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# Principal Component Analysis (PCA)
```Principal Component Analysis (PCA)
1.1
Introduction to PCA
Principal component analysis (PCA) is a statistical procedure that uses an orthogonal
transformation to convert a set of observations of possibly correlated variables into a set of
values of linearly uncorrelated variables called principal components.
This transformation is defined in such a way that the first principal component has the largest
possible variance (that is, accounts for as much of the variability in the data as possible), and
each succeeding component in turn has the highest variance possible under the constraint that
it is orthogonal to the preceding components.
PCA is sensitive to the relative scaling of the original variables.
By projecting our data into a smaller space, we’re reducing the dimensionality of our feature
space… but because we’ve transformed our data in these different “directions,” we’ve made
sure to keep all original variables in our model!
1.2
Limitations of PCA
PCA is a linear algorithm. It will not be able to interpret complex polynomial relationship
between features.
A major problem with, linear dimensionality reduction algorithms is that they concentrate on
placing dissimilar data points far apart in a lower dimension representation. But in order to
represent high dimension data on low dimension, non-linear manifold, it is important that
similar data points must be represented close together, which is not what linear
dimensionality reduction algorithms do.
Internal
1.3
Eigenvectors and Eigenvalues
Eigenvectors and eigenvalues come in pairs, i.e. every eigenvector has a corresponding
eigenvalue.
Eigenvector: direction
Eigenvalue: value that describes the variance in the data in that direction
The eigenvector with the highest eigenvalue is the principal component.
The total number of eigenvector & eigenvalue sets is equal to the total number of dimensions.
2
(
2
12
3
3
3
)×( )= ( )= 4×( )
8
1
2
2
Eigenvector
1.4
Eigenvalue
Covariance
Covariance is a measure of how much two dimensions vary from their respective means with
respect to each other. The covariance of a dimension with itself is the variance of that
dimension.
𝑐𝑜𝑣(𝑋, 𝑌) =
∑𝑛𝑖=1(𝑋𝑖 − 𝑋̅)(𝑌𝑖 − 𝑌̅)
𝑛−1
The exact value of covariance is not as important as the sign of the covariance.
Internal
+ve covariance: Both dimensions increase together.
-ve covariance: Both dimensions decrease together.
1.5
Algorithm
1.5.1 Step 1 – Standardisation
For PCA to work properly, the data should be standardised/scaled. An example is to subtract
the mean from each of the data dimensions. This produces a dataset with a mean of 0.
If the importance of features is independent of the variance of the features, then divide each
observation in a column by that column’s standard deviation.
1.5.2 Step 2 – Calculate covariance matrix
𝑐𝑜𝑣(𝑥, 𝑥)
𝐶 = (𝑐𝑜𝑣(𝑦, 𝑥)
𝑐𝑜𝑣(𝑧, 𝑥)
= 𝑍𝑇 𝑍
𝑐𝑜𝑣(𝑥, 𝑦)
𝑐𝑜𝑣(𝑦, 𝑦)
𝑐𝑜𝑣(𝑧, 𝑦)
𝑐𝑜𝑣(𝑥, 𝑧)
𝑐𝑜𝑣(𝑦, 𝑧))
𝑐𝑜𝑣(𝑧, 𝑧)
1.5.3 Step 3 – Calculate the eigenvectors and eigenvalues of the covariance matrix
The eigenvectors should be unit eigenvectors, i.e. their lengths are 1. This is done by
dividing the eigenvector by its length.
This step is also known as eigendecomposition of 𝑍 𝑇 𝑍 into 𝑃𝐷𝑃−1 , where P is the matrix of
eigenvectors and D is the diagonal matrix with eigenvalues on the diagonal and values 0
elsewhere.
1.5.4 Step 4 – Choosing components and forming a feature vector
The eigenvectors are ordered according to eigenvalue, from highest to lowest. This gives the
components in the order of significance.
Now, if you like, you can decide to ignore the components of lesser significance. You do lose
some information, but if the eigenvalues are small, you don’t lose much. If you leave out
some components, the final data set will have less dimensions than the original.
Internal
```
Colossal statues
38 Cards
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# Search by Topic
#### Resources tagged with Interactivities similar to On Target:
Filter by: Content type:
Age range:
Challenge level:
### There are 154 results
Broad Topics > Information and Communications Technology > Interactivities
### A Square of Numbers
##### Age 7 to 11 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
##### Age 5 to 11 Challenge Level:
Place six toy ladybirds into the box so that there are two ladybirds in every column and every row.
### Difference
##### Age 7 to 11 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
### Code Breaker
##### Age 7 to 11 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### One to Fifteen
##### Age 7 to 11 Challenge Level:
Can you put the numbers from 1 to 15 on the circles so that no consecutive numbers lie anywhere along a continuous straight line?
##### Age 7 to 11 Challenge Level:
Three beads are threaded on a circular wire and are coloured either red or blue. Can you find all four different combinations?
### Arrangements
##### Age 7 to 11 Challenge Level:
Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....?
### Combining Cuisenaire
##### Age 7 to 11 Challenge Level:
Can you find all the different ways of lining up these Cuisenaire rods?
### Which Symbol?
##### Age 7 to 11 Challenge Level:
Choose a symbol to put into the number sentence.
### Magic Potting Sheds
##### Age 11 to 14 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
### Junior Frogs
##### Age 5 to 11 Challenge Level:
Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible?
### First Connect Three for Two
##### Age 7 to 11 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Seven Flipped
##### Age 7 to 11 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Coded Hundred Square
##### Age 7 to 11 Challenge Level:
This 100 square jigsaw is written in code. It starts with 1 and ends with 100. Can you build it up?
### Countdown
##### Age 7 to 14 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### More Transformations on a Pegboard
##### Age 7 to 11 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
### Multiples Grid
##### Age 7 to 11 Challenge Level:
What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
##### Age 7 to 11 Challenge Level:
If you have only four weights, where could you place them in order to balance this equaliser?
### Winning the Lottery
##### Age 7 to 11 Challenge Level:
Try out the lottery that is played in a far-away land. What is the chance of winning?
### Fault-free Rectangles
##### Age 7 to 11 Challenge Level:
Find out what a "fault-free" rectangle is and try to make some of your own.
### Teddy Town
##### Age 5 to 14 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### Domino Numbers
##### Age 7 to 11 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
### One Million to Seven
##### Age 7 to 11 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Factor Lines
##### Age 7 to 14 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Coordinate Tan
##### Age 7 to 11 Challenge Level:
What are the coordinates of the coloured dots that mark out the tangram? Try changing the position of the origin. What happens to the coordinates now?
### Rod Ratios
##### Age 7 to 11 Challenge Level:
Use the Cuisenaire rods environment to investigate ratio. Can you find pairs of rods in the ratio 3:2? How about 9:6?
### Tetrafit
##### Age 7 to 11 Challenge Level:
A tetromino is made up of four squares joined edge to edge. Can this tetromino, together with 15 copies of itself, be used to cover an eight by eight chessboard?
### Red Even
##### Age 7 to 11 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
### Building Stars
##### Age 7 to 11 Challenge Level:
An interactive activity for one to experiment with a tricky tessellation
### Triangle Pin-down
##### Age 7 to 11 Challenge Level:
Use the interactivity to investigate what kinds of triangles can be drawn on peg boards with different numbers of pegs.
### Triangles All Around
##### Age 7 to 11 Challenge Level:
Can you find all the different triangles on these peg boards, and find their angles?
### Nine-pin Triangles
##### Age 7 to 11 Challenge Level:
How many different triangles can you make on a circular pegboard that has nine pegs?
### Noughts and Crosses
##### Age 7 to 11 Challenge Level:
A game for 2 people that everybody knows. You can play with a friend or online. If you play correctly you never lose!
### Overlapping Circles
##### Age 7 to 11 Challenge Level:
What shaped overlaps can you make with two circles which are the same size? What shapes are 'left over'? What shapes can you make when the circles are different sizes?
### 100 Percent
##### Age 7 to 11 Challenge Level:
An interactive game for 1 person. You are given a rectangle with 50 squares on it. Roll the dice to get a percentage between 2 and 100. How many squares is this? Keep going until you get 100. . . .
### Shapely Tiling
##### Age 7 to 11 Challenge Level:
Use the interactivity to make this Islamic star and cross design. Can you produce a tessellation of regular octagons with two different types of triangle?
### Counters
##### Age 7 to 11 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remove them. The winner is the last one to remove a counter. How you can make sure you win?
### Train for Two
##### Age 7 to 11 Challenge Level:
Train game for an adult and child. Who will be the first to make the train?
### Part the Piles
##### Age 7 to 11 Challenge Level:
Try to stop your opponent from being able to split the piles of counters into unequal numbers. Can you find a strategy?
### Rabbit Run
##### Age 7 to 11 Challenge Level:
Ahmed has some wooden planks to use for three sides of a rabbit run against the shed. What quadrilaterals would he be able to make with the planks of different lengths?
### Round Peg Board
##### Age 5 to 11 Challenge Level:
A generic circular pegboard resource.
### Odds or Sixes?
##### Age 7 to 11 Challenge Level:
Use the interactivity or play this dice game yourself. How could you make it fair?
### Times Tables Shifts
##### Age 7 to 11 Challenge Level:
In this activity, the computer chooses a times table and shifts it. Can you work out the table and the shift each time?
### Train
##### Age 7 to 11 Challenge Level:
A train building game for 2 players.
### Have You Got It?
##### Age 11 to 14 Challenge Level:
Can you explain the strategy for winning this game with any target?
### Dominoes Environment
##### Age 5 to 11 Challenge Level:
These interactive dominoes can be dragged around the screen.
### Spot Thirteen
##### Age 7 to 11 Challenge Level:
Choose 13 spots on the grid. Can you work out the scoring system? What is the maximum possible score?
### Sorting Symmetries
##### Age 7 to 11 Challenge Level:
Find out how we can describe the "symmetries" of this triangle and investigate some combinations of rotating and flipping it.
### Carroll Diagrams
##### Age 5 to 11 Challenge Level:
Use the interactivities to fill in these Carroll diagrams. How do you know where to place the numbers?
### Dotty Circle
##### Age 7 to 11 Challenge Level:
Watch this film carefully. Can you find a general rule for explaining when the dot will be this same distance from the horizontal axis?
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https://assignmentgrade.com/find-the-sum-using-fraction-bars-or-a-paper-and-pencil-write-your-answer-in-simplest-form-56-39-398743/
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# Find the sum using fraction bars or a paper and pencil. Write your answer in simplest form. 5/6 + 3/9
QUESTION POSTED AT 17/01/2020 - 01:14 AM
Find the LCD = 18
5/6 = 15/18
3/9 = 6/18
15/18 + 6/18 = 21/18 = 7/5 or 1 2/5
## Related questions
### The population of a certain geographic region is approximately 103 million and grows continuously at a relative growth rate of 1.67%. What will the population be in 5 years? Compute the answer to three significant digits.
The population ( now ) : 103 million.
In 5 years:
103 * ( 1.0167 )^5 =
= 103 * 1.08633586 =
= 111.89259 ≈ 111.893
The population will be 111.893 million.
ANSWERED AT 21/02/2020 - 10:10 AM
QUESTION POSTED AT 21/02/2020 - 10:10 AM
### In mixed fractions, which fraction is equivalent to 16/3
16/3, 32/6, 48/9, 64/12, 80/15, 96/18, 112/21, 128/24, 144/27, 160/30,
ANSWERED AT 21/02/2020 - 10:08 AM
QUESTION POSTED AT 21/02/2020 - 10:08 AM
### Sorry, here's the real questions below. 70+ Points for Best 5 Answers! 1. A triangle has an area of 150 cm². What is the area of the triangle if the height of the triangle is reduced to of its original length? A. 25 cm² B. 75 cm² C. 50 cm² D. 10 cm² 2. A right rectangular prism has a volume of 96 cm³. What is the new volume of the prism if only the length of the prism is doubled? A. 588 cm³ B. 192 cm³ C. 96 cm³ D. 48 cm³ 3. The surface of a cube is 600 cm² What is the new surface area of the cube if the length of each edge of the cube is tripled? A. 5400 cm² B. 1800 cm² C. 200 cm² D. 900 cm² 4. What is the surface area to volume ratio of the rectangular prism? [SEE ATTACHMENT] A. B. C. D. 5. A rectangle has a perimeter of 30 cm. What would be the new perimeter if the length and width of the rectangle is doubled in size? A. 18 cm B. 60 cm C. 36 cm D. 120 cm
1) 75
2) 192 cm^3
3) 1800 cm^2
4) 19/20
5) 60 cm
ANSWERED AT 21/02/2020 - 09:17 AM
QUESTION POSTED AT 21/02/2020 - 09:17 AM
### In simplest form, which mixed number is equivalent to 22\8
2 3/4
Because 8 goes into 22 2 times with a remainder of 6. That gives you 2 6/8
6/8 can be reduced to 3/4
ANSWERED AT 20/02/2020 - 11:19 PM
QUESTION POSTED AT 20/02/2020 - 11:19 PM
### In 4 years, harry’s age will be the same as jim’s age is now. in 2 years, jim will be twice as old as harry will be. find their ages now.
The answer to this question would be:
Jim =6 years
Harry =2 years
To answer this question, you need to translate the description into a mathematical equation. If H= Harry's ages and J= Jim's ages, then the equation should be look like this:
1. H+4= J
2. (J+2)= 2(H+2)
J+2= 2H+4
J= 2H +2
If you put the 1st and 2nd equation together, the result would be:
H+4= J
H+4= (2H +2)
H-2H= 2-4
-H= -2
H=2
After that insert the H value to either 1st or 2nd equation.
H+4= J
2+4=J
J= 6
ANSWERED AT 20/02/2020 - 11:19 PM
QUESTION POSTED AT 20/02/2020 - 11:19 PM
### If you eat 4 medium strawberries, you get 48% of your daily recommended amount of vitamin c. what fraction of your daily amount of vitamin c do you still need.
Step-by-step explanation:
ANSWERED AT 20/02/2020 - 11:19 PM
QUESTION POSTED AT 20/02/2020 - 11:19 PM
### If y = 3x -5 find the value of y when x = -1
Plug in negative 1 for x and do order of operations and u should get negative 8
ANSWERED AT 20/02/2020 - 11:18 PM
QUESTION POSTED AT 20/02/2020 - 11:18 PM
### If x-y=8,find the value of y when x=3
3-y=8
-y=8-3
y=-5
so just try to re arrange the equation to find the value for the variable
ANSWERED AT 20/02/2020 - 11:18 PM
QUESTION POSTED AT 20/02/2020 - 11:18 PM
### How do i find 2/4of36?
Two forths is the same as one half ...so just take half of 36 which is 18
ANSWERED AT 20/02/2020 - 11:15 PM
QUESTION POSTED AT 20/02/2020 - 11:15 PM
### In the form f(x)=a(x-h)^2+k
Vertex is (h,k)
se see vertex is (1,-4)
y=a(x-1)^2-4
find a
another point is (4,-2)
-2=a(4-1)^2-4
-2=a(3)^2-4
-2=9a-4
2=9a
divide b 9
2/9=a
ANSWERED AT 20/02/2020 - 11:14 PM
QUESTION POSTED AT 20/02/2020 - 11:14 PM
### Find the value of 2a 5b when a=-6 and b=2
2a = -12
5b = 10
Hope this helps
:)
ANSWERED AT 20/02/2020 - 11:14 PM
QUESTION POSTED AT 20/02/2020 - 11:14 PM
### Find the nth term of this sequence -2 3 8 13 18
The nth term is 5n-7
ANSWERED AT 20/02/2020 - 11:14 PM
QUESTION POSTED AT 20/02/2020 - 11:14 PM
### At is 5/9 4/6 in simplest form?
5/9 is already in it's simplest form
4/6 can be reduced to 2/3
ANSWERED AT 20/02/2020 - 11:11 PM
QUESTION POSTED AT 20/02/2020 - 11:11 PM
### Paige and her family went to the movies. They bought 4 tickets and paid $12 for popcorn. They spent$40. How much did each ticket cost? Choose two answers: one for the equation that models this situation and one for the correct answer. A. Answer: $3 B. Answer:$7 C. Equation: 12 + 4x = 40 D. Equation: 4 + 12x = 40
B. And C. Is the correct answer
ANSWERED AT 20/02/2020 - 11:11 PM
QUESTION POSTED AT 20/02/2020 - 11:11 PM
### What is the answer to 2/3 1/8 1/12?
Your question seem to me non-understandable. Are they adding up together ?
or multiplying ?
ANSWERED AT 20/02/2020 - 11:10 PM
QUESTION POSTED AT 20/02/2020 - 11:10 PM
### What is 32.15 in unit form??
3 tens, 2 ones, 1 tenths and 5 hundredths.
Step-by-step explanation:
We are asked to write 32.15 in unit form.
To write 32.15 in unit form, we need to write each digit as its place value.
Ones: 2
Tens: 3
Tenths: 1
Hundredths: 5
Therefore, 32.15 in unit form would be 3 tens, 2 ones, 1 tenths and 5 hundredths.
ANSWERED AT 20/02/2020 - 11:09 PM
QUESTION POSTED AT 20/02/2020 - 11:09 PM
### What is 1/2 divided by 3/2 equals out to be in fraction form?
When dividing fractions you have to place the fraction being divided on top of the fraction being used to divide it. So you would have 1/2 / 3/2. Then you multiply the top and bottom numbers to get the new numerator and subsequently the middle two numbers to get the denominator.
= 2/6 = 0.333
ANSWERED AT 20/02/2020 - 11:08 PM
QUESTION POSTED AT 20/02/2020 - 11:08 PM
### What improper fraction can be written as a whole number. 10/6 12\6 14\6 16\6?
12/6 can be written as 2 because 6 can go into 12 two times.
ANSWERED AT 20/02/2020 - 11:07 PM
QUESTION POSTED AT 20/02/2020 - 11:07 PM
### 1/2 x 3/4 in simplest form
Simply multiply both top numbers to get the new numerator (1x3 = 3)
Multiply both bottom numbers to get the new denominator (2x4 = 8)
Therefore the new fraction is 3/8
ANSWERED AT 20/02/2020 - 11:05 PM
QUESTION POSTED AT 20/02/2020 - 11:05 PM
21+33+15+20= 89 you started the day with $89 ANSWERED AT 20/02/2020 - 11:03 PM QUESTION POSTED AT 20/02/2020 - 11:03 PM ### 8.4×3.7 please show your work on a piece of paper 8.4 times 3.7 = 31.08 is the answer!!! ANSWERED AT 20/02/2020 - 11:00 PM QUESTION POSTED AT 20/02/2020 - 11:00 PM ### What is the answer to the first one? T=0.9 You solve this by dividing 60 on each side 54 divided by 60= 0.9 ANSWERED AT 20/02/2020 - 10:59 PM QUESTION POSTED AT 20/02/2020 - 10:59 PM ### How to find the y-intercept and graph the equation y=4x + 3 The y intercept would be 3, from there go up 4 and right 1 . ANSWERED AT 20/02/2020 - 10:57 PM QUESTION POSTED AT 20/02/2020 - 10:57 PM ### Two friends cut a large candy bar into equal pieces. Harriet ate 1/4 of the pieces. Nisha ate 1/2 of hte remaining pieces. Six pieces were left over. How many piees was the candy bar originally divided into? Harriet at 1/4 of all pieces 4/4-1/4=3/4 nisha ate 1/2 of remaining (3/4) 1/2 of 3/4=3/8 eaten ok, so eaten=1/4+3/8=2/8+3/8=5/8 eaten 8/8-5/8=3/8 left 3/8=6 pices 1/8=2 pieces 8/8=16 pices origainally divided into 16 pieces ANSWERED AT 20/02/2020 - 10:56 PM QUESTION POSTED AT 20/02/2020 - 10:56 PM ### The sum of twice a number and seven is twenty five 9 Because 9x2 is 18+7=25 ANSWERED AT 20/02/2020 - 10:55 PM QUESTION POSTED AT 20/02/2020 - 10:55 PM ### Simplify each expression, then explain what property was used:1. 6s + 2Simplified expression:Explanation:2. s + 2s + 9 + sSimplified expression:Explanation:3. 4(2s + r)Simplified expression:Explanation:Combine Like Terms:4. -3r - 3 + r + 75. 5s + 2s^ - 5s + 4s^8. Use the model below to find the area: 9. Use the model below to find the side lengths: 1. I belive this is in simplest form already. 2. 4s+9 3. 8s+4r 4. -2r+4 5. 6s Hope that helps a little bit! ANSWERED AT 20/02/2020 - 10:55 PM QUESTION POSTED AT 20/02/2020 - 10:55 PM ### Find the amount paid for the loan.$1500 at 9% for 2 years
Assuming simple interest
I=PRT
Interset=principal times rate times time
principal=1500
rate=9%=0.09
time=2
I=1500 times 0.09 times 2
I=1500 times 0.18
I=270
interest=270
total is 1500+270=1770 total paied
ANSWERED AT 20/02/2020 - 10:53 PM
QUESTION POSTED AT 20/02/2020 - 10:53 PM
### Find area of a rectangle with vertices -9,3 -9,-7 -6,-7 and -6,-3
4 times 3 equals 12.
ANSWERED AT 20/02/2020 - 10:53 PM
QUESTION POSTED AT 20/02/2020 - 10:53 PM
First question: simple possible combinations. So multiply the quantity of unique items in each set together. In this case 6 * 8 * 4 = 192.
Second questions: since there are 192 possible outcomes, and this is one possible unique outcome. The chance is 1/192.
ANSWERED AT 20/02/2020 - 10:52 PM
QUESTION POSTED AT 20/02/2020 - 10:52 PM
### Write a conjecture to save 30$I can save$30 by putting a 1 penny in a jar everyday for 3000 days.
ANSWERED AT 20/02/2020 - 10:52 PM
QUESTION POSTED AT 20/02/2020 - 10:52 PM
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## Related Rates: Ball Drop & Shadow
Posted: March 13, 2016 in Calculus: An Introduction, Differential Calculus, Differential Equations, Related Rates
Tags: ,
Here’s problem involving a falling object and the speed at which its shadow travels along the ground. As usual, in related rates, once a relationship between the variables involved has been established, the calculus required to reach its conclusion is very straight forward.
In order to make efficient use of time, these problems provide students the opportunity to practice simple differentiation procedures. In addition, the graphs provided below open the door to a discussion on the Mean Value Theorem of differential calculus, serving to either introduce or reinforce that concept.
Falling Ball
Click on the link provided here to interact with the falling ball and its shadow.
The ball’s displacement from its release point was provided in the image above. As a review (since integral calculus has already been introduced), that displacement formula is once again derived through basic differential equations; this is shown directly below.
Acceleration, Velocity and Displacement
I’ve included solutions for t=1 and t=2 below. In keeping with my belief that students can learn effectively through comparison and contrast, three varied methods are shown.
Solutions
The main subject of this entry was originally planned as an optimization problem involving differential calculus only; its been slightly modified. This more interesting approach provides the derivative up front, presenting students with three separate tasks to pursue from that point. As a consequence, students are reintroduced to differential equations and curve sketching.
A talking point emerges as well: Is there a difference between derivatives and differential equations?
Inscribed Triangle of Maximum Area
Click on the link provided here to explore area of the inscribed triangle.
Thanks for looking.
## Related Rates (continued)
Posted: March 1, 2016 in Calculus: An Introduction, Differential Calculus, Differential Equations, Implicit Differentiation, Related Rates, The Derivative
Tags:
In the spirit of consistency, several additional examples supplementing my introduction to related rates are included here. As seen in the first example below, past concepts are revisited and, in turn, connected to newly introduced ones. Mixing things up from that point on provides students with a variety of perspectives on the same theme that serves to further cement understanding.
Separable Differential Equation
Three Approaches to Related Rates
Application to Conical Container
Rate of Change of Arc Length
The approach to solving each of these related rate problems is the same; identify what is known, what is desired and then connect the two.
Surface areas of curved 3-dimensional solids tend to be much more difficult for students to conceptualize than those whose sides do not stray from a “level” plane. These will eventually be addressed but we will first discover how to calculate lengths of curves.
The circle will once again be called upon to initiate this exploration; the image below illustrates, in part, the method of exhaustion that Archimedes utilized to arrive at his estimate for π.
Click on the link here to interact with what Archimedes revealed.
……..and now this. Was Archimedes wrong???
Source: math.stackexchange.com
If the fellow above had joined pairs of points at each successive corner with a line segment (hypotenuse) and based his calculation for circumference on the sum of those, he would have found that Archimedes was correct all along.
Arc Length Formula
As Δs approaches zero in this exploration, its length becomes a more accurate estimate for the arc length near the “point” of tangency (there are always two points in very close proximity). The end result through the limiting process shown directly below is the formula for arc length.
Its always beneficial for students to work through several examples to cement their understanding of new concepts and related procedures; my preference is to provide examples that are already familiar to them. Calculating arc length (in this case) can then serve as a verification and acceptance of the new concept is achieved with confidence. The offering directly below and the link that follows connects the arc length formula to the Pythagorean theorem.
The formula for arc length is based on the Pythagorean theorem; it is therefore not surprising that they produce the same lengths on linear functions.
The real power of the formula for arc length lies in its applications to curves. Since students have known the circle’s circumference for several years, it is appropriate to now derive 2πr using our new tool. This is shown below and once again brings trigonometric substitution into play.
Circumference of the Circle
Once the circle’s circumference has been established using the arc length formula, the integration process can be further solidified by using arc length to once again calculate the circle’s area.
The formula for arc length will once again be employed in deriving the formulas for surface area of the sphere and cone.
## Differential Equations (comparisons)
Posted: February 19, 2016 in Differential Equations, Integral Calculus
Tags: ,
Having side-by-side comparisons can be valuable experience as similarities (and differences) are more readily apparent; students can gain a deeper understanding of the nuances from one to the next. For example, its worth noting that the rate of change of a circle’s area with respect to its radius is equal to that circle’s circumference; a similar relationship exists between the volume and surface area of a sphere.
Investing some thought into these and other subtleties can go a long way towards increasing one’s intuitive feel for, and enjoyment of this discipline.
Another comparison worth checking out is quadratic vs exponential growth.
## Setting the Stage for Logistic Growth
Posted: February 15, 2016 in Calculus: An Introduction, Differential Equations
Tags: ,
In order to achieve some degree of continuity, I continually strive to weave together concepts, not only within my own “area of influence”, but across other disciplines as well. Physics is naturally folded into the fabric of calculus for obvious reasons; others disciplines, not so much.
I wanted to raise awareness in students of how calculus appears in applications relating to Biology and Chemistry; logistic growth is the obvious choice for the former and is relatively straight forward once students have a feel for differential equations.
The notes directly below make clear (I hope) the distinction between two types of growth from the context of differential equations. The exponential growth model below will be expanded upon to eventually derive the well-known formula for logistic growth.
The application to Chemistry that was alluded to earlier will require First Order Differential Equations, another “diversion” that can be pursued when a change of pace is needed. We will hopefully be afforded the time to develop an adequate understanding of this before semester’s end.
## Separable Differential Equations and Slope Fields
Posted: February 15, 2016 in Differential Calculus, Differential Equations, Slope Fields
Tags: , ,
A basic understanding of differential equations has already been established through our introduction to integration. In addition, rates of change have also been linked to our brief study of differential calculus. Students now require a period of time to work through basic problem solving scenarios relating to both differentiation and integration to develop an acceptable degree of fluency.
Every now and then, a diversion from the “daily grind” can be well-received. Separable differential equations and an introduction to slope fields will be one such “diversion” and will be shared with students when deemed appropriate. Since students are very familiar with quadratic functions, will we begin there.
The equation dy/dx=x will be presented and analyzed from a “rate of change” perspective at various positions on the Cartesian plane. Once this slope field has been sketched, the shape of the parent function becomes readily apparent; the need for initial conditions arises to uniquely define each member from the family.
The image directly below sees through the completion of the scenario described above. This procedure is then repeated for other basic separable differential equations, all of which produce slope fields that are recognizable to students; these appear below our parabola example.
Differential Equation: dy/dx=kx, where k=2
Constant of Proportionality Given
Differential Equation: dy/dx=kx^2, where k=3
Constant of Proportionality Given
Constant of Proportionality Not Given
Differential Equation: dy/dx=-x/y
For the circle above, I’ve included two acceptable treatments.
The indefinite integrals require that initial conditions be substituted in after the fact to solve for the constant of integration. The second version has the initial conditions included as bounds of integration, resulting in definite integrals; same result.
Solving these differential equations tie together quite nicely the two sides of calculus to which students have been introduced. These examples also set up other such equations and problems that will be presented in the not too distant future, such as Quadratic vs Exponential Growth.
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# Free Fall
Topics: 3, Exponentiation, General relativity Pages: 4 (415 words) Published: March 12, 2015
2/12/2013
Lab 1430
Free Fall
The difference of the outline procedure and the actual procedure is the use of the brass screw was not working in our set up. So we had to improvise and use our hand as the release mechanism as what we had seen this didn’t make difference from others results. Drop Distance 50(cm)
Drop
Time(sec)
1
.306179
2
.310800
3
.304614
4
.311203
5
.298986
Drop Distance 100(cm)
Drop
Time(sec)
1
.419258
2
.417368
3
.420589
4
.416400
5
.430646
Drop Distance 150(cm)
Drop
Time(sec)
1
.516188
2
.504206
3
.495936
4
.515523
5
.502310
Drop Distance 200(cm)
Drop
Time (sec)
1
.623696
2
.616600
3
.618880
4
.628058
5
.602976
H(m)
.093756
.5
.177116
1
.256879
1.5
.38183
2
Stander Deviation
H(m)
Deviation
.5
.3063564
1
.4208522
1.5
.5068326
2
.618042
Equation Used
Percentage Error = (Abs(measured value-calculated value)/calculate value)*100% What we can see from the results and the theory of the idea of the ball dropping is that the time it takes form 1 meter and 2 meters aren’t twice as large. What we can see is that it is an exponential increase in a small amount. In theory this is proven that the time is not double just because the distance is double. And that the acceleration without air resistance will always be constant -9.81 m/s squared In question number 2 by ignoring air resistance would this tend to cause the measures value of g in this experiment to be larger or smaller. This question may be miss leading because in earth the value of g is a constant and does not change for the reason that it’s a constant it can’t be change to what in our will. In question number 3 it’s asking in what point in time the ball value of acceleration will be the smallest. I believe that at the highest point of the ball the magnitude the balls acceleration will have the smallest. I believe that this is right because it will have no acceleration so the slop will be zero and will not be...
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# Upsample data using FFTs. How is this exactly done?
I am basing my question on this post here, because I would like additional details on it, as I have not had any success in re-creating it.
I would like to simply learn, in no uncertain terms, how to do upsampling via the FFT. There seem to be a lot of details regarding if the signal is even or odd, where exactly to zero-pad, how much exactly, etc, and I am not getting those correct numbers for some reason.
Question: So let us say I have an even or odd length signal, of length $N$. I would like to upsample it to, say, $2N$, using FFTs.
How is this done exactly?
What I tried:
I tried following the post and others on the web but there seems to be very little consistency. It basically amounts to "zero-padding the frequency domain and then IFFTing" but I do not get proper results. In my case, I have a signal of length $100$, and I would like to upsample it to length $200$. So I do the following command:
ifft(fftshift([zeros(1,50) fftshift(fft(signal, 100)) zeros(1,50)]));
This however gives me complex data back, and not a signal that is simply upsampled by a factor of $2$. I am not sure where the problem is.
Thank you.
• In short, you have been misinterpreting how the spectrum is stored. Thanks for the answer, Peter K. – user7358 Mar 10 '14 at 19:37
• @user7358 Ok, but why the downvote? – TheGrapeBeyond Mar 10 '14 at 20:27
One trick, for even-length signals, is what to do with the "middle" sample. Here, I've split it half and half between each side of the FFT.
The other trick is to ensure that you have the right amplitudes in the resampled signal. Here's it's a factor of 2.
Try this in scilab:
x = rand(1,100,'normal');
X = fft(x);
XX = 2*[X(1:50) X(51)/2 zeros(1,99) X(51)/2 X(52:100)];
xx = ifft(XX);
clf;
plot([0:199]/200,xx)
plot([0:199]/200,xx,'go')
plot([0:99]/100,x,'r.')
which appears to generate the right thing. The red dots are the original samples, and the green circles (and blue connecting lines) are the resampled data.
UPDATE
For the case of an odd number of samples, there are fewer fiddly bits (no splitting the last coefficient):
x2 = rand(1,101,'normal');
X2 = fft(x2);
XX2 = 2*[X2(1:51) zeros(1,101) X2(52:101)];
xx2 = ifft(XX2);
clf;
plot([0:201]/202,xx2)
plot([0:201]/202,xx2,'gx')
plot([0:100]/101,x2,'r.')
• Thanks Peter. May you also please place an addition to your answer for the odd-cases for completion, based on the same example? Much appreciated! – TheGrapeBeyond Mar 10 '14 at 14:02
• @TheGrapeBeyond:Done! – Peter K. Mar 10 '14 at 15:21
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### Winning at a Russian Roulette
Q:Three bullets are inserted into a gun which has six slots at random positions. The trigger is pulled, and no shot is fired. What is the probability a bullet will fire if the trigger is pulled once again? What is the probability of a fire in the 3rd pull if it is pulled twice and both are blanks?
A: Let us work through the cases. There are four possible configurations the revolver could be in. These are shown in the picture below.
There are a total of 6C3 configurations possible. Each of the first three configurations can happen in 6 ways, while the last configuration can happen in just 2 ways. Given that the first shot was a blank, the probability of firing a bullet in each of the configurations are the following
The total probability (as the above cases are mutually exclusive) is the weighted sum given as
$$P(Fire) = \frac{6}{20}\times\frac{1}{3} + \frac{6}{20}\times\frac{2}{3}\times 2 + \frac{2}{20}\times1 = \frac{3}{5}$$
If two pulls on the trigger goes blank only configuration 1 has a scenario where it could go blank on the third pull. I leave it to the reader to work out that case and probability.
If you are looking to buy some books in probability here are some of the best books to learn the art of Probability
Here are a few
Fifty Challenging Problems in Probability with Solutions (Dover Books on Mathematics)
This book is a great compilation that covers quite a bit of puzzles. What I like about these puzzles are that they are all tractable and don't require too much advanced mathematics to solve.
Introduction to Algorithms
This is a book on algorithms, some of them are probabilistic. But the book is a must have for students, job candidates even full time engineers & data scientists
Introduction to Probability Theory
An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd Edition
The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (and Everyone Else!)
Introduction to Probability, 2nd Edition
The Mathematics of Poker
Good read. Overall Poker/Blackjack type card games are a good way to get introduced to probability theory
Let There Be Range!: Crushing SSNL/MSNL No-Limit Hold'em Games
Easily the most expensive book out there. So if the item above piques your interest and you want to go pro, go for it.
Quantum Poker
Well written and easy to read mathematics. For the Poker beginner.
Bundle of Algorithms in Java, Third Edition, Parts 1-5: Fundamentals, Data Structures, Sorting, Searching, and Graph Algorithms (3rd Edition) (Pts. 1-5)
An excellent resource (students/engineers/entrepreneurs) if you are looking for some code that you can take and implement directly on the job.
Understanding Probability: Chance Rules in Everyday Life A bit pricy when compared to the first one, but I like the look and feel of the text used. It is simple to read and understand which is vital especially if you are trying to get into the subject
Data Mining: Practical Machine Learning Tools and Techniques, Third Edition (The Morgan Kaufmann Series in Data Management Systems) This one is a must have if you want to learn machine learning. The book is beautifully written and ideal for the engineer/student who doesn't want to get too much into the details of a machine learned approach but wants a working knowledge of it. There are some great examples and test data in the text book too.
Discovering Statistics Using R
This is a good book if you are new to statistics & probability while simultaneously getting started with a programming language. The book supports R and is written in a casual humorous way making it an easy read. Great for beginners. Some of the data on the companion website could be missing.
### The Best Books to Learn Probability
If you are looking to buy some books in probability here are some of the best books to learn the art of Probability The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (and Everyone Else!) A good book for graduate level classes: has some practice problems in them which is a good thing. But that doesn't make this book any less of buy for the beginner. An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd Edition This is a two volume book and the first volume is what will likely interest a beginner because it covers discrete probability. The book tends to treat probability as a theory on its own Discovering Statistics Using R This is a good book if you are new to statistics & probability while simultaneously getting started with a programming language. The book supports R and is written in a casual humorous way making it an easy read. Great for beginners. Some of the data on the companion website could be missing. Fifty Cha
### The Best Books for Linear Algebra
The following are some good books to own in the area of Linear Algebra. Linear Algebra (2nd Edition) This is the gold standard for linear algebra at an undergraduate level. This book has been around for quite sometime a great book to own. Linear Algebra: A Modern Introduction Good book if you want to learn more on the subject of linear algebra however typos in the text could be a problem. Linear Algebra (Dover Books on Mathematics) An excellent book to own if you are looking to get into, or want to understand linear algebra. Please keep in mind that you need to have some basic mathematical background before you can use this book. Linear Algebra Done Right (Undergraduate Texts in Mathematics) A great book that exposes the method of proof as it used in Linear Algebra. This book is not for the beginner though. You do need some prior knowledge of the basics at least. It would be a good add-on to an existing course you are doing in Linear Algebra. Linear Algebra, 4th Ed
### Fun with Uniform Random Numbers
Q: You have two uniformly random numbers x and y (meaning they can take any value between 0 and 1 with equal probability). What distribution does the sum of these two random numbers follow? What is the probability that their product is less than 0.5. The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists A: Let z = x + y be the random variable whose distribution we want. Clearly z runs from 0 to 2. Let 'f' denote the uniform random distribution between [0,1]. An important point to understand is that f has a fixed value of 1 when x runs from 0 to 1 and its 0 otherwise. So the probability density for z, call it P(z) at any point is the product of f(y) and f(z-y), where y runs from 0 to 1. However in that range f(y) is equal to 1. So the above equation becomes From here on, it gets a bit tricky. Notice that the integral is a function of z. Let us take a look at how else we can simply the above integral. It is easy to see that f(z-y) = 1 when (
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### Home > CCA2 > Chapter Ch10 > Lesson 10.3.1 > Problem10-145
10-145.
Use Pascal’s Triangle to expand $(x + y)^7$.
Recall Pascal's Triangle. Pascal's triangle, up to n, = 5, where the first row is, n, = 0. Last row is 1, 5, 10, 10, 5, 1.
What row is needed to expand $(x + y)^7$?
Use the pattern for $(a + b)^7$Added to the triangle, row 7, 1, 7, 21, 35, 35, 21, 7, 1.
$a^7 + 7a^6b + 21a^5b^2 + 35a^4b^3 + 35a^3b^4 + 21a^2b^5 + 7ab^6 + b^7$
Now substitute $x$ for $a$ and $y$ for $b$.
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You dont have javascript enabled! Please enable it! MCQ in Engineering Mathematics Part 9 | ECE Board Exam
MCQ in Engineering Mathematics Part 9 | ECE Board Exam
This is the Multiples Choice Questions in Engineering Mathematics Part 9 of the Series. In Preparation for the ECE Board Exam make sure to expose yourself and familiarize each and every questions compiled here taken from various sources including past Board Exam Questions, Engineering Mathematics Books, Journals and other Engineering Mathematics References. In the actual board, you have to answer 100 items in Engineering Mathematics within 5 hours. You have to get at least 70% to pass the subject. Engineering Mathematics is 20% of the total 100% Board Rating along with Electronic Systems and Technologies (30%), General Engineering and Applied Sciences (20%) and Electronics Engineering (30%).
Continue Practice Exam Test Questions Part 9 of the Series
Choose the letter of the best answer in each questions.
401. A circular water main 4 meter in diameter. is closed by a bulkhead whose center is 40 m below the surface of the water in the reservoir. Find the force on the bulkhead.
a. 3419 kN
b. 4319 kN
c. 4931 kN
d. 5028 kN
Solution:
402. A plate in the form of parabolic segment is 12 m in height and 4 m deep and is partly submerged in water so that its axis is parallel to end 3 m below the water surface. Find the force acting on the plate.
a. 899.21 kN
b. 939.46 kN
c. 933.17 kN
d. 993.26 kN
Solution:
403. A cistern in the form of an inverted right circular cone is 20 m deep and 12 m diameter at the top. If the water is 16 m deep in the cistern, find the work done in Joules in pumping out the water. The water is raised to a point of discharge 10 m above the top cistern.
a. 54883992 Joules
b. 61772263 Joules
c. 68166750 Joules
d. 76177640 Joules
Solution:
404. A bag containing originally 60 kg of flour is lifted through a vertical distance of 9 m. While it is being lifted, flour is leaking from the bag at such rate that the number of pounds lost is proportional to the square root of the distance traversed. If the total loss of flour is 12 kg find the amount of work done in lifting the bag.
a. 4290 Joules
b. 4591 Joules
c. 5338 Joules
d. 6212 Joules
Solution:
405. According to Hookeโs law, the force required to stretch a helical spring is proportional to the distance stretched. The natural length of a given spring is 8 cm. a force of 4 kg will stretch it to a total length of 10 cm. Find the work done in stretching it from its natural length to a total length of 16 cm.
a. 4.65 Joules
b. 5.32 Joules
c. 6.28 Joules
d. 7.17 Joules
Solution:
406. The top of an elliptical conical reservoir is an ellipse with major axis 6 m and minor axis 4 m. it is 6 m deep and full of water. Find the work done in pumping the water to an outlet at the top of the reservoir.
a. 473725 Joules
b. 493722 Joules
c. 554742 Joules
d. 593722 Joules
Solution:
407. A bag of sand originally weighing 144 kg is lifted at a rate of 3 m/min. the sand leaks out uniformly at such rate that half of the sand is lost when the bag has been lifted 18 m. find the work done in lifting the bag of sand at this distance.
a. 6351 Joules
b. 4591 Joules
c. 5349 Joules
d. 5017 Joules
Solution:
408. A cylindrical tank having a radius of 2 m and a height of 8 m is filled with water at a depth of 6 m. Compute the work done in pumping all the liquid out of the top of the container.
a. 2 934 942 Joules
b. 3 698 283 Joules
c. 4 233 946 Joules
d. 5 163 948 Joules
Solution:
409. A right cylindrical tank of radius 2 m and a height 8 m is full of water. Find the work done in pumping the tank. Assume water to weigh 9810 N/m3.
a. 3945 kN . m
b. 4136 kN . m
c. 2846 kN . m
d. 5237 kN . m
Solution:
410. A conical vessel 12 m across the top and 15 m deep. If it contains water to a depth of 10 m find the work done in pumping the liquid to the top of the vessel.
a. 12 327.5 kN . m
b. 14 812.42 kN . m
c. 24 216.2 kN . m
d. 31 621 kN . m
Solution:
411. A hemispherical vessel of diameter 8 m is full of water. Determine the work done in pumping out the top of the tank in Joules.
a. 326 740 pi
b. 627 840 pi
c. 516 320 pi
d. 418 640 pi
Solution:
412. A spring with a natural length of 10 cm is stretched by 1/2 cm by a Newton force. Find the work done in stretching from 10 cm to 18 cm. Express your answer in joules.
a. 6.29 Joules
b. 7.13 Joules
c. 7.68 Joules
d. 8.38 Joules
Solution:
413. A 5 lb. monkey is attached to a 20 ft. hanging rope that weighs 0.3 lb/ft. the monkey climbs the rope up to the top. How much work has it done?
a. 160
b. 165
c. 170
d. 180
Solution:
414. A bucket weighing 10 Newton when empty is loaded with 90 Newton of sand and lifted at 10 cm at a constant speed. Sand leaks out of a hole in a bucket at a uniform rate and one third of sand is lost by the end of the lifting process in Joules.
a. 800 Joules
b. 850 Joules
c. 900 Joules
d. 950 Joules
Solution:
415. A conical vessel is 12 m across the top and 15 m deep. If it contains water to a depth of 10m find the work done in pumping the liquid to a height 3 m above the top of the vessel.
a. 560 pi w N.m
b. 660 pi w N.m
c. 520 pi w N.m
d. 580 pi w N.m
Solution:
416. A small in the sack of rice cause some rice to be wasted while the sack is being lifted vertically to a height of 30 m. The weight lost is proportional to the cube root of distance traversed. If the total loss was 16 kg, find the work done in lifting the said sack of rice which weighs 110 kg.
a. 2369 kg.m
b. 2409 kg.m
c. 2940 kg.m
d. 3108 kg.m
Solution:
417. A hemispherical tank of diameter 20 ft. is full of oil weighing 20 pcf. The oil is pumped to a height of 10 ft. above the top of the tank by an engine of 1/2 horsepower. How long will it take the engine to empty the tank?
a. 1 hr. 15.47 min
b. 1 hr. 24.27 min
c. 1 hr. 44.72 min
d. 2 hrs.
Solution:
418. A full tank consists of a hemisphere of radius 4 m surmounted by a circular cylinder of the same radius of altitude 8 m. Find the work done in pumping the water to an outlet of the top of the tank.
a. (2255/3) pi w
b. (2527/3) pi w
c. (2752/3) pi w
d. (5722/3) pi w
Solution:
419. Determine the differential equation of a family of lines passing thru (h, k).
a. (y – k) dx โ (x – h) dy = 0
b. (x – h) + (y – k) = dy/dx
c. (x – h) dx โ (y – k) dy = 0
d. (x + h) dx โ (y – k) dy = 0
Solution:
420. What is the differential equation of the family of parabolas having their vertices at the origin and their foci on the x-axis
a. 2x dy โ y dx = 0
b. x dy + y dx = 0
c. 2y dx โ x dy = 0
d. dy/dx โ x = 0
Solution:
421. Find the differential equations of the family of lines passing through the origin.
a. ydx โ xdy = 0
b. xdy โ ydx = 0
c. xdx + ydy = 0
d. ydx + xdy = 0
Solution:
422. The radius of the moon is 1080 miles. The gravitation acceleration of the moons surface is 0.165 miles the gravitational acceleration at the earthโs surface. What is the velocity of escape from the moon in miles per second?
a. 2.38
b. 1.47
c. 3.52
d. 4.26
Solution:
423. Find the equation of the curve at every point of which the tangent line has a slope of 2x.
a. x = -y2 + C
b. y = -x2 + C
c. x = y2 + C
d. y = x2 + C
Solution:
424. The radius of the earth is 3960 miles. If the gravitational acceleration of earth surface is 31.16 ft/sec2, what is the velocity of escape from the earth in miles/sec?
a. 3.9266
b. 5.4244
c. 6.9455
d. 7.1842
Solution:
425. Find the velocity of escape of the Apollo spaceship as it is projected from the earthโs surface that is the minimum velocity imparted to it so that it will never return. The radius of the earth is 400 miles and the acceleration of the spaceship is 32.2 ft/sec2.
a. 30426 kph
b. 50236 kph
c. 40478 kph
d. 60426 kph
Solution:
426. The rate of population growth of a country is proportional to the number of inhabitants. If a population of a country now is 40 million and expected to double in 25 years, in how many years is the population be 3 times the present?
a. 39.62 yrs.
b. 28.62 yrs.
c. 18.64 yrs.
d. 41.2 yrs.
Solution:
427. From the given differential equation xdx + 6y5dy = 0 solve for the constant of integration when x = 0, y = 2.
a. 27x dx + 4y2 dy = 0
b. 58
c. 48
d. 64
Solution:
428. Find the equation of the curve which passes through points (1, 4) and (0, 2) if d2 y/ dx2 = 1
a. 2y = x2 + 3x + 4
b. 4y = 2x2 + x + 4
c. 5y = x2 + 2x + 2
d. 3y = x2 + x + 4
Solution:
429. The rate of population growth of a country is proportional to the number of inhabitants. If a population of a country now is 40 million and 50 million in 10 years time, what will be its population 20 years from now?
a. 56.19
b. 71.29
c. 62.18
d. 59.24
Solution:
430. The Bureau of Census record in 1980 shows that the population in the country doubles compared to that of 1960. In what year will the population trebles assuming that the rate of increase in the population is proportional to the population?
a. 34.60
b. 31.70
c. 45.65
d. 38.45
Solution:
431. A tank contains 200 liters of fresh water. Brine containing 2 kg/liter of salt enters the tank at the rate of 4 liters per min, and the mixture kept uniform by stirring, runs out at 3 liters per min. Find the amount of salt in the tank after 30 min.
a. 196.99 kg
b. 186.50 kg
c. 312.69 kg
d. 234.28 kg
Solution:
432. In a tank are 100 liters of brine containing 50 kg total of dissolved salt. Pure water is allowed to run into the tank at the rate of 3 liters per minute. Brine runs out of the tank at rate of 2 liters per minute. The instantaneous concentration in the tank is kept uniform by stirring. How much salt is in the tank at the end of 1 hour?
a. 20.50
b. 18.63
c. 19.53
d. 22.40
Solution:
433. Determine the general solution of xdy + ydx = 0.
a. xy = c
b. ln xy = c
c. ln x + ln y = c
d. x + y = c
Solution:
434. The inverse Laplace transform of s/[(square) + (w square)] is:
a. sin wt
b. w
c. (e exponent wt)
d. cos st
Solution:
435. The Laplace transform of cos wt is:
a. s/[(square) + (w square)]
b. w/[(square) + (w square)]
c. w/s + w
d. s/s + w
Solution:
436. K divided by [(s square) + (k square)] is inverse Laplace transform of:
a. cos kt
b. sin kt
c. (e exponent Ky)
d. 1.0
Solution:
437. Find the inverse transform of [2/(s + 1)] โ [(4/(s + 3)] is equal to:
a. [2 e (exp โ t) โ 4e (exp โ 3t)]
b. [e (exp โ 2t) + e (exp โ 3t)]
c. [e (exp โ 2t) โ e (exp – 3t)]
d. [2e (exp โ t) โ 2e (exp – 2t)]
Solution:
438. What is the Laplace transform of e(-4t)
a. 1/ (s + 1)
b. 1/ (s + 4)
c. 1/ (s โ 4)
d. 1/ (s + t)
Solution:
439. Determine the Laplace transform of I(S) = 200 / [(s2) + 50s + 10625]
a. I(S) = 2e(-25t) sin100t
b. I(S) = 2te(-25t) sin100t
c. I(S) = 2e(-25t) cos100t
d. I(S) = 2te(-25t) cos100t
Solution:
440. Determine the inverse Laplace transform of (s + a) / [(s + a)2 + w2]
a. e(-at) cos wt
b. te(-at) cos wt
c. t sin wt
d. e(-at) sin wt
Solution:
441. Determine the inverse Laplace transform of 100/ [(S + 10) (S + 20)]
a. 10e(-10t) + 20e(-20t)
b. 10e(-10t) โ 20e(-20t)
c. 10e(-10t) โ 10e(-20t)
d. 20e(-10t) + 10e(-20t)
Solution:
442. A thin heavy uniform iron rod 16 m long is bent at the 10 m mark forming a right angle L โ shaped piece 6 m by 10 m of bend. What angle does the 10 m side make with the vertical when the system is in equilibrium?
a. 28ยฐ 12โ
b. 19ยฐ 48โ
c. 24ยฐ 36โ
d. 26ยฐ 14โ
Solution:
443. Three men carry a uniform timber. One takes hold at one end and the other two carry by means of a crossbar placed underneath. At what point of timber must the bar be placed so that each man may carry one third of the weight of the weight of the timber? The timber has a length of 12 m.
a. 4 m
b. 5 m
c. 2.5 m
d. 3 m
Solution:
444. A painters scaffold 30 m long and a mass of 300 kg, is supported in a horizontal position by a vertical ropes attached at equal distances from the ends of the scaffold. Find the greatest distance from the ends that the ropes may be attached so as to permit a 200 kg man to stand safely at one end of scaffold.
a. 8 m
b. 7 m
c. 6 m
d. 9 m
Solution:
445. A cylindrical tank having a diameter of 16 cm weighing 100 kN is resting on a horizontal floor. A block having a height of 4 cm is placed on the side of the cylindrical tank to prevent it from rolling. What horizontal force must be applied at the top of the cylindrical tank so that it will start to roll over the block? Assume the block will not slide and is firmly attached to the horizontal floor.
a. 57.74 kN
b. 58.36 kN
c. 68.36 kN
d. 75.42 kN
Solution:
446. Two identical sphere weighing 100 kN are each place in a container such that the lower sphere will be resting on a vertical wall and a horizontal wall and the other sphere will be resting on the lower sphere and a wall making an angle of 60 degrees with the horizontal. The line connecting the two centers of the spheres makes an angle of 30 degrees with the horizontal surface. Determine the reaction between the contact of the two spheres. Assume the walls to be frictionless.
a. 150
b. 120
c. 180
d. 100
Solution:
447. The 5 m uniform steel beam has a mass of 600 kg and is to be lifted from the ring B with two chains, AB of length 3 m, and CB of length 4 m. Determine the tension T in chain AB when the beam is clear of the platform.
a. 2.47 kN
b. 3.68 kN
c. 5.42 kN
d. 4.52 kN
Solution:
448. A man attempts to support a stack of books horizontally by applying a compressive force of F = 120 N to the ends of the stack with his hands, determine the number of books that can be supported in the stack if the coefficient of friction between any two books is 0.40.
a. 15 books
b. 20 books
c. 10 books
d. 12 books
Solution:
449. Two men are just to lift a 300 kg weight of crowbar when the fulcrum for this lever is 0.3m from the weight and the man exerts their strengths at 0.9 m and 1.5 m respectively from the fulcrum. If the men interchange positions, they can raise a 340 kg weight. What force does each man exert?
a. 25 kg, 40 kg
b. 30 kg, 50 kg
c. 35 kg, 45 kg
d. 40 kg, 50 kg
Solution:
450. A man exert a maximum pull of 1000 N but wishes to lift a new stone door for his cave weighing 20 000 N. if he uses lever how much closer must the fulcrum be to the stone than to his hand?
a. 10 times nearer
b. 20 times farther
c. 10 times farther
d. 20 times nearer
Solution:
Online Question and Answer in Engineering Mathematics Series
Following is the list of multiple choice questions in this brand new series:
MCQ in Engineering Mathematics
PART 1: MCQ from Number 1 โ 50 Answer key: PART 1
PART 2: MCQ from Number 51 โ 100 Answer key: PART 2
PART 3: MCQ from Number 101 โ 150 Answer key: PART 3
PART 4: MCQ from Number 151 โ 200 Answer key: PART 4
PART 5: MCQ from Number 201 โ 250 Answer key: PART 5
PART 6: MCQ from Number 251 โ 300 Answer key: PART 6
PART 7: MCQ from Number 301 โ 350 Answer key: PART 7
PART 8: MCQ from Number 351 โ 400 Answer key: PART 8
PART 9: MCQ from Number 401 โ 450 Answer key: PART 9
PART 10: MCQ from Number 451 โ 500 Answer key: PART 10
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This is a Clilstore unit. You can .
# 1st SESSION: What are Polynomials?
You can solve the exercises proposed after you have seen the video.
# TRANSCRIPTION
0:06
Hi, I’m Rob. Welcome to Math Antics.
0:09
In this video, we’re going to learn about Polynomials.
0:11
That’s a big math word for a really big concept in Algebra, so pay attention.
0:16
Now before we can understand what polynomials are, we need to learn about what mathematicians call “terms”.
0:22
In Algebra, terms are mathematical expressions that are made up of two different parts:
0:28
a number part and a variable part.
0:31
In a term, the number part and the variable part are multiplied together,
0:36
but since multiplication is implied in Algebra,
0:38
the two parts of a term are usually written right next to each other with no times symbol between them.
0:44
The number part is pretty simple… it’s just a number, like 2 or 5 or 1.4
0:49
And the number part has an official name… it’s called the “coefficient”.
0:53
Now there’s another cool math word that you can use to impress your friends at parties!
0:57
[party music, crowd noise]
0:59
…and then I said, “That’s not my wife… that’s my coefficient!”
1:04
[silence / crickets chirping]
1:07
The variable part of a term is a little more complicated.
1:11
It can be made up of one or more variables that are raised to a power.
1:14
Like… the variable part could be 'x squared'. That’s a variable raised to a power.
1:19
Or, the variable part could be just ‘y’.
1:22
If you remember what we learned in our last video, you’ll realize that that also qualifies as a variable raised to a power.
1:28
‘y’ is the same as ‘y’ to the 1st power.
1:31
But since the exponent ‘1’ doesn’t change anything, we don’t need to actually show it.
1:37
Or… the variable part of a term could be some tricky combination of variables that are raised to powers,
1:42
like ‘x squared’ times ‘y squared’.
1:44
…or ‘a’ times ‘b squared’ times ‘c cubed’.
1:48
Terms can have any number of variables like that, but the good news is that most of the time,
1:53
you’ll only need to deal with terms that have one variable. …or maybe two in complicated problems.
1:59
Oh, and there’s one thing I should point out before we move on…
2:02
if you have a term like 6y, even though it would be fine to do the multiplication the other way around and write y6,
2:09
it’s conventional to always write the number part of the term first and the variable part of the term second.
2:15
Okay, so that’s the basic idea of a term.
2:18
But there’s a little more to terms that we’ll learn in a minute.
2:20
First, let’s see how this basic idea of a term helps us understand the basic idea of a polynomial.
2:27
A polynomial is a combination of many terms.
2:30
It’s kind of like a chain of terms that are all linked together using addition or subtraction.
2:35
The terms themselves contain multiplication, but each term in a polynomial must be joined by either addition or subtraction.
2:44
And polynomials can be made from any number of terms joined together,
2:48
but there are a few specific names that are used to describe polynomials with a certain number of terms.
2:53
If there’s only one term (which isn’t really a chain) then we call it a “monomial” because the prefix “mono” means “one”.
3:01
If there are just two terms, then we call it a “binomial” because the prefix “bi” means “two”,
3:06
and if there are three terms, then we call it a “trinomial” since the prefix “tri” means “three”.
3:12
Beyond three terms, we usually just say “polynomial” since “poly” means “many”,
3:17
and in fact, it’s common to simply use the term “polynomial” even when there are just 2 or 3 terms.
3:23
Okay, so that’s the basic idea of a polynomial.
3:26
It’s a series of terms that are joined together by addition or subtraction.
3:31
Now, let’s see a typical example of a polynomial that will help us learn a little more about terms: 3 ‘x squared’ plus ‘x’ minus 5
3:41
How many terms does this polynomial have?
3:44
Well, based on what we’ve learned so far, you’re probably not quit sure.
3:48
If the terms are the parts that are joined together by addition or subtraction, then this should have three terms,
3:54
but it looks like there’s something missing with the last two terms.
3:57
This middle term is missing its number part, and this last term is missing its variable part.
4:03
That doesn’t seem to fit with our original definition of a term. What’s up with that?
4:08
Well, the middle term is easy to explain.
4:10
There really is a number part there, but it’s just ‘1’.
4:13
Do you remember how ‘1’ is always a factor of any number?
4:17
But, since multiplying by ‘1’ has no effect on a number or variable, we don’t need to show it.
4:24
So, if you see a term in a polynomial that has only a variable part, you know that the number part (or coefficient) of that term is just ‘1’.
4:33
4:37
Well, that’s a little trickier. Do you remember in our last video about exponents in Algebra,
4:43
we learned that any number or variable that’s raised to the 0th power just equals ‘1’?
4:49
That means we can think of this last term as having a variable ‘x’ that’s being raised to the 0th power.
4:56
Since that would always just equal ‘1’, it’s not really a variable in the true sense of the word,
5:00
and it has no effect on the value of the term.
5:03
But it makes sense, especially if you remember the other rule from the last video.
5:07
That rule says that any number raised to the 1st power is just itself,
5:13
which helps us see that this middle term is basically the same as ‘1x’ raised to the 1st power.
5:19
Now do you see the pattern?
5:21
Each term has a number part and each term has a variable part that is raised to a power: 0, 1 and 2.
5:28
But since ‘x’ to the ‘0’ is just ‘1’,
5:31
and ‘x’ to the ‘1’ is just ‘x’,
5:33
and anything multiplied by ‘1’ is just itself,
5:37
the polynomial gets simplified so that it no longer looks exactly like the pattern it comes from.
5:43
Oh, and this last term… the one that doesn’t have a truly variable part…
5:47
it’s called a CONSTANT term because its value always stays the same.
5:52
Alright… Now that you know what a Polynomial is, let’s talk about an important property of terms and polynomials called their “degree”.
6:01
Now that might sound like the units we use to measure temperature or angles, but the degree we’re talking about here is different.
6:07
The degree of a term is determined by the power of the variable part.
6:12
For example, in this term, since the power of the variable is 4, we say that the degree of the term is 4, or that it’s a 4th degree term.
6:20
And in this term, the power of the variable is 3, so it’s a 3rd degree term.
6:25
Likewise, this would be a 2nd degree term and this would be a 1st degree term.
6:30
Oh, and I suppose you could call a term with no variable part a “zero degree” term,
6:35
but it’s usually just referred to as a “constant term”.
6:38
Things are a little more complicated when you have terms with more that one variable.
6:43
In that case, you add up the powers of each variable to get the degree of the term.
6:47
Since the powers in this term are 3 and 2, it’s a 5th degree term because 3 + 2 = 5.
6:54
Okay, but why do we care about the degree of terms?
6:58
Well, it’s because polynomials are often referred to by the degree of their highest term.
7:04
If a polynomial contains a 4th degree term (but no higher terms), then it’s called a “4th degree” polynomial.
7:10
But if its highest term is only a 2nd degree term, then it’s called a “2nd degree” polynomial.
7:17
Another reason that we care about the degree of the terms is that it helps us decide the arrangement of a polynomial.
7:22
We arrange the terms in a polynomial in order from the highest degree to the lowest.
7:28
…ya know, cuz, mathematicians like to keep things organized…
7:37
[mumbeling] …nice… let’s see…double check…
7:45
Perfect!
7:47
For example, this polynomial (which has 5 terms)
7:50
should be rearranged so that the highest degree term is on the left, and the lowest degree term is on the right.
7:57
But of course, not every polynomial has a term of every degree.
8:01
This is a 5th degree polynomial, but it only has 3 terms.
8:05
We should still put them in order from highest to lowest, even though it has terms that are missing.
8:11
So, the “4x to the fifth” should come first.
8:14
And then the “minus 10x”.
8:17
And finally, the “plus 8”.
8:19
By the way, it’s totally fine for a polynomial to have “missing” terms like that.
8:24
And it’s sometimes helpful to think of those missing terms as just having coefficients that are all zeros.
8:30
If the coefficient of a term is zero, then the whole term has a value of zero so it wouldn’t effect the polynomial at all.
8:38
And speaking of coefficients…
8:39
What if we need to re-arrange this polynomial so that its terms are in order from highest degree to lowest degree?
8:45
The highest degree term is ‘5x squared’ but before we just move it to the front of the polynomial,
8:52
it’s important to notice that it’s got a minus sign in front of it.
8:55
Normally when we see a minus sign, we think of subtraction, but when it comes to polynomials,
9:01
it’s best to think of a minus sign as a NEGATIVE SIGN that means the term right after it has a negative value (or a negative coefficient).
9:09
In fact, instead of thinking of a polynomial as having terms that are added OR subtracted,
9:15
it’s best to think of ALL of the terms as being ADDED,
9:19
but that each term has either a POSITIVE or a NEGATIVE coefficient which is determined by the operator right in front of that term.
9:27
For example, if you have this Polynomial, you should treat it as if all of the terms are being added together,
9:33
and use the sign that’s directly in front of each term to tell you if it’s a positive or a negative term.
9:39
This first term has a coefficient of ‘negative 4’, so it’s a negative term.
9:44
The next term has a coefficient of ‘positive 6’, so it’s positive.
9:49
The next term has a coefficient of ‘negative 8’, so it’s negative.
9:54
And the constant term is just ‘positive 2’.
9:57
And recognizing positive and negative coefficients helps us a lot when
10:01
rearranging polynomials that have a mixture of positive and negative terms like our example here.
10:07
If you think of the negative sign in front of the ‘5x squared’ term as part of its coefficient,
10:13
then you’ll realize that when we move it to the front of the polynomial, the negative sign has to come with it.
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It has to come with it because it’s really a NEGATIVE term.
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If we don’t bring the negative sign along with it, we’ll be changing it into a positive term
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which would actually change the value of the polynomial.
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And in addition to helping us re-arrange them,
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treating a polynomial as a combination of positive and negative terms will be very helpful when we need to simplify them,
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which just so happens to be the subject of our next basic Algebra video.
10:46
Alright, we’ve learned a LOT about polynomials in this video,
10:50
and if you’re a little overwhelmed, don’t worry… it might just take some time for it all to make sense.
10:55
Remember, you can always re-watch this video a few times,
10:58
and doing some of the practice problems will help it all sink in.
11:01
As always, thanks for watching Math Antics, and I’ll see ya next time.
Short url: http://multidict.net/cs/5373
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2.3: Computing the F statistics
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Computing the $$F$$-value
Fisher’s ANOVA is very elegant in my opinion. It starts us off with a big problem we always have with data. We have a lot of numbers, and there is a lot of variation in the numbers, what to do? Wouldn’t it be nice to split up the variation into to kinds, or sources. If we could know what parts of the variation were being caused by our experimental manipulation (i.e., the independent variable we choose as researchers), and what parts were being caused by sampling error, we would be making really good progress. We would be able to know if our experimental manipulation was causing more change in the data than sampling error, or chance alone. If we could measure those two parts of the total variation, we could make a ratio, and then we would have an $$F$$ value. This is what the ANOVA does. It splits the total variation in the data into two parts. The formula is:
Total Variation = Variation due to Manipulation + Variation due to sampling error
This is a nice idea, but it is also vague. We haven’t specified our measure of variation. What should we use?
Remember the sums of squares that we used to make the variance and the standard deviation? That’s what we’ll use. Let’s take another look at the formula, using sums of squares for the measure of variation:
$SS_\text{total} = SS_\text{Effect} + SS_\text{Error} \nonumber$
SS Total
The total sums of squares, or $$SS\text{Total}$$ is a way of thinking about all of the variation in a set of data. It’s pretty straightforward to measure. No tricky business. All we do is find the difference between each score and the grand mean, then we square the differences and add them all up.
Let’s imagine we had some data in three groups, A, B, and C. For example, we might have 3 scores in each group. The data could look like this:
groups scores diff diff_squared
A 20 13 169
A 11 4 16
A 2 -5 25
B 6 -1 1
B 2 -5 25
B 7 0 0
C 2 -5 25
C 11 4 16
C 2 -5 25
Sums 63 0 302
Means 7 0 33.5555555555556
The data is organized in long format, so that each row is a single score. There are three scores for the A, B, and C groups. The mean of all of the scores is called the Grand Mean. It’s calculated in the table, the Grand Mean = 7.
We also calculated all of the difference scores from the Grand Mean. The difference scores are in the column titled diff. Next, we squared the difference scores, and those are in the next column called diff_squared.
Remember, the difference scores are a way of measuring variation. They represent how far each number is from the Grand Mean. If the Grand Mean represents our best guess at summarizing the data, the difference scores represent the error between the guess and each actual data point. The only problem with the difference scores is that they sum to zero (because the mean is the balancing point in the data). So, it is convenient to square the difference scores, which gets rid of the negative signs (or values) and turns all of them into positive numbers. The size of the squared difference scores still represents error between the mean and each score. And, the squaring operation exacerbates the differences as the error grows larger (squaring a big number makes a really big number, squaring a small number still makes a smallish number).
OK fine! We have the squared deviations from the grand mean, we know that they represent the error between the grand mean and each score. What next? SUM THEM UP!
When you add up all of the individual squared deviations (difference scores) you get the sums of squares. That’s why it’s called the sums of squares (SS).
Now, we have the first part of our answer:
$SS_\text{total} = SS_\text{Effect} + SS_\text{Error} \nonumber$
$SS_\text{total} = 302 \nonumber$
and
$302 = SS_\text{Effect} + SS_\text{Error} \nonumber$
What next? If you think back to what you learned about algebra, and solving for X, you might notice that we don’t really need to find the answers to both missing parts of the equation. We only need one, and we can solve for the other. For example, if we found $$SS_\text{Effect}$$, then we could solve for $$SS_\text{Error}$$.
SS Effect
$$SS_\text{Total}$$ gave us a number representing all of the change in our data, how all the scores are different from the grand mean.
What we want to do next is estimate how much of the total change in the data might be due to the experimental manipulation. For example, if we ran an experiment that causes causes change in the measurement, then the means for each group will be different from other. As a result, the manipulation forces change onto the numbers, and this will naturally mean that some part of the total variation in the numbers is caused by the manipulation.
The way to isolate the variation due to the manipulation (also called effect) is to look at the means in each group, and calculate the difference scores between each group mean and the grand mean, and then sum the squared deviations to find $$SS_\text{Effect}$$.
Consider this table, showing the calculations for $$SS_\text{Effect}$$.
groups scores means diff diff_squared
A 20 11 4 16
A 11 11 4 16
A 2 11 4 16
B 6 5 -2 4
B 2 5 -2 4
B 7 5 -2 4
C 2 5 -2 4
C 11 5 -2 4
C 2 5 -2 4
Sums 63 63 0 72
Means 7 7 0 8
Notice we created a new column called means. For example, the mean for group A was 11. You can see there are three 11s, one for each observation in row A. The means for group B and C happen to both be 5. So, the rest of the numbers in the means column are 5s.
What we are doing here is thinking of each score in the data from the viewpoint of the group means. The group means are our best attempt to summarize the data in those groups. From the point of view of the mean, all of the numbers are treated as the same. The mean doesn’t know how far off it is from each score, it just knows that all of the scores are centered on the mean.
Now that we have converted each score to it’s mean value we can find the differences between each mean score and the grand mean, then square them, then sum them up. We did that, and found that the $$SS_\text{Effect} = 72$$.
$$SS_\text{Effect}$$ represents the amount of variation that is caused by differences between the means. I also refer to this as the amount of variation that the researcher can explain (by the means, which represent differences between groups or conditions that were manipulated by the researcher).
Notice also that $$SS_\text{Effect} = 72$$, and that 72 is smaller than $$SS_\text{total} = 302$$. That is very important. $$SS_\text{Effect}$$ by definition can never be larger than $$SS_\text{total}$$.
SS Error
Great, we made it to SS Error. We already found SS Total, and SS Effect, so now we can solve for SS Error just like this:
$SS_\text{total} = SS_\text{Effect} + SS_\text{Error} \nonumber$
switching around:
$SS_\text{Error} = SS_\text{total} - SS_\text{Effect} \nonumber$
$SS_\text{Error} = 302 - 72 = 230 \nonumber$
We could stop here and show you the rest of the ANOVA, we’re almost there. But, the next step might not make sense unless we show you how to calculate $$SS_\text{Error}$$ directly from the data, rather than just solving for it. We should do this just to double-check our work anyway.
groups scores means diff diff_squared
A 20 11 -9 81
A 11 11 0 0
A 2 11 9 81
B 6 5 -1 1
B 2 5 3 9
B 7 5 -2 4
C 2 5 3 9
C 11 5 -6 36
C 2 5 3 9
Sums 63 63 0 230
Means 7 7 0 25.5555555555556
Alright, we did almost the same thing as we did to find $$SS_\text{Effect}$$. Can you spot the difference? This time for each score we first found the group mean, then we found the error in the group mean estimate for each score. In other words, the values in the $$diff$$ column are the differences between each score and it’s group mean. The values in the diff_squared column are the squared deviations. When we sum up the squared deviations, we get another Sums of Squares, this time it’s the $$SS_\text{Error}$$. This is an appropriate name, because these deviations are the ones that the group means can’t explain!
Degrees of freedom
Degrees of freedom come into play again with ANOVA. This time, their purpose is a little bit more clear. $$Df$$s can be fairly simple when we are doing a relatively simple ANOVA like this one, but they can become complicated when designs get more complicated.
Let’s talk about the degrees of freedom for the $$SS_\text{Effect}$$ and $$SS_\text{Error}$$.
The formula for the degrees of freedom for $$SS_\text{Effect}$$ is
$$df_\text{Effect} = \text{Groups} -1$$, where Groups is the number of groups in the design.
In our example, there are 3 groups, so the df is 3-1 = 2. You can think of the df for the effect this way. When we estimate the grand mean (the overall mean), we are taking away a degree of freedom for the group means. Two of the group means can be anything they want (they have complete freedom), but in order for all three to be consistent with the Grand Mean, the last group mean has to be fixed.
The formula for the degrees of freedom for $$SS_\text{Error}$$ is
$$df_\text{Error} = \text{scores} - \text{groups}$$, or the number of scores minus the number of groups. We have 9 scores and 3 groups, so our $$df$$ for the error term is 9-3 = 6. Remember, when we computed the difference score between each score and its group mean, we had to compute three means (one for each group) to do that. So, that reduces the degrees of freedom by 3. 6 of the difference scores could be anything they want, but the last 3 have to be fixed to match the means from the groups.
Mean Squares
OK, so we have the degrees of freedom. What’s next? There are two steps left. First we divide the $$SS$$es by their respective degrees of freedom to create something new called Mean Squared deviations or Mean Squares. Let’s talk about why we do this.
First of all, remember we are trying to accomplish this goal:
$\text{F} = \frac{\text{measure of effect}}{\text{measure of error}} \nonumber$
We want to build a ratio that divides a measure of an effect by a measure of error. Perhaps you noticed that we already have a measure of an effect and error! How about the $$SS_\text{Effect}$$ and $$SS_\text{Error}$$. They both represent the variation due to the effect, and the leftover variation that is unexplained. Why don’t we just do this?
$\frac{SS_\text{Effect}}{SS_\text{Error}} \nonumber$
Well, of course you could do that. What would happen is you can get some really big and small numbers for your inferential statistic. And, the kind of number you would get wouldn’t be readily interpretable like a $$t$$ value or a $$z$$ score.
The solution is to normalize the $$SS$$ terms. Don’t worry, normalize is just a fancy word for taking the average, or finding the mean. Remember, the SS terms are all sums. And, each sum represents a different number of underlying properties.
For example, the SS effect represents the sum of variation for three means in our study. We might ask the question, well, what is the average amount of variation for each mean…You might think to divide SS effect by 3, because there are three means, but because we are estimating this property, we divide by the degrees of freedom instead (# groups - 1 = 3-1 = 2). Now we have created something new, it’s called the $$MS_\text{Effect}$$.
$MS_\text{Effect} = \frac{SS_\text{Effect}}{df_\text{Effect}} \nonumber$
$MS_\text{Effect} = \frac{72}{2} = 36 \nonumber$
This might look alien and seem a bit complicated. But, it’s just another mean. It’s the mean of the sums of squares for the effect. It shows the change in the data due to changes in the means (which are tied to the experimental conditions).
The $$SS_\text{Error}$$ represents the sum of variation for nine scores in our study. That’s a lot more scores, so the $$SS_\text{Error}$$ is often way bigger than than $$SS_\text{Effect}$$. If we left our SS error this way and divided them, we would almost always get numbers less than one, because the $$SS_\text{Error}$$ is so big. What we need to do is bring it down to the average size. So, we might want to divide our $$SS_\text{Error}$$ by 9, after all there were nine scores. However, because we are estimating this property, we divide by the degrees of freedom instead (scores-groups) = 9-3 = 6). Now we have created something new, it’s called the $$MS_\text{Error}$$.
$MS_\text{Error} = \frac{SS_\text{Error}}{df_\text{Error}} \nonumber$
$MS_\text{Error} = \frac{230}{6} = 38.33 \nonumber$
Calculate F
Now that we have done all of the hard work, calculating $$F$$ is easy:
$\text{F} = \frac{\text{measure of effect}}{\text{measure of error}} \nonumber$
$\text{F} = \frac{MS_\text{Effect}}{MS_\text{Error}} \nonumber$
$\text{F} = \frac{36}{38.33} = .939 \nonumber$
Once we have the F statistics, we can find the corresponding significance or p value (the statistics program you use for calculation will likely present this value in the output automatically), and compare it to the pre-determined p critical value to make a decision about the null hypothesis.
This page titled 2.3: Computing the F statistics is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Yang Lydia Yang via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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Proof of the day
Englund
TS Contributor
In this thread we post one (or more if you can't wait) proof a day. I'll start by proving the that
$$V[b|X]=\sigma^2(X'X)^{-1}$$
in the linear regression model. We can write the beta estimator as
$$(X'X)^{-1}X'y=(X'X)^{-1}X'(X\beta+\epsilon)=$$$$(X'X)^{-1}X'X\beta+(X'X)^{-1}X'\epsilon=\beta+(X'X)^{-1}X'\epsilon$$.
Then we have that
$$V[b|X]=E[(b-\beta)(b-\beta)'|X]=E[(\beta+(X'X)^{-1}X'\epsilon-\beta)(\beta+(X'X)^{-1}X'\epsilon-\beta)'|X]=$$$$E[((X'X)^{-1}X'\epsilon)((X'X)^{-1}X'\epsilon)'|X]=E[(X'X)^{-1}X'\epsilon\epsilon'X(X'X)^{-1}|X]=$$$$(X'X)^{-1}X'E[\epsilon\epsilon'|X]X(X'X)^{-1}=$$$$(X'X)^{-1}X'\sigma^2IX(X'X)^{-1}=\sigma^2(X'X)^{-1}X'X(X'X)^{-1}=\sigma^2(X'X)^{-1}$$.
Where
$$E[\epsilon\epsilon'|X]=\sigma^2I$$
follows from one of the assumptions of the classical linear model: the spherical disturbances assumption.
spunky
Smelly poop man with doo doo pants.
me likes this. most of my proofs would come from the field of psychometrics or quantitative psychology though (mostly factor analysis and stuctural equation modelling).
here i'm doing the (rather simple) proof of how the linear factor analysis model can be parameterised as a covariance structure model. it's relevant because as a linear factor model it is unsolvable, but as a covariance structure model it is possible to obtain parameter estimates.
let the obseverd score $$x$$ be defined as the linear factor model $$x = \Lambda F+\epsilon_{i}$$ since it is known that (in the case of multivariate normality) $$E(xx')=\Sigma$$ it trivially follows that:
$$xx' = (\Lambda F+\epsilon)(\Lambda F+\epsilon)'$$
$$xx' = (\Lambda F+\epsilon)(F'\Lambda' + \epsilon')$$
$$xx' = \Lambda FF'\Lambda' + \epsilon F'\Lambda'+ \Lambda F \epsilon' + \epsilon\epsilon'$$
so taking the expectation of both sides:
$$E(xx') = E(\Lambda FF'\Lambda') + 0 + 0 + E(\epsilon\epsilon')$$
which happens because the erros are random and assumed uncorrelated with the Factors and estimated loadings. Now by linearity of expectation and substituting the covariance matrix of the Factors and of the errors we can see that:
$$E(xx') = \Lambda E(FF')\Lambda' + E(\epsilon\epsilon')$$
$$\Sigma=\Lambda \Phi \Lambda' + \Psi$$
which is known as the fundamental equation of Factor Analysis.
Englund
TS Contributor
Okay, since this day is soon over (at least according to Swedish time) and no one posted a proof yet today, I'll post another proof. I'll give a very simple, and possibly boring, proof this time. I'll prove that $$\bar{x}$$ is the value that minimizes the sum $$\sum_{i=1}^n{(x_i-a)^2}$$ (1).
By taking the first derivative with respect to a and setting it equal to zero, we get $$\sum_{i=1}^n{-2(x_i-a)}=0 \Leftrightarrow -2\sum_{i=1}^n{x_i}+2na=0 \Leftrightarrow \sum_{i=1}^n{x_i}=na \Leftrightarrow \bar{x}=a$$.
By checking the second order condition we see that it's equal to 2n, which is always positive, so now we know that $$\bar{x}$$ is at least a local minimum. By investigating (1) it is easily seen that it is also a global minimum.
Dason
I prefer the version that doesn't require the use of calculus.
$$\sum (x_i - a)^2 = \sum (x_i - \bar{x} + \bar{x} - a)^2 = \sum (x_i - \bar{x})^2 + (\bar{x} - a)^2 + 2(x_i - \bar{x})(\bar{x} - a)$$
$$= \sum (x_i - \bar{x})^2 + \sum(\bar{x} - a)^2 + 2\sum(x_i - \bar{x})(\bar{x} - a)$$
Now consider the last summation. Note that in the sum both $$a$$ and $$\bar{x}$$ are constant so we can pull them out
$$= 2(\bar{x} - a) \sum (x_i - \bar{x})$$
We know that that sum is equal to 0 so this shows the third summation disappears.
We are left with
$$\sum (x_i - a)^2 = \sum (x_i - \bar{x})^2 + (\bar{x} - a)^2$$
The first summation we can't control and the second sum is always non-negative so the minimum would occur if we can make it equal to 0 - which happens when $$a=\bar{x}$$.
Now clearly I need a few more details to make it more rigorous but I like that version a little bit more because it also gives hints at what we do in ANOVA when decomposing the sums of squares.
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spunky
Smelly poop man with doo doo pants.
a while ago (before Englund became an MVC) I posted a proof about another result in factor analysis. I thought it would be nice to resurrect it (briefly) and add it here to our small (but growing) compendium of proofs. the original thread is here
and the proof goes like this:
Let $$\bf{S}$$ be a covariance matrix with eigenvalue-eigenvector pairs ($$\lambda_1, \mathbf{e}_1$$), ($$\lambda_2, \mathbf{e}_2$$), ..., ($$\lambda_p, \mathbf{e}_p$$), where
$$\lambda_1 \ge \lambda_2 \ge ... \ge \lambda_p$$. Let $$m<p$$ and define:
$$\bf{L} = \{l_{ij}\} = \left[\sqrt{\lambda_1 }\mathbf{e}_1\ |\ \sqrt{\lambda_2} \mathbf{e}_2\ |\ ...\ |\ \sqrt{\lambda_m} \mathbf{e}_m \right]$$
and:
$$\ \mathbf\Psi = \left( \begin{array}{cccc} \psi_1 & 0 & ... & 0 \\ 0 & \psi_2 & ... & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & ... & \psi_p \\ \end{array} \right) \text{ with } \psi_i = s_{ii} - \sum_{j=1}^{m} l_{ij}^2$$
Then, PROVE:
$$\text{Sum of squared entries of } (\mathbf{S} - (\mathbf{LL'} + \mathbf{\Psi})) \le \lambda_{m+1}^2 + \cdots + \lambda_p^2$$
Spunky's attempt of a proof:
By definition of $$\psi_i$$, we know that the diagonal of $$(\mathbf{S} - (\mathbf{LL'} + \mathbf{\Psi}))$$ is all zeroes. Since
$$(\mathbf{S} - (\mathbf{LL'} + \mathbf{\Psi})))$$ and $$(\mathbf{S} - \mathbf{LL'})$$ have the same elements except on the diagonal, we know that
$$\text{(Sum of squared entries of } (\mathbf{S} - (\mathbf{LL'} + \mathbf{\Psi}))) \leq \text{ Sum of squared entries of } (\mathbf{S} - \mathbf{LL'})$$
Since $$\mathbf{S} = \lambda_1 \mathbf{e}_1 \mathbf{e}'_1 + \cdots + \lambda_p \mathbf{e}_p \mathbf{e}'_p$$
and $$\mathbf{LL'} = \lambda_1 \mathbf{e}_1 \mathbf{e}'_1 + \cdots + \lambda_m \mathbf{e}_m \mathbf{e}'_m$$, then it follows that
$$\mathbf{S} - \mathbf{LL'} = \lambda_{m+1} \mathbf{e}_{m+1} \mathbf{e}'_{m+1} + \cdots + \lambda_p \mathbf{e}_p \mathbf{e}'_p$$
Writing it in matrix form, this is saying $$\mathbf{S} - \mathbf{LL'} = \mathbf{P}_2 \mathbf{\Lambda}_2 \mathbf{P}'_2$$ where
$$\mathbf{P}_2 = [ \mathbf{e}_{m+1} | \cdots | \mathbf{e}_p ]$$ and $$\mathbf{\Lambda}_2 = Diag(\lambda_{m+1}, \cdots, \lambda_{p})$$
Then, the following is true:
$$\text{Sum of squared entries of }(\mathbf{S}- \mathbf{LL'})= \text{tr}((\mathbf{S} - \mathbf{LL'}) (\mathbf{S} - \mathbf{LL'})')=$$
$$\text{tr} (( \mathbf{P}_2 \mathbf{\Lambda}_2 \mathbf{P}'_2)( \mathbf{P}_2 \mathbf{\Lambda}_2 \mathbf{P}'_2)')=\text{tr}( \mathbf{P}_2 \mathbf{\Lambda}_2\mathbf{\Lambda}_2 \mathbf{P}'_2)$$
$$tr(\mathbf{\Lambda}_2\mathbf{\Lambda}_2)=\lambda_{m+1}^2 + \cdots + \lambda_p^2.$$
All the $$\bf{P}_2$$ disappear because by the definition of $$\bf{P}_2$$ we know that $$\bf{P}_2 '\bf{P}_2=\bf{I}$$
Last edited by a moderator:
Englund
TS Contributor
a while ago (before Englund became an MVC)
Time wasn't even defined before I became MVC, so that's per definition impossible
I posted a proof about another result in factor analysis. I thought it would be nice to resurrect it (briefly) and add it here to our small (but growing) compendium of proofs.
and the proof goes like this:
Very nice. If you keep posting stuff on FA I'll be forced to get more familiar with it, which is good
spunky
Smelly poop man with doo doo pants.
If you keep posting stuff on FA I'll be forced to get more familiar with it, which is good
i don't quite understand why but pretty much NO ONE in the Statistics world even touches on Factor Analysis. when it comes to dimension reduction techniques almost all of the undergrad stats textbooks i've seen that deal with intro to multivariate analysis stop at principal components. there may be like some small subsection in some namless appendix that says something about Factor Analysis... but that's it!
WHY!??!
gjay822
New Member
Here, it is under MATH 31 so called statistics. I am still having problem solving in this field.
Dason
We shouldn't let this thread get buried. I'm gonna sticky it.
René
New Member
Nice thread so I make my debut here: The derivation of the Ridge-Estimator in the linear Regression Model.
$$\mathbf{y}=\mathbf{X}\boldsymbol{\beta}+\mathbf{u}, \quad \mathbf{u} \sim N_n(\mathbf{0},\sigma_u^2\mathbf{I}_n)$$
with strong correlation patterns among the vectors within the data matrix $$\mathbf{X} \in Mat_{n,p}(\mathbb{R})$$. The problem with multicollinearity is that single components within the vector of parameters $$\boldsymbol{\beta} \in \mathbb{R}^k$$ can take absurdly large values. So the general idea is to restrict the length of said vector to a prespecified positve real number. Let this restriction been noted by $$\left\| \boldsymbol{\beta} \right\|_2^2=c$$, whereas $$\left\|\cdot \right\|_2$$ is just the euclidian norm on $$\mathbb{R}^n$$.
Eventually one faces the restricted least squares problem
$$Q_n(\boldsymbol{\beta},\lambda) := \left\|\mathbf{y}-\mathbf{X}\boldsymbol{\beta}\right\|_2^2 + \lambda (\left\|\boldsymbol{\beta}\right\|_2^2-c) \rightarrow \min_{\boldsymbol{\theta} \in \mathbf{\Theta}}$$
whereas the Lagrange parameter is assumed to be positive and $$\mathbf{\Theta} \subseteq \mathbb{R}^k \times \mathbb{R}_{>0}$$ is the associated parameter space. The optimization problem is equivalent to
$$Q_n(\boldsymbol{\beta},\lambda):= (\mathbf{y}-\mathbf{X}\boldsymbol{\beta})'(\mathbf{y}-\mathbf{X}\boldsymbol{\beta}) + \lambda (\boldsymbol{\beta}'\boldsymbol{\beta}-c) \rightarrow \min_{\boldsymbol{\theta} \in \mathbf{\Theta}}$$
Taking the derivative with respect to $$\boldsymbol{\beta}$$ yields
$$\displaystyle \frac{\partial}{\partial \boldsymbol{\beta}} Q_T(\boldsymbol{\beta},\lambda) = -2\mathbf{X}'(\mathbf{y}-\mathbf{X}\hat{\boldsymbol{\beta}})+ 2\lambda \hat{\boldsymbol{\beta}}$$
This leads to the first order condition (note you can set the hats already due to the fact that the potential minimizers of the problem above are already given as an implicit function)
$$-\mathbf{X}'\mathbf{y} + \mathbf{X}'\mathbf{X}\hat{\boldsymbol{\beta}} + \lambda \hat{\boldsymbol{\beta}} = \mathbf{0}$$
Arranging terms leads to the modified normal equations
$$(\mathbf{X}'\mathbf{X}+ \lambda \mathbf{I}_k)\hat{\boldsymbol{\beta}} = \mathbf{X}'\mathbf{y}$$
Since $$\mathbf{X}'\mathbf{X}$$ is at least positive semi definite and $$\lambda \mathbf{I}_k$$ is positive definite one yields that*
$$det(\mathbf{X}'\mathbf{X}+ \lambda \mathbf{I}_k) \geq det(\mathbf{X}'\mathbf{X})+det(\lambda\mathbf{I}_k) = det(\mathbf{X}'\mathbf{X}) + \lambda^n >0$$
so that $$(\mathbf{X}'\mathbf{X}+ \lambda \mathbf{I}_k)$$ is an invertible matrix even if the data matrix is of less than full column rank. This finally yields the ridge estimator in its known form
$$\hat{\boldsymbol{\beta}} = (\mathbf{X}'\mathbf{X}+ \lambda \mathbf{I}_k)^{-1}\mathbf{X}'\mathbf{y}$$
Also this is the unique global minimizer of $$Q_n$$ due to the fact that the problem under consideration is just a sum auf convex functions and $$\hat{\boldsymbol{\beta}}$$ is the only local minimizer, so one doesn't need to check the second order conditon and the associated hessians.
*One can find a good proof for that inequality in Magnus, J.R. & Neudecker, H. (1999). Matrix Differential Calculus. Wiley and Sons on page 227 theorem 28.
Last edited:
Dragan
Super Moderator
The Pearson product-moment coefficient of correlation can be interpreted as the cosine of the angle between variable vectors in $$n$$ dimensional space. Here, I will show the relationship between the Pearson and Spearman (rank-based) correlation coefficients for the bivariate normal distribution through the following series:
$$\sum_{n=1}^{\infty }\frac{\cos nx}{n}$$.
If we let $$z=\cos x+i\sin x$$, then
$$\sum_{n=1}^{m}y^{n-1}z^{n}=\frac{z\left \{ 1-\left ( yz \right )^{m} \right \}}{1-yz}$$
where it follows for $$\left | y \right |<1$$,
$$\sum_{n=1}^{\infty }y^{n-1}\left ( \cos nx+i\sin nx \right )=\frac{\cos x+i\sin x}{1-y\cos x-yi\sin x}$$
$$=\frac{\left ( \cos x-y \right )+i\sin x}{1-2y+y^{2}}$$, so that
$$\sum_{n-1}^{\infty }\cos nx=\frac{\cos x-y}{1-2y\cos x+y^{2}}$$.
This series is uniformly convergent for all values of $$y$$ and for $$\left | y \right |\leq p<1$$. Hence, integrating with respect to $$y$$, where $$0<y<1$$ gives
$$\sum_{n=1}^{\infty }y^{n}\frac{\cos nx}{n}$$
$$=\int_{0}^{y}\frac{\cos x-t}{1-2t\cos x+t^{2}}dt$$
$$=-\frac{1}{2}\ln \left ( 1-2y\cos x+y^{2} \right )$$.
Suppose that $$x$$ is neither zero nor a multplei of $$2\pi$$.
Then the series $$\sum_{n=1}^{\infty }\frac{\cos nx}{n}$$ is convergent, and, for $$0\leq y\leq 1$$, $$y^{n}$$, is positive, monotonic, decreasing and bounded. As such the series:
$$\sum_{n=1}^{\infty }y^{n}\frac{\cos nx}{n}$$
is therefore uniformly convergent on the interval $$0\leq x\leq 1$$.
Subsequently letting $$x\rightarrow 1$$, then it follows that if $$x$$ is neither $$0$$ nor a multiple of $$2\pi$$ we have
$$\sum_{n=1}^{\infty }\frac{\cos nx}{n} =-\frac{1}{2}\ln \left ( 2-2\cos x \right )$$
$$=-\ln \left ( 2\sin \frac{1}{2} x\right )$$.
Setting $$x=\frac{\pi }{3}r_{s}$$ and exponentiating $$e^{-1}$$ gives the relationship (for large sample sizes) between the Pearson and Spearman correlation coefficients as:
$$r_{p}=2\sin\left ( \frac{\pi }{6}r _{s}\right )$$
for the bivariate normal distribution.
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# Mad Minute Math Worksheets For Second Grade
Angelina Elisabeth November 15, 2020 Worksheet
Many people believe that you have to have a knack for math in order to do well in it. However, understanding the basic principles of math does not need any innate talent, or a genius intellect. What
The present generation seems to be blessed immensely with intellect and the benefits of mastering math are something worth considering. It is a well-known fact that math is not a subject that one learns by simply reading the problems and its solutions. In order to master the subject, earnest practice on multiple problems is the best way to go. However, not every person is bestowed with required materials like math worksheets to receive adequate amount of practice.
Most of even beginning algebra depends on being able to do two things–one, doing multiplication quickly and accurately in your head, two, knowing how to add, subtract, multiply, and divide fractions. You might remember a concept in algebra called ”factoring.” Factoring means breaking up into parts that are multiplied together to give you the whole. You can factor numbers. For instance, 6 factors into 2 and 3–2×3 =6. In elementary algebra we learn to factor expressions such as x^2+4x+4. This particular expression is easily factorable into (x+2)^2. If this doesn’t make any sense to you, don’t worry about it. Just trust me, if you don’t know your multiplication tables, you can’t factor. If you can’t factor, you won’t do well at all in algebra, geometry, or trigonometry.
No matter what materials you choose, it is most important that you supervise your child constantly so that mistakes get caught rather than practiced. I learned this particular lesson the hard way. When my daughter was young, she did something that needed ”attention.” I no longer remember what it was that she did, but I told her to write the sentence ”I will not disobey my parents again” 50 times. I should have known better, but I didn’t check on her at the beginning and then I got busy. So, sometime later, she brought me 50 sentences of ”I will not disobey my parents agen.” She had just practiced misspelling ”again” as ”agen” — 50 times! I’m not certain that we ever really got that fixed.
The primary problem with most math worksheets is that the problems are already written out and the child need only write the answers. For learning and practicing the basic skills of addition, subtraction, multiplication, and division, it is much more beneficial for the child to write out the entire fact and say the entire fact out loud. A child will learn a multiplication fact much faster if they are writing out 6 x 8 = 48 at the same time they are saying ”six times eight is forty-eight” than if they just see 6 x 8 = ___ and only have to supply the 48.
At the grassroots level, teachers in schools are given a packed curriculum for the year. Schools try to teach the students a number of procedures without delving much into its finer details. Hence, the student is left in a confounding position as to when a particular procedure must be used. The key ingredient to understanding math is constant practice and math assignment help. Unfortunately, this is not a common scenario among the popular math classes.
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Question
# Let WW be a subset of a vector space VV. By justifying your answer, determine whether...
Let WW be a subset of a vector space VV. By justifying your answer, determine whether WW is a subspace of VV.
(a) [5 marks] W={(x1,x2,x3,x4):x1x4=0}W={(x1,x2,x3,x4):x1x4=0} and V=R4V=R4.
(b) [5 marks] W={A:|A|≥1}W={A:|A|≥1} and V=M3,3V=M3,3, where |A||A| is the determinant of AA.
(c) [10 marks] W={p(x)=a0+a1x+a2x2+a3x3:a0=a1anda2=a3}W={p(x)=a0+a1x+a2x2+a3x3:a0=a1anda2=a3} and V=P3V=P3.
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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• Index
• » Help Me !
## #1 2008-10-22 02:47:02
Danbee
Member
Offline
(x+5)² +(y+2)²=169
xy-21=0
## #2 2008-10-26 10:35:18
justlookingforthemoment
Moderator
Offline
If you think of this problem visually, you are trying to find the points on a graph where a hyperbola and a circle intersect.
1 -- (x+5)² +(y+2)²=169
2 -- xy - 21=0
xy - 21 = 0
xy = 21
y = 21/x
substitute y = 21/x into equation 1:
(x+5)² +(21/x+2)²=169
and expand:
x² + 10x + 25 + 441/(x²) + 84/x + 4 = 169
which gives:
x^4 + 10x³ - 140x² + 84x + 441 = 0
(x - 7)(x³ + 17x² - 21x - 63) = 0
solve for x:
∴ x ≈ -17.9734 or x ≈ -1.44775 or x ≈ 2.42112 or x = 7
• Index
• » Help Me !
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Home » single variable calculus vs calculus
# single variable calculus vs calculus
if you are interested to know the differences between single variables and calculus this is the right post for you.
## the difference between single variable calculus vs calculus
To know the difference between a single variable and calculus, you have to understand that calculus is two parts: single variable and multivariable.
students begin with single variable calculus to finally be able to study and understand multivariable calculus.
At the beginning of calculus courses, students take single variable calculus courses which are Calculus 1 and calculus 2
Before that students take precalculus courses in high school or in college to prepare for college math courses. Especially single variable calculus that leads finally to multivariable calculus.
After finishing the precalculus course students go to the next step which is Calculus 1. It is the first course of single variable calculus you will study. You would be learning the famous topics of calculus like:
After finishing the Calculus 1 course you would be going to the next step which is taking calculus 2. In other words, you will be studying the same topics you’ve discovered in Calculus 1 like integrals and derivatives in an advanced and more detailed way.
This is why calculus 2 is more difficult and comes after calculus 1.
Until this moment you’re still studying single variable calculus. In other words, you are applying derivatives and integrals to single variable functions which mean functions that have one single variable like this example below :
The first step you will have in multivariable calculus is when you take calculus 3 or multivariable calculus, they are the same.
Then going further by taking differential equations which are always considered multivariable calculus, not a single variable calculus like Calculus 1 and calculus 2.
In multivariable calculus will be studying multivariable functions like the example below:
but which is easier, single variable or multivariable calculus?
this is what will be responding to in the next paragraph.
### Which is easier, Multivariable calculus or single variable calculus
As you’ve seen in the preview images you can notice a big difference between multivariable calculus function and single variable calculus function. Single variable calculus functions are easier and not more complex than multivariable ones.
For instance, if we have a single variable calculus function and want to apply a derivative it would be easier to do that. It requires less amount of time and effort, we wouldn’t notice this simple example below:
while if you’ve been asked to make a derivative of multivariable calculus they will require a lot of time and effort to finish. This is another simple to understand the difference between the previous example.
So as you can see there is no doubt, that multivariable calculus is more difficult than single variable calculus.
Moreover, it is impossible to study multivariable calculus before taking single-variable courses. Because all the basics that you will need in multivariable calculus are taught in single-variable calculus courses like Calculus 1 and calculus 2.
To simplify more multivariable calculus is just the group of single variable calculus applications that you apply together in a function.
In other words, instead of making one single derivative operation in a single variable calculus function. You make 2 or 3 derivatives combined on the same function in multivariable calculus.
For this reason, they have classified calculus courses gradually from Calculus 1 to calculus 2 then finally calculus 3.
## Is single variable calculus hard
Single variable calculus is the easiest calculus course you would be taking as a stem major. it includes Calculus 1 and calculus 2 courses that are based on applying derivatives and integrals to single variable functions.
It just requires solid basics or foundations in a pre-calculus course to succeed and not struggle, especially in calculus 2 Where a lot of people fail
To succeed in single variable calculus you have to focus on the three following topics:
• Limits and continuity it already included in the pre-calculus course that you should normally take at a high school. If you haven’t taken yet precalculus courses you should focus a lot on this topic.
• the second topic that you should focus on is derivatives. Typically derivatives are taught in the Calculus 1-course college calculus course. If you are still in high school you could study derivatives in AP calculus courses like AP calculus 1 and calculus 2 in addtion to honors calculus.
you could read this related article if you are interested to know what are AP classes and the difference between AP Calculus classes and honors calculus.
• The last subject you should have to understand or succeed in single variable calculus is integral. after finishing or studying limits and continuity in audition derivatives it is possible to go to the next step and learn Integrals.
It is impossible to understand or study integrals and you have not yet learned the basics about limits and continuity. In addition to derivatives that have solid Basics to integrals.
## difference between single variable and calculus 1
There is no difference between single variable calculus and calculus 1. Calculus 1 is considered a single variable calculus in addition to calculus 2. While other topics like differential equations and multivariable calculus are considered multivariable calculus.
So single variable calculus is the same as Calculus 1 and calculus 2 they are all based on applying calculus topics like derivatives and integrals I’m single variable functions.
If you are a Stem major or thinking to major in engineering, mathematics, or science branches you would be required to complete single variable calculus and multivariable calculus both.
but for all the majors like business economics and finance, they will not only be required to complete a single variable calculus and even less complicated calculus courses like applied calculus add survey calculus.
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# Exploring Genetic Algorithms with Kotlin
Jun 7, 2016 · 4 min read
This story is about chimps trying to land on Mars. Yes, really.
There are a lot and some of them can succeed. We use a genetic algorithm to select the best chimps of each generation until the selection is down to the hero who will land safely on Mars with the fullest fuel tank possible.
Disclaimer: This post is about my first experiments on genetic algorithms. I’m not an expert.
# Landing on Mars
The algorithm answers CodinGame’s first puzzle on landing on Mars.
The Lander evolves in two dimensions. The chimp must control it using angle and thrust power commands. The gravity is -3,711 m/s² (see details on CodinGame’s website).
The challenge is to find the correct commands to have the lander hit the ground on the landing zone (horizontal) with a vertical speed lower than 40 m/s and an horizontal speed under 20 m/s.
Genetic Algorithms try to mimic the natural selection in order to resolve problems where solutions are hard to find.
The steps are :
1. generate a population of random genomes,
2. convert theses genome into input parameters of our simulation,
3. play the simulation,
4. evaluate the results of each genome,
5. select the best genomes, recombine and mutate them to produce the next generation,
6. iterate until a solution is found.
# Representing the model
We must have a simulator representing precisely the model.
For that purpose, I split my code into a simplified cartesian system, small physics classes and the Lander.
This is the code of my simplified cartesian system.
The main idea of these classes is to simplify the manipulation of points, vector and lines.
Points and Vector will help us managing the position, the speed, the acceleration of the lander. The rotate function of Vector will be the only one containing trigonometry.
The line is here to represent Mars surface and check if the lander hits the ground.
Kotlin operator overloading is interesting to manipulate these concepts and simplify our domain. This will allow us to write code like:
`val endPoint = startPoint + moveVectorval composedVector = vector1 + vector2`
One more file for physics :
We now have all the basics to implement the Lander simulator:
The state class holds the Lander state at one moment : it’s position, angle, power, speed and fuel.
The ControlCmd holds the command that can be asked : an angle and a thrust power.
Given a lander at an initial state, we can simulate its complete trajectory after having applied a list of commands. The simulation should respect exactly the target domain constraints if we want our algorithm to be relevant. In our case, we check that applying a list of static commands ( 100 × (15°,power 1) for example) to an initial state results in having the Lander at the exact same state on CodinGame simulator.
Ok, we have our simulator. Now, let’s code the genetic algorithm.
# The Genetic Algorithm
So here is the code of a generic Genetic Algorithm in kotlin:
The Gene class represents a single gene implemented by a random double. We provide a method to simplify its use as an Int.
A Genome is an array of Genes.
We have a utility class GenomeAndResult which keeps together a genome and the corresponding object created from it. This class is generic and can be reused for any domain.
The entry point of the algorithm is the findBestChimp function. This function takes two parameters that are also functions:
1. A create function that builds a object from a genome.
2. A fitness function that evaluates an object according to the goal of the algorithm.
The genetic algorithm creates a population from random genomes. The population is sorted with the fitness function. The selection process then randomly picks individuals, the best ones having more chances to procreate.
The creation of new individuals is done through a crossover combination of parent genomes. After the crossover, the genome is altered by mutations that help exploring new areas.
The last effort is now on implementing the function that creates a Lander from a genome and the fitness function. In fact it is the trickiest part. The performance of the genetic algorithm (it’s ability to rapidly converge to a solution) is defined by the quality of the creation and fitness functions.
Even if the lander crashed we should be able to compare the crashes to allow the algorithm to give more chances to the «best crashes» genomes.
Here is my code for the first CodinGame puzzle. The Lander has just a vertical trajectory and we don’t have to manage it’s angle.
Running the program we can see that the generations are rapidly converging to the best solution.
Using the commands found by the algorithm shows a perfect landing.
All the code of the MarsLander 1 is available on this github repository.
Implementing the create and fitness functions for Mars Lander 2 requires 40 lines of code. Look them up for some nice touchdowns:
Written by
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# Central Limit Theorem Application
• Dec 1st 2011, 05:09 PM
bjnovak
Central Limit Theorem Application
A die is continuously rolled 65 -1 times. What is the probability that the total sum of all rolls does not exceed 225?
I know that the expected value for each roll is 7/2 so this means that the mean sum of all the rolls would be 64*(7/2) = 224.
Also, the variance for each roll is 35/12. How do I use this to find the std deviation in all the rolls?
• Dec 2nd 2011, 04:14 AM
SpringFan25
Re: Central Limit Theorem Application
The variance of the sum is $\displaystyle n\sigma^2$ where $\displaystyle \sigma^2$ is the variance of 1 roll.
However to apply the central limit theorum you should be thinking about the properties of the sample mean. if the total does not exceed 225, what can you say about the sample mean?
Spoiler:
$\displaystyle P(total< 225) = P(sample~mean < \frac{225}{64})$
by CLT, sample mean approximately follows $\displaystyle N(3.5,\frac{\sigma^2}{n})$
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# Thread: Quick equivalence relation problem
1. ## Quick equivalence relation problem
define a relation R on $R^R$, the set of functions from R to R by f R g if f(0) = g(0). prove that R is an equivalence relation on $R^R$. let f(x) = x for all $x \in R$. describe [f].
2. Originally Posted by Projectt
define a relation R on $R^R$, the set of functions from R to R by f R g if f(0) = g(0). prove that R is an equivalence relation on $R^R$. let f(x) = x for all $x \in R$. describe [f].
Well, whoever gave you this problem expects you to know the definition of "equivalence relation"!
An equivalence relation, xRy, on a set is an equivalence relation if
1) It is "reflexive": xRx for every x in the set.
Suppose f is a function from R to R. Is fRf? That is, is f(0)= f(0)?
2) It is "symmetric ": if xRy then yRx.
Suppose f and g are such functions and fRg (that is, f(0)= g(0)). Is gRf (is g(0)= g(0))?
3) It is "transitive": if xRy and yRz then xRz.
Suppose f, g, and h are such functions, fRg (that is, f(0)= g(0)) and gRh (that is, g(0)= h(0)). Is fRh (is f(0)= h(0))?
[f] is the "equivalence class" of f, the set of all functions that are equivalent to f(x)= x. If g(x) is equivalent to that, if xRg, what must be true about g(x)?
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# Summation with combinations
Prove that $n$ divides $$\sum_{d \mid \gcd(n,k)} \mu(d) \binom{n/d}{k/d}$$ for every natural number $n$ and for every $k$ where $1 \leq k \leq n.$ Note: $\mu(n)$ denotes the Möbius function.
I have tried numerous values for this summation and the result seems to hold true. For example, if $n = 20, k = 12$
$$\sum_{d \mid 4} \mu(d) \binom{20/d}{12/d} = \mu(1) \binom{20}{12}+\mu(2) \binom{10}{6}+\mu(4)\binom{5}{3} = \binom{20}{12}-\binom{10}{6}=125760,$$ which is divisible by $20$. Similarly if we tried it for any $k$ with $1 \leq k\leq 20$, we would have $20$ divide the expression.
How do I prove this result in the general case? That is, given any positive integer $n$, for all $k$ with $1 \leq k \leq n$ $$n \mid \sum_{d \mid \gcd(n,k)} \mu(d) \binom{n/d}{k/d}.$$
#### Solutions Collecting From Web of "Summation with combinations"
A Combinatorial Approach (Hint)
Let $G:=C_n$ be the cyclic group of order $n$ generated by $g$ acting on the left on the set $X:=\{0,1\}^n$ by rotation:
$$g\cdot\left(x_1,x_2,\ldots,x_{n-1},x_n\right)=\left(x_n,x_1,x_2,\ldots,x_{n-1}\right)$$
for all $\left(x_1,x_2,\ldots,x_n\right)\in X$. Then, if $Y\subseteq X$ consisting of $n$-tuples with $k$ occurrences of $1$’s, then $G$ also acts on $Y$. An asymmetric $G$-orbit in $Y$ is a $G$-orbit in $Y$ that consists of exactly $n$ elements. Prove that the number of asymmetric $G$-orbits in $Y$ is precisely $\displaystyle \frac{1}{n}\,\sum_{d\mid\gcd(n,k)}\,\mu(d)\,\binom{n/d}{k/d}$. You may also show that the number of all $G$-orbits in $Y$ is $\displaystyle \frac{1}{n}\,\sum_{d\mid \gcd(n,k)}\,\phi(d)\,\binom{n/d}{k/d}$, where $\phi$ is Euler’s totient function.
Even more generally, the number of ways to arrange $n=k_1+k_2+\ldots+k_r$ objects on a circle, where there are $r$ mutually distinct types and the $j$-th type consists of $k_j$ identical objects, is $$\frac{1}{n}\,\sum_{d\mid\gcd\left(k_1,k_2,\ldots,k_n\right)}\,\phi(d)\,\binom{n/d}{k_1/d,k_2/d,\ldots,k_r/d}\,.$$
There are exactly
$$\frac{1}{n}\,\sum_{d\mid\gcd\left(k_1,k_2,\ldots,k_n\right)}\,\mu(d)\,\binom{n/d}{k_1/d,k_2/d,\ldots,k_r/d}$$
asymmetric arrangements. For $r=n$ and $k_1=k_2=\ldots=k_n=1$, we retrieve the number of ways to arrange $n$ mutually distinct objects on a circle: $$\frac{1}{n}\,\binom{n}{\underbrace{1,1,\ldots,1}_{n\text{ ones}}}=(n-1)!\,.$$
P.S. Just saw that arctic tern also posted an essentially the same solution.
This proof uses symmetry (so, group actions) and a combinatorial interpretation of $v_{n,k}$.
Any group $G$ acts regularly on itself by (say left) multiplication. If $G$ acts on a set $X$, then there is an induced action of $G$ on the collection of $k$-subsets of $X$ (i.e. subsets of $X$ of size $k$). Let $v_{n,k}$ be the number of $k$-subsets of $\mathbb{Z}/n\mathbb{Z}$ with “no symmetry,” i.e. no stabilizer with respect to this induced action. This is the collection of subsets $A$ for which $r+A=A$ implies $r=0$ in $\mathbb{Z}/n\mathbb{Z}$.
First, we prove the following:
$$\sum_{d\mid (n,k)}v_{n/d,k/d}=\binom{n}{k}.$$
Suppose $A$ is any subset of $\mathbb{Z}/n\mathbb{Z}$, and that its symmetry group $S$ has size $d=|S|$. (Note that $|S|=d$ is equivalent to $S=\frac{n}{d}\mathbb{Z}/n\mathbb{Z}$.) Since $S$ acts freely on $A$ we know $|S|$ divides $|A|$, and from Lagrange’s theorem we know $|S|$ divides $|\mathbb{Z}/n\mathbb{Z}|$, i.e. $d$ divides $n$ and $k$. That $S$ acts freely on $A$ also implies that $A$ is a union of cosets of $S$.
The $k$-subsets of $\mathbb{Z}/n\mathbb{Z}$ with symmetry group of size $d$ correspond bijectively to $\frac{k}{d}$-subsets of $\mathbb{Z}/\frac{n}{d}\mathbb{Z}$ with trivial stabilizer. The correspondence is given in one direction by applying the projection $\mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/\frac{n}{d}\mathbb{Z}$ and in the other direction by taking preimages. So if $M(n,k,d)$ denotes the collection of $k$-subsets of $\mathbb{Z}/n\mathbb{Z}$ with symmetry group of size $d$ we get
$$\binom{n}{k}=\sum_d |M(n,k,d)|=\sum_{d\mid n,k}|M(\frac{n}{d},\frac{k}{d},1)|=\sum_{d\mid \gcd(n,k)}v_{n/d,k/d}.$$
Hence my $v_{n,d}$s are the same as your $u_{n,d}$s. Finally, notice $\mathbb{Z}/n\mathbb{Z}$ acts freely on $M(n,d,1)$, so we find that $v_{n,d}$ is divisible by $n$.
I would like to present the algebraic aspects of this problem to
facilitate understanding. Suppose we have $r$ types of objects
(e.g. colors) with $k_1 + k_2 + \cdots + k_r = n$ objects total where
$k_q$ gives the number of objects of color $q$ and we ask about the
number of necklaces we can form with these (rotational symmetry as
opposed to dihedral symmetry).
Applying the Polya Enumeration Theorem (PET) we have the cycle
index of the cyclic group
$$Z(C_n) = \frac{1}{n}\sum_{d|n} \varphi(d) a_d^{n/d}.$$
PET now yields for the generating function of necklaces using at most
$r$ colors
$$q_n = \frac{1}{n}\sum_{d|n} \varphi(d) (A_1^d + A_2^d +\cdots+A_r^d)^{n/d}.$$
We now introduce the concept of primitive necklaces $p_n$ i.e.
necklaces on at most $r$ colors not having any rotational
symmetry. Observe that an ordinary necklace is formed by concatenating
$d$ copies of a primitive necklace of size $n/d.$ (In fact it does not
matter where we open the primitive necklace ($n/d$ possibilities)
because when we arrange the copies of the opened necklace we always
get the same necklace regardless of where we opened the primitive
necklace.)
We will use a variety of Moebius inversion which is
Inclusion-Exclusion on the divisor poset in order to compute $p_n$ (a
generating function) and extract the desired coefficient. The possible
symmetries that can occur correspond to the divisors $f$ of $n$
($f|n$).
Using the variable $f$ we obtain as explained a segment of length
$n/f$ being repeated $f$ times, copies being placed next to each
other, thus creating $n/f$ cycles of length $f.$. These segments are
themselves necklaces of length $n/f.$ This means that the maximal
symmetry (smallest size of the constituent cycles) is a divisor of
$n/f$ because the segment could itself be a concatenation of repeated
segments. Ordering these in a poset by division yields an upside-down
instance of the divisor poset of $n$. Note that the generating
function for the contribution from $f$ is not
$$\frac{1}{n/f}\sum_{d|n/f} \varphi(d) (A_1^d + A_2^d + \cdots + A_r^d)^{n/f/d}.$$
but rather
$$\frac{1}{n/f}\sum_{d|n/f} \varphi(d) (A_1^{df} + A_2^{df} + \cdots + A_r^{df})^{n/f/d}.$$
which represents the $f$ copies of the source segment.
We thus obtain by Inclusion-Exclusion
$$\sum_{f|n} \mu(f) \frac{f}{n}\sum_{d|n/f} \varphi(d) (A_1^{df} + A_2^{df} + \cdots + A_r^{df})^{n/f/d}.$$
We put $fd=k$ so that $d=k/f$ to get
$$\sum_{f|n} \mu(f) \frac{f}{n}\sum_{k/f|n/f} \varphi(k/f) (A_1^{k} + A_2^{k} + \cdots + A_r^{k})^{n/k} \\ = \frac{1}{n} \sum_{f|n} f\mu(f) \sum_{k|n \wedge f|k} \varphi(k/f) (A_1^{k} + A_2^{k} + \cdots + A_r^{k})^{n/k} \\ = \frac{1}{n} \sum_{k|n} (A_1^{k} + A_2^{k} + \cdots + A_r^{k})^{n/k} \sum_{f|k} f\mu(f) \varphi(k/f).$$
There are several ways to simplify the term
$$\sum_{f|k} f\mu(f) \varphi(k/f).$$
E.g. note that if
$$L_1(s) = \sum_{n\ge 1} \frac{n\mu(n)}{n^s} = \prod_p \left(1-\frac{p}{p^s}\right) = \frac{1}{\zeta(s-1)}$$
and
$$L_2(s) = \sum_{n\ge 1} \frac{\varphi(n)}{n^s} = \frac{\zeta(s-1)}{\zeta(s)} \quad\text{because}\quad \sum_{n\ge 1} \frac{1}{n^s} \sum_{d|n} \varphi(d) = \zeta(s-1)$$
then
$$L_1(s) L_2(s) = \frac{1}{\zeta(s)} \quad\text{and hence}\quad \sum_{f|k} f\mu(f) \varphi(k/f) = \mu(k).$$
Substitute this into the formula to obtain
$$\frac{1}{n} \sum_{k|n} \mu(k) (A_1^{k} + A_2^{k} +\cdots + A_r^{k})^{n/k}.$$
We seek
$$[A_1^{k_1} A_2^{k_2} \cdots A_r^{k_r}] \frac{1}{n} \sum_{k|n} \mu(k) (A_1^{k} + A_2^{k} +\cdots + A_r^{k})^{n/k}.$$
Now observe that the term in the variables only produces powers that
are multiples of $k$ so we get the condition that
$$k|\gcd(n, k_1, k_2, \ldots k_r)$$
(we see that this produces a divisor of $n$) in which case we obtain a
contribution of (using $d$ for $k$ for readability)
$${n/d\choose k_1/d, k_2/d,\ldots k_r/d}$$
for an end result of
$$\frac{1}{n}\sum_{d|\gcd(k_1, k_2, \ldots k_r)} \mu(d) {n/d\choose k_1/d, k_2/d,\ldots k_r/d}.$$
We now conclude by inspection that the sum is indeed a multiple of
$n.$
A similar problem appeared at this
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# Exercise 29 - Chapter 4
edited June 2018
1) Justify each of the four steps $$(=, \le, \le, =)$$ in Eq. (4.27).
2) In the case $$\mathcal{V} = \textbf{Bool}$$, we can directly show each of the four steps in Eq. (4.27) is actually an equality. How?
Equation 4.27
Φ(p, q) I ⊗ Φ(p, q) ≤ P(p, p) ⊗ Φ(p, q) ≤ Ü P(p, p 1 ) ⊗ Φ(p 1 , q) (U P .Φ)(p, q).
• Options
1.
1. $$\Phi(p,q)=I\otimes\Phi(p,q)$$ holds because $$\mathcal{V}$$ is a symmetric monoidal preorder (in particular, this is property b) in the definition of a symmetric monoidal structure on a preorder). $$I\otimes\Phi(p,q)\leq\mathcal{P}(p,p)\otimes\Phi(p,q)$$ holds since $$\mathcal{P}$$ is a $$\mathcal{V}-$$category (in particular, property a) in the definition of a $$\mathcal{V}-$$category and property a) in the definition of a monoidal structure on a preorder). $$\mathcal{P}(p,p)\otimes\Phi(p,q)\leq\bigvee_{p_1\in\mathcal{P}}\mathcal{P}(p,p_1)\otimes\Phi(p_1,q)$$ is true since $$p\in\mathcal{P}$$, by the definition of join (joins always exist since $$\mathcal{V}$$ is a quantale). $$\bigvee_{p_1\in\mathcal{P}}\mathcal{P}(p,p_1)\otimes\Phi(p_1,q)=(U_\mathcal{P}\circ\Phi)(p,q)$$ by the definition of the unit profunctor and the definition of profunctor composition.
2. If $$\Phi(p,q)=\mathrm{true}$$ then, since $$\mathrm{true}$$ is the top element of Bool, all of the other profunctors must also evaluate to $$\mathrm{true}$$, since there is nothing larger for them to be. So all the inequalities are equalities. The only other possible value for $$(U_\mathcal{P}\circ\Phi)(p,q)$$ is $$\mathrm{false}$$, which is the bottom element of Bool, so the other profunctors must also be $$\mathrm{false}$$, as there is nothing smaller for them to be. So, again all the inequalities are equalities. Since these two possibilities cover all possible values of the profunctors in each part of the chain of equalities and inequalities, all of the relations in the chain are equalities when $$\mathcal{V}=\textbf{Bool}$$.
Comment Source:1. \$$\Phi(p,q)=I\otimes\Phi(p,q)\$$ holds because \$$\mathcal{V}\$$ is a symmetric monoidal preorder (in particular, this is property b) in the definition of a symmetric monoidal structure on a preorder). \$$I\otimes\Phi(p,q)\leq\mathcal{P}(p,p)\otimes\Phi(p,q)\$$ holds since \$$\mathcal{P}\$$ is a \$$\mathcal{V}-\$$category (in particular, property a) in the definition of a \$$\mathcal{V}-\$$category and property a) in the definition of a monoidal structure on a preorder). \$$\mathcal{P}(p,p)\otimes\Phi(p,q)\leq\bigvee\_{p\_1\in\mathcal{P}}\mathcal{P}(p,p\_1)\otimes\Phi(p\_1,q)\$$ is true since \$$p\in\mathcal{P}\$$, by the definition of join (joins always exist since \$$\mathcal{V}\$$ is a quantale). \$$\bigvee\_{p\_1\in\mathcal{P}}\mathcal{P}(p,p\_1)\otimes\Phi(p\_1,q)=(U\_\mathcal{P}\circ\Phi)(p,q)\$$ by the definition of the unit profunctor and the definition of profunctor composition. 2. If \$$\Phi(p,q)=\mathrm{true}\$$ then, since \$$\mathrm{true}\$$ is the top element of **Bool**, all of the other profunctors must also evaluate to \$$\mathrm{true}\$$, since there is nothing larger for them to be. So all the inequalities are equalities. The only other possible value for \$$(U\_\mathcal{P}\circ\Phi)(p,q)\$$ is \$$\mathrm{false}\$$, which is the bottom element of **Bool**, so the other profunctors must also be \$$\mathrm{false}\$$, as there is nothing smaller for them to be. So, again all the inequalities are equalities. Since these two possibilities cover all possible values of the profunctors in each part of the chain of equalities and inequalities, all of the relations in the chain are equalities when \$$\mathcal{V}=\textbf{Bool}\$$.
• Options
2.
In the current version of Seven Sketches (in Compositionality), there is a third part:
3 Justify each of the three steps $$(=,\leq,\leq)\newcommand{\cat}[1]{\mathcal{#1}}$$ in \eqref{SecondHalfLeftUnitLaw}. $$\label{SecondHalfLeftUnitLaw} \tag{4.29} \cat{P}(p,p_1)\otimes\Phi(p_1,q)=\cat{P}(p,p_1)\otimes\Phi(p_1,q)\otimes I\leq\cat{P}(p,p_1)\otimes\Phi(p_1,q)\otimes\cat{Q}(q,q)\leq\Phi(p,q).$$(Which the text says follows from Definitions 2.46 and 4.8.)
These follow from $$\mathcal{V}$$ being a symmetric monoidal preorder (same property as in the above comment), $$\mathcal{Q}$$ being a $$\mathcal{V}-$$category (same as above, which is Definition 2.46), and exercise 4.9 (which immediately follows definition 4.8 and is one of the equivalent ways of defining $$\mathcal{V}-$$profunctors).
Comment Source:In the current version of Seven Sketches (in Compositionality), there is a third part: 3 Justify each of the three steps \$$(=,\leq,\leq)\newcommand{\cat}[1]{\mathcal{#1}}\$$ in \eqref{SecondHalfLeftUnitLaw}. $$\label{SecondHalfLeftUnitLaw} \tag{4.29} \cat{P}(p,p_1)\otimes\Phi(p_1,q)=\cat{P}(p,p_1)\otimes\Phi(p_1,q)\otimes I\leq\cat{P}(p,p_1)\otimes\Phi(p_1,q)\otimes\cat{Q}(q,q)\leq\Phi(p,q).$$(Which the text says follows from Definitions 2.46 and 4.8.) These follow from \$$\mathcal{V}\$$ being a symmetric monoidal preorder (same property as in the above comment), \$$\mathcal{Q}\$$ being a \$$\mathcal{V}-\$$category (same as above, which is Definition 2.46), and exercise 4.9 (which immediately follows definition 4.8 and is one of the equivalent ways of defining \$$\mathcal{V}-\$$profunctors).
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# Four kilograms of water are placed in an inclosed volume of 1 m^3. Heat is added until the temperature is 150 C.Find the pressure, mass and volume of the vapor.
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
We'll calculate the volume of 4 kg of saturated vapor, at the temperature of 150 C.
0.3928*4 = 1.5712 m^3
The resulted volume is bigger than the given volume of 1 m^3.
If the given volume is less than the resulted volume, the state is in the quality region, whose pressure is P = 475.8 KPa.
Now, we'll calculate the mass of the vapor. First, we'll have to determine the quality.
The quality of the mixture is given by the ratio:
x = mass of saturated vapor/total mass
v = 1/4
v = 0.25 m^3/Kg
The total volume of the mixture is the sum of the volume occupied by the liquid and the volume occupied by the vapor.
v = v1 + x(v2-v1)
0.25 = 0.00109 + x(0.3928 - 0.00109)
We'll remove the brackets and we'll have:
0.25 = 0.00109 + 0.39171x
0.39171x = 0.24891
x = 0.24891/0.39171
x = 0.6354
Mass of saturated vapor = Total mass*x
m = 4* 0.6354
m = 2.542 Kg
The volume of the vapor is:
V = 0.3928*2.542
V = 0.998 m^3
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# Proofs with the "Purified Pigeonhole Principle". [closed]
In his EWD980 and EWD1094, Dijkstra provides a formulation of the Pigeonhole Principle as such:
"For a non-empty, finite bag of numbers, the maximum value is at least the average (and the minimum is at most the average)"
One of the examples he provides where this is useful is in a proof of the following statement:
On a ranch, each cowboy is the exclusive owner of at least one horse. Show that the number of cowboys $$\leq$$ the number of horses.
The proof is simple if we realise that the minimum number of horses owned by a cowboy is one and thus the average number of horses owned by a cowboy is greater than or equal to 1, which gives the expression:
$$\frac{\text{number of horses}}{\text{number of cowboys}} \geq 1$$
I thought that this proof was quite neat and would like to ask if there are any other problems which can be solved quite elegantly using this "Purified" Pigeonhole Principle".
My intention is to source various examples whereby this formulation principle is used in an elegant manner, especially when the usual formulation makes the argument much more complicated.
• FYI I know this as the "third principle of PGP". (I believe I got that name from Koh Khee Meng's book, but I don't have it available as a reference.) $\quad$ Note that it's also true of real numbers (instead of just integers that PGP usually requires). Commented Jul 19 at 4:01
• EWD980, which you mention, is cs.utexas.edu/~EWD/transcriptions/EWD09xx/… and has a couple of examples. Commented Jul 19 at 7:30
• There is also an example at protocols.netlab.uky.edu/~calvert/classes/275/… Commented Jul 19 at 7:34
Solution: The optimal packing of the plane by disks–the hexagonal packing–has a packing density of about $$90.7\%$$. (The important fact is that this number is strictly larger than $$90\%$$.) So choose a random optimal packing. Then each of the $$10$$ points has about a $$9.3\%$$ chance of not being covered by the discs in this packing, so by the union bound there is at least about a $$7\% = 100 - 10(9.3)$$ chance that a random tiling covers the $$10$$ points. Since this chance is positive, there must be some tiling that covers all 10 points. For such a tiling, there are at most $$10$$ disks touching a point, so this is our solution.
To phrase it in terms of your "purified pigeonhole principle," let $$X$$ be a random variable that samples a random optimal tiling of the plane (equivalent to choosing a point in a hexagonal region uniformly at random for the center of one of the disks) and outputs $$1$$ if the corresponding tiling covers all $$10$$ points, $$0$$ otherwise. We've shown that $$E(X) > 0$$, therefore $$X = 1$$ with positive probability.
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The Unit Circle and Trigonometry
The Unit Circle
The unit circle is a circle with radius $1$. We centre the unit circle at the origin ($(0,0)$) so that we can use it to talk about lengths and angles.
Sines, Cosines and Tangents, and the Unit Circle
For the past few months, I have been teaching some American students who are preparing for their college entrance exams. They seem to base their knowledge of trigonometry on the unit circle. I haven't seen trigonometry taught this way in Australia, but the approach does point out some interesting connections between the unit circle and trigonometric functions. It's well worth having a look at.
We can draw a line from the centre of the unit circle to each point $(x,y)$ on its circumference. If $\theta$ is the angle that the line makes with the positive $x$-axis, then the $x$-coordinate of point $(x,y)$ is equal to $\cos \theta$, and the $y$-coordinate is equal to $\sin \theta$. The tangent of angle $\theta$ is equal to $\tan \theta = \dfrac{y}{x}$. So, we can read off these trigonometric ratios from the perpendicular distances between the point $(x,y)$ and the axes.
If the angle $\theta$ that the line from $(0,0)$ to $(x,y)$ makes with the positive $x$-axis is $0^\circ$, then the $x$-coordinate of point $(x,y)$ is equal to $\cos 0 = 1$, and the $y$-coordinate is equal to $\sin 0 = 0$. The tangent of $0^\circ$ is equal to $\tan 0 = 0$. So, the point has coordinates $(1,0)$.
If the angle $\theta$ that the line from $(0,0)$ to $(x,y)$ makes with the positive $x$-axis is $90^\circ$, then the $x$-coordinate of point $(x,y)$ is equal to $\cos 90 = 0$, and the $y$-coordinate is equal to $\sin 90 = 1$. The tangent of $90^\circ$ is undefined. So, the point has coordinates $(0,1)$.
Note: We know that the coordinates of different points around the unit circle will have positive $x$-coordinates, or negative $x$-coordinates, and similarly for their $y$-coordinates. This lines up nicely with the sine and cosine values of different angles having positive and negative values. We'll talk more about this later.
Pythagoras' Theorem, The Unit Circle and Trigonometric Identities
The equation of the unit circle centred at the origin is $x^2 + y^2 = 1$. This corresponds to Pythagoras' theorem (the square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides) for a right-angled triangle with side lengths $x$ and $y$, and hypotenuse $1$:
\begin{align*} x^2 + y^2 &= 1^2\\ x^2 + y^2 &= 1 \end{align*}
We also know that $x = \cos\theta$ and $y = \sin \theta$, so we can use this relationship to establish the trigonometric identity:
\begin{align*} \cos^2 \theta + \sin^2 \theta &= 1 \end{align*}
This identity (equation that is always true, regardless of the value of $\theta$) is sometimes called the Pythagorean Trigonometric Identity. If you've never seen it before, it's a really good idea to memorise it. It's probably one of the most important items in your mathematical toolI kit: I've lost count of the number of times I've used it.
Exact Angle Ratios
This is a slight digression to talk about the trig ratios for some important angles. My American students have been told to just memorise the ratios of the sides of these triangles. I think we're better off knowing how to work them out. So, we'll do just that in this section. I have never memorised these ratios, but I can very quickly work them out.
Don't worry, I'm not really going off on a tangent (apologies, I couldn't resist) - we'll need these ratios very soon.
The first one is the simplest to remember. Draw yourself an isosceles, right-angled triangle of side length one. By Pythagoras' theorem, the square of the hypotenuse is $1^2 + 1^2 = 2$, so the hypotenuse has length $\sqrt{2}$.
The triangle is isosceles, so its base angles are equal. As the other angle is $90^\circ$, they are each equal to $45^\circ$.
Now we can find our trig ratios (you might remember the mnemonic sohcahtoa):
• $\sin 45^\circ = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{1}{\sqrt{2}}$
• $\cos 45^\circ = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{1}{\sqrt{2}}$
• $\tan 45^\circ = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{\sqrt{2}}{\sqrt{2}} = 1$
Now draw yourself an equilateral triangle of side length two. Chop it in half using the perpendicular bisector of the base. This will also bisect the top angle of the triangle and form a right-angled triangle with hypotenuse $2$ and one of its side-lengths equal to $1$. We can find the length of the third side using Pythagoras' theorem. If it has length $a$, then by Pythagoras' theorem, the square of the hypotenuse is $1^2 + a^2 = 2^2$, so $a^2 = 3$ and $a = \sqrt{3}$. The angles of the right-angled triangle are $60^\circ$ (all angles of an equilateral triangle are equal to $60^\circ$) and $30^\circ$ (we bisected the other angle).
Now we can find our trig ratios (you might remember the mnemonic sohcahtoa):
• $\sin 30^\circ = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{1}{2}$
• $\cos 30^\circ = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{\sqrt{3}}{2}$
• $\tan 30^\circ = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{1}{\sqrt{3}}$
• $\sin 60^\circ = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{\sqrt{3}}{2}$
• $\cos 60^\circ = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{1}{2}$
• $\tan 60^\circ = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{\sqrt{3}}{1} = \sqrt{3}$
These are the exact angle ratios that I use, but you can sometimes lose marks if you write them in this form. It is, perhaps better, to rationalise the denominators of these ratios. The table gives each of these ratios after the denominators have been rationalised.
Angle Sine Cosine Tangent
$0^\circ$ $0$ $1$ $0$
$30^\circ$ $\dfrac{1}{2}$ $\dfrac{\sqrt{3}}{2}$ $\dfrac{\sqrt{3}}{3}$
$45^\circ$ $\dfrac{\sqrt{2}}{2}$ $\dfrac{\sqrt{2}}{2}$ $1$
$60^\circ$ $\dfrac{\sqrt{3}}{2}$ $\dfrac{1}{2}$ $\sqrt{3}$
$90^\circ$ $1$ $0$ undefined
Plotting the Exact Coordinates On the Unit Circle
The $x$-coordinates and Cosine
As we move around the unit circle from the point $(1,0)$ in an anticlockwise direction, the size of the angle $\theta$ that a line joining our point to the origin makes with the positive $x$-axis increases. Our $x$-coordinate gets smaller as more of the length of the radius is taken up by moving upwards. So, the cosine of $\theta$ gets smaller as we move towards $90^\circ$, where it is equal to zero. This means that $\cos 30^\circ > \cos 45^\circ > \cos 60^\circ$, and so, if we don't remember the values of $\cos 30^\circ$ and $\cos 60^\circ$, we can get around the problem by remembering that $\cos 30^\circ$ is bigger, so it must be the one that's equal to $\dfrac{\sqrt{3}}{2}$, and $\cos 60^\circ$ is smaller, so it must be the one equal to $\dfrac{1}{2}$.
As we move around the unit circle from the point $(1,0)$ in an anticlockwise direction, the size of the angle $\theta$ that a line joining our point to the origin makes with the positive $x$-axis increases. Our $y$-coordinate gets larger as the angle gets bigger. So, the sine of $\theta$ gets larger as we move towards $90^\circ$, where it is equal to 1. This means that $\sin 30^\circ Filling in the Rest of the Circle This is one of the really cool things about this approach to trigonometry. Usually we have to remember that the \(xy$-plane is divided up into four quadrants. In the first quadrant sine, cosine and tangent are all positive. In the second quadrant sine is positive, and the others are negative. In the third quadrant, only tangent is positive, and in the fourth quadrant only cosine is positive. You've probably used a mnemonic like "All Stations to Central", or "All Seniors Turn Crazy" or "All Students Take Calculus" to remember this. If you haven't, you have a few options for remembering it now!
In the unit circle approach, we just use the properties of coordinates in the $xy$-plane, remembering that the $x$-coordinate corresponds to cosine, and the $y$-coordinate corresponds to sine. So, in the first quadrant, both coordinates are positive, so sine and cosine must both be positive. This means that tangent is also positive. In the second quadrant, the $x$-coordinate is negative, so cosine is negative, and the $y$-coordinate is positive, so sine is positive. This means that tangent negative. In the third quadrant, both $x$ and $y$ are negative. This means that both sine and cosine are negative, but tangent (being their ratio) is positive. Finally, in the fourth quadrant, $x$ is positive and $y$ is negative. So, cosine is positive, but both sine and tangent are negative.
The way the two approaches line up is nice, isn't it?
We mustn't forget that mathematicians really prefer to use radians. Remember the conversion: $180^\circ = \pi \text{ radians}$. We can label the angles in our unit circle using radians as well. The unit circle, with all the exact values filled in looks like this:
Let's finish by using the unit circle to find the sines and cosines of a few angles.
Example
What is the sine of $315^\circ$?
$315^\circ$ is equal to $360^\circ - 45^\circ$, but it is in the fourth quadrant, so its sine is negative. Therefore, $\sin 315^\circ = - \sin 45^\circ = - \dfrac{\sqrt{2}}{2}$.
Example
What is the sine of $300^\circ$?
$300^\circ$ is equal to $360^\circ - 60^\circ$. It is further around the circle, so its sine is the larger value, but it is in the fourth quadrant, so its sine is negative. Therefore, $\sin 300^\circ = - \sin 60^\circ = - \dfrac{\sqrt{3}}{2}$.
Example
What is the cosine of $\dfrac{7\pi}{6}$ radians?
$\dfrac{7\pi}{6}$ radians is equal to $\pi + \dfrac{\pi}{6}$ radians. Remember that $\dfrac{\pi}{6}$ is equal to $30^\circ$, so we have the larger distance from the $y$-axis. The angle is in the third quadrant, so its cosine is negative. Therefore, $\cos \dfrac{7\pi}{6} = - \cos \dfrac{\pi}{6} = - \dfrac{\sqrt{3}}{2}$.
Description
coordinate geometry is a branch of geometry where the position of the points on the plane is defined with the help of an ordered pair of numbers also known as coordinates. In this tutorial series, you will learn about vast range of topics such as Cartesian Coordinates, Midpoint of a Line Segment etc
Audience
year 10 or higher, several chapters suitable for Year 8+ students.
Learning Objectives
Explore topics related to Coordinates Geometry
Author: Subject Coach
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# How do you solve -1+11=\frac{1}{3}x?
Jan 16, 2018
$x = 30$
#### Explanation:
$\text{note that } \frac{1}{3} x = \frac{x}{3}$
$- 1 + 11 = \frac{x}{3}$
$\Rightarrow 10 = \frac{x}{3}$
$\text{multiply both sides by 3}$
$\left(3 \times 10\right) = \cancel{3} \times \frac{x}{\cancel{3}}$
$\Rightarrow x = 30 \text{ is the solution}$
Jan 16, 2018
x = 30
#### Explanation:
-1 + 11 = $\frac{1}{3} x$
10 = $\frac{1}{3} x$
Multiply each side by 3
$10 \cdot 3 = 3 \cdot \frac{1}{3} x$
$30 = \frac{1 \cancel{3}}{\cancel{3}} x$
$30 = x$
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# Solve a Literal Equation for a Specific Variable
I have a literal equation that needs to be solved for: $$r_{a}$$
Here is the equation:
$$\frac{ T_{a} }{ T_{b} } = \left( \frac{ r_{a} }{ r_{b} } \right)^\frac{3}{2}$$
The answer I was able to achieve was:
$$\frac{( T_{a} )^ \frac{2}{3}\,. r_{b} }{( T_{b} )^ \frac{2}{3} } = r_{a}$$
I originally believed my answer was correct but now a friend of mine is convinced it can be further simplified. Could anyone help me with this problem? Thanks in advance!
-
Your friend is correct in saying that it can be further simplified down. Remember that $(\frac{a}{b})^n=\frac{a^n}{b^n}$
$$r_b\cdot\left(\frac{T_a}{T_b}\right)^\frac{2}{3}=r_a$$
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# Knight's Tour
Consider the knight's tour problem. Can a knight travel around the chess board, landing on each square exactly once? ♞ This would be incredibly tedious to work out by hand, but a simple computer program finds a solution in a few minutes. I'll print the solution here, in algebraic notation, so you can verify it. If you started playing chess in the 60's or earlier, you might remember the old clunky notation: rook to queen's bishop 5, or rqb5, but even that wasn't sufficient if there were two rooks that could land on the same square. So they switched to algebraic in the 70's, which is basically battleship, a through h (rook to rook), and 1 through 8 up the board. Here is the knight's tour, starting at the lower left and ending at the upper left. Grab a chess board and a knight and give it a whirl.
a1 c2 e3 g4 h6 g8 e7 c8 a7 b5 d6 f7 h8 g6 f8 d7 b8 a6 c7 e8 f6 h7 g5 e6 g7 f5 d4 c6 d8 b7 a5 b3 c5 a4 b2 d1 c3 b1 a3 c4 e5 d3 e1 g2 h4 f3 h2 f1 d2 e4 f2 h1 g3 h5 f4 h3 g1 e2 c1 a2 b4 d5 b6 a8
Here is the program that finds the above solution. Pass it the arguments 8 8 0 0 for an 8 by 8 board starting at the lower left.
## Height 3
Ok, all programming and no math is pretty dull, so let's explore some smaller boards from first principles. Can a knight travel around a 3 by 3 grid? No, because if it starts in the center it has no place to go, and if it starts on the edge it can never get to the center. A 4 by 3 board, 4 across and 3 high, works, I'll leave that one to you. Hint: start at the lower left and end at the upper right.
Place two such boards together, end to end, giving an 8 by 3 field. Cover the first board as above, then jump to the lower left of the second board, and cover it as above. You can chain n such boards together, hence a 4n by 3 board works.
A 7 by 3 board has this tour, from lower left to an interior square at the right: a1 c2 a3 b1 c3 a2 c1 b3 d2 f3 g1 e2 g3 f1 e3 g2 e1 d3 b2 d1 f2. Put additional 4 by 3 boards on the left, and chain them together, and every 4n+3 by 3 board works (n positive).
An n by 2 board never works, because if the knight starts at the lower left, it keeps moving to the right forever.
How about a 4 by 4 board? One corner might start, and one corner might end, but at least 2 corners are in the middle of the tour. Assume one of these is the lower left, thus the knight jumps from c2 to a1 to b3. d4 cannot be in the middle of the tour, for that would also consume c2 and b3, whence only those 4 squares are in the circuit. Thus d4 starts or ends the tour. We can safely say the tour starts: d4 c2 a1 b3. In the same way, the tour ends with b2 d1 c3 a4, or these four possibly in a different order. The 8 moves in the middle of the tour consume the 8 side squares. If you jump from b3 to c1, you are forced to run a circuit of four, clockwise or counterclockwise, e.g. c1 d3 b4 a2 c1. The same thing happens if you jump from b3 to d2. Thus we cannot tour the 4 by 4 board.
A 5 by 5 board works: a1 c2 e3 d5 b4 a2 c3 d1 b2 a4 c5 e4 d2 b1 a3 b5 d4 e2 c1 b3 a5 c4 e5 d3 e1. If there are 13 black squares and 12 white, the tour has to start and end on black. Thus the knight cannot jump from the last square to the first, completing a circuit. We can run a tour, but not a circuit. An odd number of squares never allows for a circuit.
With the tricks at hand, you should be able to prove the 5 by 3 board is impossible.
By the black and white argument, you can't start or end on a2. The tour runs … c1 a2 c3 … Similarly, e2 is in the middle of the tour. That builds the 4 cycle c1 a2 c3 e2 c1.
Here is 9 by 3, which is, happily, from lower left to upper right.
a1 c2 a3 b1 d2 b3 c1 a2 c3 e2 g3 i2 g1 h3 i1 g2 e3 f1 h2 f3 e1 d3 b2 d1 f2 h1 i3
Chain additional 4 by 3 boards on the end, and 4n + 9 by 3 works.
How about a 6 by 3 board? This one is tricky. If a2 and e2 are both in the middle of the tour, then we have the 4 cycle: a2 c1 e2 c3 a2. Thus either a2 or e2 starts or ends the tour. In the same way, either b2 or f2 starts or ends the tour. Each of the 4 corners is in the middle of the tour. The tour includes b1 a3 c2 a1 b3 (or its reverse), and e1 f3 d2 f1 e3 (or its reverse). We can't start or end at b1, so connect it to c3. Then connect b3 to c1. c1 and c3 cannot start or end the tour, so one connects to a2 and the other to e2. We have a subtour of length 9, and it cannot connect to the other subtour of length 9.
Here is the 10 by 3 board, again, from lower left to upper right.
a1 c2 a3 b1 d2 b3 c1 a2 c3 e2 g3 i2 g1 f3 e1 g2 i3 j1 h2 f1 e3 d1 b2 d3 f2 h1 j2 h3 i1 j3
This extends to 4n+10 by 3, and that completes all boards of height 3. A board works if its length is 4, or ≥ 7.
## Chess Board
Return to the 8 by 8 chess board. A rook can start at a1 and tour the entire board, and even return to a1. (I'll leave that one to you.) A queen or king could do the same thing, and even a pawn if you let the pawn turn into a queen when it reaches the eighth row. A bishop can't tour since she always has to stay on her own color. A bishop can't tour all the black squares. It's a relatively easy proof; I'll leave it to you. Hint: it would have to start at one corner and end at the other.
Can a knight tour the chess board and then return to start, a circuit if you will? Run the program with arguments 8 8 - -, the dashes indicating a circuit, and the answer is yes.
a1 c2 e3 g4 h6 g8 e7 c8 a7 b5 d6 f7 h8 g6 f8 d7 b8 a6 c7 a8 b6 a4 c5 e6 g7 e8 f6 h7 g5 e4 c3 d5 b4 a2 c1 e2 d4 f5 h4 g2 e1 d3 b2 d1 f2 h1 g3 h5 f4 h3 g1 f3 h2 f1 d2 b1 a3 c4 e5 c6 d8 b7 a5 b3
## Height 4
We analyzed all boards of height 3; how about boards of height 4? Boards of width 1 and 2 fail trivially, 3 works, and 4 fails, as shown earlier. 5 works: a1 c2 e1 d3 b4 a2 c1 e2 d4 b3 d2 e4 c3 b1 a3 c4 e3 d1 b2 a4.
Recall the strategy we used for boards of height 3. We chained boards of width 4 together to analyze boards of widths 4n, 4n+1, 4n+2, and 4n+3. You might want to do the same thing here, perhaps using blocks of width 5, but you can't. In fact there is no way to chain blocks together, of any width. Color arguments will force us down a different path.
Color the top and bottom stripes red, and the middle two stripes green. The knight always has to move from red to green. Write the tour as a sequence of colors rgrggrgggrgr… At the end, there are equal numbers of greens and reds. If, at any point in the sequence, green exceeds red by 2, then there will always be more green than red, all the way to the end. This is a contradiction. If you start with g then the next color is r, then grgrgrgrgr all the way to the end. We never jump from green to green. If you start with red, then you can have one gg jump. After that, red and green alternate, ending in red. Those are the only possibilities. The 5 by 4 solution shown above looks like this. rgrgrgrgrggrgrgrgrgr
Next, overlay a white and black checkerboard on the green stripes. The red stripes do not participate. Look at an alternating sequence like grgrgrg… If the first square is white then all the g squares are white. If your tour starts with g, then all the g squares are white, and that is impossible. Therefore the tour starts with r.
For a while, all the g squares are white, until you run into the one and only gg jump. After that all the g squares are black. At the end of the sequence, black equals white. Therefore there is one gg jump, and it occurs exactly in the middle. The tour can be summarized: rgrgr…gg…rgrgr.
We start on red, and end on red. A circuit is not possible, since you cannot jump back from r to r. Also, a chain of two blocks is not possible. The first block ends on r, and the second block starts on r, and we cannot jump from r to r. Thus chaining does not work. Each x by 4 board is one big tour.
Start the tour in the lower left, with a component like this. a1 c2 b4 a2 c1 b3. From here we can jump to d4, and start the next component upside down. d4 f3 e1 d3 f4 e2. From here we can jump to g1 and start the next component right side up. This continues for multiples of 3. Notice there are no gg jumps.
Let the tour end at the upper left, designated a4. Walk backwards from the end of the tour, building the upside down component a4 c3 b1 a3 c4 b2. This leads to the next component, and the next, and so on. After some multiple of 3, we have the start of the tour ending at, say, t3, and the end of the tour starting at t2. In fact, t3 must jump to v4, and v1 jumps to t2. Thus we need a tour from v1 to v4, and that completes the board. We saw this earlier with a block of width 5, from columns v to z for example. Thus there is a solution to the board of width 26, or more generally, 3n+5.
If we can find tours for the boards of width 6 and 7, starting at a1 and ending at a4, then we are done with height 4. Once again the computer helps us out here.
a1 c2 b4 a2 c1 e2 f4 d3 e1 f3 d4 b3 d2 f1 e3 c4 a3 b1 c3 e4 f2 d1 b2 a4
a1 c2 d4 b3 c1 e2 g1 f3 e1 g2 f4 d3 b4 a2 c3 e4 g3 f1 d2 b1 a3 c4 e3 g4 f2 d1 b2 a4
## Height 5
For boards of height 5, chaining works again. Use this block of width 4. Then chain blocks together for boards of width 4n.
a1 c2 d4 b5 a3 b1 d2 c4 a5 b3 c5 a4 b2 d1 c3 d5 b4 a2 c1 d3
Next, we need to solve boards of width 5 6 and 7. Once again the computer provides solutions. The 5 by 5 solution starts at the lower left and ends at the lower right. Reflect the 4 by 5 board above, and put it next to this board, and chain, giving a solution to the 9 by 5 board. Keep chaining, thus covering 4n+5 by 5.
The solutions for 6 by 5 and 7 by 5 also support chaining to the right.
## Chain and Stack
Sometimes we can chain blocks together to create a row, then stack rows on top of each other to make a larger board. Recall the tour of the 7 by 3 board. It starts at a1 and ends at f2. Chain additional boards on the left, to tour a 4n+7 by 3 board. Reflect this pattern, and place it on top of the first, and the knight can tour the 4n+7 by 6 board. In general we have a solution to every board that is 4n+3 by 3m, n and m positive.
Here is a tour of 8 by 3 that goes from an interior square to lower right. Use this to chain, and stack, and tour any board that is 4n+4 by 3m, n and m positive.
b2 d3 c1 a2 c3 b1 a3 c2 a1 b3 d2 f1 h2 f3 e1 g2 e3 d1 f2 h3 g1 e2 g3 h1
There is a 9 by 3 tour, from one corner to an interior piece at the other end, and that is enough to chain and stack. Boards that are 4n+5 by 3m, n and m positive, are covered.
a3 b1 d2 f3 e1 d3 b2 d1 f2 h3 i1 g2 i3 h1 g3 i2 g1 e2 c3 a2 c1 b3 a1 c2 e3 f1 h2
With a little more work, you might be able to write a program that, given the dimensions of a rectangular board, cranks out a knights tour on that board in polynomial time, or declares the tour impossible, as on a 3 by 3 board. That would be a fun project.
## Summary
There is more to explore here, but I'll stop for now. None of this helps feed the hungry, but it is good training in logic and formal proofs. If you're a high school teacher, and you are able to take a break from algebra or trigonometry, your students might enjoy this recreational diversion.
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# Boolean Algebra
Discussion in 'Electronic Basics' started by Beanq, Oct 19, 2003.
1. ### BeanqGuest
How can I simplify this? I need to implement it using NAND gates only. I
thought maybe Demorgan, but I don't see how.
2. ### CFoley1064Guest
How can I simplify this? I need to implement it using NAND gates only. I
Hi, Bean. Straight homework help here. Cool.
Beginning digital logic classes are given three tools -- truth table, Karnaugh
mapping, and Demorgan, right? First thing I did was just draw a truth table.
Inspection shows there are only five cases where F is not true, so I decided to
do this inverted -- that is, solve for what's not true, rather than what's
true.
Let Q = (A! AND B! AND C!)
Let R = (A! AND B! AND D!)
Let S = (A! AND C! AND D!)
Let T = (B! AND C! AND D!)
It should be obvious looking at the table that F! = Q + R + S + T (with A! AND
B! AND C! AND D! being the redundant case which is true for all). Now, you can
Demorganize that, and come up with what you have below, with the last NAND
being used to invert again from F! to F. View in fixed font (like M\$
.---.
A!----| |
B!----| |o----------
C!----|& | |
| | |
'---' |
.---. |
A!----| | | .-----.
B!----| |o------- | | |
D!----|& | | '-- | | __
| | | | | |-----| |
'---' '----- | |o------| |& |o-
.---. | & | ------|__|
A!----| | .------| |
C!----| |o------- | |
D!----|& | .---| |
| | | '-----'
'---' |
.---. |
B!----| | |
C!----| |o----------
D!----|& |
| |
'---'
created by Andy´s ASCII-Circuit v1.22.310103 Beta www.tech-chat.de
This is pretty similar to factoring quadratics back in high school -- your
ability improves with practice. Try solving problems in different ways, and
just do what works.
Good luck.
Chris
3. ### petrus bitbyterGuest
Looks like schoolwork. But nevertheless:
It's (almost) written in NANDS and there is nothing to simplify. (Just draw
a Karnaugh map to see it.) Just use De Morgan:
= /(/(a*b) * /(a*c) * /(a*d) * /(b*c) * /(b*d) * /(c*d))
So you need six two inputs NAND and one six inputs NAND.
(N)AND that's it.
pieter
4. ### SHAUNGuest
chris wrote:
Shaun writes: great answer!, even though I don't understand one word you
wrote, but it did sound great
5. ### BaphometGuest
It must be those cheesy ASCII representations Shaun ;-) I don't think
anybody understands them with the possible exception of the poster!
6. ### Roger JohanssonGuest
You are using Outlook Express and it is not easy to see the ASCII
schematics in it.
You need to right-click the message in the message list, choose
"properties", details, source code, maximize window. then you will see
the message in notepad, with a fixed font, and these ASCII schematics
will make sense to you.
7. ### BaphometGuest
Thanks for the tip Roger -
I guess I'm just old fashioned. I still prefer a good schematic but I'll try
8. ### Roger JohanssonGuest
It works, if you only try to do it step by step.
First rightclick the message, and choose properties at the bottom of
the menu which appears, then choose the "details tab" at the top of
that window, then use the button named source (there is only one
button there so it doesn't matter what it says really), and maximize
that new popup window. There is the raw message and it starts with a
lot of headers, scroll down to see the real message.
9. ### BaphometGuest
I'm still too stupid to figure out how to do this in O.E., but I did look at
the post in Google and it made perfect sense, even to a dolt like me...DUH
10. ### Lord GarthGuest
In OE, one clicks 'view' then 'text size' then 'fixed'.
One can also set the default font to couier ....
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Miscellaneous Calculations - Leveraging Excel Functions - Excel 2016 Formulas (2016)
# Excel 2016 Formulas (2016)
### Chapter 10 Miscellaneous Calculations
In This Chapter
· Converting between measurement units
· Formulas that demonstrate various ways to round numbers
· Formulas for calculating the various parts of a right triangle
· Calculations for area, surface, circumference, and volume
· Matrix functions to solve simultaneous equations
· Useful formulas for working with normal distributions
This chapter contains reference information that may be useful to you at some point. Consider it a cheat sheet to help you remember the stuff you may have learned but have long since forgotten.
Unit Conversions
You know the distance from New York to London in miles, but your European office needs the numbers in kilometers. What’s the conversion factor?
Excel’s CONVERT function can convert between a variety of measurements in the following categories:
§ Area
§ Distance
§ Energy
§ Force
§ Information
§ Magnetism
§ Power
§ Pressure
§ Speed
§ Temperature
§ Time
§ Volume (or liquid measure)
§ Weight and mass
The CONVERT function requires three arguments: the value that you want to convert, the from-unit, and the to-unit. For example, if cell A1 contains a distance expressed in miles, use this formula to convert miles to kilometers:
=CONVERT(A1,"mi","km")
The second and third arguments are unit abbreviations, which are listed in the Excel Help system. Some of the abbreviations are commonly used, but others aren’t. And, of course, you must use the exact abbreviation. Furthermore, the unit abbreviations are case sensitive, so the following formula returns an error:
=CONVERT(A1,"Mi","km")
The CONVERT function is even more versatile than it seems. When using metric units, you can apply a multiplier. In fact, the first example we presented uses a multiplier. The actual unit abbreviation for the third argument is m for meters. We added the kilo-multipler—k—to express the result in kilometers.
Sometimes you need to use a bit of creativity. For example, if you need to convert 100 km/hour into miles/sec, the formula requires two uses of the CONVERT function:
=CONVERT(100,"km","mi")/CONVERT(1,"hr","sec")
Figure 10.1 shows a conversion table for area measurements. The CONVERT argument codes are in column B and duplicated in row 3. Cell A1 contains a value. The formula in cell C4, which was copied down and across, is
Figure 10.1 A table that lists all the area units supported by the CONVERT function.
=CONVERT(\$A\$1,\$B4,C\$3)
The table shows, for example, that 1 square foot is equal to 144 square inches. It also shows that one square light year can hold a lot of acres.
By the way, this formula is also a good example of when to use an absolute reference and mixed references in a formula.
On the Web
The workbook shown in Figure 10.1 is available at this book’s website. The filename is area conversion table.xlsx.
Figure 10.2 shows part of a table that lists all the conversion units supported by the CONVERT function. The table can be sorted and filtered and indicates which of the units support the metric prefixes.
Figure 10.2 A table that lists all the units supported by the CONVERT function.
On the Web
This workbook is available at this book’s website. The filename is conversion units table.xlsx.
If you can’t find a particular unit that works with the CONVERT function, perhaps Excel has another function that will do the job. Table 10.1 lists some other functions that convert between measurement units.
Table 10.1 Other Conversion Functions
Function Description ARABIC* Converts an Arabic number to decimal BASE* Converts a decimal number to a specified base BIN2DEC Converts a binary number to decimal BIN2OCT Converts a binary number to octal DEC2BIN Converts a decimal number to binary DEC2HEX Converts a decimal number to hexadecimal DEC2OCT Converts a decimal number to octal DEGREES Converts an angle (in radians) to degrees HEX2BIN Converts a hexadecimal number to binary HEX2DEC Converts a hexadecimal number to decimal HEX2OCT Converts a hexadecimal number to octal OCT2BIN Converts an octal number to binary OCT2DEC Converts an octal number to decimal OCT2HEX Converts an octal number to hexadecimal RADIANS Converts an angle (in degrees) to radians
* Function available in Excel 2013 and higher versions.
Need to convert other units?
The CONVERT function, of course, doesn’t handle every possible unit conversion. To calculate other unit conversions, you need to find the appropriate conversion factor. The Internet is a good source for such information. Use any web search engine and enter search terms that correspond to the units you use. Likely, you’ll find the information that you need.
Also, you can download a copy of Josh Madison’s popular (and free) Convert software. This excellent program can handle just about any conceivable unit conversion that you throw at it. The URL is http://joshmadison.com/convert-for-windows.
You can use the program to find the conversion factor and then use that value in your formulas.
Rounding Numbers
Excel provides quite a few functions that round values in various ways. Table 10.2 summarizes these functions.
Warning
It’s important to understand the difference between rounding a value and formatting a value. When you format a number to display a specific number of decimal places, formulas that refer to that number use the actual value, which may differ from the displayed value. When you round a number, formulas that refer to that value use the rounded number.
Table 10.2 Excel Rounding Functions
Function Description CEILING.MATH Rounds a number up to the nearest specified multiple DOLLARDE Converts a dollar price expressed as a fraction into a decimal number DOLLARFR Converts a dollar price expressed as a decimal into a fractional number EVEN Rounds a number up (away from zero) to the nearest even integer FLOOR.MATH Rounds a number down to the nearest specified multiple INT Rounds a number down to make it an integer MROUND Rounds a number to a specified multiple ODD Rounds a number up (away from zero) to the nearest odd integer ROUND Rounds a number to a specified number of digits ROUNDDOWN Rounds a number down (toward zero) to a specified number of digits ROUNDUP Rounds a number up (away from zero) to a specified number of digits TRUNC Truncates a number to a specified number of significant digits
Cross-Ref
Chapter 6, “Working with Dates and Times,” contains examples of rounding time values.
The following sections provide examples of formulas that use various types of rounding.
Basic rounding formulas
The ROUND function is useful for basic rounding to a specified number of digits. You specify the number of digits in the second argument for the ROUND function. For example, the formula that follows returns 123.4. (The value is rounded to one decimal place.)
=ROUND(123.37,1)
If the second argument for the ROUND function is zero, the value is rounded to the nearest integer. The formula that follows, for example, returns 123.00:
=ROUND(123.37,0)
The second argument for the ROUND function can also be negative. In such a case, the number is rounded to the left of the decimal point. The following formula, for example, returns 120.00:
=ROUND(123.37,-1)
The ROUND function rounds either up or down. But how does it handle a number such as 12.5, rounded to no decimal places? You’ll find that the ROUND function rounds such numbers away from zero. The formula that follows, for instance, returns 13.0:
=ROUND(12.5,0)
The next formula returns –13.00. (The rounding occurs away from zero.)
=ROUND(-12.5,0)
To force rounding to occur in a particular direction, use the ROUNDUP or ROUNDDOWN functions. The following formula, for example, returns 12.0. The value rounds down:
=ROUNDDOWN(12.5,0)
The formula that follows returns 13.0. The value rounds up to the nearest whole value:
=ROUNDUP(12.43,0)
Rounding to the nearest multiple
The MROUND function is useful for rounding values to the nearest multiple. For example, you can use this function to round a number to the nearest 5. The following formula returns 135:
=MROUND(133,5)
The second argument for MROUND can be a fractional number. For example, this formula rounds the value in cell A1 to the nearest one-eighth:
=MROUND(A1,1/8)
Rounding currency values
Often, you need to round currency values. For example, you may need to round a dollar amount to the nearest penny. A calculated price may be something like \$45.78923. In such a case, you’ll want to round the calculated price to the nearest penny. This may sound simple, but there are actually three ways to round such a value:
§ Round it up to the nearest penny.
§ Round it down to the nearest penny.
§ Round it to the nearest penny. (The rounding may be up or down.)
The following formula assumes that a dollar-and-cents value is in cell A1. The formula rounds the value to the nearest penny. For example, if cell A1 contains \$12.421, the formula returns \$12.42:
=ROUND(A1,2)
If you need to round the value up to the nearest penny, use the CEILING function. The following formula rounds the value in cell A1 up to the nearest penny. For example, if cell A1 contains \$12.421, the formula returns \$12.43:
=CEILING(A1,0.01)
To round a dollar value down, use the FLOOR function. The following formula, for example, rounds the dollar value in cell A1 down to the nearest penny. If cell A1 contains \$12.421, the formula returns \$12.42:
=FLOOR(A1,0.01)
To round a dollar value up to the nearest nickel, use this formula:
=CEILING(A1,0.05)
You’ve probably noticed that many retail prices end in \$0.99. If you have an even-dollar price and you want it to end in \$0.99, just subtract .01 from the price. Some higher-ticket items are always priced to end with \$9.99. To round a price to the nearest \$9.99, first round it to the nearest \$10.00 and then subtract a penny. If cell A1 contains a price, use a formula like this to convert it to a price that ends in \$9.99:
=(ROUND(A1/10,0)*10)-0.01
For example, if cell A1 contains \$345.78, the formula returns \$349.99.
A simpler approach uses the MROUND function:
=MROUND(A1,10)-0.01
Working with fractional dollars
The DOLLARFR and DOLLARDE functions are useful when working with fractional dollar values, as in stock market quotes.
Consider the value \$9.25. You can express the decimal part as a fractional value (\$9 1/4, \$9 2/8, \$9 4/16, and so on). The DOLLARFR function takes two arguments: the dollar amount and the denominator for the fractional part. The following formula, for example, returns 9.1 (and is interpreted as “nine and one-quarter”):
=DOLLARFR(9.25,4)
This formula returns 9.2 (interpreted as “nine and two-eighths”):
=DOLLARFR(9.25,8)
Warning
In most situations, you won’t use the value returned by the DOLLARFR function in other calculations. To perform calculations on such a value, you need to convert it back to a decimal value by using the DOLLARDE function.
The DOLLARDE function converts a dollar value expressed as a fraction to a decimal amount. It also uses a second argument to specify the denominator of the fractional part. The following formula, for example, returns 9.25:
=DOLLARDE(9.1,4)
Working with feet and inches
A problem that has always plagued Excel users is how to work with measurements that are in feet and inches. Excel doesn’t have any special features to make it easy to work with this type of measurement units, but in some cases, you can use the DOLLARDE and DOLLARFR functions.
If you enter 5'9" into a cell, Excel interprets it as a text string—not five feet, nine inches. But if you enter the value as 5.09, you can work with it as a value by using the DOLLARDE function. The following formula returns 5.75, which is the decimal representation of five feet nine inches expressed as fractional feet:
=DOLLARDE(5.09,12)
If you don’t work with fractional inches, you can even create a custom number format. Entering the following custom format will display values as feet and inches.
#" ft". 00" in."
On the Web
This workbook is available at this book’s website. The filename is feet and inches.xlsx.
Using the INT and TRUNC functions
On the surface, the INT and TRUNC functions seem similar. Both convert a value to an integer. The TRUNC function simply removes the fractional part of a number. The INT function rounds a number down to the nearest integer, based on the value of the fractional part of the number.
In practice, INT and TRUNC return different results only when using negative numbers. For example, the following formula returns –14.0:
=TRUNC(-14.2)
The next formula returns –15.0 because –14.3 is rounded down to the next lower integer:
=INT(-14.2)
The TRUNC function takes an additional (optional) argument that’s useful for truncating decimal values. For example, the formula that follows returns 54.33 (the value truncated to two decimal places):
=TRUNC(54.3333333,2)
Rounding to an even or odd integer
The ODD and EVEN functions are provided when you need to round a number up to the nearest odd or even integer. These functions take a single argument and return an integer value. The EVEN function rounds its argument up to the nearest even integer. The ODD function rounds its argument up to the nearest odd integer. Table 10.3 shows some examples of these functions.
Table 10.3 Results Using the EVEN and ODD Functions
Number EVEN Functiwon ODD Function –3.6 –4 –5 –3.0 –4 –3 –2.4 –4 –3 –1.8 –2 –3 –1.2 –2 –3 –0.6 –2 –1 0.0 0 1 0.6 2 1 1.2 2 3 1.8 2 3 2.4 4 3 3.0 4 3 3.6 4 5
Rounding to n significant digits
In some cases, you may need to round a value to a particular number of significant digits. For example, you might want to express the value 1,432,187 in terms of two significant digits: that is, as 1,400,000. The value 9,187,877 expressed in terms of three significant digits is 9,180,000.
If the value is a positive number with no decimal places, the following formula does the job. This formula rounds the number in cell A1 to two significant digits. To round to a different number of significant digits, replace the 2 in this formula with a different number:
=ROUNDDOWN(A1,2-LEN(A1))
For nonintegers and negative numbers, the solution gets a bit trickier. The formula that follows provides a more general solution that rounds the value in cell A1 to the number of significant digits specified in cell A2. This formula works for positive and negative integers and nonintegers:
=ROUND(A1,A2-1-INT(LOG10(ABS(A1))))
For example, if cell A1 contains 1.27845 and cell A2 contains 3, the formula returns 1.28000 (the value, rounded to three significant digits).
Note
Don’t confuse this technique with significant decimal digits, which are the number of digits displayed to the right of the decimal points. It’s an entirely different concept. Also, these formulas make the values less accurate.
Solving Right Triangles
A right triangle has six components: three sides and three angles. Figure 10.3 shows a right triangle with its various parts labeled. Angles are labeled A, B, and C; sides are labeled Hypotenuse, Base, and Height. Angle C is always 90 degrees (or PI/2 radians). If you know any two of these components (excluding Angle C, which is always known), you can use formulas to solve for the others.
Figure 10.3 A right triangle’s components.
The Pythagorean theorem states that
Height^2 + Base^2 = Hypotenuse^2
Therefore, if you know two sides of a right triangle, you can calculate the remaining side. The formula to calculate a right triangle’s height (given the length of the hypotenuse and base) is as follows:
=SQRT((hypotenuse^2) - (base^2))
The formula to calculate a right triangle’s base (given the length of the hypotenuse and height) is as follows:
=SQRT((hypotenuse^2) - (height^2))
The formula to calculate a right triangle’s hypotenuse (given the length of the base and height) is as follows:
=SQRT((height^2)+(base^2))
Other useful trigonometric identities are
SIN(A) = Height/Hypotenuse
SIN(B) = Base/Hypotenuse
COS(A) = Base/Hypotenuse
COS(B) = Height/Hypotenuse
TAN(A) = Height/Base
SIN(A) = Base/Height
Note
Excel’s trigonometric functions assume that the angle arguments are in radians. To convert degrees to radians, use the RADIANS function. To convert radians to degrees, use the DEGREES function.
If you know the height and base, you can use the following formula to calculate the angle formed by the hypotenuse and base (angle A):
=ATAN(height/base)
The preceding formula returns radians. To convert to degrees, use this formula:
=DEGREES(ATAN(height/base))
If you know the height and base, you can use the following formula to calculate the angle formed by the hypotenuse and height (angle B):
=PI()/2-ATAN(height/base)
The preceding formula returns radians. To convert to degrees, use this formula:
=90-DEGREES(ATAN(height/base))
On the Web
This book’s website contains a workbook —solve right triangle.xlsm— with formulas that calculate various parts of a right triangle, given two known parts. These formulas give you some insight on working with right triangles. The workbook uses a simple VBA macro to enable you to specify the known parts of the triangle.
Figure 10.4 shows a workbook containing formulas to calculate the various parts of a right triangle.
Figure 10.4 This workbook is useful for working with right triangles.
Area, Surface, Circumference, and Volume Calculations
This section contains formulas for calculating the area, surface, circumference, and volume for common two- and three-dimensional shapes.
Calculating the area and perimeter of a square
To calculate the area of a square, square the length of one side. The following formula calculates the area of a square for a cell named side:
=side^2
To calculate the perimeter of a square, multiply one side by 4. The following formula uses a cell named side to calculate the perimeter of a square:
=side*4
Calculating the area and perimeter of a rectangle
To calculate the area of a rectangle, multiply its height by its base. The following formula returns the area of a rectangle, using cells named height and base:
=height*base
To calculate the perimeter of a rectangle, multiply the height by 2 and then add it to the width multiplied by 2. The following formula returns the perimeter of a rectangle, using cells named height and width:
=(height*2)+(width*2)
Calculating the area and perimeter of a circle
To calculate the area of a circle, multiply the square of the radius by (π). The following formula returns the area of a circle. It assumes that a cell named radius contains the circle’s radius:
The radius of a circle is equal to one-half of the diameter.
To calculate the circumference of a circle, multiply the diameter of the circle by (π). The following formula calculates the circumference of a circle using a cell named diameter:
=diameter*PI()
The diameter of a circle is the radius times 2.
Calculating the area of a trapezoid
To calculate the area of a trapezoid, add the two parallel sides, multiply by the height, and then divide by 2. The following formula calculates the area of a trapezoid, using cells named parallel side 1, parallel side 2, and height:
=((parallel side 1+parallel side 2)*height)/2
Calculating the area of a triangle
To calculate the area of a triangle, multiply the base by the height and then divide by 2. The following formula calculates the area of a triangle, using cells named base and height:
=(base*height)/2
Calculating the surface and volume of a sphere
To calculate the surface of a sphere, multiply the square of the radius by (π) and then multiply by 4. The following formula returns the surface of a sphere, the radius of which is in a cell named radius:
To calculate the volume of a sphere, multiply the cube of the radius by 4 times (π) and then divide by 3. The following formula calculates the volume of a sphere. The cell named radius contains the sphere’s radius:
Calculating the surface and volume of a cube
To calculate the surface area of a cube, square one side and multiply by 6. The following formula calculates the surface of a cube using a cell named side, which contains the length of a side of the cube:
=(side^2)*6
To calculate the volume of a cube, raise the length of one side to the third power. The following formula returns the volume of a cube, using a cell named side:
=side^3
Calculating the surface and volume of a rectangular solid
The following formula calculates the surface of a rectangular solid using cells named height, width, and length:
=(length*height*2)+(length*width*2)+(width*height*2)
To calculate the volume of a rectangular solid, multiply the height by the width by the length:
=height*width*length
Calculating the surface and volume of a cone
The following formula calculates the surface of a cone (including the surface of the base). This formula uses cells named radius and height:
To calculate the volume of a cone, multiply the square of the radius of the base by (π), multiply by the height, and then divide by 3. The following formula returns the volume of a cone, using cells named radius and height:
Calculating the volume of a cylinder
To calculate the volume of a cylinder, multiply the square of the radius of the base by (π) and then multiply by the height. The following formula calculates the volume of a cylinder, using cells named radius and height:
Calculating the volume of a pyramid
Calculate the area of the base, multiply by the height, and then divide by 3. This formula calculates the volume of a pyramid. It assumes cells named width (the width of the base), length (the length of the base), and height (the height of the pyramid).
=(width*length*height)/3
Solving Simultaneous Equations
This section describes how to use formulas to solve simultaneous linear equations. The following is an example of a set of simultaneous linear equations:
3x + 4y = 8
4x + 8y = 1
Solving a set of simultaneous equations involves finding the values for x and y that satisfy both equations. For this set of equations, the solution is as follows:
x = 7.5
y = –3.625
The number of variables in the set of equations must be equal to the number of equations. The preceding example uses two equations with two variables. Three equations are required to solve for three variables (x, y, and z).
The general steps for solving a set of simultaneous equations follow. See Figure 10.5, which uses the equations presented at the beginning of this section.
1. Express the equations in standard form. If necessary, use simple algebra to rewrite the equations such that all the variables appear on the left side of the equal sign. The two equations that follow are identical, but the second one is in standard form:
2. 3x –8 = –4y
3x + 4y = 8
2. Place the coefficients in an n × n range of cells, where n represents the number of equations. In Figure 10.5, the coefficients are in the range I2:J3.
3. Place the constants (the numbers on the right side of the equal sign) in a vertical range of cells. In Figure 10.5, the constants are in the range L2:L3.
4. Use an array formula to calculate the inverse of the coefficient matrix. In Figure 10.5, the following array formula is entered into the range I6:J7. (Remember to press Ctrl+Shift+Enter to enter an array formula, and omit the curly brackets.)
{=MINVERSE(I2:J3)}
5. Use an array formula to multiply the inverse of the coefficient matrix by the constant matrix. In Figure 10.5, the following array formula is entered into the range J10:J11. This range holds the solution:
{=MMULT(I6:J7,L2:L3)}
Figure 10.5 Using formulas to solve simultaneous equations.
Cross-Ref
See Chapter 14, “Introducing Arrays,” for more information on array formulas.
On the Web
You can access the workbook, simultaneous equations.xlsx, shown in Figure 10.5, from this book’s website. This workbook solves simultaneous equations with two or three variables.
Working with Normal Distributions
In statistics, a common topic is the normal distribution, also known as a bell curve. Excel has several functions designed to work with normal distributions. This is not a statistics book, so we assume that if you’re reading this section, you’re familiar with the concept.
On the Web
The examples in this section are available at this book’s website. The filename is normal distribution.xlsx.
Figure 10.6 shows a workbook containing formulas that generate the two charts: a normal distribution and a cumulative normal distribution.
Figure 10.6 Using formulas to calculate a normal distribution and a cumulative normal distribution.
The formulas use the values entered into cell B1 (name Mean) and cell B2 (named SD, for standard deviation). The values calculated cover the mean plus/minus three standard deviations. Column A contains formulas that generate 25 equally spaced intervals, ranging from –3 standard deviations to +3 standard deviations.
A normal distribution with a mean of 0 and a standard deviation of 1 is known as the “standard normal distribution.” The worksheet is set up to work with any mean and any standard deviation.
Formulas in column B calculate the height of the normal curve for each of the 25 values in column A. Cell B5 contains this formula, which is copied down the column:
=NORM.DIST(A5,Mean,SD,FALSE)
The formula in cell C5 is the same, except for the last argument. When that argument is TRUE, the function returns the cumulative probability:
=NORMDIST(A5,Mean,SD,TRUE)
Figure 10.7 shows a worksheet with 2,600 data points in column A (named Data) that is approximately normally distributed. Formulas in column D calculate some basic statistics for this data: N (the number of data points), the minimum, the maximum, the mean, and the standard deviation.
Figure 10.7 Comparing a set of data with a normal distribution.
Column F contains an array formula that creates 25 equal-interval bins that cover the complete range of the data. It uses the technique described in Chapter 7, “Counting and Summing Techniques.” The multicell array formula, entered in F2:F26 (named Bins), is
=MIN(Data)+(ROW(INDIRECT("1:25"))*(MAX(Data)-MIN(Data)+1)/25)-1
Column G also contains a multicell array formula that calculates the frequency for each of the 25 bins:
=FREQUENCY(Data,Bins)
The data in column G is displayed in the chart as columns and uses the axis on the left.
Column H contains formulas that calculate the value of the theoretical normal distribution, using the mean and standard deviation of the Data range.
The formula in cell H2, which is copied down the column, is
=NORMDIST(F2,\$D\$5,\$D\$6,FALSE)
In the chart, the values in column H are plotted as a line and use the right axis. As you can see, the data conforms fairly well to the normal distribution. There are goodness-of-fit tests to determine the “normality” of a set of data samples, but that’s beyond the scope of this book.
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Q:
# How do I calculate cumulative averages?
A:
The cumulative average is calculated by weighting each grade by the number of credits the course was worth. The calculation of a cumulative average is how you determine your grade point average in school. This requires you to have access to your transcript so that both the credits earned and the grade earned for each course are visible.
## Keep Learning
1. Record the credit hours and grades
Record the credit hours and the grades from your transcript for each course. Typically, an A is worth four points, a B is worth three points, a C is worth two points, a D is worth one point and an F isn't worth any points. In some systems, an F is worth negative points, so make sure to check the values your school uses.
2. Calculate the total
Add the total number of points from the grades you received. It may be best to calculate the totals separately and add them together. For example, if you received 8 credits worth of an A grade and six credits for a B, you would perform the following calculation: 8*4 + 6*3 = 50.
3. Divide the total by credit hours
Add up the total number of credit hours that you have taken and divide the calculation from the previous step by this number. From the previous step, you would have 14 credit hours and perform the following calculation: 50/14 = 3.57. The number that you get is your cumulative average.
Sources:
## Related Questions
• A:
Introductory algebra is a fundamental mathematics course. It is essential to master this course before moving on to more advanced material. Key concepts in introductory algebra involve the study of variables, expressions and equations.
Filed Under:
• A:
An experimental variable is something that a scientist changes during the course of an experiment. It is distinguished from a controlled variable, which could theoretically change, but the scientists keep constant.
Filed Under:
• A:
In a ninth-grade Algebra I course, students learn to solve problems using the foundational rules of mathematical computations, and they learn to solve algebraic expressions to find the values of variables. Some Algebra I problems require students to graph equations and inequalities. The study of equations in Algebra I helps prepare students for the study of functions in Precalculus and Calculus courses.
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# 2008 AMC 12B Problems/Problem 12
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
For each positive integer $n$, the mean of the first $n$ terms of a sequence is $n$. What is the $2008$th term of the sequence?
$\textbf{(A)}\ 2008 \qquad \textbf{(B)}\ 4015 \qquad \textbf{(C)}\ 4016 \qquad \textbf{(D)}\ 4030056 \qquad \textbf{(E)}\ 4032064$
## Solution
Letting $S_n$ be the nth partial sum of the sequence:
$\frac{S_n}{n} = n$
$S_n = n^2$
The only possible sequence with this result is the sequence of odd integers.
$a_n = 2n - 1$
$a_{2008} = 2(2008) - 1 = 4015 \Rightarrow \textbf{(B)}$
## Alternate Solution
Letting the sum of the sequence equal $a_1+a_2+\cdots+a_n$ yields the following two equations:
$\frac{a_1+a_2+\cdots+a_{2008}}{2008}=2008$ and
$\frac{a_1+a_2+\cdots+a_{2007}}{2007}=2007$.
Therefore:
$a_1+a_2+\cdots+a_{2008}=2008^2$ and $a_1+a_2+\cdots+a_{2007}=2007^2$
Hence, by substitution, $a_{2008}=2008^2-2007^2=(2008+2007)(2008-2007)=4015(1)=4015\implies\boxed{\textbf{B}}$
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PHYS112-Lecture 2-Connect - PHYS 112 003 Introductory Physics I Lecture 2 INSTRUCTOR MARIE-PIERRE MILETTE Assignment u There is a bonus assignment to
# PHYS112-Lecture 2-Connect - PHYS 112 003 Introductory...
This preview shows page 1 - 13 out of 34 pages.
PHYS 112 - 003 Introductory Physics I Lecture 2 INSTRUCTOR: MARIE-PIERRE MILETTE
Assignment u There is a “bonus” assignment to introduce you to Mastering Physics (extra credit).
Vectors Summary Vector ࠵? : u Arrow’s length indicate the vector’s magnitude A u Arrow’s direction indictae the vector direction ࠵? Vectors can also be represented with their scalar components: u ࠵? = ࠵? & , ࠵? ( x y ࠵? ࠵?
Today’s Outline u Introducing physics u What is physics? u Modeling u Problem solving u Why is this true? u Motion and how to describe it u Motion diagrams u Quantities to describe motion
Today’s Outline u Introducing physics u What is physics? u Modeling u Problem solving u Why is this true? u Motion and how to describe it u Motion diagrams u Quantities to describe motion
What is physics?
Modeling u A model is a simplified representation of an object, a system (group of objects), an interaction or a process. u When creating a model, scientist decides which features to include and which to neglect.
Example of modeling: Simple point-like object Car moving from point A to point B u If looking at distance the car is moving we can model it as a point like object A B x
Example of modeling: Simple point-like object Car moving from point A to point B u Car moving into a parking spot can not be modeled as a point like object
Problem solving Problem solving steps: 1. Sketch and translate 2. Simplify and diagram 3. Represent mathematically 4. Solve and Evaluate
Learning Physics u The most important strategy to help you learn physics is called interrogation
Today’s Outline u Introducing physics u What is physics? u Modeling u Problem solving u Why is this true?
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# Solving Simultaneous Equations
Students learn how to set up and solve a pair of simultaneous equations using the elimination method. Learning progresses from solving equations where the coefficients are equal to setting up a pair of equations with different coefficients from known facts.
This unit takes place in Term 2 of Year 11 and follows on from solving equations with a single unknown.
##### Revision Lessons
Prerequisite Knowledge
• Solve two simultaneous equations in two variables algebraically;
• Find approximate solutions to simultaneous equations in two variables using a graph
• Translate simple situations or procedures into algebraic expressions or formulae; derive an equation
Success Criteria
• Solve two simultaneous equations in two variables algebraically;
• Find approximate solutions using a graph
Key Concepts
• For every unknown an equation is needed.
• Students need to have a secure understanding of adding and subtracting with negatives when eliminating an unknown.
• Coefficients need to be equal in magnitude to eliminate an unknown.
• Students need to check their solutions by substituting the calculated values into the original equations.
Common Misconceptions
• Students often struggle knowing when to add or subtract the equations to eliminate the unknown. Review addition with negatives to address this.
• Equations need to be aligned so that unknowns can be easily added or subtracted. If equations are not aligned students may add or subtract with non like variables.
• Students often try to eliminate variables with their coefficients being equal
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# If $\det(\mathrm{adj}\,A) \ne 0$, then $\det(A) \ne 0$.
I'm trying to understand the reason why $A$ is invertible only if $\mathrm{adj}\,A$ is invertible.
That's what I have right now: $A\, \mathrm{adj}\,A = |A|\cdot I$.
So if we take $\det$ of both sides we get: $|A\,\mathrm{adj}\,A| = ||A|\cdot I|$
and then: $|A| \cdot |\mathrm{adj}\,A| = |A|^n$
but now I'm stuck...
-
Is $A$ a $5\times 5$ matrix? – Michael Albanese Oct 13 '12 at 14:22
What is $\mathrm{adj} A$? – Norbert Oct 13 '12 at 14:31
– anon Oct 13 '12 at 14:38
@anon thanks!${}$ – Norbert Oct 13 '12 at 14:40
Suppose $\det{(\operatorname{adj}{A})} \neq 0$ but $\det{A} = 0$. Since $A \operatorname{adj} A = 0$ and $\operatorname{adj}{A}$ is invertible, we have $A=0$, so $\operatorname{adj}{A} = 0$, giving $\det{(\operatorname{adj}{A})} = 0$ which is a contradiction.
-
For simplicity put $\,B:=adj\, A\,$ , so:
$$AB=|A|\cdot I\Longrightarrow |A||B|=|A|^n$$
We're done, since
$$|B|=0\Longrightarrow |A|^n=0\Longrightarrow |A|=0$$
-
I don't understand why did you set |B| = 0. My question may not be clear. I mean if we know that |adjA| isn't zero how can we assert that |A| isn't zero. Now in your proof you set |B|=|adjA|=0 and you show that |A|=0. how does it help me? – SyndicatorBBB Oct 13 '12 at 15:13
@user44471: You should change "equal" to "not equal" in the title then, so that the title matches the actual question. – Hans Lundmark Oct 13 '12 at 15:29
Yes my mistake... sorry. – SyndicatorBBB Oct 13 '12 at 15:48
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```Properties of Real Numbers
IN WORDS
ISN Page 32
You have learned properties of real numbers starting in 4th grade. 8th grade math pulls them all together so that we
can justify our steps when solving equations. Let’s review.
IN WORDS:
If you add the same number on both sides of
an equation, the two sides remain equal.
2) Subtraction Property of Equality:
IN WORDS:
If you subtract the same number on both sides of
an equation, the two sides remain equal.
3) Multiplication Property of Equality:
IN WORDS:
If you multiply the same number on both sides of
an equation, the two sides remain equal.
4) Division Property of Equality:
IN WORDS:
If you divide the same number on both sides of
an equation, the two sides remain equal.
IN WORDS:
The way in which addends are grouped does not
change the sum.
6) Associative Property of Multiplication:
IN WORDS:
The way in which factors are grouped does not
change the sum.
IN WORDS:
The order in which numbers are added does not
change the sum
8) Commutative Property of Multiplication:
IN WORDS:
The order in which numbers are multiplied does
not change the sum
IN WORDS:
The sum of an addend and zero
10) Identity Property of Multiplication (Multiplicative Identity)
IN WORDS:
The product of a factor and 1 is
the factor.
IN WORDS:
The sum of a number and its
12) Inverse Property of Multiplication (Multiplicative Inverse)
IN WORDS:
The product of a number and its
multiplicative inverse is 1.
13) Distributive Property: Not available in just words. We will practice on the back
14) Zero Product Property:
IN WORDS: The product of a factor and zero is zero.
There are a lot of properties to know. Use flashcards to help you remember them. Always do your assigned
homework, and complete the notes on the back.
Properties of Real Numbers
EXAMPLES
ISN Page 32
Match the property on the left with its example on the right.
1) Addition Property of Equality: Example:
A
a + (b + c) = (a + b) + c
2) Subtraction Property of Equality: Example:
B
2 + (3+4) = (2+3) + 4
a+0=a
4+0=4
3) Multiplication Property of Equality: Example:
4) Division Property of Equality: Example:
D
5) Associative Property of Addition: Example:
a
C
(b c) = (a
a
1=a
b)
c
4
1=4
2 (3 4) = (2 3)
6) Associative Property of Multiplication: Example:
7) Commutative Property of Addition: Example:
E
a
b=b
a
4
8) Commutative Property of Multiplication: Example:
a
G
4
2(x + 2) = 2x + 4
4(x – 1) = 4x - 1
F
10) Identity Property of Multiplication (Multiplicative Identity) Example:
3=3
0=0
4
0=0
12) Inverse Property of Multiplication (Multiplicative Inverse) Example:
H
a + (-a) = 0
3 + (-3) = 0
13) Distributive Property: Example:
J
14) Zero Product Property: Example:
a+b=b+a
1+4=4+1
3x = 6
3 3
K
x-2=6
x-2+2=6+2
L
x+1=5
x + 1 -1 = 5 - 1
M
N
A
a
1
a
=1
x
2 5 2
2
3
1
=1
3
4
```
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# Playing roulette
Suppose we have a roulette wheel with $38$ total slots ($18$ black/red, $2$ neither). The ball lands in one slot selected uniformly at random, independent of all previous spins. An $\$x$bet on "black" pays$\$2x$ if the ball lands on black, and $\$0$otherwise. If you bet$1 on black for 100 consecutive spins, how much money will you end up with in expectation?
You walk up to the Roulette table with $\$15$and intend to walk away with$\$16$ using the following betting strategy. You will first bet $\$1$on black. If you win you have$\$16$ and walk away, otherwise you have $\$14$and bet$\$2$ on black on the next spin. If you win that bet you walk away, otherwise you bet $\$4$on the next spin. If you win that bet you walk away, otherwise you bet$\$8$ on the next spin and walk away, either with $\$16$if you win or$\$0$ if you lose.
What's the probability you will walk away with $\$16$? How much money are you expected to walk away with? My biggest problem with these kinds of questions is figuring out how to choose the random variable to calculate the expected values. I understand the formula for calculating the expected value, but translating a problem into those terms has been giving me a hard time. - ## 2 Answers Let's first look at the first question: If you bet$1 on black for 100 consecutive spins, how much money will you end up with in expectation?
So you want to know what your final return will be, at the end of 100 spins. Call this $R$. That is just giving it a name, but what is your final return? You can see that it is the sum of the returns from each bet. So let the return on the $i$th bet be $R_i$, then note that $R = R_1 + R_2 + \dots + R_{100}$. So the expected return is $E[R] = E[R_1] + E[R_2] + \dots + E[R_{100}]$ by linearity of expectation.
So to calculate $E[R]$, we'll be done if we calculate each $E[R_i]$. Let's try to calculate a particular $E[R_i]$. You bet $1$ dollar, and you get back $2$ if the ball lands on black, and $0$ if it doesn't. In other words, you gain $1$ dollar if it lands on black, and lose $1$ dollar if it doesn't. The probability of the former is $18/38$, and that of the latter is $20/38$. In other words, $R_i$ is $1$ with probability $18/38$, and $-1$ with probability $20/38$, so the expected value of $R_i$ is $E[R_i] = \frac{18}{38}(1) + \frac{20}{38}(-1) = \frac{-2}{38}$. Now, as this is the same for each $R_i$, we have $E[R] = E[R_1] + E[R_2] + \dots + E[R_{100}] = \left(\frac{-2}{38}\right)100 \approx -5.26$.
For the second question, let the amount you walk away with be $W$. Let $p = 18/38$, the probability that your bet on black succeeds. There are $5$ possible outcomes:
• you win your first bet: probability $p$
• you lose your first bet, and win your second: probability $(1-p)p$
• you lose your first two bets, and win the third: probability $(1-p)^2p$
• you lose your first three bets, and win the fourth: probability $(1-p)^3p$
• you lose all four bets: probability $(1-p)^4$
In the first four outcomes, you walk away with $16$ dollars, so the probability of that happening (let's call it $q$) is $q = p + (1-p)p + (1-p)^2p + (1-p)^3p = 1 - (1-p)^4 = 1 - (20/38)^4 \approx 0.92$.
[More simply, you could think of it as just two outcomes: (a) that you win some bet, which has probability $q = 1 - (1-p)^4$, and (b) that you win no bet (lose all bets), which has probability $(1-p)^4$.]
In other words, $W$ is $16$ with probability $q$, and $0$ with probability $1-q$. So the expected amount of money you walk away with is therefore $E[W] = q(16) + (1-q)0 = (1-(1-p)^4)16 \approx 14.77$.
[Aside: Note that this is less than the $15$ you came in with. This shows that you can't win in expectation even with your clever betting strategy; a consequence of the optional stopping theorem.]
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Spotted a small error: It should be E[Ri] = 18/38 * 1 + 20/38 * -1 – nknj Jul 22 '15 at 15:52
@nknj: Thanks, fixed. – ShreevatsaR Jul 22 '15 at 16:15
Probability of getting a black on any particular spin is 18 in 38, ie $p = \frac{18}{38} = \frac{9}{19} = 0.474$. Let $X$ be the random variable 'number of blacks in 100 consecutive spins'. Then the expected value of X is $E(X) = 100p = \frac{900}{19} = 47.37$, that is, you expect to win 47.37 times in 100 spins. Since you get $\$$2 for each \$$1 bet, you will therefore expect to walk away with$\$2\times 47.37 = \$94.74$. For the second part of your question, the only way you can lose is if for four consecutive spins, you get non-blacks each time. So what is the probability of losing 4 times in a row? Well, for 1 spin, the probability of not getting black is$1-p = \frac{20}{38} = \frac{10}{19} = 0.526$so for 4 consecutive spins, the probability of not getting any black is$(\frac{10}{19})^{4} = \frac{10,000}{130,321} = 0.0767$. That is, you have about a 7.7% chance of losing with this betting strategy. That is, you have 92.3% chance of walking away with$\16.
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Can you elaborate on why the expected value is just $100p$? Particularly, how does the linearity of expectations apply here? If we let $X_i = 1$ if we land on black and $X_i = 0$ if we don't land on black, how can we calculate the expected value then? – user1038665 Nov 14 '12 at 10:53
Also, for the second part, how would we calculate the "amount of money you are expected to walk away with"? – user1038665 Nov 14 '12 at 10:55
Expected value of a bet = wager x probability. – theo Nov 14 '12 at 11:02
If you bet 1 dollar 100 times on black you will win about 47.37 times and thus leave the casino with 94.74 of your original 100 dollars. – Hagen von Eitzen Nov 14 '12 at 11:03
The answer given is -\$(2/38)*(100), so I believe this is incorrect. – user1038665 Nov 14 '12 at 11:20
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Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago
# Suppose a mass of m = 2 kg is suspended from a spring with spring constant k = 10 N/m. ?
Suppose a mass of m = 2 kg is suspended from a spring with spring constant k = 10
N/m. Furthermore, suppose the spring is submerged in a medium that provides a damping force that is 4 times the velocity. The mass is released from a position 1 m below equilibrium with no initial velocity. Find the equation of motion and explain whether the system underdamped, overdamped or critically damped?
Relevance
• 2 months ago
I'll get you started in the right direction.
We can set up a differential equation simply by using Newton's law ΣF = ma. First, we identify all of the forces acting on the spring.
Naturally, we have gravity acting on this suspended spring. The magnitude of the force due to gravity is given by
F_g = mg
The damping force also acts on the spring. We are given that this force is four times the velocity at any given time. Thus, the magnitude of the damping force is given by
F_d = 4v, where v is the velocity at any time t.
Finally, there is a "restorative" force which tends the mass-spring system to it's equilibrium position. According to Hooke's law, the magnitude of this force is given by
F_s = kx, where k is the spring constant and x is the displacement from the equilibrium position.
Now, if we take the down direction to be negative, we can assign signs to the forces F_g, F_d, and F_s according to which direction they will act on the mass.
Note that the gravitational force acts in the downward direction, thus we assign a negative to F_g.
Since the damping force will always oppose the direction of travel of the spring system, we may assign a negative to F_d as well( whatever sign the velocity has at time t, we negate it).
Finally, the restorative force too may be assigned a negative since it opposes the direction of travel as well.
Summing up the forces we get the following:
F_g + F_d + F_s = ma
-mg - 4v - kx = ma
Note that v = dx/dt and a = dv/dt = dx²/dt²
Thus, we may rewrite as
-mg - 4(dx/dt) - kx = mdx²/dt²
We can rearrange as
mdx²/dt² + 4dx/dt + kx = -mg
Note that this a second order ODE. I'll let you take it from here.
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# Metamath Proof Explorer
## Theorem scshwfzeqfzo
Description: For a nonempty word the sets of shifted words, expressd by a finite interval of integers or by a half-open integer range are identical. (Contributed by Alexander van der Vekens, 15-Jun-2018)
Ref Expression
Assertion scshwfzeqfzo ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\to \left\{{y}\in \mathrm{Word}{V}|\exists {n}\in \left(0\dots {N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right\}=\left\{{y}\in \mathrm{Word}{V}|\exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right\}$
### Proof
Step Hyp Ref Expression
1 lencl ${⊢}{X}\in \mathrm{Word}{V}\to \left|{X}\right|\in {ℕ}_{0}$
2 elnn0uz ${⊢}\left|{X}\right|\in {ℕ}_{0}↔\left|{X}\right|\in {ℤ}_{\ge 0}$
3 1 2 sylib ${⊢}{X}\in \mathrm{Word}{V}\to \left|{X}\right|\in {ℤ}_{\ge 0}$
4 3 adantr ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {N}=\left|{X}\right|\right)\to \left|{X}\right|\in {ℤ}_{\ge 0}$
5 eleq1 ${⊢}{N}=\left|{X}\right|\to \left({N}\in {ℤ}_{\ge 0}↔\left|{X}\right|\in {ℤ}_{\ge 0}\right)$
6 5 adantl ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {N}=\left|{X}\right|\right)\to \left({N}\in {ℤ}_{\ge 0}↔\left|{X}\right|\in {ℤ}_{\ge 0}\right)$
7 4 6 mpbird ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {N}=\left|{X}\right|\right)\to {N}\in {ℤ}_{\ge 0}$
8 7 3adant2 ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\to {N}\in {ℤ}_{\ge 0}$
9 8 adantr ${⊢}\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\to {N}\in {ℤ}_{\ge 0}$
10 fzisfzounsn ${⊢}{N}\in {ℤ}_{\ge 0}\to \left(0\dots {N}\right)=\left(0..^{N}\right)\cup \left\{{N}\right\}$
11 9 10 syl ${⊢}\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\to \left(0\dots {N}\right)=\left(0..^{N}\right)\cup \left\{{N}\right\}$
12 11 rexeqdv ${⊢}\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\to \left(\exists {n}\in \left(0\dots {N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}↔\exists {n}\in \left(\left(0..^{N}\right)\cup \left\{{N}\right\}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right)$
13 rexun ${⊢}\exists {n}\in \left(\left(0..^{N}\right)\cup \left\{{N}\right\}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}↔\left(\exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\vee \exists {n}\in \left\{{N}\right\}\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right)$
14 12 13 syl6bb ${⊢}\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\to \left(\exists {n}\in \left(0\dots {N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}↔\left(\exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\vee \exists {n}\in \left\{{N}\right\}\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right)\right)$
15 fvex ${⊢}\left|{X}\right|\in \mathrm{V}$
16 eleq1 ${⊢}{N}=\left|{X}\right|\to \left({N}\in \mathrm{V}↔\left|{X}\right|\in \mathrm{V}\right)$
17 15 16 mpbiri ${⊢}{N}=\left|{X}\right|\to {N}\in \mathrm{V}$
18 oveq2 ${⊢}{n}={N}\to {X}\mathrm{cyclShift}{n}={X}\mathrm{cyclShift}{N}$
19 18 eqeq2d ${⊢}{n}={N}\to \left({y}={X}\mathrm{cyclShift}{n}↔{y}={X}\mathrm{cyclShift}{N}\right)$
20 19 rexsng ${⊢}{N}\in \mathrm{V}\to \left(\exists {n}\in \left\{{N}\right\}\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}↔{y}={X}\mathrm{cyclShift}{N}\right)$
21 17 20 syl ${⊢}{N}=\left|{X}\right|\to \left(\exists {n}\in \left\{{N}\right\}\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}↔{y}={X}\mathrm{cyclShift}{N}\right)$
22 21 3ad2ant3 ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\to \left(\exists {n}\in \left\{{N}\right\}\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}↔{y}={X}\mathrm{cyclShift}{N}\right)$
23 22 adantr ${⊢}\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\to \left(\exists {n}\in \left\{{N}\right\}\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}↔{y}={X}\mathrm{cyclShift}{N}\right)$
24 oveq2 ${⊢}{N}=\left|{X}\right|\to {X}\mathrm{cyclShift}{N}={X}\mathrm{cyclShift}\left|{X}\right|$
25 24 3ad2ant3 ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\to {X}\mathrm{cyclShift}{N}={X}\mathrm{cyclShift}\left|{X}\right|$
26 cshwn ${⊢}{X}\in \mathrm{Word}{V}\to {X}\mathrm{cyclShift}\left|{X}\right|={X}$
27 26 3ad2ant1 ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\to {X}\mathrm{cyclShift}\left|{X}\right|={X}$
28 25 27 eqtrd ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\to {X}\mathrm{cyclShift}{N}={X}$
29 28 eqeq2d ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\to \left({y}={X}\mathrm{cyclShift}{N}↔{y}={X}\right)$
30 29 adantr ${⊢}\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\to \left({y}={X}\mathrm{cyclShift}{N}↔{y}={X}\right)$
31 cshw0 ${⊢}{X}\in \mathrm{Word}{V}\to {X}\mathrm{cyclShift}0={X}$
32 31 3ad2ant1 ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\to {X}\mathrm{cyclShift}0={X}$
33 lennncl ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \right)\to \left|{X}\right|\in ℕ$
34 33 3adant3 ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\to \left|{X}\right|\in ℕ$
35 eleq1 ${⊢}{N}=\left|{X}\right|\to \left({N}\in ℕ↔\left|{X}\right|\in ℕ\right)$
36 35 3ad2ant3 ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\to \left({N}\in ℕ↔\left|{X}\right|\in ℕ\right)$
37 34 36 mpbird ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\to {N}\in ℕ$
38 lbfzo0 ${⊢}0\in \left(0..^{N}\right)↔{N}\in ℕ$
39 37 38 sylibr ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\to 0\in \left(0..^{N}\right)$
40 oveq2 ${⊢}0={n}\to {X}\mathrm{cyclShift}0={X}\mathrm{cyclShift}{n}$
41 40 eqeq1d ${⊢}0={n}\to \left({X}\mathrm{cyclShift}0={X}↔{X}\mathrm{cyclShift}{n}={X}\right)$
42 41 eqcoms ${⊢}{n}=0\to \left({X}\mathrm{cyclShift}0={X}↔{X}\mathrm{cyclShift}{n}={X}\right)$
43 eqcom ${⊢}{X}\mathrm{cyclShift}{n}={X}↔{X}={X}\mathrm{cyclShift}{n}$
44 42 43 syl6bb ${⊢}{n}=0\to \left({X}\mathrm{cyclShift}0={X}↔{X}={X}\mathrm{cyclShift}{n}\right)$
45 44 adantl ${⊢}\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {n}=0\right)\to \left({X}\mathrm{cyclShift}0={X}↔{X}={X}\mathrm{cyclShift}{n}\right)$
46 45 biimpd ${⊢}\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {n}=0\right)\to \left({X}\mathrm{cyclShift}0={X}\to {X}={X}\mathrm{cyclShift}{n}\right)$
47 39 46 rspcimedv ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\to \left({X}\mathrm{cyclShift}0={X}\to \exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{X}={X}\mathrm{cyclShift}{n}\right)$
48 32 47 mpd ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\to \exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{X}={X}\mathrm{cyclShift}{n}$
49 48 adantr ${⊢}\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\to \exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{X}={X}\mathrm{cyclShift}{n}$
50 49 adantr ${⊢}\left(\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\wedge {y}={X}\right)\to \exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{X}={X}\mathrm{cyclShift}{n}$
51 eqeq1 ${⊢}{y}={X}\to \left({y}={X}\mathrm{cyclShift}{n}↔{X}={X}\mathrm{cyclShift}{n}\right)$
52 51 adantl ${⊢}\left(\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\wedge {y}={X}\right)\to \left({y}={X}\mathrm{cyclShift}{n}↔{X}={X}\mathrm{cyclShift}{n}\right)$
53 52 rexbidv ${⊢}\left(\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\wedge {y}={X}\right)\to \left(\exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}↔\exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{X}={X}\mathrm{cyclShift}{n}\right)$
54 50 53 mpbird ${⊢}\left(\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\wedge {y}={X}\right)\to \exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}$
55 54 ex ${⊢}\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\to \left({y}={X}\to \exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right)$
56 30 55 sylbid ${⊢}\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\to \left({y}={X}\mathrm{cyclShift}{N}\to \exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right)$
57 23 56 sylbid ${⊢}\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\to \left(\exists {n}\in \left\{{N}\right\}\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\to \exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right)$
58 57 com12 ${⊢}\exists {n}\in \left\{{N}\right\}\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\to \left(\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\to \exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right)$
59 58 jao1i ${⊢}\left(\exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\vee \exists {n}\in \left\{{N}\right\}\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right)\to \left(\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\to \exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right)$
60 59 com12 ${⊢}\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\to \left(\left(\exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\vee \exists {n}\in \left\{{N}\right\}\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right)\to \exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right)$
61 14 60 sylbid ${⊢}\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\to \left(\exists {n}\in \left(0\dots {N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\to \exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right)$
62 fzossfz ${⊢}\left(0..^{N}\right)\subseteq \left(0\dots {N}\right)$
63 ssrexv ${⊢}\left(0..^{N}\right)\subseteq \left(0\dots {N}\right)\to \left(\exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\to \exists {n}\in \left(0\dots {N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right)$
64 62 63 mp1i ${⊢}\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\to \left(\exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\to \exists {n}\in \left(0\dots {N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right)$
65 61 64 impbid ${⊢}\left(\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\wedge {y}\in \mathrm{Word}{V}\right)\to \left(\exists {n}\in \left(0\dots {N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}↔\exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right)$
66 65 rabbidva ${⊢}\left({X}\in \mathrm{Word}{V}\wedge {X}\ne \varnothing \wedge {N}=\left|{X}\right|\right)\to \left\{{y}\in \mathrm{Word}{V}|\exists {n}\in \left(0\dots {N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right\}=\left\{{y}\in \mathrm{Word}{V}|\exists {n}\in \left(0..^{N}\right)\phantom{\rule{.4em}{0ex}}{y}={X}\mathrm{cyclShift}{n}\right\}$
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Question
# Please answer both!! Suppose the national average dollar amount for an automobile insurance claim is \$805.66....
Suppose the national average dollar amount for an automobile insurance claim is \$805.66. You work for an agency in Michigan and you are interested in whether or not the state average is less than the national average. The hypotheses for this scenario are as follows: Null Hypothesis: μ ≥ 805.66, Alternative Hypothesis: μ < 805.66. You take a random sample of claims and calculate a p-value of 0.02 based on the data, what is the appropriate conclusion? Conclude at the 5% level of significance.
Question 12 options:
1) The true average claim amount is significantly higher than \$805.66.
2) The true average claim amount is higher than or equal to \$805.66.
3) The true average claim amount is significantly less than \$805.66.
4) The true average claim amount is significantly different from \$805.66.
5) We did not find enough evidence to say the true average claim amount is less than \$805.66.
It is believed that students who begin studying for final exams a week before the test score differently than students who wait until the night before. Suppose you want to test the hypothesis that students who study one week before score different from students who study the night before, giving you the following hypotheses: Null Hypothesis: μ1 = μ2, Alternative Hypothesis: μ1 ≠ μ2. A random sample of 40 students who indicated they studied early shows an average score of 90.78 (SD = 4.15) and 29 randomly selected procrastinators had an average score of 88.15 (SD = 6.977). Perform a two independent samples t-test assuming that early studiers are group 1 and procrastinators are group 2. What is the test statistic and p-value of this test? Assume the population standard deviations are the same.
Question 14 options:
1) Test Statistic: -1.957, P-Value: 0.0545
2) Test Statistic: 1.957, P-Value: 0.0545
3) Test Statistic: 1.957, P-Value: 1.9728
4) Test Statistic: 1.957, P-Value: 0.9728
5) Test Statistic: 1.957, P-Value: 0.0272
Solution :
12)
P-value = 0.02
P-value < 0.05
Reject the null hypothesis
The appropriate conclusion is :
The true average claim amount is significantly less than \$805.66.
Option 3)
14)
Test Statistic: 1.957, P-Value: 0.0545
Option 2)
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Sir Isaac Newton Law of Gravitation.
Presentation on theme: "Sir Isaac Newton Law of Gravitation."— Presentation transcript:
Sir Isaac Newton Law of Gravitation
What Really Happened with the Apple
Probably the more correct version of the story is that Newton, upon observing an apple fall from a tree, began to think along the following lines: The apple is accelerated, since its velocity changes from zero as it is hanging on the tree and moves toward the ground. Thus, by Newton's 2nd Law there must be a force that acts on the apple to cause this acceleration. Let's call this force "gravity", and the associated acceleration the "accleration due to gravity". Then imagine the apple tree is twice as high. Again, we expect the apple to be accelerated toward the ground, so this suggests that this force that we call gravity reaches to the top of the tallest apple tree.
Sir Isaac's Most Excellent Idea
Now came Newton's truly brilliant insight: if the force of gravity reaches to the top of the highest tree, might it not reach even further; in particular, might it not reach all the way to the orbit of the Moon! Then, the orbit of the Moon about the Earth could be a consequence of the gravitational force, because the acceleration due to gravity could change the velocity of the Moon in just such a way that it followed an orbit around the earth. This can be illustrated with the thought experiment shown in the following figure. Suppose we fire a cannon horizontally from a high mountain; the projectile will eventually fall to earth, as indicated by the shortest trajectory in the figure, because of the gravitational force directed toward the center of the Earth and the associated acceleration. (Remember that an acceleration is a change in velocity and that velocity is a vector, so it has both a magnitude and a direction. Thus, an acceleration occurs if either or both the magnitude and the direction of the velocity change.)
But as we increase the muzzle velocity for our imaginary cannon, the projectile will travel further and further before returning to earth. Finally, Newton reasoned that if the cannon projected the cannon ball with exactly the right velocity, the projectile would travel completely around the Earth, always falling in the gravitational field but never reaching the Earth, which is curving away at the same rate that the projectile falls. That is, the cannon ball would have been put into orbit around the Earth. Newton concluded that the orbit of the Moon was of exactly the same nature: the Moon continuously "fell" in its path around the Earth because of the acceleration due to gravity, thus producing its orbit.
By such reasoning, Newton came to the conclusion that any two objects in the Universe exert gravitational attraction on each other, with the force having a universal form:
The constant of proportionality G is known as the universal gravitational constant. It is termed a "universal constant" because it is thought to be the same at all places and all times, and thus universally characterizes the intrinsic strength of the gravitational force.
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# MASTER NOTES LAB PHYSICS Newtons’ Laws and Force Only Chapter 4
## Presentation on theme: "MASTER NOTES LAB PHYSICS Newtons’ Laws and Force Only Chapter 4"— Presentation transcript:
MASTER NOTES LAB PHYSICS Newtons’ Laws and Force Only Chapter 4
Forces and the Laws of Motion Table of Contents Section 1 Changes in Motion Section 2 Newton's First Law Section 3 Newton's Second and Third Laws Section 4 Everyday Forces
Chapter 4 Section 1 Changes in Motion Force A force is an action exerted on an object which may change the object’s state of rest or motion. Forces can cause accelerations. The SI unit of force is the newton, N. Forces can act through contact or at a distance.
Chapter 4 Force Diagrams
Section 1 Changes in Motion Force Diagrams The effect of a force depends on both magnitude and direction.Thus, force is a vector quantity. Diagrams that show force vectors as arrows are called force diagrams. Force diagrams that show only the forces acting on a single object are called free-body diagrams.
Force Diagrams, continued
Chapter 4 Section 1 Changes in Motion Force Diagrams, continued Force Diagram Free-Body Diagram In a force diagram, vector arrows represent all the forces acting in a situation. A free-body diagram shows only the forces acting on the object of interest—in this case, the car.
Chapter 4 Newton’s First Law
Section 2 Newton’s First Law Newton’s First Law An object at rest remains at rest, and an object in motion continues in motion with constant velocity (that is, constant speed in a straight line) unless the object experiences a net external force. In other words, when the net external force on an object is zero, the object’s acceleration (or the change in the object’s velocity) is zero.
Chapter 4 Section 2 Newton’s First Law Net Force Newton's first law refers to the net force on an object.The net force is the vector sum of all forces acting on an object. The net force on an object can be found by using the methods for finding resultant vectors. Although several forces are acting on this car, the vector sum of the forces is zero. Thus, the net force is zero, and the car moves at a constant velocity.
Chapter 4 Sample Problem Determining Net Force
Section 2 Newton’s First Law Sample Problem Determining Net Force Derek leaves his physics book on top of a drafting table that is inclined at a 35° angle. The free-body diagram below shows the forces acting on the book. Find the net force acting on the book.
Sample Problem, continued
Chapter 4 Section 2 Newton’s First Law Sample Problem, continued 1. Define the problem, and identify the variables. Given: Fgravity-on-book = Fg = 22 N Ffriction = Ff = 11 N Ftable-on-book = Ft = 18 N Unknown: Fnet = ?
Sample Problem, continued
Chapter 4 Section 2 Newton’s First Law Sample Problem, continued 2. Select a coordinate system, and apply it to the free-body diagram. Choose the x-axis parallel to and the y-axis perpendicular to the incline of the table, as shown in (a). This coordinate system is the most convenient because only one force needs to be resolved into x and y components. Tip: To simplify the problem, always choose the coordinate system in which as many forces as possible lie on the x- and y-axes.
Sample Problem, continued
Chapter 4 Section 2 Newton’s First Law Sample Problem, continued 3. Find the x and y components of all vectors. Draw a sketch, as shown in (b), to help find the components of the vector Fg. The angle q is equal to 180– 90 – 35 = 55. Add both components to the free-body diagram, as shown in (c).
Sample Problem, continued
Chapter 4 Section 2 Newton’s First Law Sample Problem, continued 4. Find the net force in both the x and y directions. Diagram (d) shows another free-body diagram of the book, now with forces acting only along the x- and y-axes. For the x direction: SFx = Fg,x – Ff SFx = 13 N – 11 N SFx = 2 N For the y direction: SFy = Ft – Fg,y SFy = 18 N – 18 N SFy = 0 N
Sample Problem, continued
Chapter 4 Section 2 Newton’s First Law Sample Problem, continued 5. Find the net force. Add the net forces in the x and y directions together as vectors to find the total net force. In this case, Fnet = 2 N in the +x direction, as shown in (e). Thus, the book accelerates down the incline.
Chapter 4 Section 2 Newton’s First Law Inertia Inertia is the tendency of an object to resist being moved or, if the object is moving, to resist a change in speed or direction. Newton’s first law is often referred to as the law of inertia because it states that in the absence of a net force, a body will preserve its state of motion. Mass is a measure of inertia.
Inertia and the Operation of a Seat Belt
Chapter 4 Section 2 Newton’s First Law Inertia and the Operation of a Seat Belt While inertia causes passengers in a car to continue moving forward as the car slows down, inertia also causes seat belts to lock into place. The illustration shows how one type of shoulder harness operates. When the car suddenly slows down, inertia causes the large mass under the seat to continue moving, which activates the lock on the safety belt.
Chapter 4 Section 2 Newton’s First Law Equilibrium Equilibrium is the state in which the net force on an object is zero. Objects that are either at rest or moving with constant velocity are said to be in equilibrium. Newton’s first law describes objects in equilibrium. Tip: To determine whether a body is in equilibrium, find the net force. If the net force is zero, the body is in equilibrium. If there is a net force, a second force equal and opposite to this net force will put the body in equilibrium.
net force = mass acceleration
Section 3 Newton’s Second and Third Laws Chapter 4 Newton’s Second Law The acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to the object’s mass. SF = ma net force = mass acceleration SF represents the vector sum of all external forces acting on the object, or the net force.
Chapter 4 Newton’s Third Law
Section 3 Newton’s Second and Third Laws Chapter 4 Newton’s Third Law If two objects interact, the magnitude of the force exerted on object 1 by object 2 is equal to the magnitude of the force simultaneously exerted on object 2 by object 1, and these two forces are opposite in direction. In other words, for every action, there is an equal and opposite reaction. Because the forces coexist, either force can be called the action or the reaction.
Action and Reaction Forces
Section 3 Newton’s Second and Third Laws Chapter 4 Action and Reaction Forces Action-reaction pairs do not imply that the net force on either object is zero. The action-reaction forces are equal and opposite, but either object may still have a net force on it. Consider driving a nail into wood with a hammer. The force that the nail exerts on the hammer is equal and opposite to the force that the hammer exerts on the nail. But there is a net force acting on the nail, which drives the nail into the wood.
Chapter 4 Section 4 Everyday Forces Weight The gravitational force (Fg) exerted on an object by Earth is a vector quantity, directed toward the center of Earth. The magnitude of this force (Fg) is a scalar quantity called weight. Weight changes with the location of an object in the universe.
Chapter 4 Weight, continued Calculating weight at any location:
Section 4 Everyday Forces Weight, continued Calculating weight at any location: Fg = mag ag = free-fall acceleration at that location Calculating weight on Earth's surface: ag = g = 9.81 m/s2 Fg = mg = m(9.81 m/s2)
Chapter 4 Section 4 Everyday Forces Normal Force The normal force acts on a surface in a direction perpendicular to the surface. The normal force is not always opposite in direction to the force due to gravity. In the absence of other forces, the normal force is equal and opposite to the component of gravitational force that is perpendicular to the contact surface. In this example, Fn = mg cos q.
Chapter 4 Section 4 Everyday Forces Friction Static friction is a force that resists the initiation of sliding motion between two surfaces that are in contact and at rest. Kinetic friction is the force that opposes the movement of two surfaces that are in contact and are sliding over each other. Kinetic friction is always less than the maximum static friction.
Friction Forces in Free-Body Diagrams
Chapter 4 Section 4 Everyday Forces Friction Forces in Free-Body Diagrams In free-body diagrams, the force of friction is always parallel to the surface of contact. The force of kinetic friction is always opposite the direction of motion. To determine the direction of the force of static friction, use the principle of equilibrium. For an object in equilibrium, the frictional force must point in the direction that results in a net force of zero.
The Coefficient of Friction
Chapter 4 Section 4 Everyday Forces The Coefficient of Friction The quantity that expresses the dependence of frictional forces on the particular surfaces in contact is called the coefficient of friction, m. Coefficient of kinetic friction: Coefficient of static friction:
Chapter 4 Section 4 Everyday Forces
Chapter 4 Sample Problem Overcoming Friction
Section 4 Everyday Forces Sample Problem Overcoming Friction A student attaches a rope to a 20.0 kg box of books.He pulls with a force of 90.0 N at an angle of 30.0° with the horizontal. The coefficient of kinetic friction between the box and the sidewalk is Find the acceleration of the box.
Sample Problem, continued
Chapter 4 Section 4 Everyday Forces Sample Problem, continued 1. Define Given: m = 20.0 kg mk = 0.500 Fapplied = 90.0 N at q = 30.0° Unknown: a = ? Diagram:
Sample Problem, continued
Chapter 4 Section 4 Everyday Forces Sample Problem, continued 2. Plan Choose a convenient coordinate system, and find the x and y components of all forces. The diagram on the right shows the most convenient coordinate system, because the only force to resolve into components is Fapplied. Fapplied,y = (90.0 N)(sin 30.0º) = 45.0 N (upward) Fapplied,x = (90.0 N)(cos 30.0º) = 77.9 N (to the right)
Sample Problem, continued
Chapter 4 Section 4 Everyday Forces Sample Problem, continued Choose an equation or situation: A. Find the normal force, Fn, by applying the condition of equilibrium in the vertical direction: SFy = 0 B. Calculate the force of kinetic friction on the box: Fk = mkFn C. Apply Newton’s second law along the horizontal direction to find the acceleration of the box: SFx = max
Sample Problem, continued
Chapter 4 Section 4 Everyday Forces Sample Problem, continued 3. Calculate A. To apply the condition of equilibrium in the vertical direction, you need to account for all of the forces in the y direction: Fg, Fn, and Fapplied,y. You know Fapplied,y and can use the box’s mass to find Fg. Fapplied,y = 45.0 N Fg = (20.0 kg)(9.81 m/s2) = 196 N Next, apply the equilibrium condition, SFy = 0, and solve for Fn. SFy = Fn + Fapplied,y – Fg = 0 Fn N – 196 N = 0 Fn = –45.0 N N = 151 N Tip: Remember to pay attention to the direction of forces. In this step, Fg is subtracted from Fn and Fapplied,y because Fg is directed downward.
Sample Problem, continued
Chapter 4 Section 4 Everyday Forces Sample Problem, continued B. Use the normal force to find the force of kinetic friction. Fk = mkFn = (0.500)(151 N) = 75.5 N C. Use Newton’s second law to determine the horizontal acceleration. a = 0.12 m/s2 to the right
Sample Problem, continued
Chapter 4 Section 4 Everyday Forces Sample Problem, continued 4. Evaluate The box accelerates in the direction of the net force, in accordance with Newton’s second law. The normal force is not equal in magnitude to the weight because the y component of the student’s pull on the rope helps support the box.
Chapter 4 Air Resistance
Section 4 Everyday Forces Air Resistance Air resistance is a form of friction. Whenever an object moves through a fluid medium, such as air or water, the fluid provides a resistance to the object’s motion. For a falling object, when the upward force of air resistance balances the downward gravitational force, the net force on the object is zero. The object continues to move downward with a constant maximum speed, called the terminal speed.
Chapter 4 Fundamental Forces There are four fundamental forces:
Section 4 Everyday Forces Fundamental Forces There are four fundamental forces: Electromagnetic force Gravitational force Strong nuclear force Weak nuclear force The four fundamental forces are all field forces.
Review: Newton’s 1st Law
An object in motion stays in motion in a straight line, unless acted upon by unbalanced force. A push or pull will cause object to speed up, slow down, or change direction.
Review: Forces are Balanced
Object at Rest V = zero m/s Objects in Motion V ≠ zero m/s a = 0 m/s2 a = 0 m/s2 Stay at Rest Stay in Motion (same speed and direction
Basically, objects just keep on doing whatever they are doing unless they are acted upon by an unbalanced force.
Review: Common Examples
Ketchup stays in the bottom (at rest) until you bang (outside force) on the end of the bottom. A headrest in a car prevents whiplash injuries during a rear-end collision ( your head goes forward and then jerks backward). Animation 1 – ladder truck Animation 2 – no seatbelt
Free-body diagrams Free-body diagrams are used to show the relative magnitude and direction of all forces acting on an object.
This diagram shows four forces acting upon an object
This diagram shows four forces acting upon an object. There aren’t always four forces, For example, there could be one, two, or three forces.
Problem 1 A book is at rest on a table top. Diagram the forces acting on the book.
Problem 1 In this diagram, there are normal and gravitational forces on the book.
Problem 2 An egg is free-falling from a nest in a tree. Neglect air resistance. Draw a free-body diagram showing the forces involved.
Gravity is the only force acting on the egg as it falls.
Problem 3 A flying squirrel is gliding (no wing flaps) from a tree to the ground at constant velocity. Consider air resistance. A free body diagram for this situation looks like…
Gravity pulls down on the squirrel while air resistance keeps the squirrel in the air for a while.
Problem 4 A rightward force is applied to a book in order to move it across a desk. Consider frictional forces. Neglect air resistance. Construct a free-body diagram. Let’s see what this one looks like.
Note the applied force arrow pointing to the right
Note the applied force arrow pointing to the right. Notice how friction force points in the opposite direction. Finally, there is still gravity and normal forces involved.
Problem 5 A skydiver is descending with a constant velocity. Consider air resistance. Draw a free-body diagram.
Gravity pulls down on the skydiver, while air resistance pushes up as she falls.
Problem 6 A man drags a sled across loosely packed snow with a rightward acceleration. Draw a free-body diagram.
The rightward force arrow points to the right
The rightward force arrow points to the right. Friction slows his progress and pulls in the opposite direction. Since there is not information that we are in a blizzard, normal forces still apply as does gravitational force since we are on planet Earth.
Problem 7 A football is moving upwards toward its peak after having been booted by the punter. Draw a free-body diagram.
The force of gravity is the only force described
The force of gravity is the only force described. It is not a windy day (no air resistance).
Problem 8 A car runs out of gas and is coasting down a hill.
Even though the car is coasting down the hill, there is still the dragging friction of the road (left pointing arrow) as well as gravity and normal forces.
Net Force Now let’s take a look at what happens when unbalanced forces do not become completely balanced (or cancelled) by other individual forces. An unbalanced forces exists when the vertical and horizontal forces do not cancel each other out.
Example 1 Notice the upward force of 1200 Neutons (N) is more than gravity (800 N). The net force is 400 N up.
Example 2 Notice that while the normal force and gravitation forces are balanced (each are 50 N) the force of friction results in unbalanced force on the horizontal axis. The net force is 20 N left.
Another way to look at balances and unbalanced forces
Balanced or unbalanced?
Balanced or Unbalanced?
Evaluation Complete question #9 on the Free-body Diagram Worksheet.
Special thanks to the Physics Classroom used to prepare this lesson
Chapter 4 Multiple Choice Standardized Test Prep
Use the passage below to answer questions 1–2. Two blocks of masses m1 and m2 are placed in contact with each other on a smooth, horizontal surface. Block m1 is on the left of block m2. A constant horizontal force F to the right is applied to m1. 1. What is the acceleration of the two blocks? A. C. B. D.
Chapter 4 Multiple Choice Standardized Test Prep
Use the passage below to answer questions 1–2. Two blocks of masses m1 and m2 are placed in contact with each other on a smooth, horizontal surface. Block m1 is on the left of block m2. A constant horizontal force F to the right is applied to m1. 1. What is the acceleration of the two blocks? A. C. B. D.
Multiple Choice, continued
Chapter 4 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 1–2. Two blocks of masses m1 and m2 are placed in contact with each other on a smooth, horizontal surface. Block m1 is on the left of block m2. A constant horizontal force F to the right is applied to m1. 2. What is the horizontal force acting on m2? F. m1a G. m2a H. (m1 + m2)a J. m1m2a
Multiple Choice, continued
Chapter 4 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 1–2. Two blocks of masses m1 and m2 are placed in contact with each other on a smooth, horizontal surface. Block m1 is on the left of block m2. A constant horizontal force F to the right is applied to m1. 2. What is the horizontal force acting on m2? F. m1a G. m2a H. (m1 + m2)a J. m1m2a
Multiple Choice, continued
Chapter 4 Standardized Test Prep Multiple Choice, continued 3. A crate is pulled to the right with a force of 82.0 N, to the left with a force of 115 N, upward with a force of 565 N, and downward with a force of 236 N. Find the magnitude and direction of the net force on the crate. A N at 96° counterclockwise from the positive x-axis B N at 6° counterclockwise from the positive x-axis C x 102 at 96° counterclockwise from the positive x-axis D x 102 at 6° counterclockwise from the positive x-axis
Multiple Choice, continued
Chapter 4 Standardized Test Prep Multiple Choice, continued 3. A crate is pulled to the right with a force of 82.0 N, to the left with a force of 115 N, upward with a force of 565 N, and downward with a force of 236 N. Find the magnitude and direction of the net force on the crate. A N at 96° counterclockwise from the positive x-axis B N at 6° counterclockwise from the positive x-axis C x 102 at 96° counterclockwise from the positive x-axis D x 102 at 6° counterclockwise from the positive x-axis
Multiple Choice, continued
Chapter 4 Standardized Test Prep Multiple Choice, continued 4. A ball with a mass of m is thrown into the air, as shown in the figure below. What is the force exerted on Earth by the ball? A. mballg directed down B. mballg directed up C. mearthg directed down D. mearthg directed up
Multiple Choice, continued
Chapter 4 Standardized Test Prep Multiple Choice, continued 4. A ball with a mass of m is thrown into the air, as shown in the figure below. What is the force exerted on Earth by the ball? A. mballg directed down B. mballg directed up C. mearthg directed down D. mearthg directed up
Multiple Choice, continued
Chapter 4 Standardized Test Prep Multiple Choice, continued 5. A freight train has a mass of 1.5 x 107 kg. If the locomotive can exert a constant pull of 7.5 x 105 N, how long would it take to increase the speed of the train from rest to 85 km/h? (Disregard friction.) A. 4.7 x 102s B. 4.7s C. 5.0 x 10-2s D. 5.0 x 104s
Multiple Choice, continued
Chapter 4 Standardized Test Prep Multiple Choice, continued 5. A freight train has a mass of 1.5 x 107 kg. If the locomotive can exert a constant pull of 7.5 x 105 N, how long would it take to increase the speed of the train from rest to 85 km/h? (Disregard friction.) A. 4.7 x 102s B. 4.7s C. 5.0 x 10-2s D. 5.0 x 104s
Multiple Choice, continued
Chapter 4 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 6–7. A truck driver slams on the brakes and skids to a stop through a displacement Dx. 6. If the truck’s mass doubles, find the truck’s skidding distance in terms of Dx. (Hint: Increasing the mass increases the normal force.) A. Dx/4 B. Dx C. 2Dx D. 4Dx
Multiple Choice, continued
Chapter 4 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 6–7. A truck driver slams on the brakes and skids to a stop through a displacement Dx. 6. If the truck’s mass doubles, find the truck’s skidding distance in terms of Dx. (Hint: Increasing the mass increases the normal force.) A. Dx/4 B. Dx C. 2Dx D. 4Dx
Multiple Choice, continued
Chapter 4 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 6–7. A truck driver slams on the brakes and skids to a stop through a displacement Dx. 7. If the truck’s initial velocity were halved, what would be the truck’s skidding distance? A. Dx/4 B. Dx C. 2Dx D. 4Dx
Multiple Choice, continued
Chapter 4 Standardized Test Prep Multiple Choice, continued Use the passage below to answer questions 6–7. A truck driver slams on the brakes and skids to a stop through a displacement Dx. 7. If the truck’s initial velocity were halved, what would be the truck’s skidding distance? A. Dx/4 B. Dx C. 2Dx D. 4Dx
Multiple Choice, continued
Chapter 4 Standardized Test Prep Multiple Choice, continued Use the graph at right to answer questions 8–9. The graph shows the relationship between the applied force and the force of friction. 8. What is the relationship between the forces at point A? F. Fs=Fapplied G. Fk=Fapplied H. Fs<Fapplied I. Fk>Fapplied
Multiple Choice, continued
Chapter 4 Standardized Test Prep Multiple Choice, continued Use the graph at right to answer questions 8–9. The graph shows the relationship between the applied force and the force of friction. 8. What is the relationship between the forces at point A? F. Fs=Fapplied G. Fk=Fapplied H. Fs<Fapplied I. Fk>Fapplied
Multiple Choice, continued
Chapter 4 Standardized Test Prep Multiple Choice, continued Use the graph at right to answer questions 8–9. The graph shows the relationship between the applied force and the force of friction. 9. What is the relationship between the forces at point B? A. Fs, max=Fk B. Fk> Fs, max C. Fk>Fapplied D. Fk<Fapplied
Multiple Choice, continued
Chapter 4 Standardized Test Prep Multiple Choice, continued Use the graph at right to answer questions 8–9. The graph shows the relationship between the applied force and the force of friction. 9. What is the relationship between the forces at point B? A. Fs, max=Fk B. Fk> Fs, max C. Fk>Fapplied D. Fk<Fapplied
Chapter 4 Short Response Base your answers to questions 10–12 on the
Standardized Test Prep Short Response Base your answers to questions 10–12 on the information below. A 3.00 kg ball is dropped from rest from the roof of a building m high.While the ball is falling, a horizontal wind exerts a constant force of 12.0 N on the ball. 10. How long does the ball take to hit the ground?
Chapter 4 Short Response Base your answers to questions 10–12 on the
Standardized Test Prep Short Response Base your answers to questions 10–12 on the information below. A 3.00 kg ball is dropped from rest from the roof of a building m high.While the ball is falling, a horizontal wind exerts a constant force of 12.0 N on the ball. 10. How long does the ball take to hit the ground? Answer: 6.00 s
Short Response, continued
Chapter 4 Standardized Test Prep Short Response, continued Base your answers to questions 10–12 on the information below. A 3.00 kg ball is dropped from rest from the roof of a building m high.While the ball is falling, a horizontal wind exerts a constant force of 12.0 N on the ball. 11. How far from the building does the ball hit the ground?
Short Response, continued
Chapter 4 Standardized Test Prep Short Response, continued Base your answers to questions 10–12 on the information below. A 3.00 kg ball is dropped from rest from the roof of a building m high.While the ball is falling, a horizontal wind exerts a constant force of 12.0 N on the ball. 11. How far from the building does the ball hit the ground? Answer: 72.0 m
Short Response, continued
Chapter 4 Standardized Test Prep Short Response, continued Base your answers to questions 10–12 on the information below. A 3.00 kg ball is dropped from rest from the roof of a building m high.While the ball is falling, a horizontal wind exerts a constant force of 12.0 N on the ball. 12. When the ball hits the ground, what is its speed?
Short Response, continued
Chapter 4 Standardized Test Prep Short Response, continued Base your answers to questions 10–12 on the information below. A 3.00 kg ball is dropped from rest from the roof of a building m high.While the ball is falling, a horizontal wind exerts a constant force of 12.0 N on the ball. 12. When the ball hits the ground, what is its speed? Answer: 63.6 m/s
Short Response, continued
Chapter 4 Standardized Test Prep Short Response, continued Base your answers to questions 13–15 on the passage. A crate rests on the horizontal bed of a pickup truck. For each situation described below, indicate the motion of the crate relative to the ground, the motion of the crate relative to the truck, and whether the crate will hit the front wall of the truck bed, the back wall, or neither. Disregard friction. 13. Starting at rest, the truck accelerates to the right.
Short Response, continued
Chapter 4 Standardized Test Prep Short Response, continued Base your answers to questions 13–15 on the passage. A crate rests on the horizontal bed of a pickup truck. For each situation described below, indicate the motion of the crate relative to the ground, the motion of the crate relative to the truck, and whether the crate will hit the front wall of the truck bed, the back wall, or neither. Disregard friction. 13. Starting at rest, the truck accelerates to the right. Answer: at rest, moves to the left, hits back wall
Short Response, continued
Chapter 4 Standardized Test Prep Short Response, continued Base your answers to questions 13–15 on the passage. A crate rests on the horizontal bed of a pickup truck. For each situation described below, indicate the motion of the crate relative to the ground, the motion of the crate relative to the truck, and whether the crate will hit the front wall of the truck bed, the back wall, or neither. Disregard friction. 14. The crate is at rest relative to the truck while the truck moves with a constant velocity to the right.
Short Response, continued
Chapter 4 Standardized Test Prep Short Response, continued Base your answers to questions 13–15 on the passage. A crate rests on the horizontal bed of a pickup truck. For each situation described below, indicate the motion of the crate relative to the ground, the motion of the crate relative to the truck, and whether the crate will hit the front wall of the truck bed, the back wall, or neither. Disregard friction. 14. The crate is at rest relative to the truck while the truck moves with a constant velocity to the right. Answer: moves to the right, at rest, neither
Short Response, continued
Chapter 4 Standardized Test Prep Short Response, continued Base your answers to questions 13–15 on the passage. A crate rests on the horizontal bed of a pickup truck. For each situation described below, indicate the motion of the crate relative to the ground, the motion of the crate relative to the truck, and whether the crate will hit the front wall of the truck bed, the back wall, or neither. Disregard friction. 15. The truck in item 14 slows down.
Short Response, continued
Chapter 4 Standardized Test Prep Short Response, continued Base your answers to questions 13–15 on the passage. A crate rests on the horizontal bed of a pickup truck. For each situation described below, indicate the motion of the crate relative to the ground, the motion of the crate relative to the truck, and whether the crate will hit the front wall of the truck bed, the back wall, or neither. Disregard friction. 15. The truck in item 14 slows down. Answer: moves to the right, moves to the right, hits front wall
Chapter 4 Extended Response
Standardized Test Prep Extended Response 16. A student pulls a rope attached to a 10.0 kg wooden sled and moves the sled across dry snow. The student pulls with a force of 15.0 N at an angle of 45.0º. If mk between the sled and the snow is 0.040, what is the sled’s acceleration? Show your work.
Chapter 4 Extended Response
Standardized Test Prep Extended Response 16. A student pulls a rope attached to a 10.0 kg wooden sled and moves the sled across dry snow. The student pulls with a force of 15.0 N at an angle of 45.0º. If mk between the sled and the snow is 0.040, what is the sled’s acceleration? Show your work. Answer: 0.71 m/s2
Extended Response, continued
Chapter 4 Standardized Test Prep Extended Response, continued 17. You can keep a 3 kg book from dropping by pushing it horizontally against a wall. Draw force diagrams, and identify all the forces involved. How do they combine to result in a zero net force? Will the force you must supply to hold the book up be different for different types of walls? Design a series of experiments to test your answer. Identify exactly which measurements will be necessary and what equipment you will need.
Extended Response, continued
Chapter 4 Standardized Test Prep Extended Response, continued 17. You can keep a 3 kg book from dropping by pushing it horizontally against a wall. Draw force diagrams, and identify all the forces involved. How do they combine to result in a zero net force? Will the force you must supply to hold the book up be different for different types of walls? Design a series of experiments to test your answer. Identify exactly which measurements will be necessary and what equipment you will need. Answer: Plans should involve measuring forces such as weight, applied force, normal force, and friction.
Chapter 4 Section 1 Changes in Motion Force Diagrams
Inertia and the Operation of a Seat Belt
Chapter 4 Section 2 Newton’s First Law Inertia and the Operation of a Seat Belt
Download ppt "MASTER NOTES LAB PHYSICS Newtons’ Laws and Force Only Chapter 4"
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Strand: NUMBER AND OPERATIONS IN BASE TEN (2.NBT)
Understand place value (Standards 2.NBT.1-4). They use place value understanding and properties of operations to add and subtract (Standards 2.NBT.5-9).
• Boxes and Cartons of Pencils
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• Bundling and Unbundling
In this task students determine the number of hundreds, tens and ones that are necessary to write equations when some digits are provided. The student must, in some cases, decompose hundreds to tens and tens to ones.
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• Comparisons 1
This task requires students to compare numbers that are identified by word names and not just digits. The order of the numbers described in words are intentionally placed in a different order than their base-ten counterparts so that students need to think carefully about the value of the numbers.
• Comparisons 2
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• Grade 2 Math Module 1: Sums and Differences to 100 (EngageNY)
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• Grade 2 Math Module 4: Addition and Subtraction Within 200 with Word Problems to 100 (EngageNY)
In Module 4, students develop place value strategies to fluently add and subtract within 100; they represent and solve one- and two-step word problems of varying types within 100; and they develop conceptual understanding of addition and subtraction of multi-digit numbers within 200. Using a concrete to pictorial to abstract approach, students use manipulatives and math drawings to develop an understanding of the composition and decomposition of units, and they relate these representations to the standard algorithm for addition and subtraction.
• Grade 2 Math Module 5: Addition and Subtraction Within 1,000 with Word Problems to 100 (Engage NY)
In Module 5, students build upon their mastery of renaming place value units and extend their work with conceptual understanding of the addition and subtraction algorithms to numbers within 1,000, always with the option of modeling with materials or drawings. Throughout the module, students continue to focus on strengthening and deepening conceptual understanding and fluency.
In order to assist educators with the implementation of the Common Core, the New York State Education Department provides curricular modules in Pre-K-Grade 12 English Language Arts and Mathematics that schools and districts can adopt or adapt for local purposes.
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• Grade 2 Unit 2: Becoming Fluent with Addition and Subtraction (Georgia Standards)
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• Grade 2 Unit 4: Applying Base Ten Understanding (Georgia Standards)
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http://www.uen.org - in partnership with Utah State Board of Education (USBE) and Utah System of Higher Education (USHE). Send questions or comments to USBE Specialists - Trish French or Molly Basham and see the Mathematics - Elementary website. For general questions about Utah's Core Standards contact the Director - Jennifer Throndsen.
These materials have been produced by and for the teachers of the State of Utah. Copies of these materials may be freely reproduced for teacher and classroom use. When distributing these materials, credit should be given to Utah State Board of Education. These materials may not be published, in whole or part, or in any other format, without the written permission of the Utah State Board of Education, 250 East 500 South, PO Box 144200, Salt Lake City, Utah 84114-4200.
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Dr. Mark Gardener
Available soon from
Pelagic Publishing
# Statistics for Ecologists Using R and Excel (Edition 2)
### by: Mark Gardener
Available soon from Pelagic Publishing
Welcome to the support pages for Statistics for Ecologists. These pages provide information and support material for the book. You should be able to find an outline and table of contents as well as support datafiles and additional material.
## Exercise 7.3.3
Section 7.3.3
Get exercise data here:
Wilcoxon.xlsx
Top
### 7.3.3 Use Excel to carry out a Wilcoxon matched pairs test
This exercise is concerned with matched pairs tests (Section 7.3) and in particular how to carry out the non-parametric Wilcoxon Matched Pairs test using Excel (Section 7.3.3).
#### Introduction
There is no in-built function that will carry out the Wilcoxon Matched Pairs test in Excel. However, you can rank the data and compute the rank sums you require using the RANK.AVG function.
You do need to omit zero differences from the calculations and also to separate ranks of positive differences from ranks of negative differences. In this exercise you can see how to use the IF function to help you do this separation.
You can get the sample data here: Wilcoxon.xlsx. There are two worksheets, one has the data only and the other has a completed set of calculations so you can check your progress.
Data in matched pairs
Wilcoxon matched pairs test is for non-parametric data
Top
Here's what the data look like:
Obs A B 1 8 11 2 6 8 3 12 4 4 2 9 5 3 3 6 3 5 7 1 2 8 7 2
You can see that there are 8 pairs of data.
Use IF to make a "decision" about which item is largest.
Then work out if the difference is + or - or zero
Top
#### Show the direction of the difference
Start by making a column to show the direction of the difference. Make a heading in cell D1, Dir will do.
Now in cell D2 type a formula to show if the difference between B2 and C2 is positive or negative. You're going to subtract the second column from the first. You also want to take into account possible zero differences:
=IF(B2<C2,"-",IF(B2=C2,"","+"))
So if cell C2 is larger than cell B2 you'll get a minus symbol. If the cells are equal then you get a blank. If B2 is larger than C2 you'll get a + symbol. You will use these symbols to separate the ranks later.
Copy the formula down the rest of the column to show the direction of each of the differences.
Obs A B Dir 1 8 11 - 2 6 8 - 3 12 4 + 4 2 9 - 5 3 3 6 3 5 - 7 1 2 - 8 7 2 +
Note that observation 5 (row 6 of the spreadsheet) shows a blank because the ites are the same (i.e. a zero difference).
Calculate differences as absolute magnitude, use the ABS function.
For zero differences show a blank ""
Top
#### Calculate the differences
Make a column for the difference between samples, make the heading in cell E1, D(A-B) or something similar.
You want to subtract the values in the second column of data from the first column of data (i.e. Col B - Col C). So in cell E2 type a formula to do that. You'll need to omit any zero difference, so use an IF function to place a blank "" if the difference is zero:
=IF(B2-C2=0,"",ABS(B2-C2))
You are not interested in the sign of any difference, just the magnitude, so the ABS function is needed.
Copy the result down the rest of the column.
Obs A B Dir D(A-B) 1 8 11 - 3 2 6 8 - 2 3 12 4 + 8 4 2 9 - 7 5 3 3 6 3 5 - 2 7 1 2 - 1 8 7 2 + 5
You can see that you have 7 values, with one blank (a zero difference, observation 5).
Use RANK.AVG to work out ranks of non-zero differences.
Make sure ranks are sorted ascending, the smallest difference should get the smallest rank.
Top
#### Calculate ranks of differences
Now you want to work out the ranks of the differences. That is the ranks of the absolute value of the differences that you just worked out (column E) and called D(A-B).
Make a new column label in cell F1, call it Rd or something similar.
Now in cell F2 type a formula to work out the ranks of the items from column E:
=IF(E2="","",RANK.AVG(E2,\$E\$2:\$E\$9,1))
Note that you need to take care of any possible blank cells (corresponding to zero differences). You also need to "fix" the cell range E2:E9 using \$ since you will be copying the cell down the rest of the column. The final 1 makes sure that your ranks are sorted ascending order, with the smallest difference getting the smallest rank.
Obs A B Dir D(A-B) Rd 1 8 11 - 3 4 2 6 8 - 2 2.5 3 12 4 + 8 7 4 2 9 - 7 6 5 3 3 6 3 5 - 2 2.5 7 1 2 - 1 1 8 7 2 + 5 5
You can see that there are some tied ranks (each given a value of 2.5).
Separate ranks from positive differecnes from ranks of negative differences
Top
#### Ranks of + and of - differences
Now you need to separate the ranks due to the positive differences and the ranks due to negative differences. Mae two more column headings in cells G1 and H1, R+ and R- will do fine.
In cell G2 type a formula that shows the rank if it is due to a positive difference but leaves the cell blank is not:
=IF(D2="+",F2,"")
Copy the cell down the rest of the column G.
In cell H2 type a formula that shows the rank if it is due to a negative difference but leaves the cell blank is not:
=IF(D2="-",F2,"")
Copy the cell down the rest of the column H.
Obs A B Dir D(A-B) Rd R+ R- 1 8 11 - 3 4 4 2 6 8 - 2 2.5 2.5 3 12 4 + 8 7 7 4 2 9 - 7 6 6 5 3 3 6 3 5 - 2 2.5 2.5 7 1 2 - 1 1 1 8 7 2 + 5 5 5
You should now see the ranks split according to the sign of the difference between the samples.
Sum the ranks using SUM to get the final test statistic W.
The smaller of the two rank sums is the test result.
Use COUNT to get the number of non-zero differences.
Look up the critical value and compare to your test score.
Top
#### Rank sums
Now you need to add up the ranks for the positive and negative differences (columns G & H). You can use the SUM function for this.
In cells A11 and A12 type labels for the counts and sums, n and Sum, will do nicely.
In cell B11 type a formula to work out hte number of observations:
=COUNT(B2:B9)
Copy this into cells C11 and F11. The latter being the number of non-zero differences.
In cell B12 type a formula to sum the data in column 12:
=SUM(B2:B9)
Copy the cell into cells C12, F12, G12 and H12. Cells F12:H12 contain the ranks sums; overall, +ve and -ve. Note that the overall sum of ranks should equal the two other rank sums combined.
Obs A B Dir D(A-B) Rd R+ R- 1 8 11 - 3 4 4 2 6 8 - 2 2.5 2.5 3 12 4 + 8 7 7 4 2 9 - 7 6 6 5 3 3 6 3 5 - 2 2.5 2.5 7 1 2 - 1 1 1 8 7 2 + 5 5 5 n 8 8 7 ∑ 42 44 28 12 16
The two rank sums are 12 and 16 so the 12 is the test statistic, W. You'll need the number of non-zero differences to look up the appropriate critical value (see Table 7.13 in the book).
For Nd = 7 the critical value is 2. The calculated value of W is 12, which is larger than the critical value so the result is not significant.
Top
My Publications
My Publications
See my personal pages at GardenersOwn
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# Two cars 2
Two cars started from two positions 87 km distant at the same time in opposite directions at speeds 81 km/h and 75 km/h. What was the distance between them after 2 hours 50 minutes of driving.
Result
x = 529 km
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
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The students vice adventure had a 2,367 km flight. If they travel time was 2 hours and 56 minutes, what was their average speed in kilometres per hour?
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# While controversy rages over whether the sign language taught to some
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While controversy rages over whether the sign language taught to some [#permalink]
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22 Nov 2019, 09:38
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While controversy rages over whether the sign language taught to some great apes is truly human-like speech, there is no similar dispute that our powers of communication are greater by far than that of any other animal.
(A) are greater by far than that of any other animal
(B) are far greater than that of any other animal
(C) are greater by far than any other animal
(D) are far greater than those of any other animal
(E) have been far greater than those of other animals
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Re: While controversy rages over whether the sign language taught to some [#permalink]
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22 Nov 2019, 09:51
While controversy rages over whether the sign language taught to some great apes is truly human-like speech, there is no similar dispute that our powers of communication are greater by far than that of any other animal.
(A) are greater by far than that of any other animal
Redundant
(B) are far greater than that of any other animal
When compared to "powers of communication," the underlined text above should also be plural
(C) are greater by far than any other animal
Redundant + wrong comparison
(D) are far greater than those of any other animal
Correct
(E) have been far greater than those of other animals
Wrong because there is no time marker
Re: While controversy rages over whether the sign language taught to some [#permalink] 22 Nov 2019, 09:51
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# Is 0.7 A Strong Correlation?
## What does a correlation of 0.7 mean?
Values between 0.3 and 0.7 (0.3 and −0.7) indicate a moderate positive (negative) linear relationship through a fuzzy-firm linear rule.
6.
Values between 0.7 and 1.0 (−0.7 and −1.0) indicate a strong positive (negative) linear relationship through a firm linear rule..
## Is 0.5 strong correlation?
Correlation coefficients whose magnitude are between 0.5 and 0.7 indicate variables which can be considered moderately correlated. Correlation coefficients whose magnitude are between 0.3 and 0.5 indicate variables which have a low correlation.
## What does a correlation of 0.25 mean?
When interpreting the value of the corrrelation coefficient, the same rules are valid for both Pearson’s and Spearman’s coefficient, and r values from 0 to 0.25 or from 0 to -0.25 are commonly regarded to indicate the absence of correlation, whereas r values from 0.25 to 0.50 or from -0.25 to -0.50 point to poor …
## What does R 2 tell you?
The Formula for R-Squared Is R-Squared is a statistical measure of fit that indicates how much variation of a dependent variable is explained by the independent variable(s) in a regression model.
## Why is correlation not significant?
If the P-value is bigger than the significance level (α =0.05), we fail to reject the null hypothesis. We conclude that the correlation is not statically significant. Or in other words “we conclude that there is not a significant linear correlation between x and y in the population”
## What does a correlation of 0.9 mean?
The magnitude of the correlation coefficient indicates the strength of the association. … For example, a correlation of r = 0.9 suggests a strong, positive association between two variables, whereas a correlation of r = -0.2 suggest a weak, negative association.
## What does a correlation of 0.4 mean?
This represents a very high correlation in the data. … Generally, a value of r greater than 0.7 is considered a strong correlation. Anything between 0.5 and 0.7 is a moderate correlation, and anything less than 0.4 is considered a weak or no correlation.
## How do you interpret a weak correlation?
A weak correlation means that as one variable increases or decreases, there is a lower likelihood of there being a relationship with the second variable. In a visualization with a weak correlation, the angle of the plotted point cloud is flatter. If the cloud is very flat or vertical, there is a weak correlation.
## Is Correlation good or bad?
Many folks make the mistake of thinking that a correlation of –1 is a bad thing, indicating no relationship. Just the opposite is true! A correlation of –1 means the data are lined up in a perfect straight line, the strongest negative linear relationship you can get.
## Is 0.6 A strong correlation?
Correlation Coefficient = 0.8: A fairly strong positive relationship. Correlation Coefficient = 0.6: A moderate positive relationship. … Correlation Coefficient = -0.8: A fairly strong negative relationship. Correlation Coefficient = -0.6: A moderate negative relationship.
## What does a correlation of 0.5 mean?
The strength of the relationship between X and Y is sometimes expressed by squaring the correlation coefficient and multiplying by 100. The resulting statistic is known as variance explained (or R2). Example: a correlation of 0.5 means 0.52×100 = 25% of the variance in Y is “explained” or predicted by the X variable.
## What is a strong R value?
The relationship between two variables is generally considered strong when their r value is larger than 0.7. The correlation r measures the strength of the linear relationship between two quantitative variables. Pearson r: • r is always a number between -1 and 1.
## Is 0.2 A strong correlation?
For example, a correlation coefficient of 0.2 is considered to be negligible correlation while a correlation coefficient of 0.3 is considered as low positive correlation (Table 1), so it would be important to use the most appropriate one.
## What is an example of zero correlation?
A zero correlation exists when there is no relationship between two variables. For example there is no relationship between the amount of tea drunk and level of intelligence.
## Is 0.8 A strong correlation?
A coefficient of correlation of +0.8 or -0.8 indicates a strong correlation between the independent variable and the dependent variable. An r of +0.20 or -0.20 indicates a weak correlation between the variables.
## How do you know if a correlation is significant?
To determine whether the correlation between variables is significant, compare the p-value to your significance level. Usually, a significance level (denoted as α or alpha) of 0.05 works well. An α of 0.05 indicates that the risk of concluding that a correlation exists—when, actually, no correlation exists—is 5%.
## What percentage of correlation is significant?
Values at or close to zero imply weak or no linear relationship. Correlation coefficient values less than +0.8 or greater than -0.8 are not considered significant.
## What does the correlation indicate?
Correlation coefficients are indicators of the strength of the relationship between two different variables. A correlation coefficient that is greater than zero indicates a positive relationship between two variables. A value that is less than zero signifies a negative relationship between two variables.
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# A strange phenomenon of the composition of permutations of order 2
Suppose that $f,g,h:X\rightarrow X$ are permutations such that $f^{2}=g^{2}=\textrm{Id}_{X}$ (i.e. $f,g$ have order 2). Let $F=f\circ g$. Let $D_{f,g}$ be the distribution that takes the value $n$ where $n$ is the least positive integer such that $F^{n}(x_{0})=x_{0}$ for randomly chosen $x_{0}$.
So if $f,g$ are chosen at random, then the expected value $E(D_{f,g})$ has order around $|X|$ with high probability. However, in my experience, if $f,g$ are not chosen randomly and they have some sort of relation to each other, then $E(D_{f,g})$ often falls very close to either $\sqrt{|X|},2\sqrt{|X|},4\sqrt{|X|},\frac{4}{3}\sqrt{|X|}$. Furthermore, if $f_{1},g_{1},f_{2},g_{2}$ are all closely related in some mysterious way, then $$\frac{E(D_{f_{1},g_{1}})}{E(D_{f_{2},g_{2}})}$$ is typically very close to $1$.
For example, suppose that $X=\{0,1\}^{n}\times\{0,1\}^{n}$ and $R,S:\{0,1\}^{n}\rightarrow\{0,1\}^{n}$ are randomly selected functions. Let $f(x,y)=(x,y\oplus R(x)),g(x,y)=(x\oplus S(y),y)$. Then $f,g$ are permutations of order 2 which are related and where $E(D_{f,g})$ is typically around $2^{n}$ (The phenomenon holds even when the relation between $f$ and $g$ is more complicated. It is hard to tell when the phenomenon will hold though).
1. What is the mathematical explanation for this phenomenon?
2. Is there any way to predict $E(D_{f,g})$ when $f,g$ have a simple description and are easy to compute?
3. I am wondering if this phenomenon could cause an insecurity in a symmetric cryptosystem that is constructed from permutations of order two. For example, suppose that $f,g$ are closely related permutations of order two, and suppose $h$ is another permutation of order 2. Let $\ell_{0}=f\circ g$ and $\ell_{1}=f\circ g\circ h$. Suppose that for an expanded key $a_{0}\dots a_{n-1}$, the block encryption function is $$E_{a_{0}\dots a_{n-1}}=\ell_{a_{n-1}}\circ...\circ \ell_{a_{0}}.$$ Then could there be any cryptographic weakness from the block encryption function $E$? In particular, could there be any security weakness that arises from a cryptographic hash function that arises from the Matyas–Meyer–Oseas construction from the encryption function $E$? Should I be suspicious of a possible weakness of cryptosystems of this form?
• define, maybe with an example, closely related Commented Feb 10, 2018 at 2:38
• So at cstheory.stackexchange.com/q/40417/47025, I have found another example of the low period property in the reversible cellular automaton Critters and in similar reversible cellular automata. The cellular automaton Critters is the composition of two block functions which are nearly involutions. However, the low period phenomenon for Critters is much more extreme since Critters preserves the number of on states. Commented Mar 24, 2018 at 23:40
So it turns out that there is nothing strange about the composition of random permutations of order 2. The thing is that the distribution of the size of the orbit of an element under the operation $g\circ f$ where $f,g$ are involutions strongly depends on the number of fixed points the functions $f,g$ and if $f$ or $g$ have large quantities of fixed points, then $g\circ f$ has small orbits.
When computing the iterates $(f\circ g)^{k}(x)$, when one encounters a fixed point of the function $f$ or of the function $g$ before going back to $x$, one begins to uncompute the iterates $(f\circ g)^{k}(x)$. After one uncomputes the iterates $(f\circ g)^{k}(x)$, one eventually uncomputes back to $x$, and at this point, one starts computing $(f\circ g)^{-k}(x)$. When computing $(f\circ g)^{-k}(x)$, one eventually encounters another fixed point of the function $f$ or of the function $g$. At this point, one reverses the direction of computation until one returns to $x$ in a complete orbit. The amount of time it takes to find the first fixed point follows an exponential distribution, and the amount of time it takes to find the second fixed point after crossing $x$ also follows an exponential distribution with the same parameter. Therefore, the amount of time it takes to make a complete orbit will be $X+X$ for some exponential distribution $X$. The probability density function of $X+X$ is $xe^{-x}$ up to renormalization however.
Observe that every permutation can be written as the composition of two involutions. However, some permutations have a much higher probability of arising as the composition of two involutions than other permutations.
Suppose $1\leq r\leq k$. If $h:[k]\rightarrow[k]$ is a random permutation, then the probability that the size of the orbit with respect to $h$ of a random element $x$ is $r$ is precisely $\frac{1}{k}$. Suppose that $f,g:[2k]\rightarrow[2k]$ are permutations selected at random subject to the condition that $f,g$ have no fixed points. Then let $P_{k,r}$ denote the probability that the orbit with respect to $g\circ f$ of a random element $x\in[2k]$ has cardinality $2r$. Then, after one performs an exact computation of $P_{k,r}$, by taking a limit, one obtains $\lim_{n\rightarrow\infty}n\cdot P_{n,\lfloor nx\rfloor}=\frac{1}{2\sqrt{1-x}}$ for all $x\in(0,1)$ (in other words, $x$ probably has a very large orbit).
Now suppose that $f,g:[k]\rightarrow[k]$ are randomly selected permutations subject to the conditions that $\{x\in[k]|f(x)\neq x\}=2m,\{x\in[k]|g(x)\neq x\}=2n$. Then let $P_{m,n,k,r}$ denote the probability that the orbit of a random element has length $r$.
Then for $x>0$, we have $\frac{k\cdot P_{n,n,k,v}}{k-2n}\rightarrow xe^{-x}$ as $v\cdot\frac{k-2n}{k}\rightarrow x,k\rightarrow\infty,v\rightarrow\infty$. (I may probably give more details as to why this is the case later on).
Now, take note that the expected value of the distribution with probability density function $x\cdot e^{-x}$ is $2$. Therefore, the expected value of the size of the orbit of a random element approaches $\frac{2k}{k-2n}$. In the scenario outlined in the question, the functions $f,g$ typically have $O(\sqrt{|X|})$ many fixed points which would ensure that there are around $\frac{2k}{k-2n}$ many fixed points.
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# The Extended Distributive Property
🏆Practice extended distributive property
The extended distributive property allows us to solve exercises with two sets of parentheses that are multiplied by eachother.
For example: $(a+1)\times(b+2)$
To find the solution, we will go through the following steps:
• Step 1: Multiply the first term in the first parentheses by each of the terms in the second parentheses.
• Step 2: Multiply the second term in the first parentheses by each of the terms in the second parentheses.
• Step 3: Associate like terms.
$ab+2a+b+2$
## Test yourself on extended distributive property!
$$(3+20)\times(12+4)=$$
## Exercises to practice the distributive property
$(x-4)\times (x-2) = x^2 - 2x - 4x + 8 = x^2 - 6x + 8$
$(x+3)\times (x+6) = x^2 + 6x + 3x + 18 = x^2 + 9x + 18$
The distributive property allows us to remove parentheses and simplify an expression, even if there is more than one set of parentheses.
In order to get rid of of the parentheses, we will multiply each term of the first parentheses by each term of the second parentheses, paying special attention to the addition/ subtraction signs.
For example:
$(5+8)\times (7+2)$
Using the distributive property, we can simplify the expression.
All we need to do is to multiply each of the terms in the first parentheses by each of the terms in the second parentheses:
$(5+8)\times (7+2) =$
$5\times 7+5\times 2+8\times 7+8\times 2 =$
$35+10+56+16 =$
$117$
Basic distributive property
Let's take a moment to remember our basic distributive property.
Below we can see the formula:
$a\times(b+c)=ab+ac$
Here, we have multiplied $a$ by each of the terms inside the parentheses, keeping the same order.
Extended distributive property
Now we will apply the same concept in the extended distributive property. This allows us to solve exercises with two sets of parentheses.
For example:
$(a+b)\times(c+d)=ac+ad+bc+bd$
How does the extended distributive property work?
• Step 1: Multiply the first term in the first parentheses by each of the terms in the second parentheses.
• Step 2: Multiply the second term in the first parentheses by each of the terms in the second parentheses.
• Step 3: Associate like terms.
## Example 1
Step 1: Multiply $A$ by each of the terms included in the second parentheses.
Step 2: Multiply $2$ by each of the terms included in the second parentheses.
Step 3: Order the terms and combine like terms, if any:
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## Example 2: What do we do with a minus sign?
So, what do we do when we see a minus sign in one or both of the parentheses? Do we do anything different?
The method is the same! The only difference is that we need to make sure to put out minus/ negative signs in the right places when we distribute.
It can helpful to remember that a "minus sign" is the same as saying "plus a negative number."
For example, $4-2=4+(-2)=2$
Let's look at the exercise:
Step 1: Multiply $A$ by each of the terms included inside the second parentheses.
Step 2: Multiply $5$ by each of the terms included inside the second parentheses.
Pay attention to the signs of each of the terms! For example, we will see that, $-5$ times $-3$ equals $+15$.
In this case, there are no terms that we want to combine.
## Example 3
Find the value of $X$:
$(X+2)^2=(X+5)\times(X-2)$
Let's look at the left side of the equation and simplify:
$(X+2)^2=(X+2)\times(X+2)$
Now we can use the extended distributive property on each side of the equation.
Now the equation looks like this:
$(X+2)\times(X+2)=(X+5)\times(X-2)$
After applying the distributive property:
$X^2 + 2X + 2X + 4 = X^2 – 2X + 5X – 10$
Let's reduce, combine like terms and arrange the equation.
We will get:
$X = - 14$
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Do you know what the answer is?
## Exercises using the distributive property
### Exercise 1
Assignment:
A painter has a canvas with the following dimensions:
$(23x+12)$ length
$(20x+7)$ width
What is the area the painter needs to paint?
Solution:
We multiply the length of the canvas by the width to find the area.
$(23x+12)\times(20x+7)=$
Multiply each term in the first parentheses by each term in the second parentheses.
$23x\times20x+23x\times7+12\times20x+12\times7=$
We solve accordingly
$460x^2+161x+240x+84=$
$460x^2+401x+84$
$460x^2+401x+84$
### Exercise 2
Find the area of the following rectangle:
Solution:
To find the area we multiply the width by the length.
$3y\times(y+3z)=$
Multiply 3y by each of the terms in parentheses.
$3y\times y+3y\times3z=$
Solve accordingly
$3y^2+9yz$
$3y^2+9yz$
### Exercise 3
$(3+20)\times(12+4)=$
Solution:
We multiply each of the terms in the first parentheses by the terms in the second parentheses.
$3\times12+3\times4+20\times12+20\times4=$
Solve accordingly
$36+12+240+80=$
$48+320=368$
$368$
### Exercise 4
$(12+2)\times(3+5)=$
Solution:
We multiply each of the terms in the first parentheses by the terms of the second parentheses.
$12\times3+12\times5+2\times3+2\times5=$
Solve accordingly
$36+60+6+10=$
$96+16=112$
$112$
Do you think you will be able to solve it?
### Exercise 5
$(7x+4)\times(3x+4)=$
Solution:
We multiply each of the terms of the first parentheses by the terms of the second parentheses.
$7x\times3x+7x\times4+4\times3x+4\times4=$
Solve accordingly
$21x^2+28x+12x+16=$
$21x^2+40x+16$
$21x^2+40x+16$
### Exercise 6
$(2x-3)\times(5x-7)$
We multiply each of the terms of the first parentheses by the terms of the second parentheses.
$2x\times5x+2x\times(-7)+(-3)\times5x+(-3)\times(-7)=$
Solve accordingly
$10x^2-14x-15x+21=$
$10x^2-29x+21$
$10x^2-29x+21$
## Review questions
### What is the distributive property of multiplication?
The distributive property of multiplication is a rule in mathematics that says that multiplying the sum of two (or more) numbers is the same as multiplying the numbers separately and adding/ subtracting them together.
Distributive property of multiplication over addition:
$a\times\left(b+c\right)=a\times b+a\times c$
Distributive property of multiplication over subtraction:
$a\times\left(b-c\right)=a\times b-a\times c$
### What is the distributive property of division?
Just as in the distributive property of multiplication, the distributive property of division (also over addition or subtraction) helps us to simplify an expression.
We can express it as follows:
$\left(a+b\right):c=a:c+b:c$
Do you know what the answer is?
### What is the extended distributive property?
The extended distributive property uses the same concept as the basic distributive property to simplify expressions with two sets of parentheses.
### Where is the extended distributive property used?
#### Example 1
Solve $\left(x+3\right)\left(x-8\right)=$
We will use the extended distributive property, multiplying each of the terms as follows:
$\left(x+3\right)\left(x-8\right)=x^2-8x+3x-24$
Reducing like terms we get
$\left(x+3\right)\left(x-8\right)=x^2-5x-24$
$x^2-5x-24$
#### Example 2
$\left(2x-1\right)\left(3x-5\right)=$
Using the extended distributive property we get:
$\left(2x-1\right)\left(3x-5\right)=6x^2-10x-3x+5$
Reducing like terms:
$\left(2x-1\right)\left(3x-5\right)=6x^2-13x+5$
$6x^2-13x+5$
## examples with solutions for extended distributive property
### Exercise #1
It is possible to use the distributive property to simplify the expression below?
What is its simplified form?
$(ab)(c d)$
### Step-by-Step Solution
Let's remember the extended distributive property:
$(\textcolor{red}{a}+\textcolor{blue}{b})(c+d)=\textcolor{red}{a}c+\textcolor{red}{a}d+\textcolor{blue}{b}c+\textcolor{blue}{b}d$Note that the operation between the terms inside the parentheses is a multiplication operation:
$(a b)(c d)$Unlike in the extended distributive property previously mentioned, which is addition (or subtraction, which is actually the addition of the term with a minus sign),
Also, we notice that since there is a multiplication among all the terms, both inside the parentheses and between the parentheses, this is a simple multiplication and the parentheses are actually not necessary and can be remoed. We get:
$(a b)(c d)= \\ abcd$Therefore, opening the parentheses in the given expression using the extended distributive property is incorrect and produces an incorrect result.
Therefore, the correct answer is option d.
No, $abcd$.
### Exercise #2
$(3+20)\times(12+4)=$
### Step-by-Step Solution
Simplify this expression paying attention to the order of arithmetic operations which states that exponentiation precedes multiplication and division before addition and subtraction and that parentheses precede all of them.
Therefore, let's first start by simplifying the expressions within parentheses, then we perform the multiplication between them:
$(3+20)\cdot(12+4)=\\ 23\cdot16=\\ 368$Therefore, the correct answer is option A.
368
### Exercise #3
$(12+2)\times(3+5)=$
### Step-by-Step Solution
Simplify this expression paying attention to the order of arithmetic operations which states that exponentiation precedes multiplication and division before addition and subtraction and that parentheses precede all of them.
Therefore, let's start by simplifying the expressions within parentheses, then perform the multiplication between them:
$(12+2)\cdot(3+5)= \\ 14\cdot8=\\ 112$Therefore, the correct answer is option C.
112
### Exercise #4
It is possible to use the distributive property to simplify the expression?
If so, what is its simplest form?
$(a+c)(4+c)$
### Step-by-Step Solution
We simplify the given expression by opening the parentheses using the extended distributive property:
$(\textcolor{red}{x}+\textcolor{blue}{y})(t+d)=\textcolor{red}{x}t+\textcolor{red}{x}d+\textcolor{blue}{y}t+\textcolor{blue}{y}d$Keep in mind that in the distributive property formula mentioned above, we assume that the operation between the terms inside the parentheses is an addition operation, therefore, of course, we will not forget that the sign of the term's coefficient is ery important.
We will also apply the rules of multiplication of signs, so we can present any expression within parentheses that's opened with the distributive property as an expression with addition between all the terms.
In this expression we only have addition signs in parentheses, therefore we go directly to opening the parentheses,
We start by opening the parentheses:
$(\textcolor{red}{x}+\textcolor{blue}{c})(4+c)\\ \textcolor{red}{x}\cdot 4+\textcolor{red}{x}\cdot c+\textcolor{blue}{c}\cdot 4+\textcolor{blue}{c} \cdot c\\ 4x+xc+4c+c^2$To simplify this expression, we use the power law for multiplication between terms with identical bases:
$a^m\cdot a^n=a^{m+n}$
In the next step like terms come into play.
We define like terms as terms in which the variables (in this case, x and c) have identical powers (in the absence of one of the variables from the expression, we will refer to its power as zero power, this is because raising any number to the power of zero results in 1).
We will also use the substitution property, and we will order the expression from the highest to the lowest power from left to right (we will refer to the regular integer as the power of zero),
Keep in mind that in this new expression there are four different terms, this is because there is not even one pair of terms in which the variables (different) have the same power. Also it is already ordered by power, therefore the expression we have is the final and most simplified expression:$\textcolor{purple}{4x}\textcolor{green}{+xc}\textcolor{black}{+4c}\textcolor{orange}{+c^2 }\\ \textcolor{orange}{c^2 }\textcolor{green}{+xc}\textcolor{purple}{+4x}\textcolor{black}{+4c}\\$We highlight the different terms using colors and, as emphasized before, we make sure that the main sign of the term is correct.
We use the substitution property for multiplication to note that the correct answer is option A.
Yes, the meaning is $4x+cx+4c+c^2$
### Exercise #5
$(35+4)\times(10+5)=$
### Step-by-Step Solution
We will open the parentheses using the extended distributive property and create a long addition exercise:
We multiply the first term of the left parenthesis by the first term of the right parenthesis.
Then we multiply the first term of the left parenthesis by the second term of the right parenthesis.
Now we multiply the second term of the left parenthesis by the first term of the left parenthesis.
Finally, we multiply the second term of the left parenthesis by the second term of the right parenthesis.
In the following way:
$(35\times10)+(35\times5)+(4\times10)+(4\times5)=$
We solve each of the exercises within parentheses:
$350+175+40+20=$
We solve the exercise from left to right:
$350+175=525$
$525+40=565$
$565+20=585$
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0
# How do you do three number ratios?
Wiki User
2013-07-10 23:37:29
Divide all 3 numbers by the smallest. If you end up with floating point numbers, you can either round them to the nearest whole or multiply out until you get to whole numbers and then divide the resultant numbers by a common foctor. i.e.
14:8:78
14 ÷ 8 = 1.75
8 ÷ 8 = 1
78 ÷ 8 = 9.75
So you have 1.75 : 1 : 9.75 which rounded gives 2:1:10
Multiply the numbers out to get integers gives 175:100:975
Divide by 5 = 35:20:195
Divide by 5 again = 7:4:39
If you round the original numbers to the nearest 5, you get 15:10:80
Divided by 5 = 3:2:16
If you carry out the division equation on that ratio;
3 ÷ 2 = 1.5
2 ÷2 = 1
16 ÷ 2 = 8
1.5 : 1 : 8 = approximately 2:1:10
There are loads of different ways to do ratios. I find using the division technique with a bit of rounding works the best.
So 14:8:78 = 2:1:10 = 7:4:39 = 3:2:16 approximately.
-NEW RESPONSE-
There appears to be a HUGE amount of rounding in there and I don't know why. To deal with 3 number ratios you need to figure out what the question is, if you are trying to figure out given 3 chemicals in a ratio how much you need of each given a specific amount of final substance? Or are you asking if you need 3 ingredients in a ratio, and you have a certain amount of 1 how much do you need of the others?
The first part is by adding all the numbers together, 14:8:78, gives us 100. Then taking the total amount you need at the end, divide by 100, then multiply by each of those numbers in the ratio to find out how much of that ingredient you need.
The second type, you just have to figure out what multiple you need of the original number in the ratio to get to the new amount you have. Example, if you have 20 ounces of the first chemical (14), to get 14 to 20, you just take 14 and multiply it by 20/14. Then you would multiply the 8 and 78 by 20/14 as well. That gives you the amount of each of the others you would need.
I am really not sure what the original poster of this response was talking about... DON'T ROUND LIKE THAT until the end of the problem.
Wiki User
2013-07-10 23:37:29
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# If f(x)=4[x-1] and g(x)=x+1, how do you find f(g(-1))?
Sep 11, 2015
$f \left(g \left(- 1\right)\right) = - 4$
#### Explanation:
When you have an expression:
$f \left(y\right)$ ,
this means that you should plug in whatever $y$ is into $x$ in the function $f \left(x\right)$. That is, replace every instance of $x$ with $y$.
There are two ways to find $f \left(g \left(- 1\right)\right)$
The first is to solve from the inside out.
Plug in $- 1$ into $g \left(x\right)$ to get $g \left(- 1\right)$:
$g \left(x\right) = x + 1$
$g \left(- 1\right) = - 1 + 1$
$\textcolor{b l u e}{g \left(- 1\right) = 0}$
Then, plug in $g \left(- 1\right)$ into $f \left(x\right)$:
$f \left(x\right) = 4 \left[x - 1\right]$
$f \left(g \left(- 1\right)\right) = 4 \left[0 - 1\right]$
$\textcolor{b l u e}{f \left(g \left(- 1\right)\right) = - 4}$
The second way is to get $f \left(g \left(x\right)\right)$ first, then plug in $- 1$:
$\textcolor{red}{f \left(x\right) = 4 \left[x - 1\right]}$
$\textcolor{b l u e}{g \left(x\right) = x + 1}$
$\textcolor{red}{f \left(\textcolor{b l u e}{g \left(x\right)}\right)} = \textcolor{red}{4 \left[\textcolor{b l u e}{x + 1} - 1\right]}$
Then, you can plug in $- 1$
$f \left(g \left(- 1\right)\right) = 4 \left[- 1 + 1 - 1\right]$
$\textcolor{b l u e}{f \left(g \left(- 1\right)\right) = - 4}$
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# Ch 8.2 Confidence Interval for Mean One Sample No Sigma
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### Ch 8.2 Confidence Interval for mean with one sample when sigma is not known.
Use one sample with size n, $$\bar{x}$$, s
$$\fbox { 1) point estimate of \( \mu: \bar{x}$$ } \)
$$\fbox{ 2) Interval estimate of μ: \( \bar{x} - E < \mu < \bar{x} +E$$ } \)
When $$E (EBM) = t_ {\alpha /2} \frac{s}{\sqrt{n}}$$ when σ is not given or unknown.
Use Online calculator to find confidence interval of mean when is not given.
- use online statdisk: https://www.statdisk.com/#
Analysis/Confidence Intervals/Mean one sample/.
If summary statistics ( $$\bar{x}$$ , s, n) are given, select “use Summary Statistics” tab otherwise use “use data” tab.
Enter Clevel, n, $$\bar{x}$$ , s, or select with data tab.
Do not enter population standard deviation.
Output: E is the margin of error or Error bound for Mean.
Confidence interval estimate is $$\bar{x} - E < \mu < \bar{x} +E$$
Note: the requirement for this confidence interval is CTL applies (n > 30 or X is normal). Sample is SRS.
### B) T-distribution
$$t_ {\alpha /2}$$ is the critical value with C-level in the middle.
To find $$t_ {\alpha /2}$$
Use Online Stat book Inverse-t Calculator: http://onlinestatbook.com/2/calculators/inverse_t_dist.html
the degree of freedom = n – 1, Select confidence interval from 90%, 95% or 99%. “Calculate”
The “t for confidence interval” is the critical value tα/2.
Properties of t-distributions:
1) $$t = \frac{x-\mu } {s/\sqrt{n}}$$ = no. of times of $$s/\sqrt{n}$$ $$\bar{x}$$ is from mean μ when $$\bar{x}$$ has a normal distribution.
2) Each t-distribution is different for different sample size according to the degree of freedom df = n – 1.
3) Each t-distribution has the same general symmetric bell shape as standard normal, but with more variability.
4) As sample size increase, the student t- distribution approaches the standard normal distribution.
Ex1: Patients with insomnia are treated with zopiclone. After the treatment, the 16 subjects had a mean wake time of 98.9 min. and a sample standard deviation of 42.3 min. Assume that wake times are normally distributed.
a) Construct a 98% confidence interval estimate of mean wake time for all patients treated by zopiclone.
- since X is normal, x-bar is normal, σ is not known, so use t-distribution.
- Identify C-level = 0.98, n= 16, $$\bar{x}$$ = 98.9, s = 42.3,
- Use statdisk/ Analysis/Confidence interval/ Mean one sample/
- output: E = 27.521, Confidence interval is 71.379 < μ < 126.421
interval format (71.379, 126.421), +/- format: x-bar $$\pm$$ E, 89.9 ± 27.521
b) Can we conclude the mean wake time is less than 100 min.?
No, since the interval is from 71.4 to 126.4, not all values are less than 100 min.
c) If confidence level is decreased to 95%, what will happen to the margin of error?
When confidence level decrease, $$t_{\alpha /2}$$ will decrease, so Margin of error will decrease.
### C) Interpret the Confidence interval of mean:
i) We estimate with ________ confidence that the true population mean for ______________is between ___ and ___ .
ii) 90% of confidence interval constructed in this way contains the true ______________population mean.
Make conclusion from a confidence Interval:
1) Any value in the confidence Interval can be μ.
2) If the whole interval > a, we can conclude μ > a
3) If the whole interval < a, we can conclude μ < a.
4) When two confidence interval overlap, we can conclude that the two μ may be the same. We cannot conclude one of the μ is higher.
5) Never make conclusion about population mean based solely on value of $$\bar{x}$$ .
Ex2. Listed below are amounts of arsenic (in ug, per serving) in samples of brown rice from California.
5.4, 5.6, 7.3, 4.5, 7.5, 1.5, 5.5, 9.1, 8.7, 8.4
a) Can we assume that amount of arsenic in brown rice is from a normal distribution?
Use Normal Quantile plot to access Normality.
Copy data to statdisk. Data/Normal Assessment
The points are reasonably close to a straight line. So we can assume population X is normal.
b) Construct 90% confidence interval of mean arsenic in all brown rice from California. What distribution is sued?
-Since X is normal, so $$\bar{x}$$ is normal, σ is not known, so use t distribution.
-identify Clevel = 0.9, column in statdisk.
- Statdisk Analysis/Confidence intervals/Mean One sample /use data tab.
output: E = 1.348 , 5.002 < mean < 7.698
-Interval estimate: 5.0 ug < μ < 7.7 ug
c) Find the critical value.
Since t-distribution is used when σ is not known, use online calculator inverse t-calculator.
degree of freedom = 9, C-level = 90% , critical value t0.05 = 1.833
d) Interpret the meaning of the confidence Interval:
“We estimate with 90% confidence that the mean amount of arsenic is between 5.0 ug and 7.7 ug.”
“90% of all confidence interval collected from sample of size 10 will contain the true mean arsenic.”
d) Can we conclude the mean amount is less than 8 ug?
Yes, because all values in the interval are less than 8 ug. We can conclude that the mean amount of arsenic in all brown rice are less than 9 ug. at 90% confidence.
Ch 8.2 Confidence Interval for Mean One Sample No Sigma is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
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Home > Basic Signals > Background > Signal Operations
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With signalized intersections, there is a lot of terminology that must be known before much analysis or design can take place. The best place to begin explaining this terminology is with the concept of phasing, which covers phase assignments, definitions, and phase diagrams.
Phase Assignments
Intersection with NEMA phase labels
In the illustration above, we have a four way intersection with through lanes and separate left turn lanes. The numbers on the diagram each correspond to a movement, and those numbers are set by the National Electrical Manufacturers Association (NEMA). There are also numbers assigned to right turns and pedestrians, but we are not worrying about those here.
When intersections are signalized, that is, stop lights are installed, movements are often lumped together to run at the same time. These sets of movements are known as phases. There may be more than one movement served in a phase, but at least one NEMA movement number must be assigned. Two approaches can be taken to the assignment of NEMA movement numbers.
It is common to set the phase number equal to the lowest through movement number in the phase. For example, if all of the north-south movements were given the right of way at the same time, that could be one phase (diagram at right). Since the movements involved are 1, 6, 2, and 5, and 2 and 6 are through movements, the phase would be labeled as phase 2. The movements can also be assigned to separate phases, giving the south bound and north bound traffic separate phases, giving the through and turning traffic separate phases, or even giving each movement its own phase. For simplicity, we will refer to individual movements as "movement X", and phases as "phase X"
It is important to note that only some of the movements can safely go at the same time which is why we regulate these movements with traffic signals. Using signal indications that are changed in intervals, traffic signals can prevent conflicting movements from having the right of way at the same time. A signal indication is any one of the traffic lights at a signalized intersection and an interval is a time during which none of the indications change. Right-of-way is assigned using these signal indications, where a red light means that no right-of-way is given and a green light indicates that right-of-way is given. Note that it is not quite that simple, especially when dealing with left turning traffic, where a green ball indicates that left turn traffic may proceed but they must yield to opposing through traffic. When this situation occurs, it is said that the movement is permitted.
To solidify how intervals and indications work at a signalized intersection, refer to the following table. The table shows a signal plan for the intersection at the top of the page. The time is the length of the interval, and the R (Red), G (Green), and Y (Yellow) refer to the indication during the specific interval. This table assumes that the eastbound and westbound left turn traffic has a left turn arrow and that the northbound and southbound left turn traffic does not.
Interval Time (sec) NBLT NBTH SBLT SBTH WBLT WBTH EBLT EBTH 1 12 R R R R G R G R 2 4 R R R R Y R Y R 3 1 R R R R R R R R 4 25 R R R R R G R G 5 4 R R R R R Y R Y 6 1 R R R R R R R R 7 14 G G G G R R R R 8 3 Y Y Y Y R R R R 9 1 R R R R R R R R
The first interval in this table shows that the westbound and eastbound left turn movements have green left turn indications. All of the other traffic movements have red indications.After 12 seconds, the westbound/eastbound yellow left turn indication comes on to communicate to traffic that their turn for right-of-way at the intersection is about to end.As before, all of the other traffic movements have red indications. In interval three, all of the indications are red to provide time for the intersection to clear before the next phase has the right of way. This process of going from one interval to the next repeats itself over time. When the last interval is reached the signal controller returns to interval one, repeating the sequence of intervals over again.
Definitions
Now it is time to put forth some definitions using the table, figure and discussion above.
• Cycle: A cycle is a complete sequence of intervals. In the example provided, the sequence would go from interval one to nine and then another cycle would start with interval one.
• Cycle length:A cycle length is the time it takes to complete one cycle. In the table this would be the sum of the interval times, or 65 seconds. The minimum time for a cycle length is generally 45 seconds, to limit the time lost starting and stopping traffic.
• Phase:A phase is the part of the cycle assigned to a fixed set of traffic movements. When any of these movements change, the phase changes. The first phase in the above table is comprised of intervals one, two, and three. The third interval is included even though no traffic movements have the right-of-way.
• Yellow Change interval:This is an interval in which yellow indications tell drivers in the phase with the right-of-way that their movement is about to lose its right-of-way. An example of this is interval five.
• Red Clearance interval:This describes the interval when all of the indications are red and is a safety measure designed to give the oncoming traffic enough time to clear the intersection before the next phase begins. An example of this is interval six.
• Intergreen time: This is the summation of the time allocated to the change and clearance intervals for a given phase (yellow and all red time).
Phase Diagrams
In putting together movements to make a phase, it is important to remember that not all movements can be put together. There are some movements that just cannot go at the same time. For this reason we put together phase and ring diagrams. A phase diagram is a diagram which groups movements into phases and each phase is shown in a single block. As we discussed above, this intersection has three phases. Let us consider the phases as Phase A, Phase B and Phase C as shown below.
Phase Diagram
If the phases change as a block, as is often the case in pretimed signals, the phase diagram is adequate to ensure that no conflicts arise However, the movements do not necessarily change as a block. For this reason, we have ring diagrams. Ring diagrams with barriers ensure movements do not conflict, thus ensuring driver safety. The ring and barrier structure for a signal with 8 movements is shown below.
Ring Diagram
Possible Four Phase Diagram
The two rings operate independently except that their control must cross the "BARRIER" at the same time (i.e. movements on the left side of the barrier must all be terminated before movements on the right side of the barrier can begin). All of the movements from one street must be assigned to the left side of the barrier and all of the other movements from the other street must be assigned to right side.
On both sides of the barrier there are four movements, two through movements and two left movements. A movement in a given barrier may be served simultaneously with either of the movements in the other ring, but can not be served with the movement that is in the same ring. The figure below shows the possible scenarios for both sides of the barrier.>
Compatible Movements
In summary, there are two basic rules that govern phasing using the dual ring diagram and they are listed below.
1. Movements from opposite sides of the barrier can not be served simultaneously.
2. Movements from the same ring can not be served simultaneously.
These rules can be used for defining phases for fixed time control and for governing the operations of actuated controllers, which is discussed in the actuated signals module.
Move on to Calculations
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Cody
# Problem 89. Counting in Finnish
Solution 187134
Submitted on 7 Jan 2013 by Vialla
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% a = [1 2 3 4 5 6 7 8 9 0]; b = [8 2 3 6 4 0 7 5 9 1]; out = finnishOrdering(a); assert(isequal(out, b));
l = Columns 1 through 8 1 10000 -10000 -1000 -1 100 -100 10 Columns 9 through 11 -100000 1000 -10 t = Columns 1 through 8 -100000 -10000 -1000 -100 -1 1 10 100 Columns 9 through 10 1000 10000 b = 8 2 3 6 4 10 7 5 9 1 ans = 8 2 3 6 4 0 7 5 9 1
2 Pass
%% a = [1 1 1 2 0 4 5 3 7 2 9 8 9]; b = [8 2 2 3 4 0 7 5 9 9 1 1 1]; out = finnishOrdering(a); assert(isequal(out, b));
l = Columns 1 through 8 1 10000 -10000 -1000 -1 100 -100 10 Columns 9 through 11 -100000 1000 -10 t = Columns 1 through 8 -100000 -10000 -10000 -1000 -1 1 10 100 Columns 9 through 13 1000 1000 10000 10000 10000 b = 12 4 10 8 6 5 9 7 11 13 1 2 3 ans = 8 2 2 3 4 0 7 5 9 9 1 1 1
3 Pass
%% a = [0 0 0 1 5 4 3 2 7 5 9]; b = [2 3 4 0 0 0 7 5 5 9 1]; out = finnishOrdering(a); assert(isequal(out, b));
l = Columns 1 through 8 1 10000 -10000 -1000 -1 100 -100 10 Columns 9 through 11 -100000 1000 -10 t = Columns 1 through 8 -10000 -1000 -1 1 1 1 10 100 Columns 9 through 11 100 1000 10000 b = 8 7 6 1 2 3 9 5 10 11 4 ans = 2 3 4 0 0 0 7 5 5 9 1
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What Four Consecutive Integers Have A Sum Of 66?
What four consecutive integers have a sum of 354?
The four integers are the two integers either side of, and closest to, 88.5: 87, 88, 89, and 90.
Let the smallest integer be x; then the others are x+1, x+2, and x+3..
What four consecutive odd numbers have a sum of 40?
So the second number is . The sum of 4 numbers = 40. So the average = 40/4 = 10. Since the numbers are 4 odd consecutive numbers then the numbers are 7, 9, 11 and 13.
What is the sum of two consecutive Number?
The sum of any two consecutive numbers is always odd. Example, 4 + 5 = 9; –8 + (–7) = –15.
What are two consecutive integers?
Consecutive numbers are numbers that follow each other in order. They have a difference of 1 between every two numbers. In a set of consecutive numbers, the mean and the median are Equal.
How do you find three consecutive integers whose sum is?
Three consecutive even integers can be represented by x, x+2, x+4. The sum is 3x+6, which is equal to 108. Thus, 3x+6=108.
How do you find the sum of consecutive integers?
By using Carl Gauss’s clever formula, (n / 2)(first number + last number) = sum, where n is the number of integers, we learned how to add consecutive numbers quickly. We now know that the sum of the pairs in consecutive numbers starting with the first and last numbers is equal.
What are 5 consecutive integers?
The first five consecutive, positive integers (1, 2, 3, 4, and 5) multiply together to make 120. The largest of those integers is 5. The sum of five consecutive integers is 0.
What three consecutive integers have a sum of 33?
Therefore, the three consecutive integers are 10, 11, and 12. To check that, add them up. They all equal 33 and they are consecutive, which means this is the right answer!
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LFD Book Forum Exercise 1.11
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#1
02-07-2014, 07:07 AM
tatung2112 Junior Member Join Date: Feb 2014 Posts: 4
Exercise 1.11
Thank you Prof. Yaser. Your book is really easy to follow. I have just started it for a week and I am trying to finish every exercises in the book.
About exercise 1.11, I don't know where to check the answer so I post it here. Could you please tell me whether my answers are right or wrong? Is there any place that I can check my answer on exercise by myself?
Ex 1.11:
Dataset D of 25 training examples.
X = R, Y = {-1, +1}
H = {h1, h2} where h1 = +1, h2 = -1
Learning algorithms:
S - choose the hypothesis that agrees the most with D
C - choose the hypothesis deliberately
P[f(x) = +1] = p
(a) Can S produce a hypothesis that is guaranteed to perform better than random on any point outside D?
In case that all examples in D have yn = +1
(b) Is it possible that the hypothesis that C produces turns out to be better than the hypothesis that S produces?
(c) If p = 0.9, what is the probability that S will produce a better hypothesis than C?
Answer: P[P(Sy = f) > P(Cy = f)] where Sy is the output hypothesis of S, Cy is the output hypothesis of C
+ Since yn = +1, Sy = +1. Moreover, P[f(x) = +1] = 0.9 --> P(Sy = f) = 0.9
+ We have, P(Cy = +1) = 0.5, P(Cy = -1) = 0.5, P[f(x) = +1] = 0.9, P[f(x) = -1] = 0.1
--> P[Cy = f] = 0.5*0.9 + 0.5*0.1 = 0.5
Since 0.9 > 0.5, P[P(Sy = f) > P(Cy = f)] = 1
(d) Is there any value of p for which it is more likely than not that C will produce a better hypothesis than S?
I am not sure that my answer of (a) and for (c) is not conflict.
Thank You and Best Regards,
#2
02-08-2014, 06:17 AM
yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,478
Re: Exercise 1.11
Quote:
Originally Posted by tatung2112 I am not sure that my answer of (a) and for (c) is not conflict.
Your answers to (a) and (c) are both correct. They are not in conflict since (a) is asking a deterministic question while (c) is asking a probabilistic question.
__________________
Where everyone thinks alike, no one thinks very much
#3
02-12-2014, 05:50 PM
tatung2112 Junior Member Join Date: Feb 2014 Posts: 4
Re: Exercise 1.11
Prof. Yaser, thank you very much for your replying. I will keep studying. Thank you!
#4
07-10-2014, 04:31 PM
BojanVujatovic Member Join Date: Jan 2013 Posts: 13
Re: Exercise 1.11
Quote:
Originally Posted by tatung2112 (c) If p = 0.9, what is the probability that S will produce a better hypothesis than C? Answer: P[P(Sy = f) > P(Cy = f)] where Sy is the output hypothesis of S, Cy is the output hypothesis of C + Since yn = +1, Sy = +1. Moreover, P[f(x) = +1] = 0.9 --> P(Sy = f) = 0.9 + We have, P(Cy = +1) = 0.5, P(Cy = -1) = 0.5, P[f(x) = +1] = 0.9, P[f(x) = -1] = 0.1 --> P[Cy = f] = 0.5*0.9 + 0.5*0.1 = 0.5 Since 0.9 > 0.5, P[P(Sy = f) > P(Cy = f)] = 1
Can you please elaborate more on why P(Cy = -1) = 0.5, I cannot understand that part?
Here is my reasoning for the (c) part: the event S produces a better hypothesis than C means that is smaller than , so
#5
12-17-2015, 09:57 AM
Andrew87 Junior Member Join Date: Feb 2015 Posts: 6
Re: Exercise 1.11
Hi,
according to the first post, I can't understand why the answer to the question (d) is p < 0.5.
Intuitively my answer is that there are no values of p that make probabilistically C better than S. That's why S try to minimize the error on the training data which should reflect the true distribution. In this case, C do better than S only if
(the majority of the examples are +1 GIVEN p < 0.5) OR (the majority of the examples are -1 GIVEN p > 0.5). However both the cases are less probable than the ones for which S works better. As a results, there are no value for p to reverse the situation.
Am I right ?
#6
02-02-2016, 06:47 AM
MaciekLeks Member Join Date: Jan 2016 Location: Katowice, Upper Silesia, Poland Posts: 17
Re: Exercise 1.11
Quote:
Originally Posted by Andrew87 Hi, according to the first post, I can't understand why the answer to the question (d) is p < 0.5. Intuitively my answer is that there are no values of p that make probabilistically C better than S. That's why S try to minimize the error on the training data which should reflect the true distribution. In this case, C do better than S only if (the majority of the examples are +1 GIVEN p < 0.5) OR (the majority of the examples are -1 GIVEN p > 0.5). However both the cases are less probable than the ones for which S works better. As a results, there are no value for p to reverse the situation. Am I right ?
Referring to point (d): The crucial part is the assumption that y_n=+1 (see point (b)), C always chooses h_2, S always chooses h_1.
#7
02-03-2016, 06:21 AM
MaciekLeks Member Join Date: Jan 2016 Location: Katowice, Upper Silesia, Poland Posts: 17
Re: Exercise 1.11
"(a) Can S produce a hypothesis that is guaranteed to perform better than random on any point outside D?"
Can anyone give me some tips on this part of the exercise:
(1) Should we calculate it to be sure that S guarantees/(does't guarantee) to beat random result? If so, any tip is appreciated to deal with this deterministic task.
(3) Does "any point" in this context mean "every point" or "some point"?
#8
05-29-2016, 09:44 AM
henry2015 Member Join Date: Aug 2015 Posts: 31
Re: Exercise 1.11
Quote:
Originally Posted by yaser Your answers to (a) and (c) are both correct. They are not in conflict since (a) is asking a deterministic question while (c) is asking a probabilistic question.
For part c, I thought:
Given p = 0.9, h1 is a better hypothesis than h2.
Hence, the probability that S produces a better hypothesis than C is the probability that S picks h1 essentially as C will pick the other hypothesis that S doesn't pick.
In other words, P[S produces a better hypothesis than C] = P[S picks h1 based on the 25 training examples].
S will pick h1 if 13 out of 25 training examples give +1, so we will have:
P[S picks h1]
= P[13 or more out of 25 training examples give +1]
=
= 0.9999998379165839813935344
It is quite different from tatung2112's explanation for c.
Could you comment further?
Thanks!
Last edited by henry2015; 05-29-2016 at 09:48 AM. Reason: fixing latex syntax
#9
06-04-2016, 08:41 AM
henry2015 Member Join Date: Aug 2015 Posts: 31
Re: Exercise 1.11
Quote:
Originally Posted by henry2015 For part c, I thought: Given p = 0.9, h1 is a better hypothesis than h2. Hence, the probability that S produces a better hypothesis than C is the probability that S picks h1 essentially as C will pick the other hypothesis that S doesn't pick. In other words, P[S produces a better hypothesis than C] = P[S picks h1 based on the 25 training examples]. S will pick h1 if 13 out of 25 training examples give +1, so we will have: P[S picks h1] = P[13 or more out of 25 training examples give +1] = = 0.9999998379165839813935344 It is quite different from tatung2112's explanation for c. Could you comment further? Thanks!
I just noticed that the formula in my post actually is one form of the formula in Problem 1.7...
Now, I am even more confused.
#10
06-04-2016, 05:04 PM
htlin NTU Join Date: Aug 2009 Location: Taipei, Taiwan Posts: 610
Re: Exercise 1.11
Quote:
Originally Posted by henry2015 I just noticed that the formula in my post actually is one form of the formula in Problem 1.7... Now, I am even more confused.
I think henry2015's detailed steps are the right way to go, while Yaser's old comments are just highlighting that (a) and (c) do not conflict with each other. Thanks for asking.
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# An interesting inequality
Prove that for any positive integer values of $$m$$ and $$n$$, the following inequalities are fulfilled.
$\large 2^{mn} > m^n, \quad 2^{mn} > n^m$
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Raise both sides of the first inequality by $$\frac 1 n$$ and both sides of the second inequality by $$\frac 1 m$$.
You get $$2^m>m$$ in the first inequality and $$2^n>n$$ in the second inequality which are both true.
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Data Sufficiency in Logical Reasoning
Data Sufficiency in Logical Reasoning is an important topic to prepare for competitive exams in India. It tests the logical, analytical and thinking capacity or ability of a candidate. Candidates need to practice the questions for scoring well in the examination
The following article contains various kinds of Data Sufficiency Questions. Candidates must go through the article to practice all the important questions of the topic.
Tips and Tricks to Solve Data Sufficiency Questions
1. Read the entire question carefully before attempting it. Then go through the options to find the correct answer.
2. Use the information given in the question and avoid making any assumptions.
3. Both the statements must be carefully read and then a make a conclusion.
4. Avoid any kind of confusion and answer the question.
Data Sufficiency Questions and Answers
Q1. Who is the tallest among the brothers A, B, C, D ? Statement 1 : C is shorter than only B Statement 2 : D is taller than only A
I) Statement 1 alone is sufficient
II) Statement 2 alone is sufficient
III) Both statement 1 and statement 2 together are sufficient
IV) Both statement 1 and statement 2 even together are not sufficient
Solution : If we consider statement 1 only, we can safely conclude that B is the tallest of all. If we consider statement 2 only, we can safely conclude that A is the shortest of all. Therefore, statement 1 alone is sufficient to answer the question.
Q2. What is the value of ‘x’ Statement 1 : x2 + x – 6 = 0 Statement 2 : x ≥ 0
I) Statement 1 alone is sufficient
II) Statement 2 alone is sufficient
III) Both statement 1 and statement 2 together are sufficient
IV) Both statement 1 and statement 2 even together are not sufficient
Solution : If we consider statement 1 alone, we have x2 + x – 6 = 0 => x2 + 3x – 2x – 6 = 0 => (x + 3) (x – 2) = 0 => x = -3, 2 If we consider statement 2 alone, we have x ≥ 0 Combining the two statements, we can safely say that x = 2 Therefore, both statement 1 and statement 2 together are sufficient but neither alone is sufficient.
Q3. What is the sum of the ages of John and Peter? Statement 1: John is 5 years older than Peter. Statement 2: The average of their ages is 25.
I) Statement 1 alone is sufficient
II) Statement 2 alone is sufficient
III) Both statement 1 and statement 2 together are sufficient
IV) Both statement 1 and statement 2 even together are not sufficient
Answer: III) Both statement 1 and statement 2 together are sufficient
Q4. What is the value of x in the equation x^2 + 6x + 9 = 0? Statement 1: x = -3 Statement 2: The discriminant of the equation is zero.
I) Statement 1 alone is sufficient
II) Statement 2 alone is sufficient
III) Both statement 1 and statement 2 together are sufficient
IV) Both statement 1 and statement 2 even together are not sufficient
Answer: III) Both statement 1 and statement 2 together are sufficient
Q5. Is triangle ABC an equilateral triangle? Statement 1: The length of AB is equal to the length of AC. Statement 2: The length of BC is equal to twice the length of AB.
I) Statement 1 alone is sufficient
II) Statement 2 alone is sufficient
III) Both statement 1 and statement 2 together are sufficient
IV) Both statement 1 and statement 2 even together are not sufficient
Answer: IV) Both statement 1 and statement 2 even together are not sufficient
Q6. What is the value of x in the equation 2x + 5 = 11? Statement 1: x = 3 Statement 2: x is a prime number.
I) Statement 1 alone is sufficient
II) Statement 2 alone is sufficient
III) Both statement 1 and statement 2 together are sufficient
IV) Both statement 1 and statement 2 even together are not sufficient
Answer: I) Statement 1 alone is sufficient
Q7. How much did Mr. Smith spend on his vacation? Statement 1: Mr. Smith spent \$500 on his flight ticket. Statement 2: Mr. Smith spent \$800 on his hotel stay.
I) Statement 1 alone is sufficient
II) Statement 2 alone is sufficient
III) Both statement 1 and statement 2 together are sufficient
IV) Both statement 1 and statement 2 even together are not sufficient
Answer: III) Both statement 1 and statement 2 together are sufficient
Q8. Is the given rectangle a square? Statement 1: The length of each side of the rectangle is 8 cmStatement 2: The diagonals of the rectangle are equal in length.
I) Statement 1 alone is sufficient
II) Statement 2 alone is sufficient
III) Both statement 1 and statement 2 together are sufficient
IV) Both statement 1 and statement 2 even together are not sufficient
Answer: III) Both statement 1 and statement 2 together are sufficient
Q9. What is the sum of the angles of a triangle? Statement 1: One of the angles of the triangle is 60 degrees. Statement 2: The triangle is an equilateral triangle.
I) Statement 1 alone is sufficient
II) Statement 2 alone is sufficient
III) Both statement 1 and statement 2 together are sufficient
IV) Both statement 1 and statement 2 even together are not sufficient
Answer: II) Statement 2 alone is sufficient
Q10. What is the total number of students in a class? Statement 1: There are 20 boys in the class. Statement 2: The ratio of girls to boys in the class is 2:5.
I) Statement 1 alone is sufficient
II) Statement 2 alone is sufficient
III) Both statement 1 and statement 2 together are sufficient
IV) Both statement 1 and statement 2 even together are not sufficient
Answer: III) Both statement 1 and statement 2 together are sufficient
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Salaries of Ravi and Sumit are in the ratio 2:3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40:57. What is Sumit's salary
1. Rs. 17,000
2. Rs. 20,000
3. Rs. 34,000
4. Rs. 38,000
4
Correct Answer :
Rs. 38,000
Explanation :
No Explanation available for this question
If 0.75:x::5:8, then x is equal to:
1. 1.12
2. 1.20
3. 1.25
4. 1.30
4
Correct Answer :
1.20
Explanation :
No Explanation available for this question
The sum of three numbers is 98. If the ratio of the first to second is 2:3 and that of the second to the third is 5:8, then the second number is:
1. 20
2. 30
3. 48
4. 58
4
Correct Answer :
30
Explanation :
No Explanation available for this question
If Rs. 782 be divided into three parts, proportional to ½::¾, then the first part is:
1. Rs. 182
2. Rs. 190
3. Rs. 196
4. Rs. 204
4
Correct Answer :
Rs. 204
Explanation :
No Explanation available for this question
Rs.432 is divided amongst three workers A, B and C such that 8 times A’s share is equal to 12 times B’s share which is equal to 6 times C’s share. How much did A get
1. Rs.192
2. Rs.133
3. Rs.144
4. Rs.128
4
Correct Answer :
Rs.144
Explanation :
No Explanation available for this question
If 20 men or 24 women or 40 boys can do a job in 12 days working for 8 hours a day, how many men working with 6 women and 2 boys take to do a job four times as big working for 5 hours a day for 12 days
1. 8 men
2. 12 men
3. 2 men
4. 24 men
4
Correct Answer :
2 men
Explanation :
No Explanation available for this question
P,Q and R enter into a partnership with capitals in the ratio 3:2:1. After 4 months, P leaves the business and after 4 more months Q also leaves the business and R continues till the end of the year. If R takes 10% of the profit for managing the business, then what part of the profit does R get
1. 37%
2. 36%
3. 27%
4. 30%
4
Correct Answer :
37\%
Explanation :
No Explanation available for this question
An outgoing batch of students wants to gift a PA system worth Rs 4,200 to their school. If the teachers, offer to pay 50% more than the students and an external benefactor gives three times the teacher's contribution, then how much should the teachers donate
1. Rs 600
2. Rs 840
3. Rs 900
4. Rs 1,200
4
Correct Answer :
Rs 900
Explanation :
No Explanation available for this question
Two cogged wheels of which one has 32 cogs and other 54 cogs, work into each other. If the latter turns 80 times in three quarters of a minute, how often does the other turn in 8 seconds
1. 48
2. 24
3. 38
4. None of these
4
Correct Answer :
24
Explanation :
No Explanation available for this question
The monthly incomes of A and B are in the ratio 4 : 5, their expenses are in the ratio 5 : 6. If 'A' saves Rs.25 per month and 'B' saves Rs.50 per month, what are their respective incomes
1. Rs.400 and Rs.500
2. Rs.240 and Rs.300
3. Rs.320 and Rs.400
4. Rs.440 and Rs.550
4
Correct Answer :
Rs.400 and Rs.500
Explanation :
No Explanation available for this question
Interview Questions
MongoDB Java Script Node JS PHP JQuery Python
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# ecet hwk chap2
## ecet hwk chap2
2-10 Define frequency Spectrum and bandwidth
Frequency spectrum is the wide range of frequencies found in the Frequency domain.
Bandwidth is the specific start and stop of the allowable frequency range.
2-11 Define electrical noise
Any undesirable electrical energy that falls within the pass band of a signal.
2-12 give a brief description of the following forms of electrical noise; man-made, thermal, correlated and impulse
Man-made: noise that is produced by device.
Thermal: is the rapid and random movement of electrons agitating and is present in all electronic components and communications systems.
Correlated: noise that is mutually to the signal and must not be present in a circuit.
Impulse - are high-amplitude peaks of short duration in the total noise spectrum.
2-16 Brief describe the significant of the Shannon limit for information capacity.
Shannon limit indicates that the higher the signal to noise the better the BER.
2-17 What is meant by the M-ary encoding?
M-ary comes from the term "binary", but M represents a digit concerning the number of conditions, levels, or combinations possible for a given number of binary variables.
problem chap2
Problems Chap 2
2-10
using shannon-hartley theorem
signal to noise ratio, SNR = 20 dB or 10000
that is , S/N = 10000
Bandwidth, B = 100 kHz
C = B*log(1 + S/N) [log is taken in base 2]
C = 100*log(1+10000) kbps
C = 1328785.66 bps
2-11
If by number of conditions it means the number of different combinations then,
for n bits,
number of different conditions = 2^n
for exaple for n = 2 , (00,01,10,11) that is total 2^2 = 4
using this
a. = 2^3 = 8
b. = 2^5 = 32
c. = 2^7 = 128
d. = 2^12 = 4096
2-12
M - number of different levels in signal
M = 2^no. of bits = 2^4
B - bandwidth
B = 10000 Hz
highest bit rate possible = 2*B*log(M) [log is in base 2]
highest bit rate possible = 2*10000*log(2^4) bps
highest bit rate possible = 80000 bps
2-16...
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Practice gaussian elimination method quiz, applied math quiz 111 for online learning. Free mathematics MCQs questions and answers to practice gaussian elimination method MCQs with answers. Practice MCQs to test knowledge on gaussian elimination method, simplex methods, simplex preliminaries, linear functions in maths, break even analysis calculations worksheets.
Free gaussian elimination method worksheet has multiple choice quiz question as set of stocks usually in ownership of investor is considered as, answer key with choices as row ownership , stock ownership, portfolio and column ownership problem solving to test study skills. For online learning, viva help and jobs' interview preparation tips, study systems of linear equations multiple choice questions based quiz question and answers.
Gaussian Elimination Method Quiz
MCQ. The set of stocks usually in the ownership of investor is considered as
1. row ownership
2. stock ownership
3. portfolio
4. column ownership
C
Simplex Methods Quiz
MCQ. According to algebra of simplex method, the slack variables are assigned zero coefficients because
1. no contribution in objective function
2. high contribution in objective function
3. divisor contribution in objective function
4. base contribution in objective function
A
Simplex Preliminaries Quiz
MCQ. The non negative slack variable is added to any less than or equal to constraint of simplex method is treated as
1. negative variable
3. constrained variable
4. non constrained variable
B
Linear Functions in Maths Quiz
MCQ. The raw material cost is \$6.5 USD, labor cost \$30.0 USD, units sold 2000 and total fixed cost is \$30,000 USD then total cost is
1. \$103,000
2. \$203,000
3. \$303,000
4. \$403,000
A
Break Even Analysis Calculations Quiz
MCQ. The fixed cost is \$100,000 USD, the price per unit is \$50 USD and the variable cost per unit is \$40 USD then the output level break even point is
1. 3000 units
2. 10000 units
3. 5000 units
4. 4000 units
B
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# NumPy Illustrated: The Visual Guide to NumPy
NumPy is a fundamental library that most of the widely used Python data processing libraries are built upon pandas, inspired by PyTorch, or can efficiently share data with (TensorFlow, Keras, etc). Understanding how NumPy works gives a boost to your skills in those libraries as well. It is also possible to run NumPy code with no or minimal changes on GPU.
The central concept of NumPy is an n-dimensional array. The beauty of it is that most operations look just the same, no matter how many dimensions an array has. But 1D and 2D cases are a bit special.
I took a great article, A Visual Intro to NumPy by Jay Alammar, as a starting point, significantly expanded its coverage, and amended a pair of nuances.
## Numpy: Array vs. Python List
At first glance, NumPy arrays are similar to Python lists. They both serve as containers with fast item getting and setting and somewhat slower inserts and removals of elements.
The hands-down simplest example when NumPy arrays beat lists is arithmetic:
NumPy vs Python List
Other than that, NumPy arrays are:
• more compact, especially when there’s more than one dimension
• faster than lists when the operation can be vectorized
• slower than lists when you append elements to the end
• usually homogeneous: can only work fast with elements of one type
Here O(N) means that the time necessary to complete the operation is proportional to the size of the array (see Big-O Cheat Sheet site), and `O*(1)` (the so-called "amortized" `O(1)`) means that the time does not generally depend on the size of the array (see Python Time Complexity wiki page)
## NumPy: Vectors, the 1D Arrays
### NumPy: Vector initialization
One way to create a NumPy array is to convert a Python list. The type will be auto-deduced from the list element types:
Be sure to feed in a homogeneous list, otherwise you’ll end up with dtype='object', which annihilates the speed and only leaves the syntactic sugar contained in NumPy.
NumPy arrays cannot grow the way a Python list does: No space is reserved at the end of the array to facilitate quick appends. So it is a common practice to either grow a Python list and convert it to a NumPy array when it is ready or to preallocate the necessary space with np.zeros or np.empty:
It is often necessary to create an empty array which matches the existing one by shape and elements type:
Actually, all the functions that create an array filled with a constant value have a `_like` counterpart:
There are as many as two functions for array initialization with a monotonic sequence in NumPy:
If you need a similar-looking array of floats, like [0., 1., 2.], you can change the type of the arange output: arange(3).astype(float), but there’s a better way. The arange function is type-sensitive: If you feed ints as arguments, it will generate ints, and if you feed floats (e.g., arange(3.)) it will generate floats.
But arange is not especially good at handling floats:
This 0.1 looks like a finite decimal number to us but not to the computer: In binary, it is an infinite fraction and has to be rounded somewhere thus an error. That’s why feeding a step with fractional part to arange is generally a bad idea: You might run into an off-by-one error. You can make an end of the interval fall into a non-integer number of steps (solution1) but that reduces readability and maintainability. This is where linspace might come in handy. It is immune to rounding errors and always generates the number of elements you ask for. There’s a common gotcha with linspace, though. It counts points, not intervals, thus the last argument is always plus one to what you would normally think of. So it is 11, not 10 in the example above.
For testing purposes it is often necessary to generate random arrays:
### NumPy: Vector indexing
Once you have your data in the array, NumPy is brilliant at providing easy ways of giving it back:
All of those methods including fancy indexing are mutable: They allow modification of the original array contents through assignment, as shown above. This feature breaks the habit of copying arrays by slicing them:
Another super-useful way of getting data from NumPy arrays is boolean indexing, which allows using all kinds of logical operators:
As seen above, boolean indexing is also writable. Two common use cases of it spun off as dedicated functions: the excessively overloaded np.where function (see both meanings below) and np.clip.
### NumPy: Vector operations
Arithmetic is one of the places where NumPy speed shines most. Vector operators are shifted to the c++ level and allow us to avoid the costs of slow Python loops. NumPy allows the manipulation of whole arrays just like ordinary numbers:
As usual in Python, a//b means a div b (quotient from division), x**n means xⁿ
The same way ints are promoted to floats when adding or subtracting, scalars are promoted (aka broadcasted) to arrays:
Most of the math functions have NumPy counterparts that can handle vectors:
Scalar product has an operator of its own:
You don’t need loops for trigonometry either:
Arrays can be rounded as a whole:
NumPy is also capable of doing the basic stats:
Sorting functions have fewer capabilities than their Python counterparts:
In the 1D case, the absence of the reversed keyword can be easily compensated for by reversing the result. In 2D it is somewhat trickier (feature request⁵).
### NumPy: Searching for an element in a vector
As opposed to Python lists, a NumPy array does not have an index method. The corresponding feature request⁶ has been hanging there for quite a while now.
• One way of finding an element is `np.where(a==x)[0][0]`, which is neither elegant nor fast as it needs to look through all elements of the array even if the item to find is in the very beginning.
• A faster way to do it is via accelerating `next((i[0] for i, v in np.ndenumerate(a) if v==x), -1)` with Numba⁷ (otherwise it’s way slower in the worst case than where).
• Once the array is sorted though, the situation gets better: `v = np.searchsorted(a, x); return v if a[v]==x else -1` is really fast with O(log N) complexity, but it requires O(N log N) time to sort first.
Actually, it is not a problem to speed up searching by implementing it in C. The problem is float comparisons. This is a task that simply does not work out of the box for arbitrary data.
### NumPy: Comparing floats
The function np.allclose(a, b) compares arrays of floats with a given tolerance
• np.allclose assumes all the compared numbers to be of a typical scale of 1. For example, if you work with nanoseconds, you need to divide the default atol argument value by 1e9: `np.allclose(1e-9, 2e-9, atol=1e-17) == False`.
• math.isclose makes no assumptions about the numbers to be compared but relies on a user to give a reasonable abs_tol value instead (taking the default np.allclose atol value of 1e-8 is good enough for numbers with a typical scale of 1): `math.isclose(0.1+0.2–0.3, abs_tol=1e-8)==True`.
Aside from that, `np.allclose` has some minor issues in a formula for absolute and relative tolerances, for example, for certain a,b `allclose(a, b) != allclose(b, a)`. Those issues were resolved in the (scalar) function math.isclose (which was introduced later). To learn more on that, take a look at the excellent floating-point guide and the corresponding NumPy issue on GitHub.
## NumPy: Matrices and the 2D Arrays
There used to be a dedicated matrix class in NumPy, but it is deprecated now, so I’ll use the words matrix and 2D array interchangeably.
Matrix initialization syntax is similar to that of vectors:
Random matrix generation is also similar to that of vectors:
Two-dimensional indexing syntax is more convenient than that of nested lists:
The view sign means that no copying is actually done when slicing an array. When the array is modified, the changes are reflected in the slice as well.
### NumPy: Matrix arithmetic
In addition to ordinary operators (like `+`, `-`, `*`, `/`, `//` and `**`) which work element-wise, there’s a `@` operator that calculates a matrix product:
As a generalization of broadcasting from scalar that we’ve seen already in the first part, NumPy allows mixed operations between a vector and a matrix, and even between two vectors:
### NumPy: Row vectors and column vectors
As seen from the example above, in the 2D context, the row and column vectors are treated differently. This contrasts with the usual NumPy practice of having one type of 1D arrays wherever possible (e.g., a[:,j] — the j-th column of a 2D array a — is a 1D array). By default 1D arrays are treated as row vectors in 2D operations, so when multiplying a matrix by a row vector, you can use either shape (n,) or (1, n) — the result will be the same. If you need a column vector, there are a couple of ways to cook it from a 1D array, but surprisingly transpose is not one of them:
Two operations that are capable of making a 2D column vector out of a 1D array are reshaping and indexing with `newaxis`:
Here the -1 argument tells reshape to calculate one of the dimension sizes automatically and None in the square brackets serves as a shortcut for np.newaxis, which adds an empty axis at the designated place.
So, there’s a total of three types of vectors in NumPy: 1D arrays, 2D row vectors, and 2D column vectors. Here’s a diagram of explicit conversions between those:
By the rules of broadcasting, 1D arrays are implicitly interpreted as 2D row vectors, so it is generally not necessary to convert between those two — thus the corresponding area is shaded.
### NumPy: Matrix manipulations
There are two main functions for joining the arrays:
Those two work fine with stacking matrices only or vectors only, but when it comes to mixed stacking of 1D arrays and matrices, only the vstack works as expected: The hstack generates a dimensions-mismatch error because as described above, the 1D array is interpreted as a row vector, not a column vector. The workaround is either to convert it to a row vector or to use a specialized column_stack function which does it automatically:
The inverse of stacking is splitting:
Matrix replication can be done in two ways: tile acts like copy-pasting and repeat like collated printing:
Specific columns and rows can be deleted like that:
The inverse operation is insert:
The append function, just like hstack, is unable to automatically transpose 1D arrays, so once again, either the vector needs to be reshaped or a dimension added, or column_stack needs to be used instead:
Actually, if all you need to do is add constant values to the border(s) of the array, the (slightly overcomplicated) pad function should suffice:
### NumPy: Meshgrids
The broadcasting rules make it simpler to work with meshgrids. Suppose, you need the following matrix (but of a very large size):
Two obvious approaches are slow, as they use Python loops. The MATLAB way of dealing with such problems is to create a meshgrid:
The meshgrid function accepts an arbitrary set of indices, mgrid — just slices and indices can only generate the complete index ranges. fromfunction calls the provided function just once, with the I and J argument as described above.
But actually, there is a better way to do it in NumPy. There’s no need to spend memory on the whole I and J matrices (even though meshgrid is smart enough to only store references to the original vectors if possible). It is sufficient to store only vectors of the correct shape, and the broadcasting rules take care of the rest:
Without the `indexing='ij'` argument, meshgrid will change the order of the arguments: `J, I= np.meshgrid(j, i)` — it is an 'xy' mode, useful for visualizing 3D plots (see the example from the docs).
Aside from initializing functions over a two or three dimensional grid, the meshes can be useful for indexing arrays:
Works with sparse meshgrids, too
### NumPy: Matrix statistics
Just like sum, all the other stats functions (min, max, argmin, argmax, mean, std, var) accept the axis parameter and act accordingly:
The corresponding NumPy function is np.amin to avoid name clashes with Python min
The argmin and argmax functions in 2D and above have an annoyance of returning the flattened index (of the first instance of the min and max value). To convert it to two coordinates, an unravel_index function is required:
The quantifiers all and any are also aware of the axis argument:
### NumPy: Matrix sorting
As helpful as the axis argument is for the functions listed above, it is as unhelpful for the 2D sorting:
It is just not what you would usually want from sorting a matrix or a spreadsheet: axis is in no way a replacement for the key argument. But luckily, NumPy has several helper functions which allow sorting by a column — or by several columns, if required:
#### 1. `a[a[:,0].argsort()]` sorts the array by the first column:
Here argsort returns an array of indices of the original array after sorting.
This trick can be repeated, but care must be taken so that the next sort does not mess up the results of the previous one:
``````a = a[a[:,2].argsort()]
a = a[a[:,1].argsort(kind='stable')]
a = a[a[:,0].argsort(kind='stable')]
``````
#### 2. helper function lexsort
There’s a helper function lexsort which sorts in the way described above by all available columns, but it always performs row-wise, and the order of rows to be sorted is inverted (i.e., from bottom to top) so its usage is a bit contrived, e.g.
• `a[np.lexsort(np.flipud(a[2,5].T))]` sorts by column 2 first and then (where the values in column 2 are equal) by column 5.
• `a[np.lexsort(np.flipud(a.T))]` sorts by all columns in left-to-right order.
Here flipud flips the matrix in the up-down direction (to be precise, in the axis=0 direction, same as a[::-1,...], where three dots mean "all other dimensions" — so it’s all of a sudden flipud, not fliplr, that flips the 1D arrays).
#### 3. order argument to sort
There also is an order argument to sort, but it is neither fast nor easy to use if you start with an ordinary (unstructured) array.
#### 4. Sorting in Pandas
It might be a better option to do it in pandas since this particular operation is way more readable and less error-prone there:
• `pd.DataFrame(a).sort_values(by=[2,5]).to_numpy()` sorts by column 2, then by column 5.
• `pd.DataFrame(a).sort_values().to_numpy()` sorts by all columns in the left-to-right order.
## NumPy: 3D and Above
When you create a 3D array by reshaping a 1D vector or converting a nested Python list, the meaning of the indices is (z,y,x). The first index is the number of the plane, then the coordinates go in that plane:
This index order is convenient, for example, for keeping a bunch of gray-scale images: a[i] is a shortcut for referencing the i-th image.
But this index order is not universal. When working with RGB images, the (y,x,z) order is usually used: First are two pixel coordinates, and the last one is the color coordinate (RGB in Matplotlib, BGR in OpenCV):
This way it is convenient to reference a specific pixel: a[i,j] gives an RGB tuple of the (i,j) pixel.
So the actual command to create a certain geometrical shape depends on the conventions of the domain you’re working on:
Obviously, NumPy functions like hstack, vstack, or dstack are not aware of those conventions. The index order hardcoded in them is (y,x,z), the RGB images order:
Stacking RGB images (only two colors here)
If your data is laid out differently, it is more convenient to stack images using the concatenate command, feeding it the explicit index number in an axis argument:
Stacking generic 3D arrays
If thinking in terms of axes numbers is not convenient to you, you can convert the array to the form that is hardcoded into hstack and co:
This conversion is cheap: No actual copying takes place; it just mixes the order of indices on the fly.
Another operation that mixes the order of indices is array transposition. Examining it might make you feel more familiar with 3D arrays. Depending on the order of axes you’ve decided on, the actual command to transpose all planes of your array will be different: For generic arrays, it is swapping indices 1 and 2, for RGB images it is 0 and 1:
Interesting though that the default axes argument for transpose (as well as the only a.T operation mode) reverses the index order, which coincides with neither of the index order conventions described above.
And finally, here’s a function that can save you a lot of Python loops when dealing with the multidimensional arrays and can make your code cleaner — einsum (Einstein summation):
Einstein summation
It sums the arrays along the repeated indices. In this particular example `np.tensordot(a, b, axis=1)` would suffice in both cases, but in the more complex cases einsum might work faster and is generally easier to both write and read — once you understand the logic behind it.
## NumPy: Summary
If you want to test your NumPy skills, there’s a tricky crowd-sourced set of 100 NumPy exercises on GitHub.
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# Prove that $\frac{100!}{50!\cdot2^{50}} \in \Bbb{Z}$
I'm trying to prove that :
$$\frac{100!}{50!\cdot2^{50}}$$
is an integer .
For the moment I did the following :
$$\frac{100!}{50!\cdot2^{50}} = \frac{51 \cdot 52 \cdots 99 \cdot 100}{2^{50}}$$
But it still doesn't quite work out .
Hints anyone ?
Thanks
• Think the numbers of multiple of 2, multiple of 4, multiple of 8, ..., etc in {51, 52, ..., 99, 100}.
– Flan
Jul 7, 2013 at 14:40
• Using \displaystyle in the title is depreciated. Jul 7, 2013 at 14:42
• If you make it dependent on $n$, then this is called a double factorial. Jul 8, 2013 at 8:22
$$\frac{(2n)!}{n! 2^{n}} = \frac{\prod\limits_{k=1}^{2n} k}{\prod\limits_{k=1}^{n} (2k)} = \prod_{k=1}^{n} (2k-1) \in \Bbb{Z}.$$
We have $100$ people at a dance class. How many ways are there to divide them into $50$ dance pairs of $2$ people each? (Of course we will pay no attention to gender.)
Clearly there is an integer number of ways. Let us count the ways.
We solve first a different problem. This is a tango class. How many ways are there to divide $100$ people into dance pairs, one person to be called the leader and the other the follower?
Line up the people. There are $100!$ ways to do this. Now go down the line, pairing $1$ and $2$ and calling $1$ the leader, pairing $3$ and $4$ and calling $3$ the leader, and so on.
We obtain each leader-follower division in $50!$ ways, since the groups of $2$ can be permuted. So there are $\dfrac{100!}{50!}$ ways to divide the people into $50$ leader-follower pairs to dance the tango.
Now solve the original problem. To just count the number of democratic pairs, note that interchanging the leader/follower tags produces the same pair division. So each democratic pairing gives rise to $2^{50}$ leader/follower pairings. It follows that there are $\dfrac{100!}{2^{50}\cdot 50!}$ democratic pairings.
Combinatorial argument:
The number of ways to arrange the digits $1,1,2,2,3,3,4,4,5,5,\ldots, 50,50$ in a row is $\frac{100!}{2^{50}}$.
This is clearly a multiple of $50!$, since we can perform any of the $50!$ permutations on the set of 50 elements.
• nice for combinatorial argument! Jul 7, 2013 at 15:10
• It's not so clear to me. How does a permutation of 50 relate to a permutation of 100?
– Mark
Jul 7, 2013 at 18:02
• @Mark If you jointly permute the $1$s, $2$s, and so on ... Jul 8, 2013 at 14:06
$$100!=(1\cdot 3\cdot 5\cdots 99)\cdot (2\cdot 4\cdot 6\cdots 100),$$
where
$$(2\cdot 4\cdot 6\cdots 100)=2^{50}\cdot (1\cdot 2\cdot 3\cdots 50)=2^{50}\cdot 50!$$
so
$$(50\cdot 2)! = 2^{50}\cdot 50!\cdot (1\cdot 3\cdot 5\cdots 99)$$
More generally $$(m\cdot n)! \, =\, m^{n}\ \cdot\ n!\ \cdot\ \prod_{(k,\, j) \ =\ (1,\,1)}^{(n,\, m-1)}\, (m\cdot k-j)$$ Your case is $$n=50, m=2$$ and the product on the right is a natural number.
Prove $51\cdot 52\cdots 100$ is divisible by $2^{50}$. Hint: Count how many multiples of 2, 4, 8... there are in the former expression.
From here and here, we have that the highest power of a prime $p$ dividing $n!$ is given by $$\sum_{k=1}^{\infty}\left\lfloor\dfrac{n}{p^k} \right\rfloor$$ Hence, the highest power of a prime $p$ dividing $\dfrac{n!}{m!}$ is given by $$\sum_{k=1}^{\infty}\left\lfloor\dfrac{n}{p^k} \right\rfloor - \sum_{k=1}^{\infty}\left\lfloor\dfrac{m}{p^k}\right\rfloor$$
I trust you can finish it off from here.
for the number of multiples of $2$:
$$100 = 52 + 2(n_1 - 1)$$ $$n_1 = 25$$
for the number of multiples of $4$:
$$100 = 52 + 4(n_2 - 1)$$ $$n_2 = 13$$
for the number of multiples of $8$:
$$96 = 56 + 8(n_3 - 1)$$ $$n_3 = 6$$
for the number of multiples of $16$:
$$96 = 64 + 16(n_4 - 1)$$ $$n_4 = 3$$
for the number of multiples of $32$:
$$96 = 64 + 32(n_5 - 1)$$ $$n_5 = 2$$
for the number of multiples of $64$:
$$64 = 64 + 64(n_6 - 1)$$ $$n_6 = 1$$
To get the number of $2$'s as factors, add all the $n$'s up, which yields $50$.
This should cancel the $2^{50}$ at the denominator, and thus prove that the expression is an integer.
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Web Results
## Order decimals | Comparing decimals | Decimals | Arithmetic | Khan ...
Order 3 decimals with digits up to the thousandths from least to greatest. ... Comparing decimals: 156.378 and 156.348 · Compare ... Ordering decimals 1.
## Comparing decimals 3 | Comparing decimals | Khan Academy
Sal compares decimals like 45.675 and 45.645 with greater and less than symbols. ... Compare decimals (tenths and hundredths) ... Ordering decimals 1.
## Ordering Decimals - Math is Fun
www.mathsisfun.com/ordering_decimals.html
NO, not THAT type of ordering. ... Ordering decimals can be tricky. ... Compare using the first column on the left; If the digits are equal move to the next column to ...
Jun 21, 2012 ... 5th Grade Math- Ordering & Comparing Decimals ... Up next. Ordering Decimals SONG: Estimation of Place Value Instruction - Duration: 2:25.
## Comparing Decimals - Sheppard Software
www.sheppardsoftware.com/mathgames/decimals/CompareDecimals.htm
Learn how to compare decimals the fun way with Fruit Shoot Decimal Comparisons math game.
## Balloon Pop Math - Decimal Order- Level One - Sheppard Software
www.sheppardsoftware.com/mathgames/decimals/BalloonPopDecimals1.htm
Learn how to order decimals from least to greatest with this fun math game. ... Teacher's Email. Review this decimal ordering game--tell us what you think!
## Comparing and Ordering Decimals - Video & Lesson Transcript ...
In this video lesson, you'll find out how you can look at any two decimals and immediately tell which is greater than the other. You'll also learn...
## Comparing and Ordering Decimals In this cooperative learning ...
www.cpalms.org/Public/PreviewResourceLesson/Preview/49356
In this cooperative learning activity, students will have five sets of decimal cards to sort and put in order - least to greatest. The lesson starts with a short whole ...
## Comparing and ordering decimals - GreatSchools
www.greatschools.org/gk/worksheets/comparing-and-ordering-decimals/
Comparing and ordering decimals. Which decimal is greater? This math worksheet helps your child learn how to determine which decimals to the tenths and ...
## Ordering decimals - Pearson Prentice Hall Mathematics Video
You Are Viewing an Explanation For: Ordering decimals. See Prentice Hall's Mathematics Offerings at: http://www.phschool.com/math. Selected images used ...
### Ordering decimals 1 | Comparing decimals | Khan Academy
Sal orders 5 decimals from least to greatest. ... Compare decimals (tenths and hundredths) · Comparing decimals: 9.97 and 9.798 ... Ordering decimals 2.
### Ordering Decimals - Math Goodies
www.mathgoodies.com
Now let's order these decimals from least to greatest without our place-value chart. We will do this by comparing two decimals at a time.
### IXL - Put decimal numbers in order (5th grade math practice)
www.ixl.com
Fun math practice! Improve your skills with free problems in 'Put decimal numbers in order' and thousands of other practice lessons.
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# List all the three digit numbers can be made from the digits 0,5,9 without repetition.
Given :
The given digits are 0, 5 and 9.
To do :
We have to list all the three digit numbers can be made from the digits 0,5,9 without repetition.
Solution :
Three digit numbers that can be made with the digits 0, 5 and 9 are 509, 590, 905 and 950.
Tutorialspoint
Simply Easy Learning
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Top
# Bayesian Linear Regression
The approach to linear regression in which we undertake the statistical analysis within the Bayesian inference context is known commonly as Bayesian linear regression.
Related Calculators Calculating Linear Regression linear regression correlation coefficient calculator
## Bayesian Likelihood Linear Regression
We have a general linear regression problem wherein the specification of the conditional distribution of $y_i$ for $i$ = $1,\ 2,\ …,\ n$ we are given a predictor vector $X_i$ of order $k\ \times\ 1$.
$y_i$ = $X_i(^T)\ \beta\ +\ \in_i$
here, $\beta$ represents a vector of the order $k\ \times\ 1$. Also, $\in_i$ are random variables that are identically normally distributed and independent.
$\in_i\ ~\ N$ ($0$, $\sigma^2$)
All this is corresponding to the likelihood function that is mentioned below:
$\rho$ ($y\ |\ X$, $\beta$, $\sigma^2$) $\alpha$ ($\sigma^2$)$^{(\frac{- n}{2})}$ $exp$ ($\frac{- 1}{(2 \sigma^2)}$ ($y\ –\ X\ \beta$)$^T$ ($y\ –\ X\ \beta$))
## Bayesian Multivariate Linear Regression
In this type of linear regression, the outcome that is predictable is not a vector of a single and scalar random variable, rather a vector of random variables which are correlated.
Let us consider a regression problem with a dependent variable that is to be predicted which is a vector of real number that are correlated and the vector is of length say $m$. There are $n$ observations with each of the observation consisting of $k\ –\ 1$ variables which are explanatory as well and again being grouped into a vector of length $k$ say $X_i$. We have,
$y_{(i,\ 1)}$ = $X_i^{T}\ \beta_1\ +\ \in_{(i, 1)}$
…..
…..
$y_{(i,\ m)}$ = $X_i^{T}\ \beta_m\ +\ \in_{(i,\ m)}$
It is given that the {$\in_{(i,\ 1)},\ …,\ \in_{(i,\ m)}$} that is the set of errors are correlated.
We can write it as a single regression problem as below:
$y_i^{T}$ = $X_i^{T}\ B\ +\ \in_i^{T}$
$B$ is a coefficient matrix with order $k\ \times\ m$. The coefficient vectors $\in_{(i,\ m)}$ are all stacked in $B$ horizontally.
The noise vector is also normal jointly for each $i$ in order to have correlated outcomes for a certain observation.
$\in_i\ ~\ N$ ($0$, $\sum_\in^2$)
In matrix form, the entire problem of regression can be written as
$Y$ = $X\ B\ +\ E$
Both $Y$ and $E$ are matrices of order $n\ \times\ m$. the matrix $X$ is a design matrix of the order $n\ \times\ k$ in which the observations are vertically stacked.
## Bayesian Estimation of Linear Regression
We require estimating the parameters in $\beta$ that are not known. For $\beta$, the maximum likelihood estimate is based on the likelihood by Gaussian as below:
$p\ (y\ |\ X,\ \beta; \sigma^2)$ = $\frac{1}{(2 \pi\ \beta^2)^{(\frac{n}{2})}}$ $exp$ ($\frac{-\ 1}{(2 \sigma^2)}$ $(||y\ –\ X \beta||^2))$
It is to be noted at all times that this is nothing but a product of likelihoods of the individual components of $y$.
First, we took the log of the above likelihood and then derivate it with respect to $\beta$. We then get,
$\bigtriangledown_\beta$ ln $p\ (y\ |\ X,\ \beta; \sigma^2)$ = $\frac{-\ 1}{(\sigma^2)}$ $X^T (y – X \beta)$
When this is put equal to zero, we get,
$\hat{\beta}$ = $(X^T\ X)^{- 1}$ $X^T\ y$
The maximum likelihood estimate found for $\beta$ is unbiased and Gaussian distributed:
$\hat{\beta}$ = $N\ (\beta,\ \sigma^2\ (X^T\ X)^{- 1})$
## Conjugate Priors for Basic Linear Regression
Here we are considering conjugate priors the posterior distribution of which can be derived in an analytical manner.
A prior say $\rho$ ($\beta$, $\sigma^2$) will only be conjugate to the likelihood function if with respect to $\sigma$ and $\beta$, the functional form it has is the same. The log likelihood of $\beta$ being quadratic is written again in such a manner that we get a normal likelihood ($\beta\ -\ \hat{\beta}$).
$(y\ –\ X\ \beta)^T$ $(y\ –\ X\ \beta)$ = $(y\ –\ X\ \hat{\beta})^T$ ($y\ –\ X \hat{\beta}$) $+$ $(\beta - \hat{\beta})^T$ $(X^T\ X)$ ($\beta\ -\ \hat{\beta}$)
Again, we rewrite the likelihood as below.
$\rho$ ($y\ |\ X,\ \beta, \sigma^2$) $\alpha$ $(\sigma^2)^{(\frac{-\ v}{2})}$ $exp$ ($\frac{-\ v\ s^2}{2\ \sigma^2}$) $(\sigma^2)^{(- \frac{(n\ –\ v)}{2})}$ $\times\ exp$ ($\frac {-\ 1}{(2\ \sigma^2)}$ $(\beta\ -\ \hat{\beta})^T$ $(X^T\ X)$ $(\beta\ -\ \hat{\beta})$)
Here, $v\ s^2$ = $(y\ –\ X\ \hat{\beta})^T$ ($y\ –\ X\ \hat{\beta}$), $v$ = $n\ –\ k$ and $k$ is number of the regression coefficients.
From this we get a suggestion about the prior,
$\rho$ ($\beta, \sigma^2$) = $\rho$ ($\sigma^2$) $\sigma$ ($\beta\ |\ \sigma^2$)
Here, $\rho$ ($\sigma^2$) is the distribution of type inverse gamma
$\rho$ ($\sigma^2$) $\alpha$ $(\sigma^2)^{(- \frac{v_0}{(2 + 1)})}$ $exp$ ($-\ \frac{v_0\ (s_0)^2}{(2 \sigma^2}$))
Also, $\rho$ ($\beta\ |\ \sigma^2$) is a conditional prior density and behaves like a normal distribution
$\rho$ ($\beta\ |\ \sigma^2$) $\alpha$ $(\sigma^2)^{(- \frac{k}{2})}$ $exp$ ($- \frac{1}{(2\ \sigma^2)}$ $(\beta\ -\ \mu_0)^T$ $\wedge_0$ ($\beta\ -\ \mu_0$))
In the normal distribution type notation, the condition prior distribution is given by $N$ ($\mu_0$, $\sigma^2\ \wedge_0^{-1}$).
Related Topics Math Help Online Online Math Tutor
*AP and SAT are registered trademarks of the College Board.
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Pages
Contributors: jimblom
## Hex Basics
This page covers the very basics of hex, including an overview of the digits we use to represent hex numbers and tools we use to indicate a number is a hex value. We also cover very simple "decimal-to-hex" conversion in the form of hexadecimal counting.
### The Digits: 0-9 and A-F
Hexadecimal is a base-16 number system. That means there are 16 possible digits used to represent numbers. 10 of the numerical values you're probably used to seeing in decimal numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9; those values still represent the same value you're used to. The remaining six digits are represented by A, B, C, D, E, and F, which map out to values of 10, 11, 12, 13, 14, and 15.
You'll probably encounter both upper and lower case representations of A-F. Both work. There isn't much of a standard in terms of upper versus lower case. A3F is the same number as a3f is the same number as A3f.
#### Subscripts
Decimal and hexadecimal have 10 digits in common, so they can create a lot of similar-looking numbers. But 10 in hex is a wholly different number from that in decimal. In fact hex 10 is equivalent to decimal 16. We need a way to explicitly state whether a number we're talking about is base 10 or base 16 (or base 8, or base 2). Enter base subscripts:
Hexadecimal 10, indicated by a subscript 16, is equivalent to decimal 16 (notice the subscript 10).
As you'll see further down, subscripts aren't the only way to explicitly state the base of a number. Subscripts are just the most literal system we can use.
### Counting in Hex
Counting in hex is a lot like counting in decimal, except there are six more digits to deal with. Once a digit place becomes greater than "F", you roll that place over to "0", and increment the digit to the left by 1.
Let's do some counting:
0088
1199
2210A
3311B
4412C
5513D
6614E
7715F
Once you've reached F16, just as you would roll from 910 to 1010 in decimal, you roll up to 1016:
16102418
17112519
1812261A
1913271B
2014281C
2115291D
2216301E
2317311F
And once you've reached 1F16, roll up to 2016 and keep churning the right-most digit from 0 to F.
### Hex Identifiers
"BEEF, it's what's for dinner". Am I channelling my inner Sam Elliott (McConaughey?), or expressing my hunger for the decimal number 48879? To avoid confusing situations like that, you'll usually see a hexadecimal number prefixed (or suffixed) with one of these identifiers:
IdentifierExampleNotes
0x0x47DEThis prefix shows up a lot in UNIX and C-based programming languages (like Arduino!).
##FF7734Color references in HTML and image editting programs.
%%20Often used in URLs to express characters like "Space" (%20).
\x\x0AOften used to express character control codes like "Backspace" (\x08), "Escape" (\x1B), and "Line Feed" (\x0A).
&#xΩUsed in HTML, XML, and XHTML to express unicode characters (e.g. Ω prints an Ω).
0h0h5EA prefix used by many programmable graphic calculators (e.g. TI-89).
Numeral/Text SubscriptBE3716, 13FhexThis is more of a mathematical represenatation of base 16 numbers. Decimal numbers can be represented with a subscript 10 (base 10). Binary is base 2.
There are a variety of other prefixes and suffixes that are specific to certain programming languages. Assembly languagues, for example, might use an "H" or "h" suffix (e.g. 7Fh) or a "\$" prefix (\$6AD). Consult examples if you're not sure which prefix or suffix to use with your programming language.
The "0x" prefix is one you'll see a lot, especially if you're doing any Arduino programming. We'll use that from now on in this tutorial.
In summary: DECAF? A horrible abomination of coffee. 0xDECAF? A perfectly acceptable, 5-digit hexadecimal number.
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# trouble finding derivatives
• May 30th 2012, 11:46 AM
droidus
trouble finding derivatives
I am having trouble solving these problems, and finding the derivatives. I do not need them solved, but just explain the next step (unless otherwise noted):
1) sin(2x)*cos(2x)
' = sin(2x)(-2sin2x)+cos2x(2cos(2x))
- where did the two come from? wouldn't it just be cos(2x)?
2) cot(x)/sin(x) = cos(x)/sin^2(x)
- shouldn't there be a (-) since in the identity, there is one?
3) 4sec^2(x)
- Alright, this one, I do not quite understand what they did. It looks like they took the derivative of the number (they get 8) times the sec, then times the derivative of sec^2.
• May 30th 2012, 11:57 AM
Plato
Re: trouble finding derivatives
Quote:
Originally Posted by droidus
I am having trouble solving these problems, and finding the derivatives. I do not need them solved, but just explain the next step (unless otherwise noted):
1) sin(2x)*cos(2x)
' = sin(2x)(-2sin2x)+cos2x(2cos(2x))
- where did the two come from? wouldn't it just be cos(2x)?
You are using the chain rule. Differentiate $\sin()$ then $2x.$
• May 30th 2012, 11:59 AM
HallsofIvy
Re: trouble finding derivatives
Quote:
Originally Posted by droidus
I am having trouble solving these problems, and finding the derivatives. I do not need them solved, but just explain the next step (unless otherwise noted):
1) sin(2x)*cos(2x)
' = sin(2x)(-2sin2x)+cos2x(2cos(2x))
- where did the two come from? wouldn't it just be cos(2x)?
The same place the "2" in the first term came from- the "2x" in the trig function. The derivative of sin(u), with respect to u, is cos(u) and if u= 2x, du/dx= 2.
$\frac{d(sin(2x)}{dx}= \frac{d(sin(u))}{du}\frac{du}{dx}= cos(u)(2)= 2cos(2x)$
Quote:
2) cot(x)/sin(x) = cos(x)/sin^2(x)
- shouldn't there be a (-) since in the identity, there is one?
No, why should there be? cot(x)= cos(x)/sin(x) and that sin(x) already in the denominator gives cos(x)/sin^2(x).
(You may be thinking of the "-" in the derivative of cot(x). There is NO derivative here, just an identity. If you are asking about the derivative of cot(x)/sin(x), you haven't differentiated yet. What does the quotient rule give for the derivative of cos(x)/sin^2(x)?
Quote:
3) 4sec^2(x)
- Alright, this one, I do not quite understand what they did. It looks like they took the derivative of the number (they get 8) times the sec, then times the derivative of sec^2.
No, they did not take the derivative of any "number", they used the chain rule. Letting u= sec(x), this function is $4u^2$, which has derivative, with respect to u, of 4(2u)= 8u= 8 sec(u). Now multiply that by the derivative of u with respect to x, the derivative of sec(x).
• May 30th 2012, 12:01 PM
droidus
Re: trouble finding derivatives
oh, ok - see for the first one... thanks!
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# Toronto Math Forum
## MAT334-2018F => MAT334--Tests => Quiz-5 => Topic started by: Victor Ivrii on November 02, 2018, 03:33:36 PM
Title: Q5 TUT 5201
Post by: Victor Ivrii on November 02, 2018, 03:33:36 PM
Find the power-series expansion about the given point for the given function; find the largest disc in which the series is valid:
$$\frac{z+2}{z+3}\qquad\text{about}\; z_0 = -1.$$
Title: Re: Q5 TUT 5201
Post by: Tianfangtong Zhang on November 02, 2018, 03:41:27 PM
\begin{align*}
\frac{z+2}{z+3} &= \frac{z+3-1}{z+3} \\ &= 1-\frac{1}{z+3}\\
&= 1- \frac{1}{z+1+2}\\ &= 1 - \frac{1}{2} \frac{1}{1+\frac{z+1}{2}}\\
&= 1 - \frac{1}{2} \frac{1}{1-\frac{-(z+1)}{2}}\\
&= 1 - \sum_{n=0}^{\infty}(\frac{-(z+1)}{2})^n \\
&= 1 - \sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}(z-(-1))^n
\end{align*}
Title: Re: Q5 TUT 5201
Post by: ZhenDi Pan on November 02, 2018, 04:52:51 PM
We can first rewrite $\frac{z+2}{z+3}$ as
\frac{z+2}{z+3} = \frac{z+3-1}{z+3}
\begin{align*}
\frac{z+3-1}{z+3} & = 1 - \frac{1}{z+3} \\
& = 1 - \frac{1}{z+1+2} \\
& = 1 - \frac{1}{2} \times \frac{1}{1+\frac{(z+1)}{2}} \\
& = 1 - \frac{1}{2} \times \frac{1}{1-\frac{-(z+1)}{2}}
\end{align*}
The power series expansion is
1 - \frac{1}{2} \times \frac{1}{1-\frac{-(z+1)}{2}} = 1 - \frac{1}{2} \times \sum_{n=0}^{\infty} (\frac{-(z+1)}{2})^n \\
= 1 - \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}}(z-(-1))^n = 1 - \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}}(z+1)^n
The largest disc in which the series is valid:
\mid -\frac{z+1}{2}\mid < 1 \\
\mid z+1\mid < 2
Title: Re: Q5 TUT 5201
Post by: Heng Kan on November 02, 2018, 07:46:50 PM
See the attatched scanned picture. A different way to get the expansion.
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# Four equilateral triangles are described on the four sides of a rectangle with perimeter 12 cm. If the sum of the area of four triangles is 10 cm2 ,what is the area of the rectangle? (a) 5cm2 (b) 8 cm2 (c) 9 cm2 (d) 6.75 cm2
## Question ID - 50145 :- Four equilateral triangles are described on the four sides of a rectangle with perimeter 12 cm. If the sum of the area of four triangles is 10 cm2 ,what is the area of the rectangle? (a) 5cm2 (b) 8 cm2 (c) 9 cm2 (d) 6.75 cm2
3537
Then,
Sum of areas of the four triangles
=
Therefore,
Length = 4 cm, breadth = 2 cm.
Area of rectangle = 8 sq. cm.
Next Question :
Keyboard, Monitor, Mouse
(A) a (B) b (C) c (D) d
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## Algebra 1
$y^{2}$-16y+64 Using the rule of $a^{2}$-2ab+$b^{2}$= $(a-b)^{2}$ In this case, the a= y and b= 8 as substituting these gives us the original polynomial $a^{2}$-2ab+$b^{2}$ $y^{2}$-2(y)(8)+$8^{2}$ $y^{2}$-16y+64 Thus: $y^{2}$-16y+64 = $(y-8)^{2}$ = (y-8)(y-8)
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# Physics 1 Mid-Term
## Terms
undefined, object
copy deck
An object weighs 30N on earth. A 2nd object weighs 30N on the moon.Which has the greater mass?
30-N on the moon
hypothesis
an educated guess
Which has more mass, a kg of feathers or a kg of nails
they are both the same
What approximates the best resultant of a pair of 10-unit vectors at right angles to each other?
14-units
work/time
power
a rock that is thrown straight up reaches the top of its path and is starting to fall back down. what is its acceleration?
9.8 m/s/s
force x distance
work
What is the resultant of a 4-unit and 3-unit vector at right angles to each other
5 ( think multiples when dealing with 3-4-5 triangles)
If you lift one load up 2 stories how much work do you do compared to lifting it up only one story?
2xs the work
the energy due to position
Potential energy
A rifle with a muzzle velocity of 100m/s is fired from a tower. Neglecting air resistence, where will the bullet be 1 s later?
100 m
A 10 N force and a 30 N foce act in the same direction. what is the net force?
40 N
Does the vertical velocity change the horizontal velocity?
no
how much does a kg of nails weigh?
9.8 N
give one unit of speed
k/h
The weight of a person can be represented by a vector that acts in what direction?
downward
If you push a object w/ 2xs the force and 2xs the distance, how much does work increase?
4xs
In the abscence of air resistance, an object falls at a constant what?
acceleration
unit of work
joules
A cannonball is fired at some angle into the air. In the 1st second it moves 10 meters horizontally. Assuming it doesn't hit the ground and air resistance is small, how far does it move in the next second?
10 m
What is the maximum resultant when adding a 3N force to a 8N force?
11N
speed
d/t
units of power
watts
Jose can jump vertically 1 m on his skateboard when it was at rest. How far can Jose jump?
1 m
If a woman stands on a scale w/ 2 feet and weighs 500-N, how much will she weigh if she lifts one foot?
500-N
when an object falls freely from acceleration, what increases?
its velocity
What is the minimum resultant possible when adding 3N and 8N?
5N
if you lift two loads up one story, how much work do you do compared to lifting one load up one story?
2xs the work
A ball tossed vertically upwards, reahes its highest point and then falls back to its starting point.During this time,the acceleration is always in what direction?
downward
1/2mv/v
kinetic energy
At the instant a ball is thrown horizontally with a large force, an identical ball is dropped from the same height. Which ball hits the ground 1st?
they both hit at the same time
An arrow in a bow has 70 J of PE.assuming no loss of energy due to heat, how much kinetic energy will it have after it has been shot?
70 J
what is a theory
it is observation that has not been proven wrong yet
How much farther will a car at 100 km/h skid than the same car at 50 km/h
4xs as much
If one object has twice as much mass as another object, then the object also has twice the __________.
what must an object with kinetic energy have?
momentum
An object tavels 2 meters in the 1st second of travel, 2 meters again in the 2nd second of travel and 2 meters in the 3rd second of travel. What is the acceleration
0
A woman can lift barrels a vertical distance of 1 m or can roll them up a 2 m long ramp at the same elevation. Which one will require 1/2 the force?
the 2m ramp
What is friction?
a force that acts as a force of touching objects
A person on a roof throws one ball downward and an identical ball upward at the same speed. the ball thrown downward hits the ground with 100 J of KE. Ignoring air friction, with how much kinetic energy does the second ball hit the ground?
100 J
Suppose an object is in free fall. Each second the object falls a ________ distance than the second before
farther
A job is done slowly and an identical job is done quickly. Both jobs require the same amounts of work but different amounts of what?
power
The mass of a dog that weighs 100 N weighs how much in kgs?
10
A ball is thrown into the air with 100 J of kinetic energy which is transformed to gravitational potential energy at the top of its trajectory. When it returns to its original level after encountering air resistance, what is its kinetic energy?
the same- 100J
when you look at the speedometer in a moving car, you can seethe car's ...
instanteous speed
Which requires more work: lifting a 50kg sack vertically 2m or lifting a 25kg sack up 4m vertically?
both require the same amount of work
Friction is a force that always acts in what direction?
the opposite direction
Energy is changed from one form to another with no loss or gain. What is this stament?
Always true.
physics is the most basic science because...
it is about the world around us
Which has a greater kinetic energy, a car traveling at 30 km/h or a half as massive car traveling at 60km/h?
60km/h
The force required to keep maintain an object at a constant speed in free space is equal to__.
0
If an object has linear kinetic energy then it must be what?
moving
Ten second after starting from resta car is movingat 40 m/h. What is the car's average acceleration?
4m/s/s
When bullets are fired from an airplane in the forward direction, what will happen to the momentum of the airplane?
it will decrease
You would have to travel to what planet to have the largest force if it only weighs 1-N on Earth?
Pluto
An object is dropped and falls freely to the ground with an acceleration of a 1 g. What is its acceleration if its thrown upward at angle?
still gravity - 9.8 m/s/s
Consider drops falling from a water faucet. When the drops fall they fall _________ than before
farther
At what part of the path does a projectile have minimum speed?
the top of its path
An object following a straight-line path at constant speed has__ _______ _______ __ __
no forces acting on it.
A projectile is fired horizontally in a vacuum. Whay does the projectile maintain its horizontal component speed?
because of Newton's 1st law
The hang time for most athletes are about?
less than one second
If a ball is thrown in the air at some angle. What will the ball's velocity be at the top of its path?
changing
If a an object weighs 10 kg on the earth then what does it weigh on the moon?
10 kg
A bullet fired horizontally hits the ground in 0.5 seconds. If it had been fired with a much greater speed in the same direction, it would have hit the ground when?(neglecting the earth's curvature and air restistance)
0.5 seconds
What is the average speed of a car going 100 kilometers in two hours
50 km/h
a car accelerates at 2 m/s/s. Assuming the car starts from rest., how much time does does it take to accelerate to s speed of 30m/s
15 s
If you push a car on the earth and as hard as you push it on the moon. why is this?
b/c of inertia
vector
a quantity that has magnititude and direction
What does the law of inertia apply to?
objects in motion or at rest
a horizontal component of a projectile is independent of the ....
vertical component
Equilibrium occurs when?
when the net forces equal 0
If you drop a faether and a coin at the same time in a vacuum tube, which will reach the bottom first?
both will fall at the same time
If the force of gravity suddenly stopped acting on the planets, what woul happen?
it continues in a straight line
A car accelerates at 5 m/s/s. Assuming the car starts from rest , how much time would it take to accelerate the car to 30 m/s/s?
6s
A 10-N force and a 30-N force act on an object in different directions. What is the net force of the object?
20-N
A ball is thrown sraight up its acceleration is what?
9.8 m/s b/c of velocity
A ball is thrown straight upward at 20m/s. Ideally, (no air resistance) the ball will return in the thrower's hand at what speed?
20m/s
what is a projectile?
a moving object
What does the law of inertia state?
it resists change to its state of motion (an object is lazy)
what are the components of a vector?
magnitude and d9irection
------->
A rifle with a muzzle velocity of 100 m/s is fired horizontally from a tower. Neglecting air resistance, where will the bullet be in 10 seconds?
1000 m
Why will a ball never go as far as physics ideally predicts?
because of friction
If a ball rolls down one plane how far will it go up another?
to its initial point
scalar
quantity that has magnitude
How much force is needed to keep a cannon fired into frictionless space going?
0
If a free falling object were somehow equiped with a sppedometer, its speed would increase by __ m/s
9.8
A sheet of paper can be withdrawn from under a container if milk without toppling it if the paper is jerked quickly. Why can this be done?
because of inertia
Suppose a car is moving in a straight lineand steadily increases its speed. It moves from 35 km/h to 40 km/h and from 40 km/ h the next second. What is the acceleration?
5km/h
In the abscence of air fricton, does a vertical component of a projectile's velocity change as the projectile moves?
no, it does not
A ball is thrown straight up. Its instaneous speed at the top of its path is what?
0 (but its velocity is not)
At the instant a ball is thrown horizontally with a large force an identical ball is dropped from the same height. which the ball hits the ground first?
both fall at the same time
When there is no air resistance, a ball would go the farthest if thrown at this angle
45 degrees
92
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> > > Chapter 4.6
Solutions for Chapter 4.6: Graphs of Other Trigonometric Functions
Full solutions for Precalculus | 7th Edition
ISBN: 9780618643448
Solutions for Chapter 4.6: Graphs of Other Trigonometric Functions
Solutions for Chapter 4.6
4 5 0 268 Reviews
25
5
ISBN: 9780618643448
Since 98 problems in chapter 4.6: Graphs of Other Trigonometric Functions have been answered, more than 13351 students have viewed full step-by-step solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Precalculus, edition: 7. Precalculus was written by and is associated to the ISBN: 9780618643448. Chapter 4.6: Graphs of Other Trigonometric Functions includes 98 full step-by-step solutions.
Key Calculus Terms and definitions covered in this textbook
• Acute angle
An angle whose measure is between 0° and 90°
• Average velocity
The change in position divided by the change in time.
• Axis of symmetry
See Line of symmetry.
• Bias
A flaw in the design of a sampling process that systematically causes the sample to differ from the population with respect to the statistic being measured. Undercoverage bias results when the sample systematically excludes one or more segments of the population. Voluntary response bias results when a sample consists only of those who volunteer their responses. Response bias results when the sampling design intentionally or unintentionally influences the responses
• Commutative properties
a + b = b + a ab = ba
• Continuous function
A function that is continuous on its entire domain
• First octant
The points (x, y, z) in space with x > 0 y > 0, and z > 0.
• Inferential statistics
Using the science of statistics to make inferences about the parameters in a population from a sample.
• Law of sines
sin A a = sin B b = sin C c
• Order of an m x n matrix
The order of an m x n matrix is m x n.
The formula x = -b 2b2 - 4ac2a used to solve ax 2 + bx + c = 0.
• Quotient identities
tan ?= sin ?cos ?and cot ?= cos ? sin ?
• Random behavior
Behavior that is determined only by the laws of probability.
• Random variable
A function that assigns real-number values to the outcomes in a sample space.
• Re-expression of data
A transformation of a data set.
• Replication
The principle of experimental design that minimizes the effects of chance variation by repeating the experiment multiple times.
• Solve by elimination or substitution
Methods for solving systems of linear equations.
• Sphere
A set of points in Cartesian space equally distant from a fixed point called the center.
• Sum of a finite arithmetic series
Sn = na a1 + a2 2 b = n 2 32a1 + 1n - 12d4,
• Window dimensions
The restrictions on x and y that specify a viewing window. See Viewing window.
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# Eigen value , eigen vector
• Jan 4th 2010, 07:52 PM
flintstone
Eigen value , eigen vector
Find the eigen value and eigen vector of $\displaystyle \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix}$
Solution :
$\displaystyle |A- \lambda I |$ = $\displaystyle \begin{bmatrix} (1 - \lambda) & 2 & 3 \\ 3 & (1 - \lambda) & 0 \\ -2 & 0 & (1 - \lambda) \end{bmatrix}$ =
$\displaystyle (1-\lambda)[(1 - \lambda)(1-\lambda) - 0]-2[3(1 - \lambda)-0]+3[0-(-2)(1 - \lambda)]$
= $\displaystyle (1-\lambda)(1 - 2 \lambda + 2)- 6 + 6 \lambda+ 6 - 6 \lambda$
$\displaystyle = 1 -2\lambda+\lambda^2-\lambda+2\lambda^2-\lambda^3$
$\displaystyle =(1-\lambda)(1-2\lambda+\lambda^2)$
$\displaystyle (1-\lambda)(\lambda-1)(\lambda-1)$
Eigen value $\displaystyle \lambda =1$
putting $\displaystyle \lambda=1$ in the matrix eq
$\displaystyle \begin{bmatrix} 0 & 2 & 3 \\ 3 & 0 & 0 \\ -2 & 0 & 0 \end{bmatrix}$ . $\displaystyle \begin{bmatrix}x \\ y \\z \end{bmatrix}$=$\displaystyle \begin{bmatrix} 0 \\ 0 \\0 \end{bmatrix}$ $\displaystyle \Rightarrow$ $\displaystyle \begin{cases}2y + 3z = 0 \\ 3x = 0 \\ -2x = 0 \end{cases}$
Stuck here !!!!!!!
• Jan 4th 2010, 09:20 PM
Shanks
$\displaystyle x=0,y=c,z=\frac{-2c}{3}$
the solution set is a subspace generated by $\displaystyle (0,1,\frac{-2}{3})$.
• Jan 5th 2010, 12:04 AM
flintstone
thanks
• Jan 5th 2010, 04:00 AM
HallsofIvy
Quote:
Originally Posted by flintstone
Find the eigen value and eigen vector of $\displaystyle \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix}$
Solution :
$\displaystyle |A- \lambda I |$ = $\displaystyle \begin{bmatrix} (1 - \lambda) & 2 & 3 \\ 3 & (1 - \lambda) & 0 \\ -2 & 0 & (1 - \lambda) \end{bmatrix}$ =
$\displaystyle (1-\lambda)[(1 - \lambda)(1-\lambda) - 0]-2[3(1 - \lambda)-0]+3[0-(-2)(1 - \lambda)]$
= $\displaystyle (1-\lambda)(1 - 2 \lambda + 2)- 6 + 6 \lambda+ 6 - 6 \lambda$
$\displaystyle = 1 -2\lambda+\lambda^2-\lambda+2\lambda^2-\lambda^3$
$\displaystyle =(1-\lambda)(1-2\lambda+\lambda^2)$
$\displaystyle (1-\lambda)(\lambda-1)(\lambda-1)$
Eigen value $\displaystyle \lambda =1$
putting $\displaystyle \lambda=1$ in the matrix eq
$\displaystyle \begin{bmatrix} 0 & 2 & 3 \\ 3 & 0 & 0 \\ -2 & 0 & 0 \end{bmatrix}$ . $\displaystyle \begin{bmatrix}x \\ y \\z \end{bmatrix}$=$\displaystyle \begin{bmatrix} 0 \\ 0 \\0 \end{bmatrix}$ $\displaystyle \Rightarrow$ $\displaystyle \begin{cases}2y + 3z = 0 \\ 3x = 0 \\ -2x = 0 \end{cases}$
Stuck here !!!!!!!
Both second and third equations just tell you that x= 0. The first equation tells you that 2y= -3z so y= -(3/2)z. Any eigenvector is of the form <0, -(3/2)z, z>= z<0, -3/2, 1>. If z= 2, that is <0, -3 , 2> so the space of eigenvectors of this matrix is spanned by that single vector (which is, of course, a multiple of the one Shanks gave).
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# HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board
## 1 Gaj in Square Centimeter in Telangana
In this article, you will learn how to convert 1 Gaj in Square Centimeter in Telangana. Gaj is one of the most commonly used measurement units used in almost every state of India. People in Indian states refer to gaj for measurement of land. Here in this page you will learn how to calculate 1 Gaj in Square Centimeter unit.
## How to Convert from 1 Gaj to Square Centimeter?
You can easily convert gaj to Square Centimeter or the reverse with a simple method. You can multiply the figure in gaj by 8281 to determine the Square Centimeter value. This is all you need to do for undertaking the conversion procedure of Gaj to Square Centimeter in Telangana.
## Relationship between Gaj and Square Centimeter
It is not difficult to work out the actual relationship between gaj to Square Centimeter in most cases. You should know the proper calculations existing between gaj to Square Centimeter before you venture to conversion procedures. 1 Gaj to Square Centimeter calculations will be 8281 Square Centimeter.
Once you know these calculations, it will not be hard for you to calculate gaj in an Square Centimeter. One gaj can be worked out to various other units as well.
## Formula for Converting Gaj to Square Centimeter
The formula to convert Gaj to Square Centimeter is the following-
1 Gaj = 8281 Square Centimeter
Conversely, you can also use this formula-
Gaj = Square Centimeter * 8281
This is all you need to know for undertaking the conversion procedure on your part. You can also use online calculators or conversion tools for this purpose.
### Other Plot Areas in Square Centimeter in Telangana
• 1 Gaj in Square Centimeter in Telangana = 8281
• 2 Gaj in Square Centimeter in Telangana = 16562
• 3 Gaj in Square Centimeter in Telangana = 24843
• 4 Gaj in Square Centimeter in Telangana = 33124
• 5 Gaj in Square Centimeter in Telangana = 41405
• 6 Gaj in Square Centimeter in Telangana = 49686
• 7 Gaj in Square Centimeter in Telangana = 57967
• 8 Gaj in Square Centimeter in Telangana = 66248
• 9 Gaj in Square Centimeter in Telangana = 74529
• 10 Gaj in Square Centimeter in Telangana = 82810
• 15 Gaj in Square Centimeter in Telangana = 124215
• 20 Gaj in Square Centimeter in Telangana = 165620
• 25 Gaj in Square Centimeter in Telangana = 207025
• 30 Gaj in Square Centimeter in Telangana = 248430
• 35 Gaj in Square Centimeter in Telangana = 289835
• 40 Gaj in Square Centimeter in Telangana = 331240
• 45 Gaj in Square Centimeter in Telangana = 372645
• 50 Gaj in Square Centimeter in Telangana = 414050
• 55 Gaj in Square Centimeter in Telangana = 455455
• 60 Gaj in Square Centimeter in Telangana = 496860
• 65 Gaj in Square Centimeter in Telangana = 538265
• 70 Gaj in Square Centimeter in Telangana = 579670
• 75 Gaj in Square Centimeter in Telangana = 621075
• 80 Gaj in Square Centimeter in Telangana = 662480
• 85 Gaj in Square Centimeter in Telangana = 703885
• 90 Gaj in Square Centimeter in Telangana = 745290
• 95 Gaj in Square Centimeter in Telangana = 786695
• 100 Gaj in Square Centimeter in Telangana = 828100
## FAQs About 1 Gaj in Square Centimeter in Telangana
#### Q: Ek Gaj main Kitne Square Centimeter in Telangana Hote hain?
Ek Gaj main 8281 Square Centimeter in Telangana hote hain.
#### 1 Gaj main kitna Square Centimeter in Telangana Hota Hain?
1 Gaj main 8281 Square Centimeter in Telangana hote hain
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# I need helo
Use the given degree of confidence and sample data to find CI for the population standard deviation σ. Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation.
To find the standard deviation of the diameter of wooden dowels, the manufacters measures 19 randomly selected dowels and finds the standard deviation of the sample to be s= 0.16. Find the 95% CI for the population standard deviation σ
reshown Apr 18, 2015
Step 1:
The number of samples are 19.
The point estimate for standard deviation is s = 0.16.
Confidence level is 95%.
Find the 95% confidence interval for the standard deviation.
Confidence interval for standard deviation is .
Where n is number of samples,
are critical values correspond to the given level of confidence.
Step 2:
Now find the critical values .
Area to the right of .
Area to the left of .
Use the Chi-square distribution table to find the critical values.
Step 3:
Substitute in Confidence interval.
Therefore the 95% confidence interval for the standard deviation is .
Solution :
The 95% confidence interval for the standard deviation is .
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# Function uniqueness
kenewbie
Function "uniqueness"..
Ok, pardon the complete lack of terminology here.
I can define a function with one parameter such that no two different inputs give the same output. Example:
f(x) = x + 1
No value of x gives the same result as another value of x.
I believe that it is impossible to define a function which accepts two parameters, yet retains this property of "uniqueness". Am I correct? Is there a name for what I am trying to describe?
I'm pretty sure I am correct, but I have no idea how to prove it.
Lets say I have a function
f(a,b) = k
where k is some sort of calculation.
if k is arithmetic, i can break the uniqueness by doing the calculation outside f, and then feeding 0 as one of the parameters
f(a,b) = k = f(k,0)
if k is geometric I can do the same as above, only pass 1 instead of 0. Outside of these two classes of functions (and the trivial one where a or b is omitted in the calculation of the function) I can't seem to find a good way of proving myself.
Proving this for more and more "classes of calculations" seem futile, I'm not even sure everything can be categorized as neatly as arithmetic and geometric functions can. There might be a better way of going about this?
All feedback is welcome.
k
Last edited:
There is no "nice" way to define a two parameter function with a unique value. However, there are messy looking ways. Example consider the decimal expansions of x and y:
0.x1x2x3x4x5x6x7..... and 0.y1y2y3y4y5y6y7....
Then define z=f(x,y) by:
0.x1y1x2y2x3y4x4y4......
Note that this works from the unit square to the unit interval.
To do the whole real line, to get x and y use an arctan transformation (properly normalized) and the reverse process (tan) to go from z to the whole real line.
Normalization involves shifting the interval and dividing or multiplying by pi.
mistermath
What you're looking for is called a function that is one to one. It is also called injection/injective.
A typical example is: f(x,y) = a^x * b^y where a, b are primes that don't equal each other.
Ex: let f(x,y) = 2^x * 3 ^ y
It should be easy to see that f(3,4) = 2^3 * 3^4 is a unique number or the fundumental theorem of mathematics (sometimes called of arithmetic) would be contradicted.
What you're looking for is called a function that is one to one. It is also called injection/injective.
A typical example is: f(x,y) = a^x * b^y where a, b are primes that don't equal each other.
Ex: let f(x,y) = 2^x * 3 ^ y
It should be easy to see that f(3,4) = 2^3 * 3^4 is a unique number or the fundumental theorem of mathematics (sometimes called of arithmetic) would be contradicted.
This is good as long as x and y are integers. However, for real x and y, you would lose uniqueness.
kenewbie
Thanks a lot to both of you, I was clearly wrong in assuming that it cannot be done.
k
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Welcome to Test-paper.info
Sunday, May 28 2017 @ 03:36 AM CDT
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Forum Index > Test Paper Related > Primary 5 Matters catholic high 2011 p5 maths sa 2
| Printable Version
By: tictactoe (offline) Thursday, October 04 2012 @ 04:52 AM CDT (Read 1576 times)
tictactoe
Q4) In the number line below, the value of X is 1/12 while the value of Z is 1/8. Y is the midpoint between X and Z. Find the value of Y. (ans : 5/48)
(pls help with working solution)
X Y Z
|_____________|_____________|
1/12 ? 1/8
Newbie
Registered: 07/28/12
Posts: 8
By: tictactoe (offline) Thursday, October 04 2012 @ 05:29 AM CDT
tictactoe
Quote by: tictactoe
Q4) In the number line below, the value of X is 1/12 while the value of Z is 1/8. Y is the midpoint between X and Z. Find the value of Y. (ans : 5/48)
(pls help with working solution)
X Y Z
|_____________|_____________|
1/12 ? 1/8
Newbie
Registered: 07/28/12
Posts: 8
By: GIM (offline) Thursday, October 04 2012 @ 08:38 AM CDT
GIM
Difference X & Z----> 1/8 - 1/12 = 3/24 - 2/24 = 1/24
Difference X & Y----> 1/24 รท 2 = 1/24 x 1/2 = 1/48
Value of Y ------------> 1/12 + 1/48 = 4/48 + 1/48 = 5/48 (answer)
Regular Member
Registered: 11/07/10
Posts: 70
By: jo sarah (offline) Monday, October 08 2012 @ 09:37 AM CDT
jo sarah
Whenever you need to find a mid-value, just add the two ends and divide the result by 2.
So, (1/12) + (1/8) = 5/24
Now, divide that by 2, you get 5/48
Regular Member
Registered: 03/20/12
Posts: 111
By: tictactoe (offline) Thursday, October 11 2012 @ 08:55 AM CDT
tictactoe
thanks for your explanation jo sarah.
tictactoe
Newbie
Registered: 07/28/12
Posts: 8
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# C Program: Power of a Number
When $n$ is a positive integer, an integer $a$ multiplied to itself $n$ times is represented by $a^{n}$. $a$ is called the base and $n$ is known as exponent.
The result of the product is known as power.
When $n$ is a whole number,
$a^{2} = a \times a$
$a^{3} = a \times a \times a$
$a^{n} = a \times a \times a ... \times a$
When $n = 0$,
$a^{0} = 1$
In the below C program, we create a function called power() which computes the power of an entered number to some desired exponent by recursion.
#include <stdio.h>
long power (int, int);
int main() {
int exp, n;
printf("Number: ");
scanf("%d", &n);
printf("Exponent: ");
scanf("%d", &exp);
printf("%d^%d = %ld \n", n, exp, power(n, exp));
return 0;
}
long power (int number, int exponent) {
if (exponent) {
return (number * power(number, exponent - 1));
} else {
return 1;
}
}
We run the above program to find $3^{4}$, which gives the result as follows:
$./a.out Number: 3 Exponent: 4 3^4 = 81 We can also get the power of a number using the C function pow(), which takes two arguments of type double, where the first argument is the number and the second argument is the exponent. So, pow(2,3) computes$2^{3}\$ and gives 8.
The program below computes the power of a number using the pow() function.
#include <stdio.h>
#include <math.h>
int main() {
double n, exponent, result;
printf("Number: ");
scanf("%lf", &n);
printf("Exponent: ");
scanf("%lf",&exponent);
printf("%.1lf^%.1lf = %.2lf \n", n, exponent, pow(n,exponent));
return 0;
}
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# Rational Numbers
#### Statement
In mathematics, a rational number is any number that can be expressed as the quotient or fraction p/q of two integers, with the denominator q not equal to zero. Since q may be equal to 1, every integer is a rational number. Furthermore a set S is called countable if there exists an injective function f from S to the natural numbers N* = {1, 2, 3, ...}.
The set of all real numbers is not countable, but the set of all rational numbers is infinite and countable. In order to prove that let's put all rational numbers in a table with the following rules: in the first row are placed all integer numbers ordered by their absolute value, and placed after each natural number (including zero) its oposite:
0, 1, -1, 2, -2, ..., n, -n, ...,
In the second row all irreducible fractions whose denominator is 2 are placed, ordered by their absolute value and again after each positive number its oposite follows:
1/2, -1/2, 3/2, -3/2, 5/2, -5/2, ...
In general, in the nth row all irreducible fraction with denominator n are placed, ordered by their absolute value and after each positive number its oposite follows. The rational number table that has been obtained possess an infinite number of rows and columns:
Clearly every rational number is placed in the table. Let's enumerate the elements of the table according to the following schematic (arrows indicate the direction of the increasing numeration):
As a result, every rational number receives an unique natural number. Can you tell which rational number occupies the ith position in the table?
#### Input:
There is a single integer T in the first line of input (1 ≤ T ≤ 100). It stands for the number of test cases to follow. Each test case consist of a single line, which contains an integer number i (1 ≤ i ≤ 108).
#### Output:
For each test case print a single line with the ith rational number ri. For ri being an integer number print ri in the usual single integer representation, otherwise (ri is not an integer number) ri must be printed in the format p/q as an irreducible fraction (Greatest Common Divisor of p and q is 1 and q>0). Check the example output for clarification.
```4
1
2
3
9
```
```0
1
1/2
-1/3
```
#### Time and memory limit:
• 1 seconds
• 64MB
Problem source: COJ
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# Is there a physical quantity which is the reciprocal/multiplicative inverse of time?
Is there a physical quantity which is the reciprocal/multiplicative inverse of time?
If time =distance/speed what is speed/distance. Please forgive my ignorance if there is a really simple explanation.
• In a certain sense, frequency is opposite (or conjugate) to time via the Fourier transform. – Alfred Centauri Feb 1 '14 at 23:02
In a mathematical sense ${ speed \over distance }$ would give you a unit of ${time^{-1}}$. For example. Say we are in a car driving $5 {m \over s}$ and in a certain time we move $10m$. If we take this information and divide distance by speed we have: $${ 5 { m \over s} \over 10m } = 0.5s^{-1}$$ This leaves us with a ratio that (in this case) says that half of the distance was traveled per second. This unit, $s^{-1}$ is considered the inverse of the unit seconds. Physically, the inverse of time represents "something" per second (or minute or hour or whichever time unit you are working in). In the example I presented it represents a portion of distance traveled per second.
• $s^{-1}$ is the unit of frequency also caled $Hz$ hertz – user288447 Feb 1 '14 at 23:24
• That depends on the context. A $bq$ (Becquerel) is also $s^{-1}$. – pfnuesel Feb 1 '14 at 23:52
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# Limits
• Feb 27th 2011, 08:42 PM
ewqewq
Limits
I am having trouble with this limit, and I would appreciate anyone's imput.
limit (1+(1/3X))^(7+2X)
as x->0
I have tried to manipulate this into the 2 indeterminate forms that work with L'hopitals rule to no avail.
I am leaning toward the limit approaching 0 , since the limit is undefined at x=0, if we approach x from 0+, the limit goes to infinity^7 , which means 1/[(inifinity)^-7)] = 0
• Feb 27th 2011, 08:56 PM
Prove It
Is this $\displaystyle \displaystyle \lim_{x \to 0}\left[1 + \left(\frac{1}{3x}\right)^{7 + 2x}\right]$ or $\displaystyle \displaystyle \lim_{x \to 0}\left[\left(1 + \frac{1}{3x}\right)^{7+2x}\right]$, or some other form I haven't written?
• Feb 27th 2011, 08:59 PM
ewqewq
It is the second one . I am sorry mine was not clear.
• Feb 27th 2011, 09:17 PM
Prove It
$\displaystyle \displaystyle \lim_{x \to 0}\left[\left(1 + \frac{1}{3x}\right)^{7+2x}\right] = \lim_{x \to 0}\left\{e^{\ln{\left[\left(1+\frac{1}{3x}\right)^{7+2x}\right]}}\right\}$
$\displaystyle \displaystyle = \lim_{x \to 0}\left[e^{(7+2x)\ln{\left(1 + \frac{1}{3x}\right)}}\right]$
$\displaystyle \displaystyle = e^{\lim_{x \to 0}\left[(7+2x)\ln{\left(1 + \frac{1}{3x}\right)}\right]}$.
It doesn't look like you need to use L'Hospital's Rule at all...
• Feb 27th 2011, 09:28 PM
ewqewq
Thanks for your help, however I still do not understand what would happen to 1/(3x), since it would be undefined
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# durham college
posted by on .
the pont du gard near Nimes,france, is a roman aqueduct.an observer in a hot air balloon some distance away from the aqueduct determines that the angle of depression to each end is 54 degrees and 71 degrees,respectively.the horizontal distance between the balloon and the closest end of the aqueduct is 270.0m. calculate the length of the aqueduct to the nearest tenth of a metre
• physics?? - ,
.
• no, trig - ,
No, trig :)
Draw a sketch, h is height of balloon over bridge
tan 71 h/270
so
h = 270 tan 71 = 784
then
tan 54 = h/x where x is hor dist to other end from balloon
x = h/tan 54 = 570
so total length = 270 + 570 = 840 meters
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# Axial section
Axial section of the cone is equilateral triangle with area 208 dm2. Calculate volume of the cone.
Result
V = 188.1 dm3
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
#### To solve this verbal math problem are needed these knowledge from mathematics:
Tip: Our volume units converter will help you with converion of volume units. Pythagorean theorem is the base for the right triangle calculator. See also our trigonometric triangle calculator.
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The cube is inscribed in a sphere with volume 4114 cm3. Determine the length of the edges of a cube.
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Question
# If x2−ax+b=0 and x2−px+q=0 have one root common and the second equation has equal roots, then b+q is equal to
A
p+a
B
p2
C
ap2
D
a2
Solution
## The correct option is C ap2Let the root of x2−px+q=0 be β ⇒2β=p, β2=q ⇒β=p2=±√q ⋯(1) Now for x2−ax+b=0, let the roots be α and β. α+β=a, αβ=b From (1), α=2a−p2, 2a−p2⋅p2=b p2(2a−p)2=b ⇒4b=2ap−p2 ⇒4b+p2=2ap ⇒4b+4q=2ap [∵p2=±√q] ⇒b+p=ap2Mathematics
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11
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# A farmer buys a goat and a sheep for Rs 3500. He sold the sheep at a profit of 20 percent and the goat at a loss of 10 percent. If he sold both the animals at the same price, then the cost price (in Rs) of the cheaper animal was?
A) 2000 B) 1500 C) 1750 D) 2250
Explanation:
Q:
A man bought 2 articles for ₹3050 each. He sold one article at 10% loss and another at 20% profit. The total profit/loss percentage he earned is:
A) 10% loss B) 5% loss C) 5% profit D) 10% profit
Explanation:
0 93
Q:
There was 25% off on shirt. A lady bought that shirt and got an additional 20% discount for paying in cash and further 10% discount for being a loyal customer. She paid ₹324. What was the price tag (in ₹) on the shirt ?
A) 750 B) 600 C) 650 D) 725
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A person purchased a vehicle for ₹5,90,828 and sold it for ₹6,52,920. What is the profit percent he earned on this vehicle (correct to two decimal places) ?
A) 9.51% B) 9.55% C) 10.51% D) 11.55%
Explanation:
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There was 25% off onshirt. A lady boughta shirt and got an additional 20% discount for paying in cash and further 10% discount for being a loyal customer. She paid ₹405. What was the price tag (in ₹) on the shirt?
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# Algebra Question: Simplifying Ration Questions
Alright It's been a while since I did algebra and now I'm getting back into it in honors algebra 2 (shakes head) [highschool]. Anyways I got a question on factoring numerators.
$$6x^{2} + 7x + 2$$
-------------
$$4x^{2} - 1$$
I don't know where to start, if someone can be a guidance this would be much appreciated I was trying to http://www.analyzemath.com/Rational_expressions/Simp_rat_expre.html" [Broken] but the walk through didn't really show me how they got to the next step.
Last edited by a moderator:
symbolipoint
Homework Helper
Gold Member
Try factoring. If you find a binomial which occurs in both the numerator and the denominator then you can eliminate as a factor of 1.
For the numerator, start tries using:
(2x + ?)(3x + ?), and if that does not help, try (6x + ?)(x + ?);
but what you really will rely on is the presence of (2x+1) in numerator AND denominator. Otherwise, you will not be able to simplify the given expression.
For the bottom, it's just the difference of squares factorization. a^2 - b^2 = (a-b)(a+b). The top is a bit trickier. But if you're new to factoring, maybe this will help.
Set the numerator equal to (rx + s)(tx + v). Now multiply out this expression and equate its terms to the terms of the numerator. You should be able to find suitable r,s,t,v and then you'll have the factorization. Don't forget to multiply out to check.
tiny-tim
Homework Helper
Welcome to PF!
$$6x^{2} + 7x + 2$$
-------------
$$4x^{2} - 1$$
I don't know where to start,
Hi Lamonte ! Welcome to PF!
Useful tip: This isn't real life, it's an exam question, so it probably has an easy answer.
You know that they want you find a common factor, and you can probably see exactly what the two factors of the denominator are.
So just try both of them in the numerator … if it's a sensible exam question, then one of them should work!
HallsofIvy
Homework Helper
To add to tiny-tim's answer- even in "real life" where problems don't always have easy answers, since you are trying to simplify the fraction, the only reason you want to factor is to be able to cancel. You should be able to see quickly that 4x2- 1 factors as (2x-1)(2x+ 1). Then try each of those as a factor of 6x2+ 7x+ 2. If it is a factor, good, you can factor and cancel. If neither is a factor, also good. You know that cannot be simplified.
As have been stated, the denominator should be easily factored using the following shortcuts (which you should have learned, but might have forgotten if it was a while ago):
$$(a-b)^2 = a^2 - 2ab + b^2$$
$$(a+b)^2 = a^2 + 2ab + b^2$$
$$(a+b)(a-b) = a^2 -b^2$$
These things come up over and over, you need to practice so that you spot them easily.
As for the discriminant: First, I check to see if the expression is a complete square. If it is not, then I factor it using the quadratic equation. You can also factor by completing the square, but that is just more writing. It is good to know both ways though, since when you work with only variables it makes more sense to complete (as a way to derive the quadratic equation from Ax^2 + Bx + C, for example).
I assume you remember what the terms I used mean, if you don't then you probably need to backtrack a bit to freshen your memory on earlier topics.
k
Thanks for the help guys will give it a try and return with an answer
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Physics MCQs Questions class 9 Force and laws of Motion with answers
Ace the MCQ on Force and laws of Motion with ease by getting all your lingering doubts cleared effectively. Also, if you are looking for a more detailed questions on Force and laws of Motion Class 9 MCQs, enroll into our free live online classes and avail detailed study notes along with the answers to MCQs on Force and laws of Motion.
Important points to remember about Force and laws of Motion
• motion of an object along a straight line in terms of its position, velocity and acceleration.
• We saw that such a motion can be uniform or non-uniform
• First law of motion: An object continues to be in a state of rest or of uniform motion along a straight line unless acted upon by an unbalanced force.
• The natural tendency of objects to resist a change in their state of rest or of uniform motion is called inertia.
• The mass of an object is a measure of its inertia. Its SI unit is kilogram (kg).
• Force of friction always opposes motion of objects.
• Second law of motion: The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
• The SI unit of force is kg m s–2. This is also known as newton and represented by the symbol N.
• A force of one newton produces an acceleration of 1 m s–2 on an object of mass 1 kg.
• The momentum of an object is the product of its mass and velocity and has the same direction as that of the velocity. Its SI unit is kg m s–1
• Third law of motion: To every action, there is an equal and opposite reaction and they act on two different bodies.
• In an isolated system (where there is no external force), the total momentum remains conserved.
• The law of conservation of momentum has been deduced from large number of observations and experiments.
• This law was formulated nearly three centuries ago. It is interesting to note that not a single situation has been realised so far, which contradicts this law.
• Several experiences of every-day life can be explained on the basis of the law of conservation of momentum
MCQs On Force and laws of Motion pdf
1. According to the third law of motion, action and reaction
(a) always act on the same body
(b) always act on different bodies in opposite directions
(c) have same magnitude and directions
(d) act on either body at normal to each other
2. The inertia of an object tends to cause the object_______
(a) to increase its speed
(b) to decrease its speed
(c) to resist any change in its state of motion
(d) to decelerate due to friction
3. An object of mass 2 kg is sliding with a constant velocity of 4 ms-1 on a frictionless horizontal table. The force required to keep the object moving with the same velocity is_______
(a) 32 N
(b) 0 N
(c) 2 N
(d) 8 N
4. Force is the rate of change of _______.
(a) momentum
(b) velocity
(c) displacement
(d) acceleration
5. The average force acted on the object is ______.
(a) 0.1 N
(b) 0.003 N
(c) 3.3 N
(d) 3 N
6. In SI system, the gravitational unit of force is ________.
(a) kg m s-2
(b) kgf
(c) m/s
(d) metre
7. Which one of the following is not a vector quantity?
(a) Speed
(b) Momentum
(c) Impulse
(d) Force
8. If the velocity-time graph representing the motion of a body is parallel to time-axis, then the net force acting on the body is _____
(a) 10 N
(b) 100 N
(c) 1 N
(d) None of these
9. Newton’s first law introduces the concept of________
(a) Momentum
(b) Inertia
(c) Conservation of energy.
(d) Action and reaction.
10. Momentum of the object is depend on.
(a) Mass
(b) Velocity
(c) Mass and velocity
(d) Force.
11. Quantitative expression of force is given by________
(a) Newton’s second law of motion.
(b) Newton’s third law of motion.
(c) Newton’s first law of motion.
(d) Newton’s law of gravitation.
12. The relation between acceleration, mass and force is given by_______
(a) a x F = m
(b) F = m/a
(c) F x m = a
(d) a = F/m
13. A man throws a ball weighing 200 g vertically upwards with a speed of 10m/s. Its momentum at the highest point of its flight will be_________
(a) 2 kg. m/s
(b) 2000 kg.m/s
(c) Insufficient data to find the momentum.
(d) zero.
14. A mass M breaks into two pieces in the ratio 1 : 3 while at rest. If the heavier has a speed of v, the speed of the lighter is__________
(a) v
(b) 2v
(c) 3v
(d) 4v
15. A carpenter exerts a force of magnitude 1.5 N at right angles to the surface of a nail of mass 3 gram to drive it into the wood. If wood offers a resistive force of 0.6 N and the time of interaction of hammer and the nail 0.01 s, the depth through which the nail penetrates is_______
(a) 15m
(b) 1.5 cm
(c) 15 cm
(d) 1.5 mm
16. A water tanker filled up to 2/3 of its height is moving with a uniform speed. On sudden application of the brake, the water in the tank would_________
(a) move backward
(b) move forward
(c) be unaffected
(d) rise upwards
17. A passenger in a moving train tosses a coin which falls behind him. It means that motion of the train is_______
(a) accelerated
(b) uniform
(c) retarded
(d) along circular tracks
18. Due to an acceleration of 2 m/s2, the velocity of a body increases from 20 m/s to 30 m/s in a certain period. Find the displacement (in m) of the body in that period.
(a) 650
(b) 125
(c) 250
(d) 325
19. Upon catching a ball, a cricket fielder swings his hands backwards. The concept behind this is explained by________
(a) Newton’s first law of motion
(b) Newton’s second law of motion
(c) Newton’s third law of motion
(d) The law of inertia
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## Week 12 Topics
• Recursive functions
## Monday
### Finishing Sorting
Before moving on to new content, let’s wrap up a few things about sorting from last week.
### Recursion
Any function that sometimes calls itself is known as a recursive function. In most cases, recursion can be considered an alternative to iteration (loops). Instead of defining a loop structure, recursion defines a problem as a smaller problem until it reaches an end point. There are three requirements for a successful recursive function:
1. Base case: each function must have one or more base cases where the function does not make a recursive call to itself. Instead, the problem has reached its smallest point, and we can begin returning back the small solutions.
2. Recursive case: the function makes one (or more) calls to itself. The call must use a different argument to the function (or else we keep making the same recursive call over and over again!). Typically, the problem gets 1 smaller.
3. All possible recursive calls must be guaranteed to hit a base case. If you miss one, your program is susceptible to an infinite recursion bug!
#### Example: Print a Message
The `printmessage.py` program prints a given message `n` times. We’ve already seen how to solve this problem with a loop (this is known as an iterative solution, one that uses a loop to repeat). However, let’s explore how we can solve it without a loop using recursion.
To think about how to define printing a message `n` times in a recursive way, we want to define printing in terms of printing. In other words we want to think of a recursive definition of printing n things, and our recursive function’s structure will follow this with it recursive function calls:
``````printmessage(msg, num_times) is equivalent to:
1. print(msg)
2. printmessage(msg, num_times-1) # here is our recursive definition``````
To understand what’s happening with recursion, it’s often helpful to draw a stack diagram. The key concept is that Python treats the recursive function call the same as any other function call — it places a new stack frame on top and begins executing the new function. The only difference is the function in this case has the same name.
#### Example: Summing Integers
In `recursivefuncs.py` we are going to write some iterative and recursive versions of the same function. Iterative versions use loops, recursive versions do not use loops, and instead contain calls to themselves but on a smaller problem.
I’ve given you the code for an algorithm we have seen before — a `for` loop that sums all integers between `1` and `n`:
``````def iterative_sum(n):
sum = 0
for i in range(1, n+1):
sum = sum + i
return sum``````
Together, we’ll write a recursive function that accomplishes the same task. Here are some questions to consider:
1. How can you define Sum of `n` values recursively (define Sum in terms of Sum)?
2. What does the recursive call look like?
3. What is the base case?
Test out your solution on different values calling both the iterative and your recursive version in `main`. You can print out their return values to check that they return the same values for the same passed argument value. (and remember that `CNTRL-C` will kill a program stuck in an infinite loop (or stuck in infinite recursion)).
After getting your `recursiveSum` to work, next try implementing an iterative (with a loop) and a recursive version of a function that computes $n!$ (n factorial).
$n! = n \cdot (n-1) \cdot (n-2) \cdot \ldots \cdot 1 \\ 3! = 3 \cdot 2 \cdot 1 = 6 \\ 5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120 \\$
Add some calls in `main` to your functions to test them out.
### Command line arguments
Often time programers write programs that have command line arguments, values that are included in the command line to run the program. For example,
``$python3 cmdlineargs.py 13 4.5 Freya`` In this example, the program `13`, `4.5`, and `"Freya"` are values that are passed into the program `cmdlineargs.py`. In the program can use the values of these command line arguments by getting their values through the system variable `sys.argv` which is a list of string values, one per command line argument. ``````sys.argv[0] is "cmdlineargs.py" # the name of the program is always element 0 sys.argv[1] is "13" sys.argv[2] is "4.5" sys.argv[3] is "Freya"`````` Let’s look at `cmdlineargs.py` and see what it is doing. Then try running it with different command line arguments. ## Wednesday ### Tips for Recursion For any recursive problem, think of these questions: 1. What is the base case? In what instances is the problem small enough that we can solve the problem directly? Sometimes you may need more than one base case, but rarely do you need several. 2. What is the recursive call? Make one or calls in the function to itself. It is very important to use different arguments (usually "smaller") in the recursive call otherwise you obtain infinite recursion. 3. Do all possible chains of recursion lead to a base case? 4. If a recursive function is supposed to return a value, make sure all cases return a value and that recursive calls do something with the return value of the smaller problem before returning. ### Recursion on Strings and Lists We are going to look at recursive functions on strings and lists. #### Slicing (creating substrings and sublists) Before moving on to more recursion examples, we’ll briefly remind you about a Python operator called slicing that can be particularly useful when working with recursion. Slicing lets you easily look at subsets of strings or lists. The general syntax is `mystr[i:j]`. This creates a new string that is the substring of `mystr` starting at index `i` and running up to but not including index `j`. ```$ python3
>>> hw = "hello world"
>>> hw[1]
'e'
>>> hw[7:10]
'orl'
>>> hw[0:5]
'hello'
>>> hw[6:]
'world'
>>> hw[:5]
'hello'```
You can also omit `i` or `j`. `mystr[i:]` creates the substring starting at index `i` and running to the end of that string. `mystr[:j]` gives you the substring up to but not including the `j-th` character.
Note: slicing can also be used with lists e.g. if `lst = [1,2,3,4,5]`, then `lst[2:4]` creates a new list `[3,4]`:
``````lst = [1,2,3,4,5]
sublst = lst[2:4] # sublist refers to new list [3,4]``````
Let’s look at `string_rec.py` at the `print_string` function to see what it is doing.
### Exercise: String Length
Try implementing `string_len_rec` function that recursively computes the length of the passed string values (this could be how Python implements the `len` function).
### Exercise: Number of Even Values
Next, we’ll use recursion to work with lists. In `numevens_rec.py`, write an iterative and a recursive version of a function that takes a list of int values and returns the number of values in the list that are even.
Remember that the mod operator (`%`) can be used to test for evenness"
`````` 10 % 2 # is 0 the remainder of dividing 10 by 2
11 % 2 # is 1 the remainder of dividing 11 by 2
12 % 2 # is 0 the remainder of dividing 12 by 2
...``````
## Friday
### Recursions on Lists (and strings) the better way)
Recursion that we have done so far on lists and strings, creates new sublists or substrings on every recursive call. For large-sized lists, this uses a lot of space (e.g., starting with a list with 1,000 element, first recursive call creates a new sublist with with 999 elements, the next create a new sublist with 998 elements, …, last call creates an empty list).
In addition, because list elements can be modified in-place, passing a sublist to a recursive function mean that the function cannot directly modify the elements in the list that is passed to it as an argument.
To solve both of these problems, we can modify the recursive function to recurse in-place on the list (and we can recurse in-place on strings too in a similar way).
A recursive function that recurses in-place on a passed list (or string), takes one additional parameter: the next index value into the list that specifies the list element on which to perform the function’s action.
The first call to this type of recursive function typically passes in `0` or `n-1` (where `n` is the length of the list), and then either increments or decrements the index value passed to recursive calls (depending on which order the list elements are processed (low to high or high to low)). The recursion typically stops when the index is out of bounds.
For example, to compute the sum of the element in a list using in-place recursion, the function definition looks like:
``def rec_sum(lst, index):``
In the file, `list_rec.py` let’s look at this example and try it out. Draw the stack to see what it is doing.
#### Exercise
Add a recursive that squares all the values in a passed list.
``````def rec_square(lst, index):
"""
squares all the values in the list of values
lst: list of numeric values
index: the index of the element to square next
returns: None (this function has side-effects)
"""``````
Add some calls to your recursive function from `main` to test it out.
A Code Design Aside
This is not important for code you write in this class, however, often times when designing code that may be part of a library of functions, a programmer wants to hide the details of an in-place recursive implementation of the function from the user (i.e., the programmer using our library should not need to pass in a starting index value…that just seems weird).
To implement a cleaner interface to users of our function, while still implementing it using in-place recursion on a list, we implement a wrapper-function around our recursive function. The wrapper-function is the nice interface to the user, that then just makes a call to the recursive function that does the actual work. For example:
``````def square_list(lst):
"""
A library function that squares all the values in the passed list
lst: a list of int values
returns: None
"""
return rec_square(lst, 0) # hides the recursive function implementation``````
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# If `lim_(x to a) f(x)=lim_(x to a) [f(x)] and f(x) ` is non-constant continuous function, where [.] denotes the greatest integer function, then
Updated On: 16-1-2020
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Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Text Solution
A
f(x) is an integer
B
f(x) is not an integer
C
f(x) has a local maximum at x=a
D
f(x) has a local minimum at x=a
A::D
Solution
We observe that
exists when f(x) increases (or decreases) in the left neighbourd of x=a and decreases (or increases) in the right neighbourhood of x=a <br>
<br>
<br>
f(x) decreases in the left neighborhood of x=a and increases in the right neighborhood of x=a and f(a) is an integer. <br> Hence,
f(x) is an integer and f(x) has a local minimum at x=a.
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+0
# Find the area with work
0
76
1
Find the area with work
Guest Jun 14, 2017
#1
+4172
+2
area = (1/2) * base * height
Let's call the base 10 km .
Here, I just rotated it so that the base is at the bottom and added the pink line for the height.
sin θ = opposite / hypotenuse
sin 37º = height / 7 Multiply both sides of this equation by 7 .
7 sin 37º = height
area = (1/2) * base * height
area = (1/2) * (10) * (7 sin 37º)
area = 35 sin 37º Put this into a calculator to get that...
area ≈ 21.064 square km
hectictar Jun 14, 2017
Sort:
#1
+4172
+2
area = (1/2) * base * height
Let's call the base 10 km .
Here, I just rotated it so that the base is at the bottom and added the pink line for the height.
sin θ = opposite / hypotenuse
sin 37º = height / 7 Multiply both sides of this equation by 7 .
7 sin 37º = height
area = (1/2) * base * height
area = (1/2) * (10) * (7 sin 37º)
area = 35 sin 37º Put this into a calculator to get that...
area ≈ 21.064 square km
hectictar Jun 14, 2017
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confusion in RC circuit regarding initial conditions and capacitor voltage
Actually i was practicing circuit analysis with Differential Equations. This question has some confusing parts like-
1. In Steady State, t<0 switch is still open the 10mF capacitor is Charged to 5V and 0A current flows through it, now if a 8mF Uncharged capacitor(0V) is connected at time t=0 as show in picture then What will be Node voltage v(t) or voltage across both capacitors as they both are in parallel is it 0v or 5v? Because if the voltage at node A for t>0 is at 5v that means 8mF Cap. charged instantaneously and if the node voltage becomes same as the uncharged cap i.e. 0v then also 10mF cap goes from 5 to 0. someone please clarify
2. There are two Initial Conditions, for 10mF v(0⁻) = 5v and for 8mF v(0⁻) = 0v, and the Variable is v(t), so value should i put in the end ? and i m getting different voltage expression for each initial condition. here's my work
$$\Rightarrow \dfrac{v\left( t\right) }{8}+10m\cdot \dfrac{dv\left( t\right) }{dt}+8m\cdot\dfrac{dv\left( t\right) }{dt}-\dfrac{5}{8}=0 ,\hspace{1cm} By \space KCL$$ $$\Rightarrow 18m\dfrac{dv\left( t\right) }{dt}+\dfrac{v}{8}=\dfrac{5}{8}$$ $$v\left( t\right) =A⋅e^{\dfrac{-t}{144⋅10^{-3}}}+5$$
It would we be really helpful for me if someone will give me some hints or physical reasoning or teach me how to handle these types of circuit situation instead of Direct Answer.
• (1) it's indeterminate due to infinite current flow Feb 9 at 13:58
• This circuit can not be solved analytically because it violates our definitions of how capacitors behave and the meaning of parallel circuits. Connecting two ideal capacitors in parallel when they have different initial voltages requires infinite current for zero time. Feb 9 at 14:22
• @ElliotAlderson Sir, could you please explain the this line "requires infinite current for zero time", what does it means ? Feb 9 at 14:43
• @Andyaka, sir could you please elaborate it little bit ? and also is it solvable, Means can we find expression for i(t) and v(t) ? Feb 9 at 14:47
• The current through a capacitor is proportional to the derivative of the voltage across it. What happens to the voltages across two capacitors that initially have different voltages and then you connect them in parallel. What is $dV/dt$ in this case? Feb 9 at 15:32
Putting infinite instanteneous current between caps aside, the question is probably about charge redistributing between the two capacitors. So the question is really like "here is 5V cap of 10mF, how does the charge redistribute if we attach another 8mF cap to it.
We want to find total charge on capacitor C1, which will be the same as total charge on capacitors C1 and C2 once C2 is connected.
The formula for it says Q = C x V.
Two capacitors in parallel is just one larger capacitor. Which will have the same total charge as before.
C1 x V1 = Q = (C1 + C2) x V2
I hope it's obvious why voltage on two caps is the same when they're connected in parallel.
This charge distribution is valid strictly for time moment t=0, at t>0 the capacitors will be gaining extra charge via the resistor, where you have a typical RC-circuit, simply with starting condition that the capacitor is not at 0V. It will charge from V2 to 5V eventually. But I don't see explicit question about t>0, only about t=0.
Edit: perfect components, perfect connections assumed. "Spherical capacitors in a vacuum"
• ok so for t>0 how should i proceed? find differential equation by Kcl then solve it and the what condition should i apply ? please help Feb 10 at 9:14
• @fpsshubham it's simple RC circuit with IC formulae you can find anywhere. Remember that voltage is relative. Charging from 0V to 5V is IDENTICAL to the last letter and digit to charging from 995V to 1000V and from -34V to -29V. In this case, when at t=0 voltage on caps immediately drops, you will have a classic RC circuit with capacitor(s) charging from V2 to 5V.
– Ilya
Feb 10 at 9:16
• Here you can find detailed explanation as well as the formula: electronics-tutorials.ws/rc/rc_1.html Remember to adjust for charging not from 0 to 5V, but from V2 to 5V.
– Ilya
Feb 10 at 9:18
• ok i'll give it a try! Feb 10 at 9:21
• Let me know what your V2 is and how you apply the RC formula. I did some napkin math.
– Ilya
Feb 10 at 9:22
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## Why does length contraction only occur parallel to the direction of motion?
Maybe other answers say the same in a more technical manner, maybe not, I have not checked. In any case, I will share the explanation I usually give myself.
If observers measure different values for a given property, it is because they do it from different perspectives or, more technically, reference frames, that is to say, situations/circumstances which have an impact on the measurement process of the relevant property. This definition entails that observers should not disagree, however, if they come to measure another property with regard to which their circumstances are identical, that is to say, with respect to which their perspective or reference frame is the same.
Take the easy example of two persons standing on the ground and looking at each other from a distance. A sees B smaller and vice versa. That is because A looks at B from a distance and vice versa. They have different perspectives on each other, in this respect. But they do see the same distance between the the two of them, for they have the same perspective on this issue.
The same happens with length contraction. A and B have different states of motion, but only in one direction. Hence they have a different perspective in this respect, i.e. in terms of measuring this property, length in the direction of their relative motion, say the X axis. However, they do not have different states of motion in the Y axis, for example. In this respect they share the same perspective; for this purpose, they occupy the same reference frame; hence they measure the same values.
Ok so thankyou all for the help, I think I'm nearly there. I think I may have a clear understanding, is this a correct explination..... If I have a light clock on a fast moving spacecraft being observed from earth, with a vertical set of mirrors (perpendicular to the motion of the overall clock) and a horizontal set of mirrors. The mirrors time the pulses of light, eminating from the same source. I then apply the equations for calculating the time it takes for the light pulses to bounce between each set of mirrors using the values of c = 10, v = 6 and l = 4. I find initially that the horizontal bounces take 1.25 seconds and the vertical bounces take only 1 second. It's only when I apply the lorenz transformation to the length in the horizontal clock to give me new decreased value for l of 3.2, that I discover my time calculations for both mirrors agree on 1 second. I then conclude that the length must contract around the moving pulse of light to preserve its consistancy for all frames of reference. Is this correct?
Quote by peterspencers Ok so thankyou all for the help, I think I'm nearly there. I think I may have a clear understanding, is this a correct explination..... If I have a light clock on a fast moving spacecraft being observed from earth, with a vertical set of mirrors (perpendicular to the motion of the overall clock) and a horizontal set of mirrors. The mirrors time the pulses of light, eminating from the same source. I then apply the equations for calculating the time it takes for the light pulses to bounce between each set of mirrors using the values of c = 10, v = 6 and l = 4. I find initially that the horizontal bounces take 1.25 seconds and the vertical bounces take only 1 second. It's only when I apply the lorenz transformation to the length in the horizontal clock to give me new decreased value for l of 3.2, that I discover my time calculations for both mirrors agree on 1 second. I then conclude that the length must contract around the moving pulse of light to preserve its consistancy for all frames of reference. Is this correct?
Yes but it's insufficient: the Lorentz transformations state that there is no vertical contraction, so that is what you want to prove (or make plausible). You demonstrated that based on the starting assumptions (equal speed of light etc), the vertical contraction factor must differ from the horizontal one. However, you could assume, for example, that the vertical contraction is gamma and the horizontal contraction gamma square. Then with zero time dilation your calculation will also work.
However (in addition to other examples already given), imagine that two identical high objects collide; from SR symmetry they should have identical damage. Or alternatively, imagine a very fast bullet going through a narrow tube; it must not be possible to know which one "moves absolutely faster", and neither can it be that the bullet is smaller than the tube and also bigger than the tube when it passes through, so that a collision happens and also doesn't happen.
How will my calculation show length contraction with, zero time dillation? My calculation shows that time on the ship (ts) would be 0.8 and time from say earth (te) would be 1. Also why would I want to make vertical contraction possible? The vertical clock dosent contract, the path the photon takes is extended, as per a little pythagoras, hence the time dilation.
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Quote by peterspencers Ok so thankyou all for the help, I think I'm nearly there. I think I may have a clear understanding, is this a correct explination..... If I have a light clock on a fast moving spacecraft being observed from earth, with a vertical set of mirrors (perpendicular to the motion of the overall clock) and a horizontal set of mirrors. The mirrors time the pulses of light, eminating from the same source. I then apply the equations for calculating the time it takes for the light pulses to bounce between each set of mirrors using the values of c = 10, v = 6 and l = 4. I find initially that the horizontal bounces take 1.25 seconds and the vertical bounces take only 1 second. It's only when I apply the lorenz transformation to the length in the horizontal clock to give me new decreased value for l of 3.2, that I discover my time calculations for both mirrors agree on 1 second. I then conclude that the length must contract around the moving pulse of light to preserve its consistancy for all frames of reference. Is this correct?
I can't make sense of your setup. You haven't said what c, v and l are and you haven't said what their units are. Usually, we reserve the letter "c" to be the speed of light and "v" is a velocity. It's a little unusual for someone to set the speed of light to be 10, we usually make c = 1 to make the equations simpler. And you haven't said what equations you are using nor what l applies to.
Specifically, I can't figure out how you got 1 second for the vertical bounce.
Quote by peterspencers How will my calculation show length contraction with, zero time dillation? My calculation shows that time on the ship (ts) would be 0.8 and time from say earth (te) would be 1. Also why would I want to make vertical contraction possible? The vertical clock dosent contract, the path the photon takes is extended, as per a little pythagoras, hence the time dilation.
1. Your question is "Why does length contraction only occur parallel to the direction of motion?". Although you now suggest the contrary, I thought that you did not intend to discover what the Lorentz transformations state (contrary to valentin). As a matter of fact, they state that y=y' and z=z', so if that was your question and you did not want to prove what you assumed, then that was the answer and the end of this topic.
2. With zero time dilation instead of time dilation by a factor gamma, you can search if it is possible to obtain the same return times with your setup. Then you will find that this is possible if the length is decreased by a factor gamma square and the width and height by a factor gamma. If you don't get that, then you made a calculation error.
3. In my post I explained that that solution is nevertheless not an option if we want the PoR to hold, so that we assume that length contraction only occurs parallel to the direction of motion. Lorentz and Einstein gave other examples with the same conclusion.
I apologise for not showing my workings, here they are: I have a light clock onboard a spacecraft moving past the earth parallel to an observer. The lightclock measures the time it takes for light waves to bounce between two mirriors. I have two sets of mirriors, one on the vertical axis perpendicular to the direction of travel and also a horizontal set, in line with the direction of travel. c = 10 m/s (speed of light) v = 6 m/s (velocity) l = 4 m ( length between two mirrors) Firstly I take a time measurement onboard the craft: $ts$ = 2l/c = 0.8 Then from earth's reference frame I calculate the time by using the following two equations, for the vertical clock I use: $te$ = $\frac{ts}{√1-vv/cc}$ (please excuse my writing vv/cc, I mean v2/c2 however when I enter [sup[/sup] inside the fraction text it donsent seem to work :P im totally new to all this, trying to work it out as I go along, any help would be most kind) $te$ = 1 second Then for the horizontal clock: $te$ = $\frac{2lc}{cc-vv}$ (apologies again the bottom half should read c2-v2) $te$ = 1.25 seconds .....clearly there is something wrong here, both clocks shold agree on the time. So I apply the lorentz transformation to l in the horizontal clock: $Lo$ = the proper length (the length between the mirrors in their rest frame) $L$ = $Lo$√1-v2/c2 and end up with a value of 3.2 for l in the horizontal clock.... so I go back to $te$ = $\frac{2lc}{cc-vv}$ (apologies again the bottom half should read c2-v2) this time with 3.2 as my value for l, and then I get: $te$ = 1 second This means that the amount of distance the light can cover between each set of mirrors is equal, so even though the length is contracted in one clock, the 'light distance' is equal, as it is in both the rest frame and the moving frame. Now both clocks agree and i'm very happy :) .... I hope! Is this correct ??
Yes but it's insufficient: the Lorentz transformations state that there is no vertical contraction, so that is what you want to prove (or make plausible). You demonstrated that based on the starting assumptions (equal speed of light etc), the vertical contraction factor must differ from the horizontal one. However, you could assume, for example, that the vertical contraction is gamma and the horizontal contraction gamma square. Then with zero time dilation your calculation will also work.
Why is my above calculation insufficient to explain length contraction? I dont follow the reasoning here, please could someone explain (to a lamen) if the above quote really does apply to my above calculation.
Quote by peterspencers Why is my above calculation insufficient to explain length contraction? I dont follow the reasoning here, please could someone explain (to a lamen) if the above quote really does apply to my above calculation.
The Lorentz transformations tell you directly that there is no length contraction parallel to the direction of motion; to find that out you don't need to make such a calculation. However, your question was why length contraction only occurs parallel to the direction of motion.
I interpreted your question as "why are the Lorentz transformations the only feasible solution", while you seem to have meant "how do the Lorentz transformation work".
- Your calculation is fine to show how length contraction is a necessary element of the Lorentz transformations, and how that works.
- Your calculation doesn't show why that is the only feasible possibility based on the postulates.
Once more: if you assume that there is no time dilation but all distances contract by an additional factor γ (so that the length contracts by γ2 and the width by a factor γ), then your scenario will also work (the 'light distance' in both directions as measured with a clock is equal and the same in the rest frame and the moving frame).
Quote by peterspencers why does space time not contract uniformly in every direction around a fast moving object?
One can use the Robertson-Mansouri-Sexl test theory, which shows that three experiments are necessary to derive the following parameter of the Lorentz transformation:
$\alpha$ for time changes
$\beta$ for longitudinal length changes
$\delta$ for transverse length changes
The Michelson-Morley experiment measures the combination of $\beta$ and $\delta$.
The Kennedy-Thorndike experiment measures the combination of $\alpha$ and $\beta$.
In order to obtain the individual values, you have to measure one of those quantities directly.
For instance, the Ives-Stilwell experiment measures $\alpha$ in accordance with time dilation. Using this value of $\alpha$ with Kennedy-Thorndike shows that $\beta$ must be in accordance with relativistic length contraction in longitudinal direction; combining this value of $\beta$ with Michelson-Morley shows that $\delta$ must be zero.
Therefore, length contraction in transverse direction is experimentally excluded - only relativistic length contraction in longitudinal direction is allowed.
Quote by harrylin - Your calculation doesn't show why that is the only feasible possibility based on the postulates. Once more: if you assume that there is no time dilation but all distances contract by an additional factor γ (so that the length contracts by γ2 and the width by a factor γ), then your scenario will also work (the 'light distance' in both directions as measured with a clock is equal and the same in the rest frame and the moving frame).
In my scenario I have a rest observation and a moving one. Where the clock is being observed in motion, velocity is an obvious consideration. As soon as velocity is acting upon the clock the equations for calculating the time change and time dilation becomes an inevitable and unavoidable factor. This in turn leads to the nesicarry appliance of length contraction along the horizontal axis to then preserve equal light time between both sets of mirriors in both frames of reference.
I can only assume no time dilation in the clocks rest frame, in this instance all the prerequisites are met without the need to contract any lengths.
It sounds like you are describing a totally different situation, and have misinterpreted my scenario, in which case I apologise for my poor initial explanation. If I am still mistaken then please could you describe step by step using mathematics and my initial values, the situation you are describing. Many thanks
Quote by peterspencers In my scenario I have a rest observation and a moving one. Where the clock is being observed in motion, velocity is an obvious consideration. As soon as velocity is acting upon the clock the equations for calculating the time change and time dilation becomes an inevitable and unavoidable factor. [..]
I agree with that; however, based on what fact or assumption do you make that claim?
I clarified how you can get the same result from your scenario, based on the assumption of no time dilation (if you ask me to show how exactly, I'll gladly do that later; it seems rather obvious to me). The related transformations are of course different from the Lorentz transformations, according to which y'=y. Thus, please clarify if you agree with the part that you did not cite. What exactly did you intend with your "why" question?
Your explanation still dose not explain how 'assuming zero time dilation' is possible when observing my moving light clock. My scenario is one where we are making observations of a moving clock, therefore time dilation is an intrinsic part. Please can you show me your mathematical workings step by step using my initial values for l, c and v. Also an explination of how 'assuming zero time dilation' is possible in a situation where we are observing a moving clock. And how the prerequisites of the properties of light are met. In answer to your questions, I make the claim of TD being unavoidable where the clock is moving based on the fact that the motion of the overall clock increases the path the photon travels to complete one bounce between both mirrors. As the speed of light is constant this action theifore takes longer than it does in the rest frame, hence the time dilation. And, my initial 'why' question, I believe, has been answered through my calculations. This has also been verified in another thread (which you have picked up) so I am now attempting to understand your claim that my scenario is insufficient to completely describe the answer to my initial 'why' question.
Quote by peterspencers In my scenario 'assuming no time dilation' is not an option! My scenario is one where we are making observations of a moving clock, therefore time dilation is an intrinsic part.
Then I may have misunderstood your scenario, as I think that you describe two systems in which clocks are observed that are in rest in each system. Please check.
Please can you show me your mathematical workings step by step using my initial values for l, c and v. Also an explination of how 'assuming zero time dilation' is possible in a situation where we are observing a moving clock. [..]
Yes, I will, as I promised, after you reply my repeated question to you so as to be sure that we are "tuned to the same frequency". So, for the third time: please clarify if you agree with the part that you did not cite, here it is again:
Your question was why length contraction only occurs parallel to the direction of motion.
I interpreted your question as "why are the Lorentz transformations the only feasible solution", while you seem to have meant "how do the Lorentz transformation work". Correct?
I believe that my initial 'why' question has been answered through my calculations, this has been confirmed to me by someone else in another post. I am now attempting to vindicate that explination. My question may have been indirectly about the Lorentz transformation, I am still only aware of what it is and how it works so much as my scenario and calculations betray.
I am fascinated at the thought of there being other ways to mathematically and intuitively describe and explain the reasons for length contraction. Before I begin along that line of questioning however I must first clarify if my current understanding (based upon my scenario) is sufficient. As you have stated that it is not....
Then I may have misunderstood your scenario, as I think that you describe two systems in which clocks are observed that are in rest in each system. Please check.
No my scenario does not describe 2 clocks that are at rest, why would I have included v (velocity) in my equations if that were the case?
I hope I have answered you questions sufficiently, I must ask again, in light of my attempt at clarification... if my scenario and calculations are sufficient to explain length contraction and time dilation?
Quote by peterspencers [..] I am fascinated at the thought of there being other ways to mathematically and intuitively describe and explain the reasons for length contraction. Before I begin along that line of questioning however I must first clarify if my current understanding (based upon my scenario) is sufficient. As you have stated that it is not... [...]
It's sufficient to understand how the Lorentz transformations work. It's not sufficient to determine that for relativity the Lorentz transformations are the only solution. And apart of the relativity principle there is also a physical reason for length contraction (yes we could discuss that later!).
No my scenario does not describe 2 clocks that are at rest, why would I have included v (velocity) in my equations if that were the case?
I did not think that you describe 2 clocks that are in rest, so probably I did understand your scenario which looks to me the "standard" one.
I hope I have answered you questions sufficiently, I must ask again, in light of my attempt at clarification... if my scenario and calculations are sufficient to explain length contraction and time dilation?
Yes that's sufficient to get a basic understanding of how length contraction and time dilation work. And here's your calculation redone for an alternative solution* (I simply copy-pasted from you with a few modifications) :
A light clock onboard a spacecraft moves past the earth parallel to an observer. The lightclock measures the time it takes for light waves to bounce between two mirrors. There are two sets of mirrors, one on the vertical axis perpendicular to the direction of travel and also a horizontal set, in line with the direction of travel.
c = 10 m/s (speed of light)
v = 6 m/s (velocity)
l = 4 m ( length between two mirrors)
A time measurement onboard the craft should yield according to the craft's reference frame:
$ts$ = 2l/c = 0.8 s
Then from earth's reference frame we calculate the time (uncorrected for relativity) by using the following two equations, for the vertical clock we use:
$te$ = $\frac{ts}{√1-v^2/c^2}$
$te$ = 1 second
Then for the horizontal clock:
$te$ = $\frac{2lc}{c^2-v^2}$
$te$ = 1.25 seconds
.....clearly there is something wrong here, both clocks should agree on the time in both directions.
We could for example propose a length transformation by a factor γ2 in the horizontal clock:
$Lo$ = the proper length (the length between the mirrors in their rest frame)
$L_h$ = $Lo$(1-v2/c2)
and end up with a value of 2.56 for Lh in the horizontal clock....
so we go back to $te$ = $\frac{2lc}{c^2-v^2}$
this time with 2.56 as our value for Lh, and then we get:
$te$ = 1 second
Similarly for the vertical clock we can propose a length transformation by a factor γ:
$L_v$ = $Lo$(√1-v2/c2)
now with 3.2 as our value for Lv, and then we get:
$te$ = 1 second
This means that the 'light distance' in both directions as measured with a clock is equal and the same in the rest frame and the moving frame. And this solution works with a time dilation factor of 1 (no time dilation).
*based on that scenario there's in fact an infinite number of solutions, indicated with the multiplication factor l; see equation 1 of http://en.wikisource.org/wiki/On_the...ron_%28June%29
Quote by harrylin Then for the horizontal clock: $te$ = $\frac{2lc}{c^2-v^2}$ $te$ = 1.25 seconds
There's been a mistake here. The formula I found for $te$ is $te$ = $\frac{2l}{\sqrt{c^2-v^2}}$. It gives $te$ = 1 second for this case.
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# Illustrative Mathematics Grade 8, Unit 5, Lesson 13: The Volume of a Cylinder
Learning Targets:
• I can find the volume of a cylinder in mathematical and real-world situations.
• I know the formula for volume of a cylinder.
Related Pages
Illustrative Math
#### Lesson 13: The Volume of a Cylinder
Let’s explore cylinders and their volumes.
Illustrative Math Unit 8.5, Lesson 13 (printable worksheets)
#### Lesson 13 Summary
We can find the volume of a cylinder with radius and height using two ideas we’ve seen before:
• The volume of a rectangular prism is a result of multiplying the area of its base by its height.
• The base of the cylinder is a circle with radius r, so the base area is πr2.
Remember that is the number we get when we divide the circumference of any circle by its diameter. The value of π is approximately 3.14.
Just like a rectangular prism, the volume of a cylinder is the area of the base times the height. For example, take a cylinder whose radius is 2 cm and whose height is 5 cm.
The base has an area of 4π cm2 (since π · 22 = 4π), so the volume is 20π cm3 (since 4π · 5 = 20π). Using 3.14 as an approximation for π, we can say that the volume of the cylinder is approximately 62.8 cm3.
In general, the base of a cylinder with radius r units has area πr2 square units. If the height is h units, then the volume V in cubic units is
V = πr2h
#### Lesson 13.1 A Circle’s Dimensions
Here is a circle. Points A, B, C, and D are drawn, as well as Segments AD and BC.
1. What is the area of the circle, in square units? Select all that apply.
2. If the area of a circle is 49π square units, what is its radius? Explain your reasoning.
#### Lesson 13.2 Circular Volumes
What is the volume of each figure, in cubic units? Even if you aren’t sure, make a reasonable guess.
1. Figure A: A rectangular prism whose base has an area of 16 square units and whose height is 3 units.
2. Figure B: A cylinder whose base has an area of 16π square units and whose height is 1 unit.
3. Figure C: A cylinder whose base has an area of 16π square units and whose height is 3 units.
#### Are you ready for more?
Here are solids that are related by a common measurement. In each of these solids, the distance from the center of the base to the furthest edge of the base is 1 unit, and the height of the solid is 5 units. Use 3.14 as an approximation for π to solve these problems.
1. Find the area of the square base and the circular base.
Area of square base = √2 · √2 = 2
Area of circular base = π12 = 1π
2. Use these areas to compute the volumes of the rectangular prism and the cylinder. How do they compare?
Volume of rectangular prism = 2 · 5 = 10
Volume of cylinder = 1π · 5 = 5π = 15.71
3. Without doing any calculations, list the figures from smallest to largest by volume. Use the images and your knowledge of polygons to explain your reasoning.
4. The area of the hexagon is approximately 2.6 square units, and the area of the octagon is approximately 2.83 square units. Use these areas to compute the volumes of the prisms with the hexagon and octagon bases. How does this match your explanation to the previous question?
Volume of hexagonal prism = 2.6 · 5 = 13
Volume of octagonal prism = 2.83 · 5 = 14.15
#### Lesson 13.3 A Cylinder’s Dimensions
1. For cylinders A–D, sketch a radius and the height. Label the radius with an r and the height with an h.
2. Earlier you learned how to sketch a cylinder. Sketch cylinders for E and F and label each one’s radius and height.
#### Lesson 13.4 A Cylinder’s Volume
1. Here is a cylinder with height 4 units and diameter 10 units. a. Shade the cylinder’s base. b. What is the area of the cylinder’s base? Express your answer in terms of π. c. What is the volume of this cylinder? Express your answer in terms of π.
2. A silo is a cylindrical container that is used on farms to hold large amounts of goods, such as grain. On a particular farm, a silo has a height of 18 feet and diameter of 6 feet. Make a sketch of this silo and label its height and radius. How many cubic feet of grain can this silo hold? Use 3.14 as an approximation for π.
#### Are you ready for more?
One way to construct a cylinder is to take a rectangle (for example, a piece of paper), curl two opposite edges together, and glue them in place.
Which would give the cylinder with the greater volume: Gluing the two dashed edges together, or gluing the two solid edges together?
Gluing the two solid edges would give a cylinder with a greater volume
#### Lesson 13 Practice Problems
1. Match each set of information about a circle with the area of that circle.
2. a. Sketch a cylinder.
b. Label its radius 3 and its height 10.
c. Shade in one of its bases.
3. At a farm, animals are fed bales of hay and buckets of grain. Each bale of hay is in the shape a rectangular prism. The base has side lengths 2 feet and 3 feet, and the height is 5 feet. Each bucket of grain is a cylinder with a diameter of 3 feet. The height of the bucket is 5 feet, the same as the height of the bale.
a. Which is larger in area, the rectangular base of the bale or the circular base of the bucket? Explain how you know.
b. Which is larger in volume, the bale or the bucket? Explain how you know.
4. Three cylinders have a height of 8 cm. Cylinder 1 has a radius of 1 cm. Cylinder 2 has a radius of 2 cm. Cylinder 3 has a radius of 3 cm. Find the volume of each cylinder.
5. A one-quart container of tomato soup is shaped like a rectangular prism. A soup bowl shaped like a hemisphere can hold 8 oz of liquid. How many bowls will the soup container fill? Recall that 1 quart is equivalent to 32 fluid ounces (oz).
6. Two students join a puzzle solving club and get faster at finishing the puzzles as they get more practice. Student A improves their times faster than Student B.
a. Match the students to the Lines l and m.
b. Which student was faster at puzzle solving before practice?
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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## solution for integration of following expression.
on 25 Sep 2012
### Babak (view profile)
can any one solve this integration?
integration of ((exp(a*x))/(1+b*exp(c*x)));
on 25 Sep 2012
Do you want the indefinite integral, or the integral over a certain range, or what?
Siva Malla
### Siva Malla (view profile)
on 26 Sep 2012
I want indefinite integral
on 26 Sep 2012
Then you will have to look to Mathematica.
### Babak (view profile)
on 25 Sep 2012
A general form for the indefinite integral of your problem does not exist.
Take y = exp(a*x) and transform the integral over x to an integral over y. It will be the integral of
` 1/a* 1/(1+b*y^(c/a)) *dy`
depending on what the value of c/a is, a general form for the integral may/may not exist. For example, for c/a=1, the result is
` 1/(a*b)* log(1+b*y)`
but for c/a=2, b>0, the integral will be
` sqrt(b)/a*Arctan(sqrt(b)*y)`
So I don't think you can get a general form of the integral from the Symbolic Math Toolbox or any other Symbolic Math Software. You can use the numerical integrations methods and integrate it over a definite domain.
#### 1 Comment
on 25 Sep 2012
I am not sure that is what the OP asked for here. I asked above for clarification, but got none.
### Azzi Abdelmalek (view profile)
on 25 Sep 2012
``` syms x
% you must assign values to a b and c to find result
a=1;b=1;c=1;
y=((exp(a*x))/(1+b*exp(c*x)))
inty=int(y)```
Show 1 older comment
Azzi Abdelmalek
### Azzi Abdelmalek (view profile)
on 25 Sep 2012
` inty=log(exp(x) + 1)`
Babak
### Babak (view profile)
on 25 Sep 2012
Thanks! It confirms the result of
` 1/(a*b)* log(1+b*y)`
for the case where c/a=1 in my answer above. I don't think MATLAB can do the integral when a, b and c are all syms though...
on 25 Sep 2012
No, but Mathematica can:
Result
Apply Today
Play today
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# physical science
A spider hangs by its string at an eye level 20 cm in front of the plane mirror. You are behind the spider, 50 cm from the mirror. What is the distance betwen your eye and the image of the spider in the miror?
The spider's image is 20 m behind the mirror. Add 50 cm (your eye-mirror distance) to that.
1. 👍 0
2. 👎 0
3. 👁 180
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# 640 Fall Factor Trees and Level 5
Because it is fall, and 640 has many factors, I decided to make factor trees using fall colors. Get out your rakes!
There are many other possible factor trees for 640, but raking leaves can be a lot of work, so I only made two of them.
640 is the hypotenuse of the Pythagorean triple 384-512-640.
OEIS.org informs us that 640 = 16!!!!!!, but if you type 16!!!!!! into a calculator, you will get an error message as soon as you type !!.
16!!!!!! ≠ (((((16!)!)!)!)!)!
There are 6 !’s so 16!!!!!! = 16(16-6)(16-12) = 16 x 10 x 4.
Here is this week’s Level 5 puzzle:
Print the puzzles or type the solution on this excel file: 12 Factors 2015-10-05
—————————————————————————————————
• 640 is a composite number.
• Prime factorization: 640 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5, which can be written 640 = (2^7) x 5
• The exponents in the prime factorization are 7 and 1. Adding one to each and multiplying we get (7 + 1)(1 + 1) = 8 x 2 = 16. Therefore 640 has exactly 16 factors.
• Factors of 640: 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 320, 640
• Factor pairs: 640 = 1 x 640, 2 x 320, 4 x 160, 5 x 128, 8 x 80, 10 x 64, 16 x 40, or 20 x 32
• Taking the factor pair with the largest square number factor, we get √640 = (√64)(√10) = 8√10 ≈ 25.298221
—————————————————————————————————
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# doubt in Solving an inequality
I took upon following question-
Find the values of $$m$$ for which given equation has real roots
$$\sin^2 x-(m-3)\sin x+m=0$$
So I started by first satisfying $$D>=0$$ and found that $$m$$ should range from $$(-\infty,1] \cup [9,\infty)$$
However after this, I found roots by quadratic formula as
$$\sin x = (m-3)/2 \pm \sqrt{(m-1)(m-9)/4}$$
However this value should be within range of sin function and I'm unable to solve the inequality so formed for values of $$m$$.
How to solve the inequality( actually the $$\pm$$ term is confusing me)?
• What do you denote $D$? Commented May 22, 2021 at 9:23
• @Bernard Presumably the discriminant Commented May 22, 2021 at 9:28
• I think it means $b^2-4ac$ from the quadratic rules
– 00GB
Commented May 22, 2021 at 9:29
• $sin$ is supposed to be within the range $[-1, 1]$, so solve the inequality for that too. Commented May 22, 2021 at 9:30
• @ShubhamJohri: I guessed so, but I pointed that when one uses a non-standard notation, it has to be explained. Commented May 22, 2021 at 9:32
When $$m=1$$, both roots of the quadratic are $$-1$$ so a solution for $$x$$ exists. When $$m\in(-\infty,1)$$, the smaller root $$r_1=(m-3)/2-\frac12\sqrt{(m-1)(m-9)}<-1$$ as $$(m-3)/2<-1$$. So we should enforce that the larger root $$r_2=(m-3)/2+\frac12\sqrt{(m-1)(m-9)}\in[-1,1]$$, i.e.\begin{align*}&-1\le(m-3)/2+\frac12\sqrt{(1-m)(9-m)}\le1\\& \iff1-m\le\sqrt{(1-m)(9-m)}\le5-m\end{align*}Note that the first inequality is always true since $$9-m\ge1-m\ge0$$ and hence $$(1-m)(9-m)\ge(1-m)^2$$. The second inequality on squaring gives$$m^2-10m+9\le m^2+25-10m$$which is true. So all values of $$m\le1$$ will work.
When $$m\ge9$$, the larger root $$r_2\ge3$$. So we must enforce $$-1\le r_1\le 1$$, i.e. \begin{align*}&-1\le(m-3)/2-\frac12\sqrt{(m-1)(m-9)}\le1\\& \iff1-m\le-\sqrt{(m-1)(m-9)}\le5-m\\& \iff m-5\le\sqrt{(m-1)(m-9)}\le m-1 \end{align*} Correspondingly note that the second inequality is always true this time around. Squaring the first inequality would yield $$25\le9$$, which is never true. So no $$m\ge9$$ works.
Our final answer is $$m\le1.~\blacksquare$$
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# How do you do modulus in JavaScript?
## How do you do modulus in JavaScript?
To obtain a modulo in JavaScript, in place of a % n , use ((a % n ) + n ) % n .
## What is modulus operator in JavaScript with example?
Modulus is a related concept, but handles negative numbers differently. For example, -21 % 5 === -1 , because the remainder always takes the sign of the left number. However, a true modulus operator would always return a positive value, so 21 modulo 5 would equal 4.
What does modulo (%) do?
Modulo is a math operation that finds the remainder when one integer is divided by another. In writing, it is frequently abbreviated as mod, or represented by the symbol %. Where a is the dividend, b is the divisor (or modulus), and r is the remainder.
What is modulo operator example?
The modulo operation (abbreviated “mod”, or “%” in many programming languages) is the remainder when dividing. For example, “5 mod 3 = 2” which means 2 is the remainder when you divide 5 by 3.
### How do you get absolute value in JavaScript?
The Math. abs() function returns the absolute value of a number. That is, it returns x if x is positive or zero, and the negation of x if x is negative.
### How do you calculate modulo?
How to Do a Modulo Calculation. The modulo operation finds the remainder of a divided by b. To do this by hand just divide two numbers and note the remainder. If you needed to find 27 mod 6, divide 27 by 6.
What does modulo 8 mean?
The ring of integers mod 8. Addition and Multiplication in Mod 8. The ring of integers consists of the set {0, 1, 2, 3, 4, 5, 6, 7} where these eight numbers can be thought of being arranged in a circle as in the face of a clock: To add or multiply the numbers, simply follow the arrows and see where you land.
How do you make a number positive in JavaScript?
To convert a negative number to a positive one in JavaScript, use the abs() method in JavaScript. The method returns the absolute value of a number.
## How do you find the absolute value?
The absolute value of a number is the number’s distance from zero, which will always be a positive value. To find the absolute value of a number, drop the negative sign if there is one to make the number positive. For example, negative 4 would become 4.
## How do I return a string to a number?
7 ways to convert a String to Number in JavaScript
1. Using parseInt() parseInt() parses a string and returns a whole number.
2. Using Number() Number() can be used to convert JavaScript variables to numbers.
3. Using Unary Operator (+)
4. Using parseFloat()
5. Using Math.
6. Multiply with number.
7. Double tilde (~~) Operator.
How do you convert a negative number to positive in JavaScript?
An alternative approach is to use multiply by -1 . To convert a positive number to a negative number, multiply the number by -1 , e.g. 5 * -1 . By multiplying the number by -1 , we flip the number’s sign and get back a negative number. Copied!
How do you convert to positive?
Multiply with Minus One to Convert a Positive Number All you have to do just multiply a negative value with -1 and it will return the positive number instead of the negative.
### How do I convert a string to a number in JavaScript?
What is the use of modulus (%) in JavaScript?
The modulus (%) operator returns only the remainders so if either types of values like both positive and negative integers as well as the values like string types the attempt is made for to convert the string to the number formats.
What is 21 modulo 5 in JavaScript?
However, a true modulus operator would always return a positive value, so 21 modulo 5 would equal 4. In practice, you are unlikely to use the remainder operator on negative values, and many JavaScript developers aren’t aware of the difference.
## How to use modulo to calculate a list?
I think it is better to explain it this way: Modulo is the difference left when dividing a value. You can use this to calculate listing lists with items, for example: 10 % 10 gives you 0. When it is 0 you know there a 10 items in a list.
## What is the use of module operator in JavaScript?
Above example we have used module (%) operator in the two variables when we divide the two operands it has the remainder has 2 that value is positive so it returns the positive value as the result. How does Modulo Done in JavaScript?
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# Use the indicated substitution to evaluate the integral. \int_0^2 {\frac{{{x^2}}}{{\sqrt {16 -...
## Question:
Use the indicated substitution to evaluate the integral.
{eq}\int_0^2 {\frac{{{x^2}}}{{\sqrt {16 - {x^2}} }}dx, \quad x = 4\sin \theta } {/eq}
## Definite Integral:
The integral in which the value of the boundaries of the limits are given, and the resulting value is fixed is known as the definite integral. We will use the substitution method to solve this question.
$$I=\int_{0}^{2}\dfrac{x^{2}}{\sqrt{16-x^{2}}}\text{ d}x$$
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https://greprepclub.com/forum/michael-drives-x-miles-due-north-at-arrives-at-point-a-he-7038.html?sort_by_oldest=true
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# Michael drives x miles due north at arrives at Point A. He
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Michael drives x miles due north at arrives at Point A. He [#permalink] 01 Oct 2017, 22:24
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Question Stats:
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Michael drives x miles due north and arrives at Point A. He then heads due east for y miles. Finally, he drives z miles in a straight line till the starting point. If x, y, and z are integers, then how many miles did Michael drive if one of the initial two legs of the journey was 5 miles?
(A) 5 miles
(B) 12 miles
(C) 25 miles
(D) 30 miles
(E) Cannot be determined by the information given.
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[Reveal] Spoiler: OA
Last edited by chetan2u on 24 Nov 2018, 07:17, edited 1 time in total.
Corrected the Q
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Re: Michael drives x miles due north at arrives at Point A. He [#permalink] 01 Oct 2017, 23:53
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Bunuel wrote:
Michael drives x miles due north at arrives at Point A. He then heads due east for y miles. Finally, he drives z miles a straight line towards is starting point. If x, y, and z are integers, then how many miles did Michael drive if one of the legs of the journey was 5 miles?
(A) 5 miles
(B) 12 miles
(C) 25 miles
(D) 30 miles
(E) Cannot be determined by the information given.
Kudos for correct solution.
Since it is given that it travel north first , then east and finally meets the starting point so it constitutes a right angle triangle
since x,y,z are integers and 5 is the shortest distance.
Thus we know the sides of the right angle triangle and shortest side 5 is given by 5:12:13
SO the total distance traveled by him is 5+12+13=30. Option D
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Re: Michael drives x miles due north at arrives at Point A. He [#permalink] 02 Oct 2017, 07:14
pranab01 wrote:
Bunuel wrote:
Michael drives x miles due north at arrives at Point A. He then heads due east for y miles. Finally, he drives z miles a straight line towards is starting point. If x, y, and z are integers, then how many miles did Michael drive if one of the legs of the journey was 5 miles?
(A) 5 miles
(B) 12 miles
(C) 25 miles
(D) 30 miles
(E) Cannot be determined by the information given.
Kudos for correct solution.
Since it is given that it travel north first , then east and finally meets the starting point so it constitutes a right angle triangle
since x,y,z are integers and 5 is the shortest distance.
Thus we know the sides of the right angle triangle and shortest side 5 is given by 5:12:13
SO the total distance traveled by him is 5+12+13=30. Option D
How did you reach the 5:12:13 triple? Just remembered it as one of the primitive Pitagorean triple or are there any kind of computation to use?
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Re: Michael drives x miles due north at arrives at Point A. He [#permalink] 02 Oct 2017, 10:04
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IlCreatore wrote:
How did you reach the 5:12:13 triple? Just remembered it as one of the primitive Pitagorean triple or are there any kind of computation to use?
Yes they are pythagorean triplets
here are few more -
3 : 4 : 5
5 : 12 : 13
8 : 15 : 17
1 : 1 : √2
1 : √3: 2
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Re: Michael drives x miles due north at arrives at Point A. He [#permalink] 24 Nov 2018, 01:17
The question states one of the legs of the journey was 5 miles.
We haven't been told which one, the shorter one or the longer one.
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Re: Michael drives x miles due north at arrives at Point A. He [#permalink] 24 Nov 2018, 02:04
indiragre18 wrote:
The question states one of the legs of the journey was 5 miles.
We haven't been told which one, the shorter one or the longer one.
I agree. This wouldn't be a problem but we also have the answer option of 12, which corresponds to another P. Triple of 3-4-5. We are never told that 5 is the shortest leg of the journey or the longest. If we assume shortest, 5-12-13 works, if we assume longest, 3-4-5 works.
Now, the only possibility I can see here is that "one of the legs of the journey" in the original question directly corresponds to a triangle side that is NOT the hypotenuse, however this wouldn't really make sense as we are never outright told we have a triangle (figuring this out is part of the problem I assume), and classifying the 2 shorter sides as "legs" and the longer hypotenuse as something else also doesn't make sense from a terminology standpoint.
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Re: Michael drives x miles due north at arrives at Point A. He [#permalink] 24 Nov 2018, 07:15
Expert's post
projectoffset wrote:
indiragre18 wrote:
The question states one of the legs of the journey was 5 miles.
We haven't been told which one, the shorter one or the longer one.
I agree. This wouldn't be a problem but we also have the answer option of 12, which corresponds to another P. Triple of 3-4-5. We are never told that 5 is the shortest leg of the journey or the longest. If we assume shortest, 5-12-13 works, if we assume longest, 3-4-5 works.
Now, the only possibility I can see here is that "one of the legs of the journey" in the original question directly corresponds to a triangle side that is NOT the hypotenuse, however this wouldn't really make sense as we are never outright told we have a triangle (figuring this out is part of the problem I assume), and classifying the 2 shorter sides as "legs" and the longer hypotenuse as something else also doesn't make sense from a terminology standpoint.
You are correct in your thinking and even the language is having some typo.
correcting it. Thanks
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Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html
Re: Michael drives x miles due north at arrives at Point A. He [#permalink] 24 Nov 2018, 07:15
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