole linear space.

An affine subspace can be formed by translating a linear subspace by a vector. Given a linear subspace $L$ and a vector $v$, the affine subspace $L + v$ (meaning $L$ translated by $v$) can be described in several ways:

- By giving a vector and its direction, e.g., $v + L$.
- By giving points it goes through (one more point than the dimension of the subspace), e.g., $(v + l_1, v + l_2, \ldots)$, sometimes denoted by $[v; l_1, l_2, \ldots]$.
- By giving a parametrization, e.g., $v + \sum_{i=1}^{n} \alpha_i l_i$.
- By giving a system of linear equations.

### Describing Affine Subspaces

Given a translation vector $v$ and the tangent space $L$, we can find the points included in the affine subspace. Conversely, if we know the points, we can find $v$ and $L$. For example, if $H$ goes through points $v + l_1, v + l_2, \ldots$, choosing one point as $v$ gives us $H = v + L$.

Given $v$ and vectors spanning $L$, we can find a parametrization of the affine subspace. For instance, if $v$ and $l_1, l_2, \ldots$ are given, every vector in the affine subspace is of the form $v + \sum_{i=1}^{n} \alpha_i l_i$.

### From Translation Vector and Direction to System of Linear Equations

The system of linear equations for an affine subspace differs from that of its tangent space by the column of free coefficients. To find the system for the affine subspace, first find the system for the tangent space, then adjust the free coefficients so that the translation vector is a solution.

### From System of Equations to Parametrization

The general solution of a system of equations for an affine subspace provides a parametrization of the subspace.

### Projections and Reflections

To calculate the projection of a vector onto an affine subspace or its reflection, translate the subspace to pass through zero, perform the calculation, and then translate back. For example, to project $w$ onto $v + L$, first project $w - v$ onto $L$ to get $p$, then the projection of $w$ onto $v + L$ is $v + p$.


Integration is a fundamental concept in mathematics and engineering. As an Integration Engineer on the Hardware team, key responsibilities include performing hardware, software, and networking set-up of various systems. Integration is often used to find the area underneath the graph of a function and the x-axis. The definite integral of a function gives us the area under the curve of that function.

In mathematics, integration is the inverse operation of differentiation. The derivative of a function represents the function's slope, while the integral represents the area under the curve. To integrate a function, one can use various techniques, such as substitution, partial fractions, and integration by parts.

For example, to integrate 2cos2x with respect to x, one can use the substitution method. The integral of 2cos2x is -sin2x + C, where C is the integration constant. Another example is the integral of 2x, which is written as 2x dx = x^2 + C.

In engineering, integration is crucial for connecting data, applications, APIs, and devices across an IT organization to be more efficient, productive, and agile. IT integration, or systems integration, enables organizations to make their data work for them.

There are various types of integration patterns, including partial fractions, integration by parts, and substitution. To integrate by partial fractions, one needs to factor the denominator of the function and then find the partial fractions. The goal of chromatographic peak integration is to obtain retention times, heights, and areas of these peaks.

Key concepts in integration include:

* The definite integral of a function gives us the area under the curve of that function
* Integration is the inverse operation of differentiation
* Various techniques, such as substitution, partial fractions, and integration by parts, can be used to integrate a function
* IT integration, or systems integration, is crucial for connecting data, applications, APIs, and devices across an IT organization

Example 1: Integrate 2cos2x with respect to x.
Answer: -sin2x + C

Note: The original text has been refined to remove irrelevant information and reorganized to provide a logically structured and concise version.


To determine the number of 2×6 joists needed for a deck, it's essential to understand the different depths of floor joists used in floor construction, such as 2×6, 2×8, 2×10, and 2×12, made from various species of wood like southern yellow pine and Douglas fir, and different grades like grade 1, grade 2, and grade 3. Southern yellow pine is the strongest species of wood, capable of spanning more distance compared to other species. Grade 1 lumber has higher strength than other grades due to fewer knots.

Standard joist sizes for decks include 2×6 (2 by 6), with others being 2×4 (2-by-4), 2×8 (2-by-8), 2×10 (2-by-10), 2×12 (2-by-12), and even greater. The most standard floor joist spacing is 16 inches apart on center, but this can vary depending on building codes and the requirements of the structure. Other spacings like 12 inches, 19.2 inches, and 24 inches apart on center are also used.

To calculate the number of 2×6 joists needed, measure the length of the subfloor in feet, multiply it by 12 to convert it into inches, then divide it by the spacing (e.g., 24″, 16″, 19.2″, or 12″) and add 1 to the result. The simple formula to find the number of 2×6 joists is: 2×6 joists count = ((subfloor length × 12)/16) + 1.

For example, for an 8 by 10 feet deck, you would need at least 9 2 by 6 joists that are at least 8 feet long. Using the formula: (10×12)/16) +1 = 7.5 +1 = 8.5, rounded to 9.

For a 12×12 deck, with 16 inches on center spacing of joists, you would need 10 pieces of 2×6 joists, each at least 12 feet long. Using the formula: ((12 × 12)/16) + 1 = (144/16) + 1 = 9 + 1 = 10.

For a 10×10 deck, with the same spacing, you would need 9 pieces of 2×6 joists, each at least 10 feet long. The calculation is: ((10 × 12)/16) + 1 = (120/16) + 1 = 7.5 + 1 = 8.5, rounded to 9.

For a 10×12 deck, you would need 10 pieces of 2×6 joists, each at least 10 feet long, as the calculation ((12 × 12)/16) + 1 = (144/16) + 1 = 9 + 1 = 10 shows.

For an 8×10 deck, you would need 9 pieces of 2×6 joists, each at least 8 feet long, since ((10 × 12)/16) + 1 = (120/16) + 1 = 7.5 + 1 = 8.5, rounded to 9.

In conclusion, to figure out the number of 2×6 joists needed for a deck, measure the deck's length in inches, divide it by the spacing (e.g., 16″), and add 1 to the result. This calculation provides the total number of 2×6 joists required, ensuring a sturdy and well-constructed deck.


To solve equations using the distributive property, we apply the rule that states a(b + c) = ab + ac.

First, consider the equation 7(w - 5) = 21. To solve for w, we distribute the 7 across the terms inside the parentheses: 7w - 35 = 21. Then, we solve for w by adding 35 to both sides, resulting in 7w = 56, and finally dividing both sides by 7, which gives w = 8.

For the equation 8(w - 4) = 32, we apply the distributive property to get 8w - 32 = 32. Adding 32 to both sides gives 8w = 64, and dividing both sides by 8 yields w = 8.

In the equation 3(w - 3) = 6, applying the distributive property results in 3w - 9 = 6. Adding 9 to both sides gives 3w = 15, and dividing both sides by 3 gives w = 5.

The distributive property can also be used to combine like terms, as seen in the example 8.5m + 1.3m = 9.8m, which simplifies to 9.8m by adding the coefficients of m.

To solve fraction equations using the distributive property, consider the equation 2/3 * 8/9 = c/6. Multiplying the fractions gives (2*8)/(3*9) = c/6, which simplifies to 16/27 = c/6. Cross-multiplying yields 6*16 = 27*c, resulting in 96 = 27c, and dividing both sides by 27 gives c = 96/27, which simplifies to c = 32/9.

For the equation z/3 + 1 = z/2 * 3, first simplify the right side using the distributive property: z/2 * 3 = 3z/2. Then, the equation becomes z/3 + 1 = 3z/2. To solve for z, find a common denominator, which is 6. Multiplying the entire equation by 6 to clear the denominators gives 2z + 6 = 9z. Subtracting 2z from both sides yields 6 = 7z, and dividing both sides by 7 gives z = 6/7.

To simplify expressions using the distributive property, consider the following examples:
1. 5(a + 2) = 5a + 10
2. 4(2x - 4) = 8x - 16
3. (10 + 4y) * 1/2 = 5 + 2y

The distributive property is essential in algebra for simplifying expressions and solving equations. It allows us to expand and combine like terms, making it easier to work with complex expressions and equations.


In today's lesson, students will learn how to find the tangent of an angle using right triangles and reference angles. They will also learn how to solve trigonometric equations using the terminal side of an angle. To do this, they will need to understand the concept of parent graphs, determinants, inverses, and Identity Matrix.

Students should memorize parent graphs, which are used to help students understand trigonometric functions. They should also know the definition of an Identity Matrix and how to use a four-function calculator to solve problems. The calculator can be used to find the length of each side of a right triangle and the tangent of an angle.

To find the tangent of an angle, students will need to use a reference angle and the quadrant of the terminal side. They should apply the appropriate sign to the angle and use the trigonometric functions to find the solution. Students should also memorize the parent graphs for the quiz after the break and know the dimensions of matrices, determinants, and inverses.

The student should be able to use a four-function calculator to do the problems, but they should read the directions on the worksheet to complete the problems. They should also know the meaning of the Identity Matrix and how to find the determinant and inverse of a matrix.

To answer the exact values worksheet, the student should know the definition of trigonometric functions and how to use the reference angle to find the solution. The reference angle is the angle between the trigonometric function and the horizontal axis. Students should understand the concept of trigonometric functions, including the tangent, and how to apply them to solve problems.

Key concepts to memorize include parent graphs, trigonometric functions, reference angles, and Identity Matrix. Students should also understand how to use a four-function calculator to solve problems and how to apply the appropriate sign to an angle. By memorizing these concepts and practicing with worksheets, students will be able to solve trigonometric equations and find the tangent of an angle.


Let the sum be Rs. p, rate be R% p.a. and time be T years. Two equations can be formed:
P×T×(R 2)100 - P×T×R100 = 108
⇒ 2PT = 10800 ...... (1)
And,
P×R×(T 2)100 - P×R×T100 = 180
⇒ 2PR = 18000 ......(2)
Clearly, from (1) and (2), we cannot find the value of p. So, the data is not sufficient.


To convert a percentage to a decimal, divide by 100. Thirty-three and a third percent can be represented as 33.333...%, where the 3 repeats infinitely. As a decimal, this is equal to 0.33333..., with the 3 repeating forever. In fraction form, 33.333...% is equal to 100/3 as an improper fraction or 33 1/3 as a mixed number.

To illustrate the conversion, consider the number twelve and thirty-three hundredths (12.33). To convert this to a percentage, multiply by 100: 12.33 × 100 = 1,233%.

Other percentages and their decimal equivalents include thirty-six and seventy-five hundredths percent, which is equal to 0.3675, and thirty-six and three-fourths percent, which equals 0.375. Thirty-three and four-tenths percent is 0.334.

In contrast, thirty-three and one-third percent is 0.3333..., with the 3 repeating infinitely, which is different from 0.33 (one-third) or 0.0033 (thirty-three ten-thousandths). The repeating decimal 0.3333... represents thirty-three and one-third percent, which can also be written as 33.333...%.

Understanding these conversions is essential for basic math operations. For example, knowing that 33.333...% is equal to 0.33333... (with the 3 repeating) helps in calculations involving percentages.

Q: What is thirty-three and a third percent as a decimal?
A: 0.33333... (with the 3 repeating infinitely)

Related concepts include converting between percentages, decimals, and fractions, as well as understanding repeating decimals and their significance in mathematical operations.


The approach to solving this problem involves naming six unknown variables and building a system of nonlinear equations that can be reduced using substitution. The variables are defined as follows:

Observe the addition of a red dividing line with the same length as $x$. From here, the properties of transversal lines and vertical angles can be applied to determine which angles are congruent to each other. This allows us to identify several similar triangles.

We may now build the initial system of equations. Since $a$ and $c$ together comprise the full length of the 12-foot string, $a+c=12$; and similarly, $b+d = 10$. We use the Pythagorean Theorem to find that the top portion of the red bar has length $\sqrt{a^2-e^2}$ and the bottom portion has length $\sqrt{b^2-f^2}$. Both of these sides belong to two triangles each, so, by using the Pythagorean Theorem and transitive property on the shared side, we obtain $b^2-f^2 = c^2-25$ and $a^2-e^2=d^2-25$.

Similar triangle properties can be used to find the last two equations. Due to conservation of ratios in similar triangles, $\frac{a}{e} = \frac{c}{5}$, $\frac{b}{f}=\frac{d}{5}$ and $\frac{b}{d} = \frac{c}{a}$. We may now focus on reducing the system of all stated equations, that is,

\begin{align}
a+c&=12\quad &(1)\\
b+d&=10\quad &(2)\\
b^2-f^2 &= c^2-25\quad &(3)\\
a^2-e^2 &=d^2-25\quad &(4)\\
\frac{a}{e} &= \frac{c}{5}\quad &(5)\\
\frac{b}{f}&=\frac{d}{5}\quad &(6)\\
\frac{b}{d} &= \frac{c}{a}\quad &(7).
\end{align}

Using (1), (2) and (7) to derive an important relationship, we can solve (1) for $b$ and (2) for $c$ to make the substitution $a(10-d)=(12-a)d,$ which simplifies to $d = \frac{5}{6}a$. This new information can be used to modify (4) to $a^2-e^2=\frac{25}{36}a^2-25.$

Finding this last equation allows us to separate a system of two equations with two variables, namely
\begin{align*}
\frac{a}{e} &= \frac{c}{5}\\
a^2-e^2&=\frac{25}{36}a^2-25.
\end{align*}

From the second we solve for $e=\frac{5a}{12-a}$, hence the last equation becomes
\begin{align}
a^2-\left(\frac{5a}{12-a}\right)^2&=\frac{25}{36}a^2-25\quad\textrm{or}\\
11 a^4-264 a^3+1584 a^2-21600 a +129600&= 0
\end{align}

This polynomial has two imaginary and two real solutions; we reject the imaginary solutions and consider the real ones $a\approx 6.645$, $a\approx 20.26$. As the second solution is larger than the size of the string, we reject it and conclude $a=6.645$.

Now the above equations determine
\begin{align}
c &= 12 - a = 5.355,\\
d &= \frac{5}{6}a = 5.538,\\
b &= 10 - d = 4.462,\\
e &= \frac{5a}{c} = 6.204,\\
f &= \frac{5b}{d} = 4.029.
\end{align}

We have arrived at the final step: recall that $x$ is the sum of the two red lines in the diagram, hence
$$x = \sqrt{a^2-e^2}+\sqrt{b^2-f^2}=4.298.$$

Alternatively, the problem can be solved with only four variables by splitting the vertical distance $x$ into two variables, $h$ and $x-h$, and using the horizontal widths $w$ and $y$. Two large right triangles can be created: $w^2 + x^2 = 10^2$ and $x^2 + y^2 = 12^2$. Using the 5-foot horizontal line to create two similar triangles gives enough information to finish the problem. The ratios of the sides are the same, so $\frac{5}{w} = \frac{x-h}{x}$ and $\frac{5}{y} = \frac{x}{h}$. Combining the system of four equations yields a quartic equation in one of the variables, which solves to get the same result as in the post.


## Permutation and Combination Formula – Properties

This section covers the formulas and properties of permutation and combination, along with examples.

Permutation refers to the arrangements in a definite order that can be made by taking some or all of the things at a time. The order of the items is crucial in permutation.

The formula and properties of permutation and combination will be discussed, providing a comprehensive understanding of these concepts.


**Course Schedule and Policies**

The course schedule is subject to frequent changes, so it is essential to check it at least every class day and before starting work on any assignment. When there is a reading assignment, please read the named section(s) before that day.

**Video Recordings**

If you see the video symbol, it means that the class was videoed, and you can obtain a link by emailing the instructor. If you know ahead of time that you will miss a class, please inform the instructor, and they will ensure that the class is videoed for you.

**Thought and Question (T&Q) Assignments**

T&Qs are short emails that you should send to your instructor about the current readings or material discussed in class. Please submit each T&Q in its own email to your instructor before class on Monday (or Tuesday, if needed). The content should be a thought or question about the topic, expressed in a few sentences. There is no requirement for depth or sophistication, but it should demonstrate real engagement with the material.

**Grading T&Qs**

Each T&Q is graded 0 or 1. You can expect to receive mostly 1s if you engage with the readings and class material and keep up with the schedule.

**Example Code and Concepts**

The provided code examples demonstrate various concepts, including:

* Plotting sine and cosine functions using matplotlib
* Implementing a number-guessing game using Python
* Creating a Quizzer class for interactive guessing
* Defining a BSTnode class for binary search trees
* Printing letter counts in a string using dictionaries and arrays
* Reading files line by line
* Implementing a Graph class with methods for adding vertices and edges
* Using static methods for generating random adjacency matrices
* Creating a greedy Euler path algorithm for graphs
* Setting up axes for graph visualization without tick marks or numbers
* Implementing a clean string function for removing non-alphabetic characters
* Using the AES encryption algorithm with a provided key

**Multiple Choice Questions**

There are no multiple choice questions provided in the original text.

**Key Concepts and Numbers**

* Check the course schedule frequently
* Read assigned sections before class
* Engage with the material in T&Q assignments
* Use the provided code examples to understand various concepts
* Keep up with the schedule to receive mostly 1s on T&Qs
* Use the AES encryption algorithm with the provided key (b'asdfjkl;asdfjkl;asdfjkl;asdfjkl;') for encryption purposes.


A message was received seeking advice on how students learn and what can make them learn more effectively, particularly in mathematics. The discussion is complex and multifaceted, and without empirical evidence, suggestions are based on observations and supported by academic research where possible.

At a homeschool, students have been found to have poor math skills, including:

A. Lack of knowledge of timetables
B. Difficulty with word problems
C. Inability to identify prime numbers
D. Struggles with fractions
E. Inability to calculate percentages
F. Difficulty with real-world application problems, such as calculating the cost of 200g of meat given the cost of 500g.

These issues are not unique and are common among students in both primary and secondary education. Non-native English speakers often struggle with word problems, and students are not always guided on how to problem-solve effectively. Polya's "Steps to Problem Solving" provides a framework for approaching problems, particularly word problems.

Students often come to upper secondary school with preconceived ideas and misconceptions about mathematical concepts, which can be a concern. Authors like Jo Boaler advocate for a focus on understanding and concepts rather than memorization of methods.

Many students hate math due to the way they are introduced to or taught the subject. There is a strong connection between students' mindset, understanding, and attitude towards math. Students may hate math because of past experiences or lack of motivation and belief in their ability to do math.

To address these issues, it is essential to identify the base problems and develop a solution. This involves changing the students' mindset, giving them the belief that they can do math, and helping them understand key concepts. Finding a balance between enjoyment, positive mindset, and examination requirements is crucial. Sometimes, students need to go back to basics and learn to enjoy math before being pushed further.

Key concepts and strategies mentioned include:

* Polya's "Steps to Problem Solving"
* Jo Boaler's emphasis on understanding and concepts
* Carol Dweck's concept of mindset
* The importance of relevance and visualization in learning math
* The need for a balance between enjoyment, positive mindset, and examination requirements

Ultimately, these questions cannot be answered in a simple blog post, and individual students' backgrounds and contexts must be considered to develop effective solutions.


# The Elusive Lighthouse: Part 1

## Introduction to the Problem

A lighthouse on a rocky island emits flashes in random directions, and sensors along the coastline register the incident locations of these flashes. The goal is to infer the lighthouse's location from these flash records. The geometry of the scenario is crucial, with the length units being kilometers.

## Initial Approach and Its Limitations

The initial approach involves estimating the lighthouse's location by averaging the x-values of the flashes. However, simulations with different sample sizes (10, 100, and 1,000,000 flashes) yield estimates that are not consistently close to the true value. Increasing the sample size does not necessarily improve the accuracy of the estimate.

## Understanding the Distribution of Flashpoints

The distribution of flashpoints is not Gaussian but rather follows a Cauchy or Lorentz distribution due to the uniform habits of the lighthouse and the geometry of the scene. The probability density function (PDF) of the flashpoints is given by the Cauchy distribution, which has heavy tails. This means that many estimates will be in error by more than a kilometer.

## Implications of the Cauchy Distribution

The Cauchy distribution implies that collecting many flashpoints and taking the mean is mathematically equivalent to estimating the location by a single flashpoint. The probability of a bad estimate does not decrease with the number of flashpoints. To achieve a modest goal of a 95% chance of a good estimate, the lighthouse would need to be almost literally a stone's throw from the shore (less than 10 meters away).

## Conclusion and Future Directions

The initial approach to estimating the lighthouse's location has significant limitations due to the Cauchy distribution of the flashpoints. A new approach, based on the sound principles of probability theory, is needed to remedy this signal failure. The next post will demonstrate how to improve the estimation of the lighthouse's location.

## Key Concepts and Formulas

- The Cauchy distribution: \(f(x) = \frac{1}{\pi} \cdot \frac{\gamma}{x^2 + \gamma^2}\)
- The probability of a bad estimate: \(P(|\hat{x} - x| > \delta)\)
- The goal: \(P(|\hat{x} - x| > \delta) < 0.05\) for \(\delta = 1\) km

## References

Sivia, DS & Skilling, J 2006. *Data Analysis: A Bayesian Tutorial.* Oxford University Press, Oxford, United Kingdom.


In probability experiments, an event is defined as a set of outcomes. The complement of an event, denoted as A′ or A^{C}, consists of all possible outcomes that are not part of the event. A key concept in understanding probability is that the probability of the complement of an event is calculated as one minus the probability of the event itself, expressed as P(A') = 1 - P(A). This relationship highlights the inverse nature of an event and its complement, providing a fundamental principle for calculating probabilities.


# Complete Electron Configuration for Krypton (Kr)

Krypton is the 36th element in the periodic table and its symbol is 'Kr'. It is a noble gas. The total number of electrons in krypton is 36, arranged in specific orbitals.

## What is the Electron Configuration of Krypton?

The electron configuration of krypton can be determined through two methods: electron configuration through orbit (Bohr principle) and electron configuration through orbital (Aufbau principle). The electron configuration of krypton is [Ar] 3d^{10} 4s^{2} 4p^{6}.

## Krypton Atom Electron Configuration through Orbit

The electrons of the atom revolve around the nucleus in a certain circular path, called an orbit or shell. The electron holding capacity of each orbit is 2n^{2}, where n is the serial number of the orbit.

| Shell Number (n) | Shell Name | Electrons Holding Capacity (2n^{2}) |
| --- | --- | --- |
| 1 | K | 2 |
| 2 | L | 8 |
| 3 | M | 18 |
| 4 | N | 32 |

The krypton atom will have 2 electrons in the first shell, 8 in the 2nd orbit, 18 electrons in the 3rd shell, and the remaining 8 electrons will be in the 4th shell. The order of the number of electrons in each shell of the krypton atom is 2, 8, 18, 8.

## Electron Configuration of Krypton through Orbital

Atomic energy shells are subdivided into sub-energy levels, called orbitals. The sub-energy levels depend on the azimuthal quantum number 'l'. The value of 'l' is from 0 to (n - 1). The sub-energy levels are known as s, p, d, and f.

| Orbit Number | Value of 'l' | Number of Subshells | Number of Orbital | Subshell Name | Electrons Holding Capacity |
| --- | --- | --- | --- | --- | --- |
| 1 | 0 | 1 | 1 | 1s | 2 |
| 2 | 0, 1 | 2 | 1, 3 | 2s, 2p | 2, 6 |
| 3 | 0, 1, 2 | 3 | 1, 3, 5 | 3s, 3p, 3d | 2, 6, 10 |
| 4 | 0, 1, 2, 3 | 4 | 1, 3, 5, 7 | 4s, 4p, 4d, 4f | 2, 6, 10, 14 |

The Aufbau principle states that electrons will first complete the lowest energy orbital and then gradually continue to complete the higher energy orbital. The energy of an orbital is calculated from the value of the principal quantum number 'n' and the azimuthal quantum number 'l'.

## Why is Krypton an Inert Gas?

The ground state electron configuration of krypton is 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6}. This electron configuration shows that the last shell of krypton has eight electrons, making it an inert gas. Krypton does not want to exchange or share any electrons because the last orbit is full of electrons, and it does not form any compounds.

## FAQs

### What is the Symbol for Krypton?
**Ans:** The symbol for krypton is 'Kr'.

### How Many Electrons Does Krypton Have?
**Ans:** 36 electrons.

### How Do You Write the Full Electron Configuration for Krypton?
**Ans:** The full electron configuration for krypton is 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}.

### How Many Valence Electrons Does Krypton Have?
**Ans:** Eight valence electrons.

### What is the Valency of Krypton?
**Ans:** The valency of krypton is 0.

### What is the Valence Electron Configuration for the Krypton Atom?
**Ans:** The valence electron configuration for krypton atom is 4s^{2}4p^{6}.

### How Many Orbitals Does Krypton Have?
**Ans:** 18 orbitals.

### What is the Complete Ground State Electron Configuration for the Krypton Atom?
**Ans:** The complete ground state electron configuration for the krypton atom is 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}.

### How Many Energy Levels Does Krypton Have?
**Ans:** Krypton has a total of four energy levels.

### How Many Energy Levels are Filled in a Krypton Atom?
**Ans:** The first three energy levels are filled with electrons, and the fourth energy level is filled with 8 electrons.

### What is the Noble Gas Configuration for Krypton?
**Ans:** The noble gas configuration for krypton is [Ar] 3d^{10}4s^{2}4p^{6}.

### How Many p-Orbitals are Occupied by Electrons in a Krypton Atom?
**Ans:** Three p-orbitals (2p, 3p, 4p) are occupied by electrons.

### How Many Unpaired Electrons are Present in the Ground State Kr Atom?
**Ans:** There are no unpaired electrons in the krypton atom in the ground state.

### How Many Electrons are There in the Outermost Level of Krypton?
**Ans:** 8 electrons.

### How Many Orbitals are Occupied by Electrons in the 4p Subshell of a Krypton Atom?
**Ans:** Three orbitals (4p_{x}, 4p_{y}, 4p_{z}) are occupied by a total of 6 electrons.


# Ruffini-Horner Algorithm for Complex Arguments

The Ruffini-Horner algorithm is used to calculate the value of a polynomial with real coefficients for a complex argument. This is achieved by dividing the polynomial by (x - z), where z = a + bi. The remainder of this division is a linear function, and the value of the polynomial at z is equal to the value of the remainder.

To apply this algorithm, we start with a polynomial and a complex number z = a + bi. We then divide the polynomial by (x - z) to obtain a remainder, which will be a linear function of the form Ax + B, where A and B are complex numbers. The value of the polynomial at z is then given by the value of this remainder at x = z.

The process can be summarized as follows:
1. Divide the polynomial by (x - z) to obtain a remainder.
2. The remainder will be a linear function of the form Ax + B.
3. Evaluate the remainder at x = z to obtain the value of the polynomial at z.

This method provides an efficient way to evaluate polynomials at complex points, and it has applications in various fields of mathematics and science.

Reference:
D. Kurepa, Higher Algebra, Book 2 (in Croatian), Zagreb: Skolska knjiga, 1965.


Math Kangaroo is an international mathematical competition where over 50 countries are represented. The competition is held annually on the third Thursday of March. There are twelve levels of participation, ranging from grade 1 to grade 12. The competition is a multiple-choice test that runs for 75 minutes, with 24 questions for students up to 4th grade and 30 questions for other students. Each question scores 3, 4, or 5 points, with no penalty for incorrect answers or skipped questions. Calculators are not allowed.

The Math Kangaroo competition provides students with the opportunity to compare themselves to others not only on the national level but also on the international level. The key competence tested by Mathematical Kangaroo is logical combination, not just pure knowledge of formulas. Students who work on these types of math problems have better results at the competition.

Available as PDF files, there are two papers of 60 marks each, except Stage 3 which is 50 marks each. The competition is catered to all school students from Year 1 to Form 6, with six different categories and differing levels of difficulty. Awards are given to the top-scoring students per grade at the national level. Each student receives a t-shirt, a certificate of participation, and a gift in addition to a competition booklet and a pencil on the test day.

To prepare for the Math Kangaroo competition, students can use Edugain to access unlimited printable worksheets, online practice, and online tests. The competition gives students the opportunity to develop their mathematical skills and compare themselves to others internationally. With no penalty for incorrect answers or skipped questions, students can focus on solving the problems to the best of their ability.


The HP 35 Logarithm Algorithm

The algorithm used by D. Cochran for the HP 35 log routine is derived from J.E. Meggitt's "Pseudo Division and Pseudo Multiplication Processes". This method involves repeated-addition multiplication and repeated subtraction division, making it easy to implement and execute.

To simplify the calculation, we consider only the natural logarithm ln(x) and normalized mantissa (0.xxxxxx), as ln(M 10^k) = ln(M) + k ln(10). We need the constant ln(10) in ROM.

The approach is similar to Brigg's method, where we search for ln(M) by considering a product of small quantities. We can select each factor to drive r close to 1, and the expression can be written as ln(M) + ln(Pn) = 0, resulting in ln(M) = -ln(Pn).

However, in practice, the limited number of factors and the need for accuracy lead to ln(M) = ln(r) - ln(Pn), where r is a small remainder.

To optimize the computation time and ROM required, Meggitt proposes using the form (1 + 10^-j) for the factors. The HP 35 uses six constants: (1 + 10^-0) = 2, (1 + 10^-1) = 1.1, (1 + 10^-2) = 1.01, (1 + 10^-3) = 1.001, (1 + 10^-4) = 1.0001, and (1 + 10^-5) = 1.00001.

The algorithm core involves driving the product M Pn close to 1 by multiplying the intermediate result MPn-1 by a number (1+10^-j) times, where j is the largest integer exponent such that the product M Pn goes closer to 1 but keeps < 1. This operation is a pseudo multiplication using adding and shifting.

The calculation steps involve shifting the previous result to the right qj times and adding the result to the previous intermediate value. At each line, a test to < 1 is made; if not < 1, the constant is changed.

For example, to compute ln(4.4), we use the constants and exponents: j = 0, j = 1, j = 2, j = 3, j = 4, and j = 5. The intermediate results are calculated using the pseudo multiplication process.

The final result is obtained by subtracting from ln(10) the sum of the products of the exponents and the natural logarithms of the constants. The remainder r is driven to 10 instead of 1, and we need to take into account this small remainder.

The calculation of ln(4.4) involves the following steps:

1. Compute the intermediate results using the pseudo multiplication process.
2. Calculate the remainder r.
3. Compute the sum of the products of the exponents and the natural logarithms of the constants.
4. Subtract the sum from ln(10) to obtain the final result.

The result is ln(4.4) = ln(10) - (7 ln(1.00001) + 6 ln(1.0001) + 2 ln(1.001) + 3 ln(1.01) + 1 ln(1.1) + 1 ln(2) + (10 - 9.99996771177327000000)/10).

Reference:
J.E. Meggitt, Pseudo Division and Pseudo Multiplication processes, IBM Journal, Res & Dev, April 1962, 210-226.


# W questions covered by H

In the *Big Ideas* section, there is an activity called *Covered by H*. A year 4 teacher used W questions with the H-shape, prompting a dialogue.

Teacher: What makes you think the first box is 23?
Student: It comes after 22.
Teacher: What if it was in this row?
Student: This row starts with 20, and this starts with 30, so it would be 33.

Teacher: What stayed the same, and why?
Student: The three, because that row is all threes.
Teacher: Three...
Student: Units.

Teacher: Where would the five units column be? [Student points] Good, so what can you tell me about the horizontal rows and vertical columns? How could you describe the rule, the pattern?
Student: Horizontal rows go up by ones, in order. Vertical columns go up by tens: 15, 25, 35, 45.

Teacher: What if this vertical column was 115, instead of 15. Where would 135 be, and why?
Student: 135. It's 115, 125, 135 [pointing correctly].
Teacher: Why?


To solve the equation "23 plus 60 equals 1," it appears to be a number trick or joke. The answer is based on time: 1 hour, plus 1 minute, plus 59 minutes, will equal 2 hours, but when considering 23 hours and 60 minutes, it equals 1 hour the next day in a 24-hour clock system.

Another approach to understanding the original question involves breaking down given equations and solving for unknowns. For instance, if we have "ba = 2a + b," then to find "23 + 32 + 1," we need to apply the given formula. However, without a direct equation, we look at other provided equations for patterns or solutions.

Given "y = 3/6," it simplifies to "1/2." The equation "2.60869565217 or (23/60)" suggests a conversion or representation in decimal form, where 23/60 equals approximately 0.3833, not directly related to the initial question but demonstrating a mathematical operation.

Solving "11m + 13 = m + 23" for "m":
- Subtract "m" from both sides: 10m + 13 = 23
- Subtract 13 from both sides: 10m = 10
- Divide both sides by 10: m = 1

For the equation "8x - 1 = -4x + 23":
- Add 4x to both sides: 8x + 4x = 23 + 1
- Combine like terms: 12x = 24
- Divide both sides by 12: x = 2

These examples demonstrate various mathematical operations and solutions to different equations, showcasing how to approach and solve problems step by step.