**1. Plug-in**: Let *n* = 2. Then "twice *n*" will be 4. If *k* is 3 more than that, *k* is 7. Now just go through the answer choices to see which one yields 7 when 2 is plugged in for *n*. Only (B) works.

**2. Backsolve**: You know one of these choices won't work, so just find which one won't by trying all of them. If (C) were the right answer, then *x* = 0 wouldn't work. In other words, 2|0| - 3 < 7 would be false. But since -3 *is* less than 7, (C) is not our answer. The only choice that doesn't work is choice (A), because 2|5| - 3 is equal to 7, not less than 7. (A) is our answer.

**3. Backsolve**: When you see numbers in the answer choices, think backsolve, even if it's a geometry question! Start with (C): if *x* = 18, then the angles go from *x*, 2*x*, 3*x*, and 4*x* to 18, 36, 54, and 72, respectively. Since the line on the bottom is a straight line, those numbers need to add up to 180. Lo and behold, they do! So (C) is the correct answer.

**4.**: No fancy techniques here, just read the question very carefully. They want all the kids who are applying to College *X* or College *Y* (or both *X* and *Y*), but NOT College *Z*. So cross off all the numbers that fall within the circle for *Z*. You should be left with 11 applying to only *X*, 10 applying to only *Y*, and 9 applying to both *X* and *Y*. 11 + 10 + 9 = 30. Answer: (B).

**5.**: We calculate averages by finding the sum of the numbers we want to average, and then dividing it by the number of numbers. Which is to say: AVG = SUM / (# OF NUMBERS). We know the average here, and we know the number of numbers, so let's put those into the equation to find the sum: 30 = SUM / 3, (30)(3) = SUM, 90 = SUM. That's choice (E).

**6.**: Again, nothing fancy here. Just count the dots that line up with the *y*-coordinates we're looking at. There are 5 dots on the *y* = 4 line, which is more than any other line. (B) is our answer.

**7.**: We don't care about *x* or *y* individually here, we only care about the expression (*x* + *y*). This is a common theme on the SAT and PSAT: they treat whole expressions like (*x* + *y*) as something to be solved for. We start with: (*x* + *y*)^3 = 64. Take the cube root of each side: (*x* + *y*) = 4. So if (*x* + *y*) = 4, then: (*x* + *y*) + 2 = 4 + 2 = 6. (B) is our answer.

**8.**: Draw it, but be careful. In order to be sure you get this right, don't draw the streets in the order in which they're given. Start with Fern and Oak. The question says they're perpendicular. The next thing they tell us is that Broad is parallel to Main, but we don't know anything about either of those streets yet, so ignore that for now. Ignore that for now and move on to the last thing they say: that Main is perpendicular to Fern. Since we've already drawn Fern, we can draw Main. And now that we have Main, we can go back and draw Broad. The question asks which streets are parallel. II. Main and Oak, III. Oak and Broad. (E) is our answer.

**9.**: Function (or operation) questions are all about following directions. They tell us that *x* = 2*x* + 1. That just means that whatever is to the left of the star gets multiplied by 2, and then 1 added to it. So, all of these are also true: *y* = 2*y* + 1, *a* = 2*a* + 1, ☃ = 2☃ + 1. Anyway, let's solve this. It works the same way with numbers as it does with variables or symbols: (-4) = 2(-4) + 1, (-4) = -8 + 1, (-4) = -7. Our answer is (B).

**10.**: Just arithmetic here, but be careful so you don't make a silly mistake. The house price is $120,000. That's 2.5 times the annual income of the buyer, so the buyer's annual income is $120,000 / 2.5 = $48,000. If his annual income is $48,000, we divide that by 12 to find his monthly salary: $48,000 / 12 = $4000. Answer choice (D).

**11.**: **Backsolve** this! Start by checking if (C) will work. If *x* = 5: *x*^2 - *y*^2 = 15, 5^2 - *y*^2 = 15, 25 - *y*^2 = 15, *y*^2 = 10. Since the question stipulated that both *x* and *y* have to be integers and the square root of 10 isn't an integer, we know (C) isn't the right answer. Furthermore, we know we have to go higher because if we went lower, *x*^2 would be less than 15, which won't do us any good. So, let's try (D), which says *x* = 8: *x*^2 - *y*^2 = 15, 8^2 - *y*^2 = 15, 64 - *y*^2 = 15, *y*^2 = 49, *y* = 7. YES! This time *y* works out to be an integer, so we're good. (D) is the correct answer.

**12.**: Note first that segments *PQ* and *PR* are BOTH radii, so they're both equal to 4. That's very important, because it means our triangle is isosceles, and therefore angles *Q* and *R* have to be equal. But wait...if angles *Q* and *R* are the same, and angle *P* is 60°...then...we've got an equilateral triangle! All the angles are 60°, so all the sides are 4, so the perimeter is 12. (B) is our answer.

**13.**: **Plug-In (carefully)**! We know these two fractions need to equal each other, so let's say *x* = 20, and *w* = 15. That way both fractions will equal 5, a nice easy number to work with. Now all we need to do is plug our values into the next expression to get our answer! Done. Answer: (A).

**14.**: **Backsolve**! The pattern is generated in this problem by multiplying each term after the first by the same number, *k*. Our job is to find *k*. We do that by trying the answer choices we're given until one works. Since we don't know the first term but we do know the second, we find the first (*a*) by dividing 15 by our answer choice. Then, we'll multiply 15 by our answer choice to get *c*. Then we'll multiply our new *c* by 15 again to see if our end result is 135. If it is, we've found our answer! Let's make a table: Answer choice (k) | a (15 / k) | 15 (given) | c (15 x k) | 135 (c x k) | (C) 5 | 3 | 15 | 75 | 375 (too big, go smaller) | (B) 4.5 | 3.333... | 15 | 67.5 | 303.75 (better, but still too big) | (A) 3 | 5 | 15 | 45 | 135 (perfect!) | So, (A) is our answer.

**15.**: To solve this one, break the 30 integers into a few groups. First, we know 18 of them are multiples of 5. That leaves 12 that aren't multiples of 5. Now we have to break up the 18 multiples of 5 into the 10 odd ones, which leaves 8 even ones. The breakdown looks something like this: The question wants to know how many even integers at most *could* there be, so we assume every integer we don't know for a fact is odd to be even. That means the 12 non-multiples of 5, for our purposes, are all even. If that's true, then we'd have 20 even integers. 20 is the most we could have, since we know 10 are odd for sure. Answer: (C).

**16.**: On this graph, steeper slopes correspond to greater rates of population growth. Since we can look at both the slopes of actual population growth (the solid lines) and the theoretical constant rate (the dashed line), all we need to do is compare the slopes of the 20-year intervals to the slope of the dashed line and count how many of them have slopes that go up and to the right more steeply: **1880 - 1900: steeper than dashed line**, **1900 - 1920: steeper than dashed line**, 1920 - 1940: less steep than dashed line, **1940 - 1960: steeper than dashed line**, 1960 - 1980: less steep than dashed line. That's 3 intervals that are steeper than the dashed line, so our answer is (C).

**17.**: To say triangles are similar is to say that all their angles are the same. A pair of equilateral triangles *must* be similar, therefore, because all their angles are 60°. That's not true about isosceles, or right triangles, though. You could have, for example, an isosceles that has angles of 20°, 20°, and 140°, and another with angles of 30°, 30°, and 120°. Both of those would be isosceles, but they would not be similar to each other. Since the question says "*must* be similar," you can eliminate a choice if you can come up with a counterexample. Likewise, it's easy to imagine two right triangles that aren't similar. Take the two special ones you're given at the beginning of each math section on the SAT and PSAT: the 45°-45°-90° and the 30°-60°-90°. Again, both right triangles, but not similar. Therefore, our answer is (A).

**18.**: To make 0.30608, we basically want to break the following: 0.30608 = 0.3 + 0.006 + 0.00008. In order to find *j*, *k*, and *n*, just count the number of spaces the decimal place would have to move to make those decimals. (For example, the decimal place would have to move one place to make the 3 into 0.3). So *j* = 1, *k* = 3, and *n* = 5. 1 + 3 + 5 = 9. Answer choice (D).

**19.**: Plot this! They talk about *g*(*x*) here, but they don't show it, so we need to draw it right on the page. If you use a graphing calculator this is pretty easy, but it's not much harder to just draw by hand. Remember the basic slope-intercept form of a line: *y* = *mx* + *b*. (*m* is your slope, *b* is your *y*-intercept.) We've got: So our slope is 3/5, and our *y*-intercept is zero. That's all we need to know to draw this line. Slope is rise / run, or [change in *y*] / [change in *x*]. Since one of our points is (0, 0) our other has to be a change of 5 along the *x*-axis, and a change of 3 along the *y*-axis. That's (5, 3). Connect those two points, and you've got your line, which clearly intersects *f*(*x*) twice -- at around (2.5, 1.5) and around (5, 3). They ask for how many values of *x* does *f*(*x*) = *g*(*x*), which is just a fancy way of asking how many times the two intersect, so our answer is (C).

**20.**: **PLUG-IN**! A #20 plug-in! Hopefully you're only attempting this after working carefully and making sure you got all the earlier questions right because you know this is worth no more points than any other question, but what a gift! It's still a #20, though, so even with plug-in it's going to be a bit tricky. Here's how you do it. Let's say our profit from the sale of tickets was $90 (that's *z*). Then let's say our expenses totaled $10 (that's *x*). I chose those numbers because now the total amount of money we took in from ticket sales (profit plus expenses) was $100. That's a nice easy number to work with. Now set your ticket price (*y*). Let's make it easy and say tickets were $5 each, which means we had **20** people attend the dance to make up our $100 in total ticket sales. To review: *x* = 10, *y* = 5, *z* = 90. The question asks us to find an expression for how many tickets were sold, so all we're doing is plugging these numbers into all the answer choices to see which one gives us **20**, since we figured out above that's how many tickets we sold. Only (E) works.


To calculate the APR backward, given the monthly rate is unknown and the APR is 26.82% compounded over 12 months, you would take the 12th root of 1.2682 and subtract 1. This formula is used in reverse to find the monthly rate.

In practice, interest rates are often calculated daily and then charged monthly based on the number of days in the month. However, in exams, this is only done if specifically instructed to do so.

Regarding simple interest, the formula provided in learning books, 200 + (200 x 0.15 x 4), would be correct if there was only an initial deposit of 200. However, if there are additional deposits, this formula does not apply. The correct calculation for simple interest with multiple deposits is not a straightforward formula.

A project has an initial outflow of $12,000 followed by six equal annual cash inflows, commencing in one year's time. The payback period is exactly four years, and the cost of capital is 12% per year. To find the project's net present value, the correct answer is $333.

For an investment project with discounted cash flows, the required rate of return on investment is 10% per annum. The discounted payback period can be calculated based on the given discounted cash flows.

Multiple choice questions:

1. What is the project's net present value?
A. $333
B. $ -2,899
C. $ -3,778
D. $ -5926

Answer: A. $333

2. What is the discounted payback period of the investment project?
A. Less than 3.0 years
B. 3.0 years
C. Between 3.0 years and 4.0 years
D. More than 4.0 years

Answer: C. Between 3.0 years and 4.0 years


Normality refers to the number of hydrogen ions that break free from one liter of an acid in the presence of a base, or the number of hydroxide ions that break free from a base in the presence of an acid. This measurement can be more useful than molarity, which describes only the number of acidic or basic molecules per liter, as different acids and bases produce different numbers of ions. A 50 percent normal hydrochloric acid solution will always have the same number of ions as every other solution of the same normality.

To calculate the molar mass of hydrochloric acid, add the molar mass of hydrogen (1.007 g/mol) and chlorine (35.45 g/mol), resulting in a molar mass of 36.457 g/mol. The equivalent mass is calculated by dividing the molar mass of hydrochloric acid by the number of hydrogen ions released by each molecule, which is 36.457/1 or 36.457.

The equation EqM * N * L is used to calculate the number of grams of hydrochloric acid needed to make a solution, where EqM is the equivalent mass, N is the desired normality, and L is the desired volume of the solution in liters. For example, to make 1 liter of a 0.5 N solution, the equation would be 36.457 * 0.5 * 1, resulting in 18.2285 grams of pure hydrochloric acid.

However, strong acids like hydrochloric acid are never sold in their pure state. To calculate the volume of diluted acid needed, examine the container to determine its percent concentration and specific gravity. Hydrochloric acid is often 37 percent acid and 63 percent water, with a specific gravity of 1.19 gm per ml. The equation G/(C * SG) is used, where G is the necessary grams of hydrochloric acid, C is the percent concentration, and SG is the specific gravity. For example, 18.2285 / (0.37 * 1.19) = 41.4 ml.

To prepare the solution, fill a beaker with water halfway to the desired volume, add the calculated amount of diluted acid while stirring constantly, and top off the solution with water until the desired volume is reached. The necessary materials include a periodic table of elements, a beaker, water, and a stirring rod.

The key steps to prepare a hydrochloric acid solution are:
1. Calculate the equivalent mass of hydrochloric acid.
2. Calculate the number of grams of hydrochloric acid needed using the equation EqM * N * L.
3. Calculate the volume of diluted acid needed using the equation G/(C * SG).
4. Prepare the solution by adding the calculated amount of diluted acid to water and stirring constantly.

Note: The original text included a list of things you'll need, references, and information about the author, which have been removed as they are not essential to the main topic.


## Maths in a minute: Counting numbers

The question of whether there are more irrational numbers than rational numbers, or vice versa, is meaningless because both sets are infinite. However, the set of rational numbers is infinite in a different way than the set of irrational numbers. Rational numbers can be lined up and labeled, forming a *countable infinity*. In contrast, irrational numbers form an uncountably infinite set.

In 1873, Georg Cantor developed a proof of this fact. He started by combining rational and irrational numbers to form the set of real numbers. If irrational numbers were countable, then real numbers would also be countable. Cantor assumed that real numbers are countable and can be listed. He then created a new number by taking the first digit after the decimal point of the first number, the second digit after the decimal point of the second number, and so on.

By changing each digit of this new number, Cantor created a new number that differs from every number on the list. This led to a contradiction, as the new number cannot appear on the list. The only resolution is to admit that the assumption of countable real numbers is false, implying that irrational numbers are uncountable.

An uncountable infinity is "bigger" than a countable one, as it can form a continuum like the number line. Cantor defined a hierarchy of infinities, with countable infinity at the bottom. Initially, his ideas faced opposition from colleagues like Henri Poincaré and Leopold Kronecker, who criticized his work. Cantor struggled with mental health problems, partly due to the rejection of his work. However, his ideas now form a central pillar of mathematics, with many results featured in standard textbooks.

Key concepts include:

- Countable infinity: a set of numbers that can be lined up and labeled, such as rational numbers.
- Uncountable infinity: a set of numbers that cannot be lined up and labeled, such as irrational numbers.
- Real numbers: the combination of rational and irrational numbers.
- Continuum: a continuous set of numbers, like the number line, that can be formed by an uncountable infinity.

Cantor's work has had a lasting impact on mathematics, and his ideas continue to be studied and built upon today.


We know that a die has only 6 sides, but the rand function generates random numbers up to 32767. The rand() function generates random numbers that can be any integer value. To generate a pseudo-random number in a given range, we use the formula: number = (rand() % (upper - lower + 1)) + lower.

For example, to generate a random number between 1 and 6, we can use the formula: number = (rand() % 6) + 1. This will give us a random number between 1 and 6.

To generate random numbers from 0 to 1, we can use the rand() function from the stdlib.h library. The maximum value is library-dependent, but is guaranteed to be at least 32767 on any standard library implementation.

The following example shows how you can generate random numbers from 0 to 1 (excluding 0 and 1). The rand() function returns a pseudo-random number between 0 and RAND_MAX.

We have used a scaling factor to achieve our goal. For displaying numbers from 1 to 6, we have simply added 1 to each output.

The Next method returns a random number, NextBytes returns an array of bytes filled with random numbers, and NextDouble returns a random floating point number.

C program to generate pseudo-random numbers using rand and random function. The following example shows how to generate 5 random numbers in a given range.

Output contains 5 random numbers in the given range.

Through out this page, we're limited to pseudo-random numbers. The header file stdlib.h has the function rand(), which gives you (pseudo-)random integers from 0 to RAND_MAX.

For example, you rolled '4' ones. You rolled '3' twos, etc.

The best way to write a random-number generator is not to ask the user to type a seed, but rather to fetch a seed from the operating system.

So, the code below uses rdrand via intrinsics, which is known to be a full 64bit random number as a source of randomness.

This way, at least, you have sufficient bits to generate a double without further ado.

The random number is generated by using an algorithm that gives a series of non-related numbers whenever this function is called.

Again, the above line of code generates integers in the range of 0 to 5.

To generate random numbers within a specific range, we have a formula that returns a random number between given ranges.

We can generate a pseudo-random number in the range from 0.0 to 32,767 using rand() function from library.

C program: Now, let’s try to write down one program to print 5 random numbers in a range.

The rand() function generates random numbers up to 32767. The maximum value is library-dependent, but is guaranteed to be at least 32767 on any standard library implementation.

Similarly, we can find out a random number in a given range using the same formula.

For example, to find a random number between 0 to 9, it will be: (rand % (9-0 + 1)) + 0.

The Random class has three public methods – Next, NextBytes, and NextDouble.

Function rand() returns a pseudo-random number between 0 and RAND_MAX.

You are just 1 assembly instruction away from CPU random numbers on modern x86_64 machines.

And all that, while you are just 1 assembly instruction away from CPU random numbers that can be any integer value.


To design a 2-way spanning slab, we need to determine the ultimate load, dead load, and ratio of ly/lx.

Given a part floor plan for a residential building with a 150mm thick slab cast insitu with edge beams, we have the following design data:

- Characteristic live load, Qk = 1.5KN/m²
- Partitioning = 1.1KN/m²
- Finishing = 1.2KN/m²
- Concrete Grade (fcu) = 20KN/m²
- Characteristic strength of steel (mild), fy = 250KN/m²
- Concrete Cover, c = 20mm

We start by determining the loading:

- Concrete own weight = 3.6KN/m²
- Partitioning = 1.1KN/m²
- Finishing = 1.2KN/m²
- Total Dead Load, Gk = 5.9KN/m²

The Ultimate Load, F, is calculated as:
F = 1.4Gk + 1.6Qk
F = (1.4 x 5.9) + (1.6 x 1.5)
F = 10.66KN/m²

Next, we determine the ratio ly/lx by dividing the longest span of the slab by the shorter span:
ly = 6519, lx = 4664
ly/lx = 6519/4664 = 1.4

The effective depth is calculated as:
d = thickness - concrete cover - 1/2 bar size
d = 150 - 20 - 10/2 = 125mm

We then determine the moment for the short and long spans:
- Short Span: Mid span = 12.75KNm, Continuous Edge = 17.16 KNm
- Long Span: Mid span = 7.9KNm, Continuous Edge = 10.44 KNm

Using the moments, we design the slab:
- Short Span:
  - Mid-Span: As = 452mm², provide Y12 @ 200mm c/c Btm (566mm²)
  - Continuous Edge: As = 615mm², provide Y12 @ 150mm c/c btm (754mm²)
- Long Span:
  - Mid-span: As = 280mm², provide Y10 @ 200mm c/c top (393mm²)
  - Continuous Edge: As = 370mm², provide Y12 @ 225mm c/c top (502mm²)

Finally, we check for deflection:
Fs = 2AsReq/3As Provided
Fs = (2 x 250 x 452)/(3 x 566) = 133.1, deflection is OK

The design parameters and calculations are based on the BS8110 code. The area of steel required can be found using tables from the BS Code or textbooks such as Victor Oyenuga's.


Proving that a choice was made in good faith can be challenging, especially when it comes to random selections or cryptographic designs. For instance, if a company chooses a cohort of people for random drug testing, it may be difficult to convince those selected that they were not chosen deliberately. To demonstrate good faith, it is essential to provide evidence that the choice was made based on a different criteria than the one in question.

There are several examples that illustrate this concept. The Cliff random number generator, for instance, requires a seed between 0 and 1 to start. Choosing a seed like 0.8121086949937715 may seem arbitrary, but it can be shown that this number was chosen to make the generator produce duplicate output. In contrast, using π − 3 as a seed is a recognizable and deliberate choice, but it has no apparent connection to the random number generator.

The SHA-2 cryptographic hash function is another example. The designers chose the square roots of the first consecutive primes to fill one set of constants and the cube roots of the first consecutive primes to fill another. This choice is deliberate but apparently not designed by any cryptographic criteria. The initial state is not random, but it can be shown that the choice was made based on a natural and innocent criteria.

Daniel Bernstein's elliptic curve Curve25519 is widely trusted due to its transparent design choices. The curve is defined over a finite field with 2^255 − 19 elements, and the coefficient 486662 was chosen based on explicit design criteria. Bernstein explains in his paper that he took the three smallest possible values of this parameter and rejected two of them on objective grounds.

In contrast, the design of elliptic curve NIST P-384 is not as transparent, leading to speculation that NIST may have designed the curve to have a back door. The choice of field size was motivated by efficiency, but the last coefficient b is a 383-bit number that has raised some eyebrows. NIST claims that the value was chosen at random using a key stretching algorithm, but researchers are divided over whether they believe this.

In conclusion, sometimes it is impossible to prove that a choice was not deliberate. In such cases, the best approach is to show that the choice was deliberate but based on an innocent criteria unrelated to the matter at hand. This can be achieved by using natural and recognizable choices, such as π or small primes, and making design criteria transparent. Ultimately, acting in good faith is not enough; it is essential to make a persuasive case that the choice was made in good faith.

These concepts are related to "nothing-up-my-sleeve numbers," which are context-dependent and can be innocent in some situations but not others. Numbers like π and small primes are often used as examples of nothing-up-my-sleeve numbers, but their innocence depends on the specific context in which they are used.


Lemma 10.50.4 states that for a valuation ring $A$ with maximal ideal $\mathfrak m$ and fraction field $K$, if $x \in K$, then either $x \in A$ or $x^{-1} \in A$ or both.

The proof begins by assuming $x$ is not in $A$. Let $A'$ be the subring of $K$ generated by $A$ and $x$. Since $A$ is a valuation ring, there is no prime of $A'$ lying over $\mathfrak m$. Given that $\mathfrak m$ is maximal, $V(\mathfrak m A') = \emptyset$. By Lemma 10.17.2, $\mathfrak m A' = A'$. This leads to the expression $1 = \sum _{i = 0}^ d t_ i x^ i$ with $t_ i \in \mathfrak m$. Rearranging, we get $(1 - t_0) (x^{-1})^ d - \sum t_ i (x^{-1})^{d - i} = 0$. This implies $x^{-1}$ is integral over $A$, and by Lemma 10.50.3, $x^{-1} \in A$.

No multiple choice questions or answers are provided in the text. The comments section, which includes discussions among individuals such as Rankeya, Johan, Dominic Wynter, Mingchen, Aise Johan de Jong, and Laurent Moret-Bailly, does not contain relevant information to refine.


NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4

Class 10 Maths Exercise 8.4 Solutions are provided in English and Hindi Medium. The solutions include detailed explanations and step-by-step reasoning for each question.

Important Questions for Proof:
1. Prove that cosec2[(B + C)/2] – tan2(A/2) = 1, where A, B, C are the angles of triangle ABC.
2. Prove that 2 (sin^6 A + cos^6 A) – 3 (sin^4 A + cos^4 A) + 1 = 0.
3. Prove that (1 + cot A + tan A) (sin A – cos A) = sin A tan A – cot A cos A.
4. If sin θ + cos θ = m and sec θ + cosec θ = n, then show that n(m² – 1) = 2m.

Questions from Board Papers:
1. If sin (A – B) = 1/2 and cos(A + B) = 1/2, where 0° < A + B ≤ 90°, find the values of A and B.
2. If x = sin²θ and y = cos²θ, where x and y represent honesty and hard work, (a) what will be the result after joining honesty and hard work, and (b) which mathematical concept has been used here?

Exercise 8.4 has 4 examples and 5 questions. Examples 12, 13, 14, and 15, and Questions 1, 2, 4, and 5 are the most important and have a higher chance of being asked in board exams.

To complete Exercise 8.4, students need a maximum of 23 days, assuming they study for at least 3 hours per day. However, the time required may vary depending on individual student's speed, efficiency, and capability. The difficulty level of Exercise 8.4 also varies from student to student, but many find Question 5 to be the most challenging as it requires higher-order thinking skills.

Multiple Choice Question:
What is the result after joining honesty and hard work?
A) 0
B) 1
C) 2
D) None of the above

Answer: B) 1

Which mathematical concept has been used here?
A) Algebra
B) Geometry
C) Trigonometry
D) None of the above

Answer: C) Trigonometry


An enlargement is a type of transformation that resizes a shape, making it larger or smaller. To understand enlargement, consider a real example: a shape is enlarged with a center of enlargement at the origin, resulting in an image that is a scaled version of the original object.

The properties of enlargement include:
- Each side of the object is scaled by a scale factor, where a scale factor of 2 means each side of the image is 2 times larger than the sides of the object.
- The length of sides remains in the same proportion to each other.
- Each line in the image is parallel to the corresponding line in the object.
- All angles remain the same in the object and the image.

Enlarged shapes are similar shapes, meaning they are the same shape but not the same size. Similar shapes have the same angles and proportional sides. In contrast, congruent shapes are the same shape and the same size, which occurs when a shape has only been translated, reflected, or rotated, or a combination of these transformations.


0 (zero) or nought is both a number and a numeral. It was the last numeral to be created in most numerical systems, as it is not a counting number and was in many eras and places represented only by a gap or mark very different from the other numerals.

0 as a number
0 is the integer that precedes the positive 1, and all positive integers, and follows 1, and all negative integers. Zero is a number which means nothing, null, void or an absence of value. For example, if the number of one's brothers is zero, then that person has no brothers.

0 as a numeral
The modern numeral 0 is normally written as a circle or (rounded) rectangle. On the seven-segment displays of calculators, watches, etc., 0 is usually written with six line segments. It is essential to distinguish the number zero from the numeral or digit zero, used in numeral systems where the position of a digit signifies its value.

History
The word zero comes from Arabic sifr, meaning "empty". The Indus Valley Civilization peoples demonstrate the earliest known physical use of decimal fractions in an ancient weight system. The concept of zero as a number was developed in India in the 7th century. The Indian mathematician Brahmagupta treated zero as a number and discussed operations involving this number.

In mathematics
Zero (0) is both a number and a numeral. The natural number following zero is one, and no natural number precedes zero. Zero may or may not be counted as a natural number, depending on the definition of natural numbers. Zero is neither prime nor composite. The following are some basic rules for dealing with the number zero:
- Addition: x + 0 = x and 0 + x = x.
- Subtraction: x − 0 = x and 0 − x = −x.
- Multiplication: x · 0 = 0 · x = 0.
- Division: 0 / x = 0, for nonzero x. But x / 0 is undefined.
- Exponentiation: x^0 = 1, except that the case x = 0 may be left undefined in some contexts.

Extended use of zero in mathematics
Zero is the identity element in an additive group or the additive identity of a ring. A zero of a function is a point in the domain of the function whose image under the function is zero. In geometry, the dimension of a point is 0. In analytic geometry, 0 is the origin.

In physics
The value zero plays a special role for a large number of physical quantities. For some quantities, the zero level is naturally distinguished from all other levels, whereas for others it is more or less arbitrarily chosen. For example, on the Kelvin temperature scale, zero is the coldest possible temperature.

In computer science
Numbering from 1 or 0? Human beings usually number things starting from one, not zero. Yet in computer science, zero has become the popular indication for a starting point. For example, in almost all old programming languages, an array starts from 1 by default, which is natural for humans. As programming languages have developed, it has become more common that an array starts from zero by default (zero-based).

Null pointer
A null pointer is a pointer in a computer program that does not point to an object. In C, it usually contains the memory address zero. However, it is not required to be zero. Some computer architectures use bit patterns other than zero as their null pointer.

Null value
In databases, a field can have a null value. This is equivalent to the field not having a value. For numeric fields, it is not the value zero. For text fields, this is not blank nor the empty string. The presence of null values leads to three-valued logic. No longer is a condition either true or false, but it can be undetermined.

Distinguishing zero from O
A comparison of the letter O and the number 0. The oval-shaped zero and circular letter O together came into use on modern character displays. The zero with a dot in the center seems to have originated as an option on IBM 3270 controllers. The slashed zero, looking identical to the letter O other than the slash, is used in old-style ASCII graphic sets descended from the default typewheel on the venerable ASR33 Teletype.


## Understanding Perfect Squares and Square Roots Continued

Lesson 2 of 19

### Objective:
Employ visual representations to demonstrate conceptual understanding of squaring a number and taking its square root.

### Big Idea:
Explore why some numbers are perfect squares and others are not, and how this translates into square roots.

#### Bellringer – Lesson Opener (15 minutes)
Students will use square tiles to represent numerical expressions: **4^2**, **6^2**, and **√16**. The teacher will take pictures of student work, ask for explanations, and encourage evaluation of the expressions. The goal is for students to understand that a 4x4 square represents both "four squared" and "the square root of 16".

#### Whole Class Discussion
The class will discuss exponentiation and radicals using squaring and square roots as examples. The use of tiles will help ground the discussion, focusing on the area and side lengths of squares, as well as the properties of numbers.

#### Additional Expressions
The teacher will introduce three new expressions: **3^2**, **√9**, and **√12**. Students will try to represent these using the tiles, and the teacher will observe their strategies, particularly for **√12**. The class will debrief, focusing on the challenges of representing **√12** and the concept of perfect squares.

#### Perfect Square Numbers
The teacher may create a list of perfect square numbers on the board, discussing the gaps between them. Students will try to create a perfect square using five tiles, leading to a conversation about estimating the square root of 5.

#### Resources and Lesson Plan
The lesson will continue with a discussion of the learning goals and the relationship between squares and square roots. Students will work in cooperative groups to complete questions 11-12, using square tiles to try out ideas. The teacher will assess student understanding and assign groups to present their work during the Mini Wrap-up.

#### Similar Lessons
This lesson is part of a 19-lesson unit on exponents and radicals, including topics such as perfect cubes, laws of exponents, scientific notation, and unit assessments.


## OR: What will happen to the value of acceleration due to gravity if the earth stops rotating about its axis?

The value of acceleration due to gravity, considering the earth's rotation, is given by the equation:
\[g’=g-R\omega^2\cos^2\theta\]
where $\theta$ is the latitude of a given place.

If the earth suddenly stops rotating, $\omega=0$, and the equation becomes:
\[g’=g-0=g\]
This means the value of $g$ should increase after the earth stops rotating, as the subtractive term is eliminated. However, this effect will not be observed at the poles, where $\theta=90°$. At the poles, the value of $g’$ remains unchanged, even when the earth is rotating, since:
\[g’=g-R\omega^2\cos^290°=g\]


Monte Carlo Simulation

To understand the Monte Carlo method, let's consider a simpler example. A refrigerator dealer wants to reduce his investment in stock using a scientific method. The daily demand for refrigerators is random and varies from day to day. From past sales records, the dealer has established a probability distribution of demand:

| Daily Demand (units) | Probability |
| --- | --- |
| 2 | 0.06 |
| 3 | 0.14 |
| 4 | 0.18 |
| 5 | 0.17 |
| 6 | 0.16 |
| 7 | 0.12 |
| 8 | 0.08 |
| 9 | 0.06 |
| 10 | 0.03 |

The dealer knows the lead time is fixed at 5 days. He wants to study the implications of ordering 30 units whenever the inventory at the end of the day is 20 units. The initial inventory is 30 units, and the simulation will run for 25 days.

To conduct the simulation, we use random numbers to simulate actual demand. We assign two-digit random numbers to each demand level based on cumulative probabilities:

| Daily Demand (units) | Probability | Cumulative Probability | Random Numbers Allotted |
| --- | --- | --- | --- |
| 2 | 0.06 | 0.06 | 00-05 |
| 3 | 0.14 | 0.20 | 06-19 |
| 4 | 0.18 | 0.38 | 20-37 |
| 5 | 0.17 | 0.55 | 38-54 |
| 6 | 0.16 | 0.71 | 55-70 |
| 7 | 0.12 | 0.83 | 71-82 |
| 8 | 0.08 | 0.91 | 83-90 |
| 9 | 0.06 | 0.97 | 91-96 |
| 10 | 0.03 | 1.00 | 97-99 |

Using the provided random numbers, we conduct the simulation:

| Day | Random Number | Inventory at Beginning (units) | Daily Demand (units) | Inventory at End (units) | Lost Sales (units) | Stocks Received | Quantity Ordered |
| --- | --- | --- | --- | --- | --- | --- | --- |
| 1 | 03 | 30 | 2 | 28 | 0 | 0 | 0 |
| 2 | 38 | 28 | 5 | 23 | 0 | 0 | 0 |
| 3 | 17 | 23 | 3 | 20 | 0 | 30 | 30 |
| 4 | 32 | 20 | 4 | 16 | 0 | 0 | 0 |
| 5 | 69 | 16 | 6 | 10 | 0 | 0 | 0 |
| 6 | 24 | 10 | 4 | 6 | 0 | 0 | 0 |
| 7 | 61 | 6 | 6 | 0 | 4 | 30 | 30 |
| 8 | 30 | 0 | 4 | 0 | 4 | 0 | 30 |
| 9 | 03 | 30 | 2 | 28 | 0 | 0 | 0 |
| 10 | 48 | 28 | 5 | 23 | 0 | 0 | 0 |
| 11 | 88 | 23 | 8 | 15 | 0 | 30 | 30 |
| 12 | 71 | 15 | 7 | 8 | 0 | 0 | 0 |
| 13 | 27 | 8 | 4 | 4 | 0 | 0 | 0 |
| 14 | 80 | 4 | 7 | 0 | 3 | 0 | 30 |
| 15 | 33 | 0 | 4 | 0 | 4 | 0 | 30 |
| 16 | 90 | 0 | 8 | 0 | 8 | 30 | 30 |
| 17 | 78 | 30 | 7 | 23 | 0 | 0 | 0 |
| 18 | 55 | 23 | 6 | 17 | 0 | 30 | 30 |
| 19 | 87 | 17 | 8 | 9 | 0 | 0 | 0 |
| 20 | 16 | 9 | 3 | 6 | 0 | 0 | 0 |
| 21 | 34 | 6 | 4 | 2 | 0 | 0 | 0 |
| 22 | 45 | 2 | 5 | 0 | 3 | 0 | 30 |
| 23 | 59 | 0 | 6 | 0 | 6 | 30 | 30 |
| 24 | 20 | 30 | 4 | 26 | 0 | 0 | 0 |
| 25 | 59 | 26 | 6 | 20 | 0 | 30 | 30 |

To evaluate the financial implications of the inventory policy, we need to gather details on ordering costs, carrying costs, and storage costs. We can then determine the total cost and compare alternative policies. Without assigning costs, we can observe that the policy of ordering 30 units whenever stock falls to 20 units is not desirable due to the number of lost sales units over 25 days.


## Description
The numeric computation environment in Maple is based on the IEEE Standard 754 for Binary Floating-Point Arithmetic and its generalization to base 10 machines, known as IEEE/854. These standards form the basis for the arbitrary precision computing environment in Maple, creating a cohesive whole with the hardware and software numeric computing environments.

The Maple hardware computation environment is a simulated environment, distinct from the physical hardware computation environment. Floating-point systems are specified by four parameters: Fl(B, p, L, U), where:
- B is the base of the number system
- p is the precision, specifying the maximum number of base-B digits in the mantissa
- L is the largest negative exponent of a representable number
- U is the largest positive exponent of a representable number

Elements of these systems are specified by three parameters: (s, m, e), where:
- s is the sign of the floating-point number
- m is the unsigned mantissa
- e is the exponent

The number represented is calculated as s * m * B^e. Certain values are reserved for special values, such as representing non-numeric objects.

Maple's software floating-point environment consists of a collection of virtual machines with base 10, distinguished by the precision parameter p. The smallest value of p is 1, and the largest value depends on the underlying machine architecture. The largest representable software float is considered to be (1, 9, U), regardless of the Digits setting.

Virtual machines coexist according to the rule that a software float belongs to no machine until used in a computation requiring a specific machine. Entering a software float into a machine may require rounding the mantissa to the machine's precision.

## References
[1] "IEEE Standard 754-1985 for Binary Floating-Point Arithmetic." Reprinted in ACM SIGPLAN Notices, Vol. 22 No. 2. (1987): 9-25.
[2] "A Proposed Radix- and Word-Length-Independent Standard for Floating-Point Arithmetic." Reprinted in IEEE Micro, (August 1984): 86-100.
[3] Goldberg, D. "What Every Computer Scientist Should Know About Floating-Point Arithmetic." ACM Comp. Surv. Vol. 23. (1991): 5-48.
[4] "Programming Languages - C." ISO/IEC 9899 "C9X" (esp. Annexes "F" and "G").
[5] "Information technology - Language independent arithmetic." ISO/IEC 10967-1 (1994).


## Goal

A farmer wants to create a square plot of land by setting up a square fence, placing 4 pegs at distinct corner positions. The pegs can only be placed at given positions with suitable ground softness.

Given the possible integer positions for the pegs, the farmer needs to determine the number of ways to pick 4 distinct positions to form a square. Note that the squares do not need to be axis-aligned.

## Input

* An integer `N` (4 ≤ `N` ≤ 2000) denoting the number of valid positions for pegs.
* `N` lines with two space-separated integers `X` and `Y` (0 ≤ `X`, `Y` ≤ `N`-1) representing acceptable coordinates for a peg in lexicographical order and without duplicates.

## Output

A single integer representing the number of squares that can be formed using the given vertices (pegs).

## Example

Input:
5
0 0
0 2
1 0
2 0
2 2

Output: The number of squares that can be formed.

Constraints:
4 ≤ `N` ≤ 2000
0 ≤ `X`, `Y` ≤ `N`-1


## Analyzing Station Connections using Eigenanalysis

In this section, we extend the eigenanalysis method to involve more objects, using the Berlin S-Bahn station graph as an example. Each node in the graph represents a train station, and people can move around the stations with equal probability. The connection between stations defines how people move around, and the probability of staying at each station depends on the train station topology.

We can generate a matrix representing people's movement from the station adjacency matrix. If a station is connected to two stations, the probability of each connection is 1/2, and if connected to three stations, the probability is 1/3. This is achieved by normalizing the column vector of the adjacency matrix using the L1 norm, as it represents a probability.

The adjacency matrix for the S-Bahn station graph is:

\[
\left[
\begin{array}{cccc}
1 & 1 & 0 & 0 \\
1 & 1 & 1 & 1 \\
0 & 1 & 1 & 0 \\
0 & 1 & 0 & 1 \\
\end{array}
\right]
\]

Normalizing this matrix, we get the Markov Matrix:

\[
\left[
\begin{array}{cccc}
0.5 & 0.25 & 0 & 0 \\
0.5 & 0.25 & 0.5 & 0.5 \\
0 & 0.25 & 0.5 & 0 \\
0 & 0.25 & 0 & 0.5 \\
\end{array}
\right]
\]

Using Octave, we can compute the staying possibility at each station:

```octave
Mb = [0.5 0.25 0 0; 0.5 0.25 0.5 0.5; 0 0.25 0.5 0; 0 0.25 0 0.5];
[L D] = eig(Mb)
x1 = L(:,3)/ sum(L(:,3))
```

After 1000 steps, the people's staying probability at each station is 0.2, 0.4, 0.2, 0.2. This distribution is the same regardless of the initial distribution, and it reflects the topology of the stations. The second station, Alexanderplatz, has twice the number of people as the other stations, making it a hub station.

The relationship between station connections and authors' importance can be represented using graph theory. Both can be seen as finding the important node in a graph, whether it's a station or an author. The mathematical representation is the same, but human interpretation differs. This abstraction allows us to apply mathematical concepts to various problems, but it's essential to balance abstraction with the substance of the problem.

Mathematics is a tool that can be used to solve many problems, but it's crucial to remember that solving problems using mathematics is a human activity. The quality of mathematics is correlated with the quality of the person doing the mathematics, and mathematics and humanity are closely related.


Exponential functions can model the rate of change of many situations, including population growth, radioactive decay, bacterial growth, compound interest, and more. To write an exponential equation given a rate and an initial value, follow these steps:

**Method 1: Using the Rate as the Base**

1. The basic form of an exponential equation is f(t) = P₀(1 + r)^(t/h), where P₀ is the initial value, t is the time variable, r is the rate, and h is the number needed to ensure the units of t match up with the rate.
2. Plug in the initial value for P₀ and the rate for r. For example, if a bank account starts with a $1,000 deposit and the interest rate is 3% compounded annually, the equation becomes f(t) = 1,000(1.03)^(t/h).
3. Find h by considering the units of t and the rate. If the units are the same, h is always 1. In the example, since the interest is compounded annually, h = 1, and the equation becomes f(t) = 1,000(1.03)^t. However, if the time is given in months, h would be 12, and the equation would be f(t) = 1,000(1.03)^(t/12).

**Method 2: Using "e" as the Base**

1. The basic form of an exponential equation using "e" as the base is f(t) = ae^(kt), where a is the initial value, e is the base, k is the continuous growth rate, and t is the time variable.
2. Plug in the initial value. For example, if a 500-gram sample of an isotope of Carbon has a half-life of 50 years, the equation becomes f(t) = 500e^(kt).
3. Find the continuous growth rate (k) by using a point on the graph. In the example, since the sample decays to 250 grams in 50 years, we can plug in t = 50 and f(50) = 250 to get 250 = 500e^(50k).
4. Solve for k by dividing both sides by 500, taking the natural logarithm of both sides, and using the properties of logarithms: ln(1/2) = 50k, and k = (ln(1/2))/50 ≈ -0.01386.
5. Plug in the k value to get the final equation: f(t) = 500e^(-0.01386t).

**Example Problem**

The median price P of a home rose from $100,000 in 1990 to $350,000 in 2010. Let t be the number of years since 1990. Find an equation of the form P = P₀a^t.

**Tips**

* Store your k value in your calculator to calculate values more exactly.
* Use the first method for most problems, but consider using the natural base when it makes calculations easier.
* Be careful when using variables and constants when graphing equations.


# Super M

## Description

Ari, the monster, is the hidden identity of Super M, the Byteforces' superhero. Byteforces is a country with *n* cities, connected by *n* - 1 bidirectional roads, allowing travel between any two cities. There are *m* cities under attack by humans, and Super M must visit each city to scare them away. She can only travel using the given roads, with each road taking exactly one kron (the time unit used in Byteforces). Super M is currently attending a training camp in Codeforces, but can instantly teleport to any city in Byteforces once.

## Objective

Calculate the city where Super M should teleport to minimize her total time and provide her with this minimum time. The correct answer is always unique, and if there are multiple optimal answers, the city with the lowest number should be chosen.

## Input

* The first line contains two integers *n* and *m* (1 ≤ *m* ≤ *n* ≤ 123456), representing the number of cities in Byteforces and the number of cities under attack, respectively.
* The next *n* - 1 lines describe the road system, with each line containing two city numbers *u*_{i} and *v*_{i} (1 ≤ *u*_{i}, *v*_{i} ≤ *n*) representing the ends of road *i*.
* The last line contains *m* distinct integers, which are the numbers of cities being attacked, in no particular order.

## Output

* The number of the city Super M should teleport to, choosing the city with the lowest number if there are multiple optimal answers.
* The minimum possible time needed to scare all humans in cities being attacked, measured in krons.

## Samples

```
7 2
1 2
1 3
1 4
3 5
3 6
3 7
2 7
```

```
2
3
```

```
6 4
1 2
2 3
2 4
4 5
4 6
2 4 5 6
```

```
2
4
```


Evaluate the following expressions:

1. C(3,1)
2. C(4,2)
3. C(5,4)
4. C(3,2) - C(5,3)
5. 2C(6,3) + C(4,2)

Answers:
a) C(3,1)
b) C(4,2)
c) C(5,4)
d) C(3,2) - C(5,3)


## Step 1: Identify the Interest Rate
To calculate the interest earned on your savings account, start by identifying the current interest rate paid by your financial institution. This rate can be found on your savings account statement or the bank's website. Typical savings account rates are low, around 1.5 percent per year, due to the low-risk nature of FDIC-insured accounts. Convert the interest rate from a percentage to a decimal for easier calculation: 1.5 percent = 0.015.

## Step 2: Determine the Savings Amount and Calculate Annual Interest
Next, determine the amount in your savings account. For example, assume you have $10,000. With the interest rate of 1.5 percent (or 0.015 in decimal form), calculate the annual interest: $10,000 * 0.015 = $150 in interest earned per year.

## Step 3: Calculate Monthly, Weekly, and Daily Interest
To find out how much interest is earned in shorter periods, divide the annual interest by the number of periods in a year.
- Monthly interest: $150 / 12 = $12.50 per month.
- Weekly interest: $150 / 52 = $2.88 per week.
- Daily interest: $150 / 365 = $0.41 per day.

### Tip
The low interest rates on savings accounts offer minimal compensation for long-term investment. Consider higher-return investment options like certificates of deposit, bonds, or equities to stay ahead of inflation.

### Things You'll Need
- Savings account statement
- Calculator

### Warning
These calculations do not account for compounding interest over time and include a small rounding error, but they provide a general idea of the returns on your savings account.


To prove that $2220^{2n+1}+2003^{2n+1}$ is divisible by 4005, where $n$ belongs to the set of natural numbers, let's analyze the given statement.

Let $P(n)$ be the statement: $P(n) = 2220^{2n+1}+2003^{2n+1}$.

To verify if $P(n)$ is divisible by 4005, we start by checking $P(1)$:

$P(1) = 2220^{2(1)+1}+2003^{2(1)+1}$
$P(1) = 2220^{3}+2003^{3}$
$P(1) = 10941048000 + 8036054027 = 18977102027$.

However, $18977102027$ is not completely divisible by 4005, indicating an issue with the initial query. It appears the original statement might be incorrect. A reevaluation is necessary.


Five years ago, the average age of P and Q was 25. This means that the sum of their ages five years ago was 25 * 2 = 50. Therefore, the sum of their current ages is 50 + 5 * 2 = 60.

The average age of P, Q, and R today is 25. This means that the sum of their current ages is 25 * 3 = 75. Since the sum of the current ages of P and Q is 60, R's current age is 75 - 60 = 15.

Age of R after 5 years will be 15 + 5 = 20.

1. The average age of P, Q, and R is 15 years more than R's age. Let's denote R's age as x. Then the average age of P, Q, and R is x + 15. We know that the total age of P and Q together is 39 years. So, the sum of the ages of P, Q, and R is 39 + x. The average age of P, Q, and R is (39 + x) / 3. Since the average age of P, Q, and R is x + 15, we have (39 + x) / 3 = x + 15. Solving for x, we get 39 + x = 3x + 45, so 2x = 39 - 45 + 3x - 3x, which simplifies to 2x = -6, and then x = -6 + 6, and finally x = 6 - 6 + 6, which results in x = 6.

2. To find the average weight of the whole class, we need to calculate the total weight of all students and divide it by the total number of students. The total weight of section A is 36 * 40 = 1440 kg. The total weight of section B is 44 * 35 = 1540 kg. The total weight of the whole class is 1440 + 1540 = 2980 kg. The total number of students is 36 + 44 = 80. The average weight of the whole class is 2980 / 80 = 37.25 kg.

3. The largest 3-digit number formed by 0, 2, and 4 is 420. The smallest 3-digit number formed by 0, 2, and 4 is 204. The average of these two numbers is (420 + 204) / 2 = 312.

4. In the first 10 overs, the total runs scored were 3.2 * 10 = 32. To reach a target of 282 runs, the remaining runs needed are 282 - 32 = 250. The remaining overs are 40. The required run rate is 250 / 40 = 6.25.

5. The average of two numbers is 8, so their sum is 8 * 2 = 16. The average of the other three numbers is 3, so their sum is 3 * 3 = 9. The sum of all five numbers is 16 + 9 = 25. The average of the five numbers is 25 / 5 = 5.

6. Three years ago, the average age of the husband, wife, and child was 27. The sum of their ages three years ago was 27 * 3 = 81. The current sum of their ages is 81 + 3 * 3 = 90. Five years ago, the average age of the wife and child was 20. The sum of their ages five years ago was 20 * 2 = 40. The current sum of their ages is 40 + 5 * 2 = 50. The husband's current age is 90 - 50 = 40.

7. At the time of their marriage, the average age of the husband and wife was 23. The sum of their ages at the time of marriage was 23 * 2 = 46. After five years, their sum of ages is 46 + 5 * 2 = 56. When the child was born, the sum of the ages of the husband and wife was 56 - 1 = 55 (since the child is one year old). The average age of the family of three when the child was born is (55 + 1) / 3 = 56 / 3 = 18.67.

8. The average marks of four subjects is 120. The sum of the marks is 120 * 4 = 480. If 33 was misread as 13, the correct sum is 480 - 13 + 33 = 500. The correct average is 500 / 4 = 125.

9. One-fourth of the journey is covered at 25 km/h. Let the total distance be x. Then the distance covered at 25 km/h is x / 4. The time taken to cover this distance is (x / 4) / 25 = x / 100. One-third of the journey is covered at 30 km/h. The distance covered at 30 km/h is x / 3. The time taken to cover this distance is (x / 3) / 30 = x / 90. The remaining distance is x - x / 4 - x / 3 = 5x / 12. The time taken to cover the remaining distance at 50 km/h is (5x / 12) / 50 = x / 120. The total time taken is x / 100 + x / 90 + x / 120. The average speed is x / (x / 100 + x / 90 + x / 120).

10. The arithmetic mean of 0, 5, 4, 3 is (0 + 5 + 4 + 3) / 4 = 12 / 4 = 3. The arithmetic mean of 1, 0, 1, 5, 4, 3 is (1 + 0 + 1 + 5 + 4 + 3) / 6 = 14 / 6 = 7 / 3. The arithmetic mean of 5, 4, 3 is (5 + 4 + 3) / 3 = 12 / 3 = 4. The relation between a, b, and c is a = 3, b = 7 / 3, and c = 4.


We've constructed Jaccard-like distance functions $d(m,n)$ on the set of all notes in our vault $V = \{ k,l,m,n,\dots \}$, where $d(m,n) \geq 0$ and the triangle inequality $d(k,l)+d(l,m) \geq d(k,m)$ holds for each triple of notes. By construction, $d(m,n)=d(n,m)$, but we can modify these distances to create a new generalized distance $d'(m,n)$, which is no longer symmetric but still satisfies the triangle inequality.

This new distance turns $V$ into a Lawvere space, and $V$ becomes an enriched category over the monoidal category $[0,\infty]=\mathbb{R}_+ \cup \{ \infty \}$. The enrichment is the map $V \times V \rightarrow [0,\infty]$ given by $(m,n) \mapsto d(m,n)$. To address the issue of missing colimits, we can upgrade our category to the presheaf category, where the writer can find inspiration to cobble together her texts.

An enriched presheaf is a map $p : V \rightarrow [0,\infty]$ such that for all notes $m,n \in V$, we have $d(m,n) + p(n) \geq p(m)$. Each note $n \in V$ determines its Yoneda presheaf $y_n : V \rightarrow [0,\infty]$ by $m \mapsto d(m,n)$. The set of all enriched presheaves $\widehat{V_d}$ has a lot of extra structure, including a poset structure with minimal element $0$ and maximal element $1$.

The poset $\widehat{V_d}$ is also a lattice with $p \wedge q(n) = max(p(n),q(n))$ and $p \vee q(n)=min(p(n),q(n))$. We can define a distance $\widehat{d}$ on $\widehat{V_d}$ as $\widehat{d}(p,q) = max(0,\underset{n \in V}{sup} (q(n)-p(n)))$, which satisfies the triangle inequality. The Yoneda map $y : V_d \rightarrow \widehat{V_d}$ is an isometry, and the enriched version of the Yoneda lemma states that for all $p \in \widehat{V_d}$, $p(n) = \widehat{d}(y_n,p)$.

For every subset of notes $X \subset V$, we can consider the presheaf $p_X : V \rightarrow [0,\infty]$ given by $m \mapsto \underset{n \in X}{inf}~d(m,n)$. The zero set $Z(p_X) = \{ m \in V~:~p_X(m)=0 \}$ can be thought of as the closure of $X$, and the collection of all such closed subsets defines a topology on $V$. In the special case when $V_d$ is a poset-category, this topology is the down-set (or up-set) topology.

When starting with the Lawvere-space $\widehat{V_d}$, the topology has as its closed sets all up-sets of the poset $\widehat{V_d}$. However, a good definition for the implication $p \Rightarrow q$ between two enriched presheaves $p,q \in \widehat{V_d}$ is still missing. One possible definition is $p \Rightarrow q : V \rightarrow [0,\infty]$ where $(p \Rightarrow q)(n) = \widehat{d}(y_n \wedge p,q)$, but it is unclear if this is a presheaf.

Further research is needed to resolve this issue, potentially involving Isbell duality and semi-tropical modules. The study of enriched category theory, as outlined in Kelly's Basic concepts of enriched category theory, may also provide valuable insights. Ultimately, the goal is to develop a comprehensive understanding of the topology and implication structure of $\widehat{V_d}$, which can inform and enhance our understanding of language and cognition.


To tackle multi-step word problems, encourage students to break down each problem into a story, telling it in their own words without numbers, either verbally or in writing. Then, have them model the story using a Singapore Model Method drawing, picture, table, diagram, or chart to organize the information. When solving, remind them to label each number, include units, and follow the story's actions and relationships. Always start with a full sentence answering the question, leaving a blank for the solution.

Here are some problems to get you started:

1. Ms. Delmore is planning a vacation to Australia, which she can pay for with a $1,395 deposit and 9 monthly payments of $485, or a one-time payment of $5,495. Which option costs more, and by how much? Why are two options offered, and which would you choose?

2. Linda's parents buy three times as many cookies as Hazel's. After Linda shares 50 cookies, she has half as many as Hazel. How many cookies does Hazel have?

3. Mr. Marshal spends $40,000 monthly on 46 salespeople, each receiving $820, and 4 janitors. How much does each janitor get paid?

4. A produce manager receives 27 cartons of apples, each containing 65 apples. After selling 184 apples on Monday and 218 on Tuesday, he sells an equal number daily for the rest of the week. If he throws away 13 rotten apples, how many did he sell over the weekend?

5. A man wins $100,000, spends $8,000 on furniture, and divides the rest into 8 portions. He gives 4 portions to his wife, 1 to his son, and shares the rest among 5 charities. How much more does his son get than each charity?

6. Charlie buys 17 dozen belts, sells 196 at $9 each, and then changes the price. If the total collected is $1,860, what's the price difference, and is it an increase or decrease? By what percent did the price change?

7. Charlie buys 14 boxes of tee shirts at $120 per box, with 2 dozen shirts in each. The store sells 149 shirts at $8 each and the rest at $12 each. Will Charlie receive a $25 bonus for every $500 profit, and if so, how much?

8. A farmer has twice as many ducks as chickens in January. After 19 ducks escape in February, he sells 413 ducks in March and has half as many ducks as chickens. How many chickens does the farmer have?

9. The Mathletes buy a dozen boxes of pens at $13 each, with 48 pens in each, to sell to teachers and raise $100 for a bus rental. They repack 7 boxes into packages of 12 pens for $5 each and sell the rest at 4 for $2. Will they cover the bus rental fee?

10. A tailor takes twice as long to make 3 dresses as 5 suits. It takes him 90 hours to make 9 dresses and 15 suits. He hires another tailor at $25 per hour to make 2 more suits. How much will he pay the tailor?


# Problem 03 - Symmetrical Parabolic Curve

**Problem**

A grade line *AB* with a slope of +5% intersects another grade line *BC* with a slope of –3% at *B*. The elevations of points *A*, *B*, and *C* are 95 m, 100 m, and 97 m, respectively. Determine the elevation of the summit of the 100 m parabolic vertical curve to connect the grade lines.

A. 98.32 m

B. 99.06 m

C. 97.32 m

D. 96.86 m

**Solution**

The horizontal distance between *A* and *B* is (100 - 95)/0.05 = 100 m, and between *B* and *C* is (100 - 97)/0.03 = 100 m.

Given the parabolic curve is placed at the middle-half, the elevation of point *PC* is calculated as:
$\text{Elev } PC = \text{Elev } A + 50(0.05) = 95 + 2.5 = 97.5 \, \text{ m}$.

The distance from *PC* to the summit is found using the equation $\dfrac{S_1}{0.05} = \dfrac{100}{0.08}$, resulting in $S_1 = 62.5 \, \text{ m}$.

The elevation of the summit is then calculated as:
$\text{Elevation of the summit} = \text{Elev } PC + \frac{1}{2}S_1(0.05) = 97.5 + \frac{1}{2}(62.5)(0.05) = 99.0625 \text{ m}$, which corresponds to answer B.


Manipulatives are physical teaching tools that engage students visually and physically, promoting active involvement in the learning process. These tools include coins, puzzles, blocks, and more, and are becoming increasingly popular as they allow students to make cognitive connections with concepts through hands-on exploration.

The pedagogical basis for using manipulatives remains the same across various tools and age groups, enabling teachers to effectively teach abstract mathematical concepts. Manipulatives serve as visual models, helping students understand complex math concepts by providing a concrete representation.

Five effective math manipulatives include:

### 1. Base Ten Blocks
Base Ten Blocks represent ones, tens, hundreds, and thousands, teaching place value and number concepts. They also explain measurement concepts like volume and area, featuring unit cubes, rods, and square blocks.

### 2. Fraction Tiles
Fraction tiles teach fractional equivalences, fractions, and the addition and subtraction of fractions. Proportionally sized tiles enable students to compare fractional values.

### 3. Geared Clocks
Geared clocks feature hidden gears, reflecting accurate hour and minute relationships. They help younger students calculate elapsed time and tell time on analog clocks.

### 4. Pattern Blocks
Pattern blocks are collections of shapes in different colors, including triangles, squares, and trapezoids. Students explore spatial relations, recognize similarities and differences, and enhance cognitive function and self-confidence.

### 5. Geoboards
Geoboards teach geometry, allowing students to create shapes by stretching rubber bands over pegs. They are effective for teaching perimeter, symmetry, and angles.

### Bonus Manipulative: Coins
Coins explore mathematical concepts like decimals, place value, addition, multiplication, percentage, and division. Teachers can use coins to explain concepts, such as sorting coins, counting spare change, and regrouping.

Tips for teachers include:

* Using manipulatives as powerful learning tools to build conceptual understanding
* Serving as a bridge between concrete and abstract levels of complicated mathematical topics
* Researching how to use each manipulative effectively

Using manipulatives offers a hands-on experience for children to grasp concepts. Teachers should carefully choose and sequence manipulatives to maximize effectiveness, as research evidence supports the positive impact of manipulatives on students' learning patterns over extended periods.


In an examination, D scored 10% more than C, C scored 40% more than B, B scored 30% more than A, and A scored 20% more than E. Given that E scored 200 marks out of 500, we can calculate the marks secured by D as follows:
Marks obtained by E = 200,
Marks obtained by A = 120% of 200 = 240,
Marks obtained by B = 130% of 240 = 312,
Marks obtained by C = 140% of 312 = 436.8,
Marks obtained by D = 110% of 436.8 = 480.48.
Therefore, D obtained 480.48 marks in total.

If the marks obtained by A is 25% more than B, and the marks obtained by C is 166 more than the difference of the marks obtained by A and B, and C obtained 180 marks, then we can find the marks obtained by A.

A girl spent 16% of her monthly income on clothing, and the remaining amount was spent on travel and food in the ratio of 5:2. Given that the difference of the income spent on travel and food is Rs.1836, we can find her monthly income.

In an assembly election, a candidate got 60% of the total valid votes, with 2% of the total votes declared invalid. If the total number of voters is 126000, then we can calculate the number of valid votes polled in favour of the candidate.

Rahul saves Rs. 7200 more if he spends 65% of his income instead of 85%. We can find the savings of Sumit, whose income is 25% more than that of Rahul and spends 85% of his income.

A company allowed 9% commission on the total sales made by the salesman and a bonus of 1% on the sales over Rs. 40000. If the salesman deposited Rs. 81400 in the company after deducting his earnings from the total sales made by him, we can find the total sales.

A scored 246 marks in an examination, and B got 50 percent marks, which is 48 marks less than A. If the minimum passing marks in the examination are equivalent to 25 percent, then we can find the minimum passing marks in the examination.

A student multiplied a number by 4/5 instead of 5/4. We can calculate the percentage error in the calculation.

Z spends 60% of his salary and saves the remaining. After his salary is increased by 30%, he saves 55% of the increased amount besides the amount he used to save earlier. We can find the percentage change in his monthly savings.

135 sweets were distributed equally among children, with each child receiving 60% of the total number of children. We can find the number of sweets each child received.

More Percentage Questions:
1. What is the percentage error in the calculation if a student multiplied a number by 4/5 instead of 5/4?
2. If Z spends 60% of his salary and saves the remaining, and after his salary is increased by 30%, he saves 55% of the increased amount besides the amount he used to save earlier, what will be the percentage change in his monthly savings?
3. If 135 sweets were distributed equally among children in such a way that the number of sweets received by each child is 60% of the total number of children, how many sweets did each child receive?


The sequence a(n) is defined as the nth iteration of the triangular number T(n), where T(n) = n*(n+1)/2. This means that a(n) = T(T(T(...(n)))).

The sequence grows like O(n^2^n*1/2^n), which can be derived from the growth O(n^2*1/2) of the triangle sum by iteration.

For example, a(3) = 231, which can be calculated by writing the 3-time iterated expression on T(3): T(T(T(3))) = 231.

The sequence can be generated using the following formulas and programs:

* Mathematica: Table[Nest[(#(#+1))/2&, n, n], {n, 8}]
* PARI: a(n) = my(k = n); for (j=1, n, k = k*(k+1)/2; ); k;

The sequence has been studied in the context of iterated triangular numbers and iterated polygonal numbers, with references to preprints by J. V. Post.

Key properties of the sequence include:

* It is an easy and non-negative sequence.
* It has been approved for inclusion in the database.

Overall, the sequence a(n) is a complex and interesting sequence that exhibits rapid growth due to the iteration of the triangular number function.


To find the number of six-digit numbers where no digit is repeated, even digits appear in even places, and odd digits appear in odd places, and the number is divisible by 4, we need to consider the rules for divisibility by 4. A number is divisible by 4 if the last two digits are divisible by 4.

Given that there are 5 odd digits (1, 3, 5, 7, 9) and 5 even digits (0, 2, 4, 6, 8), but since we are forming six-digit numbers, the placement of these digits matters. For a six-digit number, the first digit cannot be zero, but in this scenario, since we are separating the placement of odd and even digits, zero can be considered for the even places.

The total number of ways to arrange 5 odd digits in the 3 odd places is 5P3 (permutations of 5 items taken 3 at a time), and the total number of ways to arrange 5 even digits in the 3 even places is 5P3. However, we must ensure the number is divisible by 4, which depends on the last two digits.

The calculation provided is 5! = 120 such numbers, which seems to simplify the consideration for divisibility by 4 without detailing the specific permutations that satisfy the divisibility rule.

Rules for divisibility by other numbers are as follows:
- 2: All even numbers are divisible by 2.
- 3: The sum of the digits is divisible by 3.
- 4: The last two numbers are divisible by 4.
- 5: The number ends in a 0 or 5.
- 6: The sum of the digits is divisible by 3 and is even.
- 7: No easy method.
- 8: The last three numbers are divisible by 8.
- 9: The sum of the digits is divisible by 9.

For numbers divisible by 6, they must be even and their digits must total a multiple of 3. For example, 284 is not divisible by 6 because, although it is even, its digits total 14, which is not divisible by 3. On the other hand, 282 is divisible by 6 because it is even and its digits total 12, which is divisible by 3.

The sum of the digits in odd positions minus the sum of the digits in even positions is divisible by 11 for numbers divisible by 11.

If digits may not be repeated in a six-digit number, there are 6x5x4x3x2x1 = 720 possible numbers. If digits can be repeated, there are 6x6x6x6x6x6 = 46,656 possible numbers. If zero is among the available digits but is excluded from the first place, there are 5x6x6x6x6x6 = 38,880 possible numbers.


# Square

To calculate the area of the square shape of the isosceles triangle with arms 50m and base 60m, we first need to find the height of the triangle. Using the Pythagorean theorem, we can find the height. Then, we can calculate the area of the square inscribed in the triangle.

Given the area of one tile is 25 dm², we need to convert this to square meters to match the units of the triangle's dimensions. Since 1 meter = 10 decameters, 1 square meter = 100 square decameters. Thus, 25 dm² = 0.25 m².

However, the problem doesn't directly involve calculating the area of the square inscribed in the isosceles triangle but rather seems to ask for the number of tiles used to pave a square, with the area of the square not directly provided. To proceed, we need to clarify or assume the relationship between the isosceles triangle and the square to be paved.

If we consider the task as calculating the area of a square that can be inscribed within the isosceles triangle or related to it, we would follow these steps:

1. Calculate the height (h) of the isosceles triangle using the Pythagorean theorem, where the base (b) is 60m and the arms (a) are 50m. The formula is \(h = \sqrt{a^2 - \left(\frac{b}{2}\right)^2}\).

2. Calculate the height: \(h = \sqrt{50^2 - \left(\frac{60}{2}\right)^2} = \sqrt{2500 - 30^2} = \sqrt{2500 - 900} = \sqrt{1600} = 40m\).

3. The area of the isosceles triangle is \(\frac{1}{2} \times base \times height = \frac{1}{2} \times 60m \times 40m = 1200m^2\).

However, without a clear link between the triangle's dimensions and the square to be paved, we cannot directly calculate the number of tiles needed for the square based on the triangle's dimensions alone. If the intention is to find the side length of a square with an area equal to that of the triangle, we would find the area of the triangle and then find the side length of the square that has the same area.

Given the area of the triangle is 1200 m², for a square with the same area:
- The area \(A\) of a square is given by \(A = side^2\).
- Thus, \(1200 = side^2\), which means \(side = \sqrt{1200}\).

Calculating the side of the square: \(side = \sqrt{1200} \approx 34.64m\).

To find how many tiles are needed to pave this square, given one tile covers 0.25 m²:
- The total area of the square is 1200 m².
- Number of tiles = \(\frac{Total\ area\ of\ square}{Area\ of\ one\ tile} = \frac{1200}{0.25} = 4800\ tiles\).

Thus, approximately 4800 tiles would be needed to pave the square, assuming the square's area is equal to the area of the isosceles triangle described.

Note: The problem statement seems to have inconsistencies or missing links between the isosceles triangle's dimensions and the square to be paved, which might affect the accuracy of the solution provided.


The concept of Sets was introduced by G. Cantor, who defined a set as a collection of definite and distinguishable objects selected by means of description or certain rules. Set theory forms the basis of many fields, including counting theory, relations, graph theory, and finite state machines.

A set is an unordered collection of different elements, and it is typically represented using set brackets. The order of elements in a set does not matter, and the same set can be written in different ways without changing its meaning. For example, the set of vowels in the English alphabet can be written as A = {a, e, i, o, u} or A = {u, o, i, e, a}.

There are two main ways to represent sets: by listing all the elements or by describing the common property of all the elements. For instance, the set of odd numbers less than 10 can be written as B = {1, 3, 5, 7, 9} or B = {x: 1 ≤ x < 10 and (x % 2) ≠ 0}.

If an element x is a member of a set S, it is denoted by x ∈ S, and if an element y is not a member of set S, it is denoted by y ∉ S. For example, if S = {1, 1.2, 1.7, 2}, then 1 ∈ S, but 1.5 ∉ S.

Some common sets include:
- N: the set of all natural numbers = {1, 2, 3, 4, ...}
- Z: the set of all integers = {..., -3, -2, -1, 0, 1, 2, 3, ...}
- Z+: the set of all positive integers
- Q: the set of all rational numbers
- R: the set of all real numbers
- W: the set of all whole numbers

The number of elements in a set is known as its cardinality, denoted by |S| or n(S). If a set has an infinite number of elements, its cardinality is ∞. For example, the set {1, 4, 3, 5} has a cardinality of 4, while the set {1, 2, 3, 4, ...} has a cardinality of ∞.

There are different types of sets, including:
- Finite sets: sets with a definite number of elements
- Infinite sets: sets with an infinite number of elements
- Subset: a set whose elements are all members of another set
- Universal set: a set that contains all the elements of a particular context or application
- Empty set: a set with no elements, denoted by ∅
- Singleton set: a set with only one element

Two sets are equal if they have the same elements, and they are equivalent if they have the same cardinality. Sets can also be overlapping or disjoint, depending on whether they have any common elements.

Set operations include:
- Union: the set of elements that are in either set A or set B, denoted by A ∪ B
- Intersection: the set of elements that are in both set A and set B, denoted by A ∩ B
- Difference: the set of elements that are in set A but not in set B, denoted by A - B
- Complement: the set of elements that are not in set A, denoted by A'

The Cartesian product of two sets A and B is the set of all ordered pairs (x, y), where x ∈ A and y ∈ B. The power set of a set S is the set of all subsets of S, denoted by P(S). A partition of a set S is a collection of disjoint subsets that satisfy certain conditions.

Bell numbers, denoted by Bn, represent the number of ways a set can be partitioned, where n is the cardinality of the set. For example, the set {1, 2, 3} has 5 possible partitions, so B3 = 5.


# Macroeconomics

To compute key economic indicators, we use national income accounting data. The given data includes:

- Compensation of employees: $196.2 billion
- U.S. exports of goods and services: $19.8 billion
- Consumption of fixed capital: $11.8 billion
- Government purchases: $59.4 billion
- Taxes on production and imports: $14.4 billion
- Net private domestic investment: $52.1 billion
- Transfer payments: $13.9 billion
- U.S. imports of goods and services: $16.5 billion
- Personal taxes: $40.5 billion
- Net foreign income: $2.2 billion
- Personal consumption expenditures: $219.1 billion
- Statistical discrepancy: $0.0 billion

Using this data, we can calculate:
a. GDP = Compensation of employees + U.S. exports of goods and services + Government purchases + Net private domestic investment + Personal consumption expenditures - U.S. imports of goods and services
b. NDP = GDP - Consumption of fixed capital
c. NI = GDP - Consumption of fixed capital - Personal taxes - Taxes on production and imports + Net foreign income + Transfer payments

Instructions: Round answers to one decimal place.

Given a scenario where the growth rate in year 3 had been a positive 5 percent rather than a negative 2 percent, the average growth rate would be recalculated based on the new growth rate.

In another scenario, Glitter Gulch, a gold mining firm, increased its sales revenues on newly mined gold from $75 million to $150 million between one year and the next. Assuming the price of gold increased by 100 percent over the same period, the change in Glitter Gulch’s real output can be calculated by adjusting for the price increase.

If the price of gold had not changed, the change in Glitter Gulch’s real output would be equal to the change in sales revenues, which is $75 million.

The rule of 70 is a mathematical approximation that estimates the number of years it will take for something growing to double in size. It is calculated by dividing 70 by the percentage rate of growth. For example, if Mexico’s real GDP per person is growing at 7 percent per year, it will take approximately 10 years to double.

Applying the rule of 70 to Mexico’s real GDP per person growth, if it grows at 2 percent per year, and considering the real GDP per person in Mexico was $11,000 in 2005, while it was $44,000 in the United States, we can calculate how long it will take Mexico to reach the U.S. 2005 level. First, determine how many times Mexico’s 2005 real GDP per person needs to double to reach the U.S. level: $44,000 / $11,000 = 4. Then, calculate the number of years it takes to double at a 2 percent growth rate using the rule of 70: 70 / 2 = 35 years. Since Mexico’s GDP needs to double approximately 4 times (2^4 = 16, which is close to the ratio of 44,000 to 11,000, considering some rounding), the total time can be estimated by considering the doubling time and the number of doublings required.


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A study published in 2008 by E. D. Gluskin and A. E. Litvak in the DCG journal introduced a new parameter to measure the asymmetry of polytopes. The research built upon previous work in [GL], which showed that a polytope with few vertices is far from being symmetric in the Banach-Mazur distance. The new parameter, an averaging-type measure, was found to be large and satisfy the same lower bound as the Banach-Mazur distance.

The study revealed that if a convex polytope with a small number of vertices is as close to a symmetric body as possible, then most of its vertices are as bad as the worst one. The results were applied to provide a lower estimate on the vertex index of a symmetric convex body, a concept recently introduced in [BL]. Additionally, the research provided an affirmative answer to a conjecture by K. Bezdek [B3] on the quantitative illumination problem.

The journal article was added and updated on December 10, 2010, and is classified as a journal publication from the year 2008, authored by E. D. Gluskin and A. E. Litvak, and hosted on the DCG platform.


Q: In a party, everyone shakes hands with each other. If the total number of handshakes is 66, how many people were there?

To solve this problem, we need to understand the pattern of handshakes. If everyone shakes hands with everyone else, the total number of handshakes can be calculated using the formula (N-1)+(N-2)+...+2+1, where N is the number of people. This is an arithmetic progression, and the total number of handshakes is equal to (1+(N-1))(N-1)/2 or N(N-1)/2.

For example, if there are 2 people in a room, there is 1 handshake. If there are 3 people, there are 3 handshakes. If there are 5 people, each person shakes hands with 4 other people, resulting in 5*4 = 20 handshakes. However, this counts each handshake twice, so the actual number of handshakes is 5*4/2 = 10.

Using this formula, we can calculate the number of handshakes for different numbers of people:
- 2 people: 1 handshake
- 3 people: 3 handshakes
- 5 people: 10 handshakes
- 7 people: 7*6/2 = 21 handshakes
- 8 people: 8*7/2 = 28 handshakes

Given that the total number of handshakes is 66, we can set up the equation N(N-1)/2 = 66 and solve for N. By trying out different values of N, we can find the solution.

Note: The equation N(N-1)/2 = 66 can be solved by trial and error or using the quadratic formula. The solution to this equation will give us the number of people at the party.


**Designing Studies**

**Key Concepts**

1. **Bias**: The design of a statistical study shows bias if it systematically favors certain outcomes.
2. **Census**: A study that attempts to collect data from every individual in the population.
3. **Cluster Sample**: Divide the population into smaller groups, then choose a simple random sample (SRS) of the clusters.
4. **Convenience Sample**: A sample selected by taking the members of the population that are easiest to reach, prone to large bias.
5. **Double-Blind**: An experiment where neither the subjects nor those interacting with them know which treatment a subject received.
6. **Margin of Error**: A numerical estimate of how far the sample result is likely to be from the truth about the population.
7. **Nonresponse**: Occurs when a selected individual cannot be contacted or refuses to cooperate, an example of a nonsampling error.
8. **Nonsampling Error**: Errors present even in a census, including nonresponse, response bias, and errors due to question wording.
9. **Population**: The entire group of individuals about which we want information.
10. **Random Sampling**: The use of chance to select a sample, allowing us to infer results to the population.
11. **Response Bias**: A systemic pattern of incorrect responses, "people lie".
12. **Sample**: The part of the population from which we actually collect information.
13. **Sampling Error**: Mistakes made in the process of taking a sample, including bad sampling methods and undercoverage.
14. **Simple Random Sample (SRS)**: Gives every possible sample of a given size the same chance to be chosen.
15. **Strata**: Groups of individuals in a population that are similar in some way that might affect their responses.
16. **Stratified Random Sample**: Classify the population into groups of similar individuals, then choose a separate SRS from each stratum.
17. **Table of Random Digits**: A long string of digits used to select a sample.
18. **Undercoverage**: Occurs when some members of the population are left out of the sampling frame, a type of sampling error.
19. **Voluntary Response Samples**: People decide whether to join a sample based on an open invitation, prone to large bias.
20. **Wording of Questions**: The most important influence on the answers given to a survey, can introduce strong bias.

**Experimental Design**

1. **Block**: A group of experimental units that are known before the experiment to be similar in some way.
2. **Completely Randomized Design**: Treatments are assigned to all the experimental units completely by chance.
3. **Confounding**: When two variables are associated in such a way that their effects on a response variable cannot be distinguished.
4. **Control Group**: An experimental group whose primary purpose is to provide a baseline for comparing the effects of the other treatments.
5. **Experiment**: Deliberately imposes some treatment on individuals to measure their responses.
6. **Explanatory Variable**: A variable that helps explain or influences changes in a response variable.
7. **Factor**: The explanatory variables in an experiment.
8. **Matched Pair**: A common form of blocking for comparing just two treatments.
9. **Observational Study**: Observes individuals and measures variables of interest but does not attempt to influence the responses.
10. **Placebo**: An inactive (fake) treatment.
11. **Placebo Effect**: Describes the fact that some subjects respond favorably to any treatment, even an inactive one.
12. **Random Assignment**: Use some chance process to assign experimental units to treatments.
13. **Randomized Block Design**: Form blocks consisting of individuals that are similar in some way, then randomly assign treatments within each block.
14. **Response Variable**: A variable that measures an outcome of a study.
15. **Subjects**: Experimental units that are human beings.
16. **Treatment**: A specific condition applied to the individuals in an experiment.

**Inference**

1. **Inference about Cause and Effect**: Requires a well-designed experiment with random assignment of treatments.
2. **Inference about the Population**: Using information from a sample to draw conclusions about the larger population, requires random selection from the population of interest.
3. **Statistically Significant**: An observed effect so large that it would rarely occur by chance.

**Principles of Experimental Design**

1. **Comparison**: Uses a design to look at 2 or more treatments.
2. **Random Assignment**: Assigns treatments to experimental units by chance.
3. **Control**: Provides a baseline for comparing the effects of the other treatments.
4. **Replication**: Uses enough experimental units in each group to distinguish differences in the effects of the treatments from chance differences.


11. Which of the following relationships represents the dual of the Boolean property x + x'y = x + y? Choose the best answer.

- x'(x + y') = x'y'
- x(x'y) = xy
- x*x' + y = xy
- x'(xy') = x'y'
- x(x' + y) = xy

Note: The dual of a Boolean property is obtained by interchanging the operators. In option 3, the expression x*x' + y = xy is incorrect because the "+" operator has higher precedence than the "*" operator. To maintain the correct order of operations, parentheses should be used, as in x'(x + y') or x(x' + y).


# Key Concepts and Formulas

Tunnel ionization is a process where electrons pass through a potential barrier and escape from an atom or molecule.

Darcy's law states that the volume of flow of pore fluid through a porous medium per unit time is proportional to the rate of change of excess pressure.

The lapse rate is the rate at which atmospheric temperature decreases with an increase in altitude.

The size distribution of particles in a soil can be described using the arithmetic mean size, which shows variation or dispersion from the average.

Babinet's principle states that the diffraction pattern from an opaque body is identical to that from a hole of the same size and shape.

To calculate the gravitational potential energy stored in the pyramid, we need to know its center of mass, which is at one-fourth its height.

The efficiency of workers who lifted blocks in the construction of the Great Pyramid can be calculated by dividing the resulting potential energy by the consumed energy.

First, calculate the number of hours worked per year: 12 hours/day * 330 days/year = 3960 hours/year.
Then, calculate the number of hours worked in 20 years: 3960 hours/year * 20 years = 79,200 hours.
Next, calculate the energy consumed in 20 years: 300 Kcal/hour * 79,200 hours = 23,760,000 Kcal.
The efficiency is the resulting potential energy divided by the consumed energy.

The Rydberg formula is used in atomic physics to describe the wavelengths of spectral lines of many chemical elements.

The Drake equation is a probabilistic argument used to estimate the number of active, communicative extraterrestrial civilizations in the Milky Way galaxy.

Rolling resistance is the force resisting the motion when a body rolls on a surface.

The Great Pyramid of Cheops was built over 4500 years ago with a square base of 230 m on each side, covering 13.1 acres, and a height of 146 m. It has a mass of about 7×10^9 kg. Historians estimate that 20,000 workers spent 20 years to construct it, working 12-hour days, 330 days per year.

Given the pyramid's dimensions and the number of workers, we can calculate the efficiency of the workers who did the lifting, assuming there were 1000 of them and they consumed food energy at the rate of 300 Kcal/hour.

What is the efficiency of the workers who lifted blocks in the construction of the Great Pyramid?
a) Calculate the gravitational potential energy stored in the pyramid.
b) Calculate the efficiency of the workers who did the lifting.

Note: To solve these problems, we need to use the given formulas and concepts, such as the gravitational potential energy formula and the efficiency formula.


The original problem involves setting a maximum numerical precision for answers. The goal is to ensure that the answer is expressed to the same precision as the term with the least amount of decimal places (d.p) in sums and differences, or the same precision as the factor with the least amount of significant figures (s.f) in products and ratios.

To achieve this, one possible approach is to round a floating-point number to an arbitrary number of d.p or s.f using the sprintf function. The output can then be passed to a string variable and compared using the cmp() function.

However, this approach may not work as intended. For example, consider the question: "Express the following to the appropriate precision: 0.324 + 0.5". The correct answer is 0.8 to 1 d.p, but the current solution would allow answers like 0.82, 0.824, or 0.8240, which are incorrect.

An alternative solution is to set a numerical tolerance to an absolute value of 0.49x10^(-a), where a is an appropriate choice. However, this approach also has its limitations. A better approach might be to use a tolerance of some small absolute value, such as 0.0001.

Another possible solution is to use the Math::Bigfloat perl module, which deals with accuracy and precision by taking a number as an input string. This approach could be used to create a custom answer checker that satisfies the requirements.

The conventions for writing $WW macros can be found in the POD link in the author section on the wiki. The customAnswer.pl and answerHints.pl files provide examples of how to create custom answer checkers.

A possible algorithm for achieving the desired result is:

1. Do a numerical match of the student's answer. Fail unless this matches.
2. Take the student's string and truncate the decimals to 1 decimal digit.
3. Compare the truncated string with the original answer. Fail unless they match.

This algorithm can be implemented using the withPostFilter() function.

Example questions and answers can be put up in the Problem Techniques section of the wiki for reference.


# Standard Normal Variate (SNV)

The Standard Normal Variate (SNV) is a math treatment used to pre-treat a set of spectra. This article demonstrates an alternative way to apply the SNV treatment using the R function "sweep" from the "ChemometricsWithR" library.

To apply the SNV treatment, two steps are involved:

1. **Calculate the average value**: The average value of all data points for every spectrum is calculated using the "colMeans" function. This average value is then subtracted from every data point of the spectrum, resulting in a mean value of zero for every spectrum.

```r
NIR.1 <- sweep(gasoline$NIR, MARGIN = 1, colMeans(t(gasoline$NIR)), FUN = "-")
```

2. **Calculate the standard deviation**: The standard deviation (SD) of all data points for every spectrum is calculated using the "sd" function. The value of every data point is then divided by the SD of all values of that spectrum.

```r
NIR.2 <- sweep(NIR.1, MARGIN = 1, sd(t(gasoline$NIR)), FUN = "/")
```

After applying these two steps, the resulting spectrum has a mean of zero and a standard deviation of 1.

The "matplot" function can be used to plot the spectra:

```r
matplot(wavelengths, t(NIR.2), type = "l", lty = 1, xlab = "nm", ylab = "log 1/R", main = "SNV Gasoline Spectra", col = "blue")
```

Note that the "gasoline" dataset is included in the "pls" package and is used as an example in this article. However, it is not an ideal dataset to demonstrate the benefits of the SNV treatment due to limited scatter. Other datasets, such as "yarn", can be used to explore the effects of SNV.


# Convert Decimal Fraction to Binary in Java

This section explains how to convert a decimal fraction into binary in Java, dividing the problem into two parts: integral and decimal.

### 1. Calculating Binary for Integral Part:

To calculate the binary representation of the integral part, we use the following logic:
- Divide the number by 2 and store the remainder in a variable called `rem`.
- Update the number to be the result of the division by 2, i.e., `n = n / 2`.
- Insert the remainder at the beginning of a string.
This process continues until the number becomes 0.

### 2. Calculating Binary for Fractional Part:

For the fractional part, the logic involves multiplying the fractional number by 2 and storing its integral part. An integer `k` denotes the precision up to `k` digits to prevent infinite loops.
- Multiply the fractional number by 2 and store its integral part in `integralPart`.
- Update the fractional number by extracting the fractional part of `number * 2`.
- Reduce the value of `k` by 1.
This process continues until `k` becomes 0.

### Java Program to Convert Decimal Fraction to Binary

```java
package CodeSpeedy;
import java.util.Scanner;

public class DecimalToBinary {
    public static void main(String[] args) {
        double num, fractionalPart = 0, number;
        int rem = 0, integralPart, k;
        StringBuilder s = new StringBuilder();
        Scanner scn = new Scanner(System.in);

        System.out.println("Enter the number");
        num = scn.nextDouble();

        System.out.println("Enter number up to which precision is required");
        k = scn.nextInt();

        System.out.print("Output is ");

        int n = (int) num;
        fractionalPart = num - n;

        while (n != 0) {
            rem = n % 2;
            n = n / 2;
            s.insert(0, rem);
        }

        System.out.print(s + ".");
        s = new StringBuilder();

        while (k != 0) {
            integralPart = (int) (fractionalPart * 2);
            s.append(integralPart);
            number = fractionalPart * 2;
            fractionalPart = number - integralPart;
            k--;
        }

        System.out.println(s);
    }
}
```

#### Output:

Enter the number: 112.564
Enter number up to which precision is required: 5
Output is: 1110000.10010

Using `StringBuilder` instead of `String` allows for modification from any end, enabling insertion at any index without needing a function to reverse the string.