url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://www.nicklevine.org/declarative/lectures/solutions/solutions-7.html | 1,506,456,104,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818696681.94/warc/CC-MAIN-20170926193955-20170926213955-00212.warc.gz | 506,895,334 | 2,510 | Solutions to exercises from "lecture notes 7"
(defun reducing-copy-list (list)
(reduce 'cons list :from-end t :initial-value nil))
(defun reducing-reverse (list)
(reduce (lambda (x y) (cons y x)) list :initial-value nil))
;;;;;;;;;;
#||
(defstruct node
data
child-1
child-2
child-3)
(defun copy-all-nodes (node)
(when node
(make-node :data (node-data node)
:child-1 (copy-all-nodes (node-child-1 node))
:child-2 (copy-all-nodes (node-child-2 node))
:child-3 (copy-all-nodes (node-child-3 node)))))
(defun node-find (data node)
(when node
(or (eql data (node-data node))
(node-find data (node-child-1 node))
(node-find data (node-child-2 node))
(node-find data (node-child-3 node)))))
||#
;;;;;;;;;;;
;; Actually I prefer the following, which does not assume the number
;; of child nodes and produces cleaner code...
(defstruct node
data
children)
(defun copy-all-nodes (node)
(make-node :data (node-data node)
:children (mapcar 'copy-all-nodes (node-children node))))
(defun node-find (data node)
(or (eql data (node-data node))
(dolist (child (node-children node))
(when (node-find data child)
(return t)))))
;;;;;;;;;;;
(defun govt-cons (car cdr)
(cons cdr car))
(defun govt-car (x)
(cdr x))
(defun govt-cdr (x)
(car x))
;; this question is difficult to interpret - I guess it means that the
;; argument to govt-list is a proper list -- otherwise if the argument
;; was already a govt-list we could say (defun govt-list (&rest x) x) ...
(defun govt-list (&rest things)
(in-govt-list things))
(defun in-govt-list (things)
(when things
(govt-cons (car things)
(in-govt-list (cdr things)))))
(defun govt-length (list)
(if list
(1+ (govt-length (govt-cdr list)))
0))
(defun govt-member (thing govt-list)
(when govt-list
(if (eql thing (govt-car govt-list))
govt-list
(govt-member thing (govt-cdr govt-list)))))
;;;;;;;;;;;
;; this looks like a total rewrite to me (sigh)
(defun lottery-distinct ()
(in-lottery-distinct nil 6))
(defun in-lottery-distinct (numbers how-many)
(if (plusp how-many)
(let* ((next (1+ (random 49))))
(if (find next numbers)
(in-lottery-distinct numbers how-many)
(in-lottery-distinct (cons next numbers) (1- how-many))))
(sort numbers '<)))
(defun lottery-distinct-iterative ()
(let* ((numbers nil)
(how-many 6))
(loop (let* ((next (1+ (random 49))))
(unless (find next numbers)
(push next numbers)
(when (zerop (decf how-many))
(return (sort numbers '<))))))))
;;;;;;;;;;;
(defun maximum-with-lambda (numbers)
(let* ((best (first numbers)))
(mapc (lambda (number)
(if (> number best)
(setf best number)))
(rest numbers))
best))
(defun dotty-with-lambda (things)
(mapc (lambda (junk)
(declare (ignore junk))
(format t "."))
things))
;;;;;;;;;;;
;; similar to the govt-cons above
;; probably don't need a my-nil (depends how it's going to be used...)
(defstruct my-cons
car
cdr)
(defun my-first (thing)
(my-cons-car thing))
(defun my-rest (thing)
(my-cons-cdr thing))
(defun my-cons (first rest) ; NB ok to have function name same as structure
(make-my-cons :car first :cdr rest))
(defun my-list (&rest things)
(in-my-list things))
(defun in-my-list (things)
(when things
(make-my-cons :car (first things)
:cdr (in-my-list (rest things)))))
(defun my-length (thing)
(if thing
(1+ (my-length (my-rest thing)))
0))
If you have tried the exercises, looked at the solutions and still do not understand what's going on, I am available for consultation at the times advertised on my office door. Bring your code with you in BOTH the following forms:
• Logbook containing printout (nothing handwritten please)
• file on floppy
Nick Levine | 975 | 3,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-39 | latest | en | 0.740069 |
https://studyres.com/doc/8526298/?page=5 | 1,580,030,640,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251687958.71/warc/CC-MAIN-20200126074227-20200126104227-00123.warc.gz | 675,064,884 | 23,126 | • Study Resource
• Explore
Survey
Thank you for your participation!
* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
Document related concepts
Law of large numbers wikipedia, lookup
Central limit theorem wikipedia, lookup
Bootstrapping (statistics) wikipedia, lookup
Fisher–Yates shuffle wikipedia, lookup
Transcript
```Statistics and Data
Analysis
Professor William Greene
Stern School of Business
IOMS Department
Department of Economics
5-1/35
Part 5: Random Variables
Statistics and Data Analysis
Part 5 – Random Variables
5-2/35
Part 5: Random Variables
Random Variable
Using random variables to organize the information
about a random occurrence.
Random Variable: A variable that will take a value
assigned to it by the outcome of a random
experiment.
Realization of a random variable: The outcome of
the experiment after it occurs. The value that is
assigned to the random variable is the realization.
X = the variable, x = the outcome
5-3/35
Part 5: Random Variables
Types of Random Variables
Discrete: Takes integer values
Continuous: A measurement.
5-4/35
Binary: Will an individual default (X=1) or not (X=0)?
Finite: How many female children in families with 4
children; values = 0,1,2,3,4
Finite: How many eggs in a box of 12 are cracked?
Infinite: How many people will catch a certain disease
per year in a given population? Values = 0,1,2,3,…
(How can the number be infinite? It is a model.)
How long will a light bulb last? Values X = 0 to ∞
How do we describe the distribution of biological
measurements?
Measures of intellectual performance
Part 5: Random Variables
Modeling Fair Isaacs:
A Binary Random Variable
Sample of Applicants for a Credit Card (November, 1992)
Experiment = One
randomly picked
application.
Let X = 0 if Rejected
Let X = 1 if Accepted
Rejected
5-5/35
Approved
X is DISCRETE
(Binary). This is
called a Bernoulli
random variable.
Part 5: Random Variables
The Random Variable Lenders Are
Really Interested In Is Default
Of 10,499 people whose application
was accepted, 996 (9.49%)
defaulted on their credit account
(loan). We let X denote the
behavior of a credit card recipient.
X = 0 if no default
X = 1 if default
This is a crucial variable for a
lender. They spend endless
5-6/35
Part 5: Random Variables
5-7/35
Part 5: Random Variables
Distribution Over a Count
Of 13,444 Applications, 2,561 had at least one
derogatory report in the previous 12 months.
Let X = the number of
reports for individuals
who have at least 1.
X = 1,2,…,>10. X is a
discrete random
variable. (There are also
in this data set who had
X=0.)
5-8/35
Part 5: Random Variables
Discrete Random Variable?
Response (0 to 10) to the question: How satisfied are you with
your health right now?
Experiment = the
response of an individual
drawn at random.
Let X = their response to
the question. X = 0,1,…,10
This is a DISCRETE
random variable, but it is
not a count.
Do women answer systematically differently from men?
5-9/35
Part 5: Random Variables
Continuous Variable – Light Bulb Lifetimes
Probability for a specific value is 0.
Probabilities are defined over intervals, such as
P(1000 < Lifetime < 2500). Needs calculus.
5-10/35
Part 5: Random Variables
Distribution of T = the lifetime of the bulb.
10,000 Hours?
Philips DuraMax Long Life “Lasts 1 Year” … “Life 1000 Hours.” Exactly?
5-11/35
Part 5: Random Variables
Probability Distribution
Range of the random variable = the set of values it can
take
Discrete: A set of integers. May be finite or infinite
Continuous: A range of values
Probability distribution: Probabilities associated with
values in the range.
5-12/35
Part 5: Random Variables
Bernoulli Random Variable
Probability Distribution
P(X=0)
P(X=1)
0.5556
0.4444
Experiment = A
randomly picked
application.
Let X = 0 if Rejected
Let X = 1 if Accepted
The range of X is [0,1]
Reject
5-13/35
Approve
Part 5: Random Variables
Probability Distribution over
Derogatory Reports
Derogatory
Reports
X P(X=x)
1 .5100
2 .2085
3 .0953
4 .0547
5 .0430
6 .0226
7 .0148
8 .0125
9 .0109
10 .0277
5-14/35
Part 5: Random Variables
Notation
5-15/35
Probability distribution = probabilities
assigned to outcomes.
P(X=x) or P(Y=y) is common.
Probability function = PX(x).
Sometimes called the density
function
Cumulative probability is
Prob(X < x) for the specific X.
Part 5: Random Variables
Cumulative Probability
Derogatory Reports
X P(X=x) P(X<x)
1 .5100
.5100
2 .2085
.7185
3 .0953
.8138
4 .0547
.8685
5 .0430
.9115
6 .0226
.9341
7 .0148
.9489
8 .0125
.9614
9 .0109
.9723
10 .0277 1.0000
The item marked 10 is actually 10 or more.
5-16/35
Part 5: Random Variables
Rules for Probabilities
1. 0 < P(x) < 1 (Valid probabilities)
2.
x all possible outcomes
P(x) 1
3. For different values of x, say A and
B, Prob(X=A or X=B) = P(A) + P(B)
5-17/35
Part 5: Random Variables
Probabilities
Derogatory Reports
X P(X=x) P(X<x)
1 .5100
.5100
2 .2085
.7185
3 .0953
.8138
4 .0547
.8685
5 .0430
.9115
6 .0226
.9341
7 .0148
.9489
8 .0125
.9614
9 .0109
.9723
10 .0277 1.0000
5-18/35
P(a < x < b) = P(a)+P(a+1)+…+P(b)
E.g., P(5 < Derogs < 8) = .0430 + .0226 + .0148 + .0125
= .0929
P(a < x < b) = P(x < b) – P(x < a-1)
E.g., P(5 < Derogs < 8) = P(Derogs < 8) – P(Derogs < 4)
= .9614 - .8685
= .0929
Part 5: Random Variables
Mean of a Random Variable
Average outcome; outcomes weighted by
probabilities (likelihood)
DenotedE[X] = i = all outcomes P(X xi ) xi
Typical value
Usually not equal to a value that the random
variable actually takes.
E.g., the average family size in the U.S. is
1.4 children.
Usually denoted E[X] = μ (mu)
5-19/35
Part 5: Random Variables
Expected Value
X = Derogs
x P(X=x)
1 .5100
2 .2085
3 .0953
4 .0547
5 .0430
6 .0226
7 .0148
8 .0125
9 .0109
10 .0277
μ=2.361
E[X] = 1(.5100) + 2(.2085) + 3(.0953) + … + 10(.0277) = 2.3610
5-20/35
Part 5: Random Variables
Expected Payoffs are Expected Values
of Random Variables
18 Red numbers
18 Black numbers
2 Green numbers (0,00)
5-21/35
Bet \$1 on a number
If it comes up, win \$35. If not, lose the \$1
The amount won is the random variable:
Win = -1 P(-1) = 37/38
+35 P(+35) = 1/38
E[Win] = (-1)(37/38) + (+35)(1/38)
= -0.053
= -5.3 cents (familiar).
Part 5: Random Variables
Buy a Product Warranty?
Should you buy a \$20 replacement warranty on a \$47.99 appliance?
What are the considerations?
Probability of product failure = P (?)
Expected value of the insurance = -\$20 + P*\$47.99 < 0 if P < 20/47.99.
5-22/35
Part 5: Random Variables
Median of a Random Variable
The median of X is the value x such that Prob(X < x) = .5.
For a continuous variable, we will find this using calculus.
For a discrete value, Prob(X < M+1) > .5 and Prob(X < M-1) < .5
X
0
1
2
3
4
5
6
7
8
9
10
Prob(X=x) Prob(X < x)
.0164
.0164
.0093
.0257
.0235
.0492
.0429
.0921
.0509
.1430
.1549
.2979
.0926
.3905
.1548
.5453
.2259
.7712
.1120
.8832
.1168
1.0000
Mean (6.8)
Median (7)
Health Satisfaction Sample Proportions.
5-23/35
Part 5: Random Variables
of the Random Outcomes
Derogatory
Reports
X P(X=x)
1 .5100
2 .2085
3 .0953
4 .0547
5 .0430
6 .0226
7 .0148
8 .0125
9 .0109
10 .0277
μ=2.361
5-24/35
The range is 1 to
10, but values
outside 1 to 5 are
rather unlikely.
Part 5: Random Variables
Variance
Variance = E[X – μ]2 = σ2 (sigma2)
2
2
P(X
x
)(x
)
Compute
i = all outcomes
i
i
The square root is usually more useful.
Standard deviation = σ
Compute i = all outcomes P(X xi ) (xi )2
5-25/35
2
P(X
x
)x
i = all outcomes
i
i
2
Part 5: Random Variables
Variance Computation
X = Derogatory Reports. μ = 2.361
x P(X=x)
x-μ
(x- μ)2 P(X=x)(x-μ)2
1 .5100 -1.361
1.85232
0.94468
2 .2085 -0.361
0.13032
0.02717
3 .0953
0.639
0.40832
0.03891
4 .0547
1.639
2.28632
0.14694
5 .0430
2.639
6.96432
0.29947
6 .0226
3.639 13.24232
0.29928
7 .0148
4.639 21.53032
0.31850
8 .0125
5.639 31.79832
0.39748
9 .0109
6.639 44.07632
0.48043
10 .0277
7.639 58.35432
1.61641
SUM
4.56928
5-26/35
σ2 = 4.56928
σ = 2.13759
Part 5: Random Variables
Common Results
for Random Variables
Concentration of Probability
What it means: For any random outcome,
5-27/35
For almost any random variable, 2/3 of the probability
lies within μ ± 1σ
For almost any random variable, 95% of the
probability lies within μ ± 2σ
For almost any random variable, more than 99.5% of
the probability lies within μ ± 3σ
An (observed) outcome more than one σ away from μ
is somewhat unusual.
One that is more than 2σ away is very unusual.
One that is more than 3σ away from the mean is so
unusual that it might be an outlier (a freak outcome).
Part 5: Random Variables
Outlier?
5-28/35
In the larger credit card data set, there was an
individual who had 14 major derogatory
reports in the year of observation. Is this
“within the expected range” by the measure
of the distribution?
The person’s deviation is (14 – 2.361)/2.137 =
5.4 standard deviations above the mean. This
person is very far outside the norm.
Part 5: Random Variables
Recall from day 2 of class
Reliable Rules of Thumb
5-29/35
Almost always, 66% of the observations in a sample will
lie in the range
[mean+1 s.d. and mean – 1 s.d.]
Almost always, 95% of the observations in a sample will
lie in the range
[mean+2 s.d. and mean – 2 s.d.]
Almost always, 99.5% of the observations in a sample
will lie in the range
[mean+3 s.d. and mean – 3 s.d.]
Part 5: Random Variables
A Possibly Useful “Shortcut”
E[X – μ]2 = E[X2] – μ2
=
5-30/35
2
2
P(X
x
)x
μ
i = all outcomes
i
i
Part 5: Random Variables
Application
PartyPlanners plans parties each day, and must order supplies for the events.
The number of requests for party plans varies day by day according to
P(X=0) = .4 P(X=1) = .3 P(X=2) = .25 P(X=3) = .05
How many parties should they expect on a given day?
E[X] = .4(0) + .3(1) + .25(2) + .05(3) = .95, or about 1.
What are the variance and standard deviation?
Var[X] = .4(02 )+ .3(12 ) + .25(22 ) + .05(3 2 ) -.952 = .8475.
0.8475 = 0.9206
If they plan for 1 party per day, it is rather likely that they will run out of materials
since 2 is only 1.1 standard deviations above the mean.
5-31/35
Part 5: Random Variables
Important Algebra
Linear Translation: For the random variable
X with mean E[X] = μ,
if Y = a+bX, then E[Y] = a + bμ
Scaling: For the random variable X with
standard deviation σX,
if Y = a+bX, then σY = |b| σX
It is not necessary to transform the original data.
5-32/35
Part 5: Random Variables
Example: Repair Costs
5-33/35
The number of repair orders per day at a body shop is distributed by:
Repairs
0
1
2
3
4
Probability
.1
.2
.35
.2
.15
Opening the shop costs \$500 for any repairs. Two people each cost \$100/repair to
do the work.
What are the mean and standard deviation of the number of repair orders?
μ = 0(.1) + 1(.2) + 2(.35) + 3(.2) + 4(.15)
= 2.10
2
2
2
2
2
2
2
σ = 0 (.1) + 1 (.2) + 2 (.35) + 3 (.2) + 4 (.15) – 2.1 = 1.39
σ = 1.179
What are the mean and standard deviation of the cost per day to run the shop?
Cost = \$500 + \$100*(2)*(Number of Repairs)
Mean = \$500 + \$200*(2.1) = \$920/day
Standard deviation = \$200(1.179) = \$235.80/day
Part 5: Random Variables
Summary
5-34/35
Random variables and random outcomes
Outcome or sample space = range of the random
variable
Types of variables: discrete vs. continuous
Probability distributions
Probabilities
Cumulative probabilities
Rules for probabilities
Moments
Mean of a random variable
Standard deviation of a random variable
Part 5: Random Variables
Application: Expected Profits and Risk
You must decide how many copies of your self published novel to print . Based on market
research, you believe the following distribution describes X, your likely sales (demand).
x
P(X=x)
25
.10
(Note: Sales are in thousands. Convert your final result to
40
.30
dollars after all computations are done by multiplying your
55
.45
final results by \$1,000.)
70
.15
Printing costs are \$1.25 per book. (It’s a small book.) The selling price will be \$3.25. Any
unsold books that you print must be discarded (at a loss of \$2.00/copy). You must decide how
many copies of the book to print, 25, 40, 55 or 70. (You are committed to one of these four – 0
is not an option.)
A. What is the expected number of copies demanded.
B. What is the standard deviation of the number of copies demanded.
C. Which of the four print runs shown maximizes your expected profit? Compute all four.
D. Which of the four print runs is least risky – i.e., minimizes the standard deviation of the
profit (given the number printed). Compute all four.
E. Based on C. and D., which of the four print runs seems best for you?
5-35/35
Part 5: Random Variables
X = Sales (Demand)
x
P(X=x)
25,000 .10
40,000 .30
55,000 .45
70,000 .15
A. Expected Value =
all values of x
x P(X=x)
= .1(25,000) + .3(40,000) + .45(55,000) + .15(70,000)
= 49,750
5-36/35
Part 5: Random Variables
B. Standard Deviation
Get the Variance First
2
all values of x
(x - E[x]) 2 P(X=x)
= .1(25,000 - 49,750)2 .3(40,000 - 49,750)2
+ .45(55,000 - 49,750)2 + .15(70,000 - 49,750)2
= 163,687,500
Standard Deviation = square root of variance.
=
163,687,500 = 12,794.041
There is a shortcut
2 all values of x x 2 P(X=x) 2
2
all values of x
(x - E[x]) 2 P(X=x)
.1(25,0002 ) .3(40,0002 ) + .45(55,0002 ) + .15(70,0002 ) - 49,7502
= 163,687,500
=
5-37/35
Part 5: Random Variables
x
P(X=x) Revenue per book = \$3.25
25,000 .10
Cost per book
= \$1.25
40,000 .30
Profit per book sold = \$2.00/book
55,000 .45
70,000 .15
Expected Profit | Print Run = 25,000 is \$2 25,000 = \$50,000
(Demand is guaranteed to be at least 25,000)
Expected Profit | Print Run = 40,000 is \$2 .9 40,000
+ .1 (\$2 25,000 - \$1.25 15,000) = \$75,125
(If print 40,000, .9 chance sell all and .1 chance sell only 25,000)
Expected Profit | Print Run = 55,000 is \$2 .6 55,000
+ .1 (\$2 25,000 - \$1.25 30,000)
+ .3 (\$2 40, 000 \$1.25 15, 000) = \$85,625
Expected Profit | Print Run=70,000 is \$2 .15 70,000
+ .1 (\$2 25,000 - \$1.25 45,000)
+ .3 (\$2 40, 000 \$1.25 30000)
+ .45 (\$2 55, 000 \$1.25 15000) = \$55,287,50
5-38/35
Part 5: Random Variables
Expected Profit Given Print Run
5-39/35
Part 5: Random Variables
Variances
Print Run = 25,000. Variance = 0. Std. Dev. = 0 Demand will be at least 25,000.
Print Run = 40,000. Variance =
.1*[(2* 25000 1.25*15000) 75,125]2
(if demand is only 25,000)
.9*[(2* 40000)
75,125)]2
Standard Deviation = square root = \$14625
Print Run = 55,000. Variance =
(if demand is 40,000)
.1*[(2* 25000 1.25*30, 000) 85, 625]2
.3*[(2* 40000) 1.25*15, 000) 85, 625] +
.6*[(2*55, 000
85,625]
(if demand is only 25,000)
(if demand is 40,000)
(if demand is 55,000)
Standard Deviation = square root = \$32,702.49
Print Run = 70,000. Variance =
5-40/35
.1* [(2* 25000 1.25* 45000) 55, 287.5]2
(if demand is only 25,000)
.3* [(2* 40000 1.25*30, 000) 55, 287.5]2 +
(if demand is 40,000)
.45*[(2*55, 000 1.25*15, 000) 55, 287.5]2 +
(if demand is 55,000)
.15*[2*70, 000
55, 287.5]2
Standard Deviation = square root = \$35,572.84
(if demand is 70,000)
Part 5: Random Variables
Run=70,000
Run=55,000
Run=40,000
Run=25,000
5-41/35
Part 5: Random Variables
Run=70,000
Run=55,000
Run=40,000
Run=25,000
5-42/35
Part 5: Random Variables
Run=70,000
?
Run=55,000
Run=40,000
Run=25,000
5-43/35
Part 5: Random Variables
```
Related documents | 5,915 | 15,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2020-05 | latest | en | 0.810867 |
http://gmatclub.com/forum/lt-spoiler-d-spoiler-40181.html?fl=similar | 1,435,925,849,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375096061.71/warc/CC-MAIN-20150627031816-00294-ip-10-179-60-89.ec2.internal.warc.gz | 100,348,758 | 45,929 | Find all School-related info fast with the new School-Specific MBA Forum
It is currently 03 Jul 2015, 04:17
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# lt;spoiler> D </spoiler>
Author Message
TAGS:
Senior Manager
Joined: 14 Aug 2006
Posts: 365
Followers: 1
Kudos [?]: 5 [0], given: 0
lt;spoiler> D </spoiler> [#permalink] 17 Dec 2006, 13:41
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
<spoiler>D</spoiler>
Attachments
p_g_2_21.GIF [ 8.62 KiB | Viewed 368 times ]
Intern
Joined: 17 Dec 2006
Posts: 22
Followers: 0
Kudos [?]: 0 [0], given: 0
Given AB=OB=CO
1. COD = 60
=> COA=120
Let BOA=x = BAO => ABO=180-2x
=> OBC = 180-(180-2x) = 2x
Since OBC=OCB => OCB=2x
Since BOA=x => COB=120-x [since COA=120]
Now COB + OBC + OCB = 180 => 120 - x + 2x + 2x = 180
=> x=20= BAO. sufficient
2. Since BCO=40 => OBC=40 => ABO=140 => BAO=20
Hence D
_________________
Senior Manager
Joined: 01 Oct 2006
Posts: 498
Followers: 1
Kudos [?]: 20 [0], given: 0
posted a lot of times answer is D
Similar topics Replies Last post
Similar
Topics:
16 If d denotes a decimal, is d>=0.5 ? 8 15 Oct 2012, 03:49
Is d greater than 1? (1) d>4-d (2) 1/d > -1 I know the 4 05 Jun 2006, 09:22
4 If d denotes a decimal, is d>=0.5 8 18 Apr 2006, 07:08
It's a D. from 1 -> d=4 ->r=2 so the area can be 1 10 Apr 2006, 12:15
1 If d denotes a decimal, is d >= 0.5? 16 11 Dec 2005, 09:54
Display posts from previous: Sort by | 708 | 1,981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2015-27 | longest | en | 0.813881 |
http://solution-dailybrainteaser.blogspot.com/2014/10/find-number-riddle.html | 1,532,331,842,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676595531.70/warc/CC-MAIN-20180723071245-20180723091245-00253.warc.gz | 342,738,815 | 1,947 | # Find The Number Riddle
Find The Number Riddle Solution - 4 October
I have thought of a number that is made up by using all the ten digits just once. Here are a few clues for you to guess my number:
First digits is divisible by 1.
First two digits are divisible by 2.
First three digits are divisible by 3.
First four digits are divisible by 4.
First five digits are divisible by 5.
First six digits are divisible by 6.
First seven digits are divisible by 7.
First eight digits are divisible by 8.
First nine digits are divisible by 9.
The number is divisible by 10.
Can you find out the number ? | 145 | 601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-30 | latest | en | 0.949084 |
https://www.tudelft.nl/en/eemcs/study/online-education/math-explained/differentiation/ | 1,603,558,243,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107884322.44/warc/CC-MAIN-20201024164841-20201024194841-00017.warc.gz | 926,487,998 | 11,285 | # Differentiation
Differentiation is very helpful to determine how fast something changes. Slope and speed are defined by differentiation.
### Context: rollercoaster
This video shows a rollercoaster as an example to explain the mathematical concept of differentiation. It introduces speed and slope, which can be defined using differentiation.
### Definition of derivative
The derivative is defined using speed and slope as examples. The process of determining the derivative is called differentiation. You will also learn about the tangent line.
### How to determine the derivative?
This video show the grand plan for calculating derivatives: First, you learn the derivatives of the standard functions. Second: you learn rules to calculate the derivative of combinations of standard functions, such as the chain rule. Then you use the derivatives of the standard functions to obtain its derivative.
### Rules of calculation for differentiation
Product rule for calculating the derivative of a function.
### Rules of calculation for differentiation - part 2
Chain rule for calculating the derivatives of a function.
### Derivatives of power functions
How do you calculate the derivative of power functions? In this video you will learn the general rule by looking at some examples.
### Derivative of the sine
What is the derivative of the standard function sine? It is the cosine! Look at the graph of the sine and see why the cosine is the derivative of the sine.
### Derivative of x^p and a^x
What is the derivative of the function x to the power p? And what is the derivative of the function a to the power x? We use the exponential function e^x to find the derivatives.
### Non-differentiable functions
Can you differentiate any function at any point? The answer is "no". Why?
### Tangent line
The derivative of a function is the slope of the tangent line at the graph of the function. In this video we will see how to find an equation of this tangent line.
### Finding minima and maxima
How can you find the minima and maxima of a function using differentiation? This can help you to optimize the solutions to a problem. How can you find global and local maxima and minima? What is a boundary point, critical point and singular point?
### First and second derivative test
To know the maximal or minimal values a function can have, you first have to find the critical points with the first derivative test. To see if a critical point is a local maximum or minimum you can use the second derivative test. This video shows you how!
### Implicit differentiation
This video will prepare you for learning implicit differentiation. A technique to find the formula for the derivative without actually determining explicit formulas for the functions first. | 543 | 2,779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2020-45 | latest | en | 0.913324 |
https://fr.scribd.com/document/324283798/Thermal | 1,571,157,864,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986660067.26/warc/CC-MAIN-20191015155056-20191015182556-00259.warc.gz | 490,297,360 | 71,856 | Vous êtes sur la page 1sur 3
# Chapter II
THERMAL CONCEPTS
Temperature is the intuitive concepts of hotness or coldness of a substance. It can be
measured by using a thermometer. For the example a body temperature is measured
by using Celsius thermometer.
While heat is energy that is transferred from one body into another as a result of
different temperatures.
Thermal Properties
A. Specific Heat Capacity (c)
If thermal energy is provided to a body, generally the temperature of the body will
increases.
Specific heat capacity is the amount of thermal energy needed to raise the temperature
of a mass of one kilogram of a substance by one Kelvin.
Therefore, to raise the temperature of a mass m by T Kelvin, the amount of thermal
energy required is
Q = mc T
(assuming that c is temperature independent).
The units of specific heat capacity are Jkg-1K-1
B. Heat Capacity (C)
Heat capacity is the amount thermal energy required to change the temperature of one
kilogram by one Kelvin. It is the result of mass multiplied by specific heat capacity of
a body.
C = mc
The units of heat capacity are JK-1
If a quantity of thermal energy Q is given to a body, then the rise in temperature T
will be found the following formula
Q=C T
C. Thermal Equilibrium
It is everyday experience that thermal energy flows from hot bodies into cold bodies
(see figure 1.1)
Hot
cold
Flow of thermal energy
Figure 1.1 in an isolated system thermal energy always flows from the hotter body to the colder
When a cold and a hot body are placed in contact, thermal energy will flow until the
temperature of both bodies is the same. This state of affairs is called thermal
equilibrium.
The amount of thermal energy lost by the hot body is equal to the amount of thermal
energy gained by the cold body. In some books this condition is called Black principle
11
D. Change of State
Ordinary matter can exist as a solid, a liquid or a gas. These three are called state of
matter.
Heating can turn ice into water and water into steam. Ice will turn into water if the
temperature of the ice is its melting temperature: 0oC. Similarly, to turn water into
steam the temperature must be 100oC.
This means that if we are given a piece of ice at a temperature of say, - 10oC , to melt
it we must first raise its temperature from -10oC to zero. After all of the ice has turned
into water, we have water at a temperature of 0oC.
Any additional thermal energy supplied will increase the temperature of the water.
When the temperature reaches 100oC, any additional thermal energy supplied is used
to turn water into steam at the same temperature of 100oC.
After all of the water has turned into steam, the temperature begins to increase again.
We thus see that when the state of matter is changing, the temperature does not
change.
Once at the melting point, any additional thermal energy supplied does not
increase the temperature.
The thermal energy required to melt a unit mass of material at its melting point
is called the specific latent heat of fusion, Lt and the thermal energy required
to vaporize a unit mass at its boiling point is called the specific latent heat of
vaporization, Lv.
Thus to melt or vaporize a quantity of mass m, we require a quantity of
thermal energy
Q = m. Lt and Q = m. Lv
The specific latent heats have units of Jkg-1
gas
o
T( C)
Liquid and gas
100
liquid
Solid and liquid
0
-x solid
Q (J)
12
Exercise
Questions
1. A body of mass 0.150kg has its temperature increased by 5.00oC when 385J of
thermal energy is provided to it. What is the bodys specific heat capacity?
2. How much ice at -10oC must be dropped into a cup containing 300 g of water
at 20oC in order for the temperature of the water to be reduced to 10 oC? The
cup itself has a mass of 150 g and is made out of aluminum, assume that not
thermal energy is lost to be surrounding.
3. A graph in figure 1.7 shows the change of ice phase, from data what is the value
of t? ( Les=80 cal/gram, ces= 0.5 cal/gramoC and cair= 1 cal/gramoC).
toC
t
2000
6000
7000
Q(cal)
## 4. Ice at 0 oC is added to 900 gram of water at 20 oC. The water temperature
cools down to 10 oC. How much ice was added( Les=80 cal/gram, ces= 0.5
cal/gramoC and cair= 1 cal/gramoC?
5. A body has mass of 0.15kg its temperature increased by 5 oC when 375J of
thermal energy is applied to it. What is the specific heat capacity of the body ?
13 | 1,110 | 4,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2019-43 | latest | en | 0.899584 |
https://www.projectguitar.com/forums/topic/32576-first-build-a-few-questions/ | 1,709,074,521,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00336.warc.gz | 943,721,085 | 30,287 | # First Build. A Few Questions
## Recommended Posts
So after a lot of searching and not finding what i want, I decided to build myself a guitar. Plus I think it would be a fun project.
I want to build the guitar around the Ibanez S style body shape, but i couldn't find any dimensions on the net. Does anybody knows them?
When you measure the scale to the which point of the floyd rose bridge do you measure.
How do you measure the radius of the fretboard. I ran into things like 400 or 430. But what does that mean?
Any help would be appreciated.
##### Share on other sites
The numbers you mentioned are fretboard radii in millimeters.
They're more commonly measured in inches:
7.5" =~ 191mm
9.5" =~ 241mm
10" = 254mm
12" =~ 305mm
14" =~ 356mm
15" = 381mm
16" =~ 406mm
20" = 508mm
Edited by Rick500
##### Share on other sites
When you measure the scale to the which point of the floyd rose bridge do you measure.
Scale length = [distance from nut edge (fingerboard side, obviously!) to centre of 12th fret wire] x 2
I would go with that instead of trying to guess which Floyd saddle was in the right place!!
##### Share on other sites
When you measure the scale to the which point of the floyd rose bridge do you measure.
Scale length = [distance from nut edge (fingerboard side, obviously!) to centre of 12th fret wire] x 2
I would go with that instead of trying to guess which Floyd saddle was in the right place!!
May I ask the opposite question (I hope it's useful info for the OP too) ??
By your definition above I know where the theoretical scale end lies on the body (it's a line parallel to the frets and nut).
Now where do you position your bridge around this ?? Are there any rules as to how to distribute saddle travel span around the theoretical "zero" for the different strings ??
TIA
##### Share on other sites
When you measure the scale to the which point of the floyd rose bridge do you measure.
Scale length = [distance from nut edge (fingerboard side, obviously!) to centre of 12th fret wire] x 2
I would go with that instead of trying to guess which Floyd saddle was in the right place!!
May I ask the opposite question (I hope it's useful info for the OP too) ??
By your definition above I know where the theoretical scale end lies on the body (it's a line parallel to the frets and nut).
Now where do you position your bridge around this ?? Are there any rules as to how to distribute saddle travel span around the theoretical "zero" for the different strings ??
TIA
Simplest: measure a guitar with the same scale and go from there.
Basically, though, you will never, ever need to have the string shorter than the true scale length (ie, twice the distance from nut to twelfth), usually needing about 3mm for the low E, between nothing and 1mm for the high E. Ish. Depends on scale, strings, and tuning. I tend to set things so the high E is at the scale length at the most forward possible point along its travel, low E about 3mm behind that.
There are discussions on this subject here, and certainly in the MIMF.com library, particularly with regard to tunomatic bridge placement.
However, if you've got a fairly standard type bridge, go to stewmac.com/fretcalc, type in the details, and the calculator tells you where to drill your bridge studs and/or screw holes. Easy, right?
##### Share on other sites
However, if you've got a fairly standard type bridge, go to stewmac.com/fretcalc, type in the details, and the calculator tells you where to drill your bridge studs and/or screw holes. Easy, right?
Yup, I just looked at that link, put in some info, and it does indeed give you a very useful reply! Nice one!
Thanks for the pointer!
DJ
##### Share on other sites
I highly recommend Melvyn Hiscocks' book, Build an Electric Guitar. Available through several outlets like Amazom.com and book stores. He covers everything in building a guitar from scratch.
Welcome, Vinny
##### Share on other sites
I highly recommend Melvyn Hiscocks' book, Build an Electric Guitar. Available through several outlets like Amazom.com and book stores. He covers everything in building a guitar from scratch.
##### Share on other sites
However, if you've got a fairly standard type bridge, go to stewmac.com/fretcalc, type in the details, and the calculator tells you where to drill your bridge studs and/or screw holes. Easy, right?
Great page that one !!
Of course, the bridge I'm using isn't there (wilkinson wrap-around), but I can surely work it out from the given info. Thanks !!
##### Share on other sites
However, if you've got a fairly standard type bridge, go to stewmac.com/fretcalc, type in the details, and the calculator tells you where to drill your bridge studs and/or screw holes. Easy, right?
Great page that one !!
Of course, the bridge I'm using isn't there (wilkinson wrap-around), but I can surely work it out from the given info. Thanks !!
Or email StewMac and ask. Bit odd, really, since they do sell the Gotoh Wilky stud tailpiece bridge unit, as well as a pigtail fully adjustable bridge. Overall, though, you can probably use the TOM info, although the pre-compensated bridges can be installed straight or nearly straight.
## Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.
× Pasted as rich text. Paste as plain text instead
Only 75 emoji are allowed.
× Your previous content has been restored. Clear editor
× You cannot paste images directly. Upload or insert images from URL.
×
×
• Home | 1,310 | 5,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-10 | latest | en | 0.9563 |
https://sciencing.com/continuity-equation-fluids-definition-forms-examples-13723387.html | 1,653,134,473,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662539101.40/warc/CC-MAIN-20220521112022-20220521142022-00284.warc.gz | 573,651,817 | 88,429 | # Continuity Equation (Fluids): Definition, Forms & Examples
Print
Consider a stream of cars driving down a segment of road with no onramps or offramps. In addition, suppose the cars can’t change their spacing at all – that they are somehow kept a fixed distance apart from each other. Then, if one car in the long line changes its speed, all of the cars would be automatically forced to change to the same speed. No car could ever be going faster or slower than the car in front of it, and the number of cars passing a point on the road per unit time would be the same along all points on the road.
But what if the spacing isn’t fixed and the driver of one car steps on their brakes? This causes other cars to slow down as well and can create a region of slower moving, closely spaced cars.
Now imagine you have observers at different points along the road whose job is to count the number of cars going by per unit time. An observer at a location where the cars are moving faster counts the cars as they go by, and because of the larger spacing between the cars, still ends up coming up with the same number of cars per unit time as an observer near the traffic jam location because even though the cars move more slowly through the jam, they are more closely spaced.
The reason the number of cars per unit time passing each point along the road remains roughly constant boils down to a conservation of car number. If a certain number of cars pass a given point per unit time, then those cars are necessarily moving on to pass the next point in approximately the same amount of time.
This analogy gets at the heart of the continuity equation in fluid dynamics. The continuity equation describes how fluid flows through pipes. Just as with the cars, a conservation principle applies. In the case of a fluid, it is conservation of mass that forces the amount of fluid passing any point along the pipe per unit time to be constant so long as the flow is steady.
## What Is Fluid Dynamics?
Fluid dynamics studies fluid motion or moving fluids, as opposed to fluid statics, which is the study of fluids that are not moving. It is closely related to the fields of fluid mechanics and aerodynamics but is narrower in focus.
The word fluid often refers to a liquid or an incompressible fluid, but it can also refer to a gas. In general a fluid is any substance that can flow.
Fluid dynamics studies patterns in fluid flows. There are two main ways in which fluids are compelled to flow. Gravity can cause fluids to flow downhill, or fluid can flow due to pressure differences.
## Equation of Continuity
The continuity equation states that in the case of steady flow, the amount of fluid flowing past one point must be the same as the amount of fluid flowing past another point, or the mass flow rate is constant. It is essentially a statement of the law of conservation of mass.
The explicit formula of continuity is the following:
\rho_1A_1v_1 = \rho_2A_2v_2
Where ρ is density, A is cross-sectional area and v is the flow velocity of the fluid. The subscripts 1 and 2 indicate two different regions in the same pipe.
## Examples of the Continuity Equation
Example 1: Suppose water is flowing through a pipe of diameter 1 cm with a flow velocity of 2 m/s. If the pipe widens to a diameter of 3 cm, what is the new flow rate?
Solution: This is one of the most basic examples because it occurs in an incompressible fluid. In this case, density is constant and can be cancelled from both sides of the continuity equation. You then only need to plug in the formula for area and solve for the second velocity:
A_1v_1 = A_2v_2 \implies \pi(d_1/2)^2v_1 =\pi(d_2/2)^2v_2
Which simplifies to:
d_1^2v_1 =d_2^2v_2 \implies v_2 = d_1^2v_1/d_2^2 = 0.22 \text{ m/s}
Example 2: Suppose a compressible gas is flowing through a pipe. In a region of the pipe with a cross-sectional area of 0.02 m2, it has a flow rate of 4 m/s and a density of 2 kg/m3. What is its density as it flows through another region of the same pipe with a cross-sectional area of 0.03 m2 at velocity 1 m/s?
Solution: Applying the continuity equation, we can solve for the second density and plug in values:
\rho_2 = \rho_1 \frac{A_1v_1}{A_2v_2}=5.33 \text{ kg/m}^3
Dont Go!
We Have More Great Sciencing Articles! | 1,033 | 4,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2022-21 | longest | en | 0.954759 |
https://ask.sagemath.org/question/10868/a-bug-with-solve-xyx/ | 1,547,716,861,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658901.41/warc/CC-MAIN-20190117082302-20190117104302-00619.warc.gz | 451,361,609 | 13,854 | A bug with solve xy=x ?
For solving the equation "xy=x", we can do:
sage: x,y=var('x,y')
sage: solve([x*y==x],[x,y])
[x == 0]
But this is not the complete solution.
We need to add the trivial equation "0==0" for having the complete solution:
sage: solve([x*y==x,0==0],[x,y])
[[x == 0, y == r1], [x == r2, y == 1]]
Is it a bug ?
edit retag close merge delete
1
Is it a bug? A good question. If we change the order of x,y to solve([x*y==x],[y,x]) , we get y==1 as solution. Obviously only the first variable is used as far as there is only one equation. IMHO that's not a bug, but it would be nice, if Sage could add the trivial equation automatically.
( 2013-12-30 21:30:41 -0600 )edit
Sort by » oldest newest most voted
x,y=var('x,y') show((x*y-x).solve(x,y))
"==" is a logic,not always equal "="
more | 260 | 817 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-04 | longest | en | 0.88203 |
https://discuss.codecademy.com/t/faq-interview-prep-problems-code-challenge-9/551114 | 1,670,190,728,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710980.82/warc/CC-MAIN-20221204204504-20221204234504-00086.warc.gz | 241,672,804 | 10,300 | # FAQ: Interview Prep Problems - Code Challenge 9
This community-built FAQ covers the “Code Challenge 9” exercise from the lesson “Interview Prep Problems”.
Paths and Courses
This exercise can be found in the following Codecademy content:
## FAQs on the exercise Code Challenge 9
There are currently no frequently asked questions associated with this exercise – that’s where you come in! You can contribute to this section by offering your own questions, answers, or clarifications on this exercise. Ask or answer a question by clicking reply () below.
If you’ve had an “aha” moment about the concepts, formatting, syntax, or anything else with this exercise, consider sharing those insights! Teaching others and answering their questions is one of the best ways to learn and stay sharp.
## Join the Discussion. Help a fellow learner on their journey.
Ask or answer a question about this exercise by clicking reply () below!
You can also find further discussion and get answers to your questions over in #get-help.
Agree with a comment or answer? Like () to up-vote the contribution!
Need broader help or resources? Head to #get-help and #community:tips-and-resources. If you are wanting feedback or inspiration for a project, check out #project.
Looking for motivation to keep learning? Join our wider discussions in #community
Learn more about how to use this guide.
Found a bug? Report it online, or post in #community:Codecademy-Bug-Reporting
Have a question about your account or billing? Reach out to our customer support team!
None of the above? Find out where to ask other questions here!
This isn’t so much a question… but isn’t the resulting column (running_total_gross) just ‘to_date_gross’? That’s what the gross to date IS.
What an odd challenge…
2 Likes
I’m struggling with how to phrase this, but I think this exercise is an error. That is, I think it wasn’t implemented properly by whoever implemented these exercises.
Using a window function with `PARTITION BY` , get the total change in `gross` for each movie up to the current `week` and display it next to the current `week` column along with the `title` , `week` , and `gross` columns.
So, specifically, the directions are to get the ‘change in gross’ and display it. But the answer to the challenge is as follows:
`SELECT title,week,gross, SUM(gross) OVER ( PARTITION BY title ORDER BY week ) AS 'running_total_gross' FROM box_office;`
The tip-off here that this is an implementation error is that the directions ask for the change in gross, and yet in the passing code example they use ‘running_total_gross’ as a column name. Additionally, one wouldn’t expect to use the SUM aggregate when trying to calculate the change from week to week. Granted I’m just learning SQL, but upon reading the directions my first instinct was to use LAG. Someone should probably review this one.
5 Likes
I agree with both of the above and reported this as a bug. The assignment and accepted answer don’t match, and the accepted answer is unnecessary as there is already a column for it.
Technically, the assignment should probably be reworded: “total change in gross” would actually just be that week’s gross which is also its own column. I think the intent behind the assignment is to find “change in weekly gross”
3 Likes
Completely agree with above, this question looks like a mistake. The answer they want is the same as the to_date_gross column, but the question suggests we should provide the change in gross each week using LAG function.
2 Likes | 764 | 3,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-49 | latest | en | 0.899983 |
https://www.jobilize.com/course/section/section-summary-3-6-normal-and-tension-forces-by-openstax?qcr=www.quizover.com | 1,621,379,794,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989874.84/warc/CC-MAIN-20210518222121-20210519012121-00386.warc.gz | 837,874,081 | 25,088 | # 0.5 3.6 normal and tension forces (Page 5/8)
Page 5 / 8
Solution
First, we need to resolve the tension vectors into their horizontal and vertical components. It helps to draw a new free-body diagram showing all of the horizontal and vertical components of each force acting on the system.
Consider the horizontal components of the forces (denoted with a subscript $x$ ):
${F}_{\text{net}x}={T}_{\text{L}x}-{T}_{\text{R}x}.$
The net external horizontal force ${F}_{\text{net}x}=0$ , since the person is stationary. Thus,
$\begin{array}{lll}{F}_{\text{net}x}=0& =& {T}_{\text{L}x}-{T}_{\text{R}x}\\ {T}_{\text{L}x}& =& {T}_{\text{R}x}.\end{array}$
Now, observe [link] . You can use trigonometry to determine the magnitude of ${T}_{\text{L}}$ and ${T}_{\text{R}}$ . Notice that:
$\begin{array}{lll}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& \frac{{T}_{\text{L}x}}{{T}_{\text{L}}}\\ {T}_{\text{L}x}& =& {T}_{\text{L}}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)\\ \text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& \frac{{T}_{\text{R}x}}{{T}_{\text{R}}}\\ {T}_{\text{R}x}& =& {T}_{\text{R}}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right).\end{array}$
Equating ${T}_{\text{L}x}$ and ${T}_{\text{R}x}$ :
${T}_{\text{L}}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)={T}_{\text{R}}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right).$
Thus,
${T}_{\text{L}}={T}_{\text{R}}=T,$
as predicted. Now, considering the vertical components (denoted by a subscript $y$ ), we can solve for $T$ . Again, since the person is stationary, Newton’s second law implies that net ${F}_{y}=0$ . Thus, as illustrated in the free-body diagram in [link] ,
${F}_{\text{net}y}={T}_{\text{L}y}+{T}_{\text{R}y}-w=0.$
Observing [link] , we can use trigonometry to determine the relationship between ${T}_{\text{L}y}$ , ${T}_{\text{R}y}$ , and $T$ . As we determined from the analysis in the horizontal direction, ${T}_{\text{L}}={T}_{\text{R}}=T$ :
$\begin{array}{lll}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& \frac{{T}_{\text{L}y}}{{T}_{\text{L}}}\\ {T}_{\text{L}y}={T}_{\text{L}}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)\\ \text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& \frac{{T}_{\text{R}y}}{{T}_{\text{R}}}\\ {T}_{\text{R}y}={T}_{\text{R}}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right).\end{array}$
Now, we can substitute the values for ${T}_{\text{L}y}$ and ${T}_{\text{R}y}$ , into the net force equation in the vertical direction:
$\begin{array}{lll}{F}_{\text{net}y}& =& {T}_{\text{L}y}+{T}_{\text{R}y}-w=0\\ {F}_{\text{net}y}& =& T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)+T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)-w=0\\ 2\phantom{\rule{0.25em}{0ex}}T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)-w& =& 0\\ 2\phantom{\rule{0.25em}{0ex}}T\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)& =& w\end{array}$
and
$T=\frac{w}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)}=\frac{\text{mg}}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)},$
so that
$T=\frac{\left(\text{70}\text{.}\text{0 kg}\right)\left(9\text{.}{\text{80 m/s}}^{2}\right)}{2\left(0\text{.}\text{0872}\right)},$
and the tension is
$T=\text{3900 N}.$
Discussion
Note that the vertical tension in the wire acts as a normal force that supports the weight of the tightrope walker. The tension is almost six times the 686-N weight of the tightrope walker. Since the wire is nearly horizontal, the vertical component of its tension is only a small fraction of the tension in the wire. The large horizontal components are in opposite directions and cancel, and so most of the tension in the wire is not used to support the weight of the tightrope walker.
If we wish to create a very large tension, all we have to do is exert a force perpendicular to a flexible connector, as illustrated in [link] . As we saw in the last example, the weight of the tightrope walker acted as a force perpendicular to the rope. We saw that the tension in the roped related to the weight of the tightrope walker in the following way:
$T=\frac{w}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)}.$
We can extend this expression to describe the tension $T$ created when a perpendicular force ( ${\mathbf{\text{F}}}_{\perp }$ ) is exerted at the middle of a flexible connector:
$T=\frac{{F}_{\perp }}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)}.$
Note that $\theta$ is the angle between the horizontal and the bent connector. In this case, $T$ becomes very large as $\theta$ approaches zero. Even the relatively small weight of any flexible connector will cause it to sag, since an infinite tension would result if it were horizontal (i.e., $\theta =0$ and $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =0$ ). (See [link] .)
## Section summary
• When objects rest on a surface, the surface applies a force to the object that supports the weight of the object. This supporting force acts perpendicular to and away from the surface. It is called a normal force, $\mathbf{\text{N}}$ .
• When objects rest on a non-accelerating horizontal surface, the magnitude of the normal force is equal to the weight of the object:
$N=\text{mg}.$
• When objects rest on an inclined plane that makes an angle $\theta$ with the horizontal surface, the weight of the object can be resolved into components that act perpendicular ( ${\mathbf{\text{w}}}_{\perp }$ ) and parallel ( ${\mathbf{\text{w}}}_{\parallel }$ ) to the surface of the plane. These components can be calculated using:
${w}_{\parallel }=w\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)=\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)$
${w}_{\perp }=w\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)=\text{mg}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(\theta \right).$
• The pulling force that acts along a stretched flexible connector, such as a rope or cable, is called tension, $\mathbf{\text{T}}$ . When a rope supports the weight of an object that is at rest, the tension in the rope is equal to the weight of the object:
$T=\text{mg}.$
## Problem exercises
Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts an average force of 1350 N horizontally. Each of the second team’s members has an average mass of 73 kg and exerts an average force of 1365 N horizontally. (a) What is magnitude of the acceleration of the two teams? (b) What is the tension in the section of rope between the teams?
1. $0.{\text{11 m/s}}^{2}$
2. $1\text{.}2×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{N}$
(a) Calculate the tension in a vertical strand of spider web if a spider of mass $8\text{.}\text{00}×{\text{10}}^{-5}\phantom{\rule{0.25em}{0ex}}\text{kg}$ hangs motionless on it. (b) Calculate the tension in a horizontal strand of spider web if the same spider sits motionless in the middle of it much like the tightrope walker in [link] . The strand sags at an angle of $\text{12º}$ below the horizontal. Compare this with the tension in the vertical strand (find their ratio).
(a) $7\text{.}\text{84}×{\text{10}}^{-4}\phantom{\rule{0.25em}{0ex}}\text{N}$
(b) $1\text{.}\text{89}×{\text{10}}^{–3}\phantom{\rule{0.25em}{0ex}}\text{N}$ . This is 2.41 times the tension in the vertical strand.
Suppose a 60.0-kg gymnast climbs a rope. (a) What is the tension in the rope if he climbs at a constant speed? (b) What is the tension in the rope if he accelerates upward at a rate of $1\text{.}{\text{50 m/s}}^{2}$ ?
Consider the baby being weighed in [link] . (a) What is the mass of the child and basket if a scale reading of 55 N is observed? (b) What is the tension ${T}_{1}$ in the cord attaching the baby to the scale? (c) What is the tension ${T}_{2}$ in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kg? (d) Draw a sketch of the situation indicating the system of interest used to solve each part. The masses of the cords are negligible.
how can chip be made from sand
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers! | 3,783 | 11,868 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 59, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2021-21 | latest | en | 0.702018 |
http://reference.wolfram.com/legacy/v6/tutorial/GeneralizedInput.html | 1,511,033,381,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805023.14/warc/CC-MAIN-20171118190229-20171118210229-00227.warc.gz | 251,880,511 | 16,027 | This is documentation for Mathematica 6, which was
based on an earlier version of the Wolfram Language.
Mathematica Tutorial Functions »
# Generalized Input
The fundamental paradigm of most computer languages, including Mathematica, is that input is given and processed into output. Historically, such input has consisted of strings of letters and numbers obeying a certain syntax.
Evaluate this input line to generate a table of output.
Out[1]=
Starting in Version 3, Mathematica has supported the use of two-dimensional typeset mathematical notations as input, freely mixed with textual input.
This also generates a table of output.
Out[2]=
Starting with Version 6, a wide range of non-textual objects can be used as input just as easily, and can be mixed with text or typeset notations.
Evaluate the following input, then move the slider and evaluate it again to see a new result.
Out[3]=
The "value" of this slider when it is given as input is determined by its position, in this case an integer between 1 and 10. It can be used anywhere in an input line that a textual number or variable name could be used.
How to create such controls are discussed in the next section, but it is worth noting first that in many cases there are better alternatives to this kind of input.
Casting this example in the form of a Manipulate allows you to see the effect of moving the slider in real time as it is moved.
Out[4]=
But there are situations where using a control inside a traditional Shift+Enter evaluated input works better. Some cases are if the evaluation is very slow, if you want complete flexibility in editing the rest of the input line around the control(s), or if the point of the code is to make definitions that will be used later, and the controls are being used as a convenient way to specify initial values.
For example, you might want to set up a color palette using ColorSetter to initialize named colors that will be used in subsequent code.
Click any color swatch to get a dialog allowing you to interactively choose a new color. These values can then be used in subsequent programming just as if they had been initialized with more traditional textually specified values.
Out[8]=
These color swatches provide an informative, more easily edited representation of the colors than would an expression consisting of numerical color values.
## How to Create Inline Controls
The most flexible and powerful way to create anything in Mathematica is to evaluate a function that returns it.
These examples use Slider, but the same principles apply to any controls. Control Objects lists all the available control objects.
You can create a slider by evaluating Slider[].
Out[9]=
The resulting slider object can be copied and pasted into a subsequent input cell just as if it were any other kind of input. (Or, you can just click in the output cell and start typing, which will cause it to be converted automatically into an input cell.)
Controls created this way are inert to evaluation.
For example, type 2+, then paste the previous slider after the + to create this input line, and then evaluate it.
Out[10]=
When evaluated, the slider remains a slider, which is not wanted in this case (though it is very useful in other situations). What is needed instead is a slider that, when evaluated as input, becomes the value it is set to, rather than a slider object.
DynamicSetting[e] an object which displays as e, but is interpreted as the dynamically updated current setting of e upon evaluation
Object that evaluates into its current setting.
When DynamicSetting is wrapped around a slider and evaluated, the new slider looks identical to the original one, but has the hidden property of evaluating into its current setting.
This displays the new slider.
Out[11]=
If the new slider is copied and pasted into an input line, the act of evaluation transforms the slider into its current value (by default 0.5 if it has not been moved with the mouse).
Out[12]=
The examples in the previous section were created using DynamicSetting in this way.
While copying and pasting can be used very effectively to build up input lines containing controls, it is often most convenient to use Evaluate in Place Ctrl+Shift+Enter (Command+Return on Mac) to transform control expressions in place, especially once you are familiar with the commands that create controls.
Ctrl+Shift+Enter evaluate a selection "in place", replacing the selection with the output
Evaluating in place.
For example, enter the following input line.
Out[13]=
Then, highlight the entire control expression. (Triple-clicking the word DynamicSetting is an especially quick way to do this.)
Type Ctrl+Shift+Enter (Command+Return on Mac), and the control expression will be transformed in place into an actual control. (Note that Ctrl+Shift+Enter is not the normal Shift+Enter used for evaluating input.)
Evaluating the input cell with Shift+Enter will then give the desired result.
Out[15]=
All the arguments of Slider can be used to change the initial value, range, and step size.
Out[16]=
Then evaluate in place (Ctrl+Shift+Enter) to transform the text command into a slider.
Out[17]=
Of course, this works with other kinds of controls as well.
Out[18]=
This is the result after evaluating in place.
Out[19]=
Note that the control expressions do not contain a dynamic reference to a variable as they normally would (see "Introduction to Dynamic"). Controls used in input expressions as described here are static, inert objects, much like a textual command. They are not linked to each other, and nothing happens when you move one, except that it moves. Basically they are simply recording their current state, for use when you evaluate the input line.
## Complex Templates in Input Expressions
It is possible to use whole panels containing multiple controls in input expressions. Constructing such panels is more complex than simply wrapping DynamicSetting around a single control, because you have to specify how the various control values should be assembled into the value returned when the template is evaluated.
The function Interpretation is used to assemble a self-contained input template object, which may contain many internal parts that interact with each other through dynamic variables. The arguments are Interpretation[variables, body, returnvalue].
The first argument gives a list of local variables with optional initializers in the same format as Module or DynamicModule. (In fact, Interpretation creates a DynamicModule (see "Introduction to Dynamic") in the output.)
The second argument is typeset to become the displayed form of the interpretation object. Typically it contains formatting constructs and controls with dynamic references to the variables defined in the first argument.
The third argument is the expression that will be used as the value of the interpretation object when it is given as input. Typically this is an expression involving the variables listed in the first argument.
Interpretation[e,expr] an object which displays as e, but is interpreted as the unevaluated form of expr if supplied as input Interpretation[{x=x0,y=y0,...},e,expr] allows local variables in e and expr
Object that displays as one expression and evaluates as another.
Evaluating the following input cell creates an output cell containing a template for the Plot command.
Out[20]=
This template can be copied and pasted into an input cell, and the values edited as you like. Shift+Enter evaluation of the input cell generates a plot.
Out[21]=
In the following more sophisticated example, the variable definite is used to communicate between the controls in the template, dimming the min and max value fields when indefinite integration is selected.
Out[1]=
This copy of the previous template gives the integral upon evaluation.
Out[23]=
As with the single controls in earlier sections, these input templates can be copied and pasted into new input cells, and they can be freely intermixed with textual input.
To test the result of integration, wrap the template with D to take the derivative and verify that the result is the same as the starting point.
Out[24]=
These examples are fairly generic: they look like dialog boxes in a lot of programs. But there are some important differences. For example, note the x2 in the input field. Input fields in Mathematica may look like those in other programs, but the contents can be any arbitrary typeset mathematics, or even graphics or other controls. (See the next section to learn how to write templates that can be nested inside each other.)
Mathematica templates (and dialog boxes) are also not restricted to using a regular grid of text fields.
Here is a simple definite integration template laid out typographically.
Out[25]=
Note that you do not need a template to evaluate integrals; they can be entered as plain typeset math formulas using keyboard shortcuts (as described in "Entering Two-Dimensional Expressions") or the Basic Input palette.
Out[26]=
Whether it is useful to make a template like this or not depends on many things, but the important point is that in Mathematica the full range of formatting constructs, including text, typeset math, and graphics, is available both inside and around input fields and templates.
## Advanced topic: Dealing with Premature Evaluation in Templates
Templates defined like those in the previous section do not work as you might hope if the variables given in initializers already have other values assigned to them (for example, if the variable x has a value in the previous section), or if template structures are pasted into the input fields. To deal with evaluation issues correctly, it is necessary to use InputField objects that store their values in the form of unparsed box structures rather than expressions. (Box structures are like strings in the sense that they represent any possible displayable structure, whether it is a legal Mathematica input expression or not.)
This defines a template.
Out[27]=
This copy of the previous template gives the integral upon evaluation.
Out[28]=
This template will work properly even under what one might consider abuse. For example, you can nest it repeatedly to integrate a function several times.
Out[29]=
Note how the InputField grows automatically to accommodate larger contents. (This behavior is controlled by the FieldSize option.)
The typographic template can also be made robust to evaluation.
Out[30]=
And it can be nested, though this kind of thing can easily get out of hand, so it is probably more fun than useful.
Out[31]=
## Graphics as Input
Graphical objects, including the output of Graphics, Graphics3D, plotting commands, and graphics imported from external image files, can all be used as input and freely mixed with textual input. There are no arbitrary limitations in the mixing of graphics, controls, typeset mathematics, and text.
Evaluate a simple plot command.
Out[32]=
Then click to place the insertion point just to the left of the plot and type "Table[".
Complete the command by clicking and typing to the right of the plot, then evaluating.
Out[2]=
Notice how the plot appears in several different sizes depending on its context. There are three standard automatic plot sizes, the largest if the plot is by itself in an output, smaller if used in a list or table output, and smallest if in a textual input line. This is mainly a convenience to make graphical input less bulky. You are always free to resize the graphic by clicking it and using the resize handles, or by including an explicit ImageSize option.
You can import a bitmap image from a file.
Out[3]=
Then copy/paste this into an input cell to do simple image processing on it.
Out[35]=
The ability to use graphics as input allows for remarkably rich input, as in this simple Manipulate example.
Out[36]= | 2,430 | 11,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-47 | latest | en | 0.903785 |
https://www.abcteach.com/directory/subjects-math-money-645-6-5 | 1,505,990,807,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687740.4/warc/CC-MAIN-20170921101029-20170921121029-00104.warc.gz | 738,942,662 | 18,245 | You are an abcteach Member, but you are logged in to the Free Site. To access all member features, log into the Member Site.
# Money Worksheets
FILTER THIS CATEGORY:
= Member Site Document
• This Counting Money 4 Math is perfect to practice counting skills. Your elementary grade students will love this Counting Money 4 Math. Realistic coins and dollars enable the students to practice counting money.
• This Word Problems - School Supplies (elementary) is perfect to practice problem solving skills. Your elementary grade students will love this Word Problems - School Supplies (elementary). "Cathy bought a bottle of glue and a box of crayons. How much did she spend?" Use the price list to solve these simple addition problems.
• This Addition - Money (set 2) Worksheet is perfect to practice addition skills. Your elementary grade students will love this Addition - Money (set 2) Worksheet. 6 pages of math addition worksheets, concertrates on adding monetary amounts together to solve each problem. Answer key included.
• This Addition - Money (set 3) Worksheet is perfect to practice addition skills. Your elementary grade students will love this Addition - Money (set 3) Worksheet. 6 pages of math addition worksheets, concertrates on adding monetary amounts together to solve each problem. Answer key included.
• This Counting Money Is Fun!!! Math Worksheet is perfect to practice money skills. Your elementary grade students will love this Counting Money Is Fun!!! Math Worksheet. Printable math worksheet, covering monetary work, counting money.
• This Counting Money 1 Math is perfect to practice counting skills. Your elementary grade students will love this Counting Money 1 Math. Realistic coins enable the students to practice counting money.
• Use this 'Book: Arthur's TV Troubles (elementary)' printable worksheet in the classroom or at home. Your students will love this 'Book: Arthur's TV Troubles (elementary)'. Book comprehension and vocabulary enhancement for this installment of Marc Brown's popular "Arthur" series. Arthur has trouble with truth in advertising.
• This Word Problems 1 (K-1) Morning Math is perfect to practice beginning math skills. Your elementary grade students will love this Word Problems 1 (K-1) Morning Math. No matter what time you work with math, our new "Morning Math" series is a great way to open students' eyes to the daily uses of math. Simple word problems for beginning math students ("Draw 5 apples. Put an X over 3 of them. How many apples do you have?") start the series. Common Core: K.OA1, K.OA2, 1.OA1
• This Place Value (set 1) Math Game is perfect to practice place value skills. Your elementary grade students will love this Place Value (set 1) Math Game. Who has 2 tens and 5 ones? Develop listening skills, practice numbers 1-99, and practice tens and ones place values with this all-class game.
• 10 pages of practice in counting U.S. money. Created with our abctools Math Worksheet Generator. CC:MMD.C.8
• This Add Up the Change! Worksheet is perfect to practice addition skills. Your elementary grade students will love this Add Up the Change! Worksheet. Roll a die with coin pictures on the faces (available on abcteach) or draw coins out of a bag. Write the value of each coin in the row, then add up the row.
• This Match the Money (USA currency) Cut and Paste is perfect to practice money skills. Your elementary grade students will love this Match the Money (USA currency) Cut and Paste. Cut out the six money squares and paste them on the matching squares on the board.
• This Addition - Money (set 1) Worksheet is perfect to practice addition skills. Your elementary grade students will love this Addition - Money (set 1) Worksheet. 6 pages of math addition worksheets, concertrates on adding monetary amounts together to solve each problem. Answer key included.
• This Place Values 1 Morning Math is perfect to practice place value skills. Your elementary grade students will love this Place Values 1 Morning Math. Quick place value practice, great for morning math. | 876 | 4,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-39 | latest | en | 0.900555 |
https://www.physicsforums.com/threads/numerical-puzzle.244080/ | 1,611,610,089,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703644033.96/warc/CC-MAIN-20210125185643-20210125215643-00110.warc.gz | 921,208,337 | 13,873 | Numerical puzzle
Homework Statement
Im having problem of even starting this question:
"Determine the positive integer N = abcdef such that 6N = defabc"
any ideas?
The Attempt at a Solution
Related Precalculus Mathematics Homework Help News on Phys.org
$$6N = defabc = abcdef = N$$
6N doesn't equal N unless it's 0. The only solution is 0 which is not a positive integer.
Or am I missing something really stupid?
actually i ended up solving it
abc = 142 and def = 857
....abc and def are just the digits in the question. but thanks!
dynamicsolo
Homework Helper
Actually, anyone who recognizes the decimal expansions of 1/7 and 6/7 might suspect immediately that this is an answer... (Ah, but is it unique?)
Feldoh, this type of puzzle is what is referred to as a cryptarithm. Letters stand in for digits and the prospective solver is supposed to "decode" the arithmetic problem. In classical form, this problem would be posed as
ABCDEF
x 6
______
DEFABC
The question does not say that 6N = N , but that 6N is a digital rearrangment of N.
Regardless I think the question could be worded better...
dynamicsolo
Homework Helper
It did strike me that the intent of the problem might be obscure to anyone not familiar with this kind of puzzle. | 307 | 1,252 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-04 | latest | en | 0.950126 |
https://folu.me/post/wbuaqpbbx-d-dpbz/blog/2024/07/06/retrograde | 1,721,749,848,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518058.23/warc/CC-MAIN-20240723133408-20240723163408-00791.warc.gz | 218,249,876 | 40,327 | # Calculating when a planet will appear to move backwards
submited by
Style Pass
2024-07-07 00:30:03
When we say that the planets in our solar system orbit the sun, not the earth, we mean that the motions of the planets is much simpler to describe from the vantage point of the sun. The sun is no more the center of the universe than the earth is. Describing the motion of the planets from the perspective of our planet is not wrong, but it is inconvenient. (For some purposes. It’s quite convenient for others.)
The word planet means “wanderer.” This because the planets appear to wander in the night sky. Unlike stars that appear to orbit the earth in a circle as the earth orbits the sun, planets appear to occasionally reverse course. When planets appear to move backward this is called retrograde motion.
Venus completes 13 orbits around the sun in the time it takes Earth to complete 8 orbits. The ratio isn’t exactly 13 to 8, but it’s very close. Five times over the course of eight years Venus will appear to reverse course for a few days. How many days? We will get to that shortly.
When we speak of the motion of the planets through the night sky, we’re not talking about their rising and setting each day due to the rotation of the earth on its axis. We’re talking about their motion from night to night. The image above is how an observer far above the Earth and not rotating with the Earth would see the position of Venus over the course of eight years. | 327 | 1,471 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-30 | latest | en | 0.92107 |
https://ir.lzu.edu.cn/handle/262010/224749 | 1,675,269,744,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499946.80/warc/CC-MAIN-20230201144459-20230201174459-00815.warc.gz | 348,029,316 | 24,623 | 三类非线性反应扩散方程的无网格配置方法 Alternative Title Meshfree collocation methods for three kinds of nonlinear reaction diffusion equations 肖艳敏 Thesis Advisor 伍渝江 2011-06-01 Degree Grantor 兰州大学 Place of Conferral 兰州 Degree Name 硕士 Keyword 无网格配置方法 广义Fisher方程 Allen-Cahn方程 奇异反应扩散方程 径向基函数插值 Abstract 自然界中许多生物, 物理, 化学问题的数学模型都可以用微分方程的形式给出, 而这些微分方程的精确解大多数是无法求出的. 随着计算机技术的迅速发展,可以利用计算机求方程的数值解或者模拟这些现象. 无网格方法是近年来迅速发展起来的一种新兴的科学计算方法, 它任意选取求解域内的坐标点构造插值基函数的组合逼近微分方程的解, 不需要或较少的需要网格剖分, 这样就可以很容易的模拟许多复杂的自然现象. 目前, 这种方法已经成为求解偏微分方程数值解的 一种重要方法. 在这篇文章中, 主要是利用基于径向基函数插值的无网格配置方法求解一些非线性反应扩散方程的数值解, 这些方程有广义Fisher方程,Allen-Cahn方程和奇异反应扩散方程. 此方法可以把非线性问题转化为线性问题, 与传统的有限元和有限差分相比不需要网格剖分, 简单灵活易推广到多维等优点. 它本身的这些优点促使它迅速发展并成为科学研究者进行科学和工程研究的重要方法. 文章最后的数值结果说明了这种方法的有效性. Other Abstract Many mathematical models of biology, physics and chemistry in nature can be represented by differential equations, but most of differential equations’ exact solutions can’t be found. With the swift development of Computer technology, We can figured out the numerical solution of differential equations by computer and simulate natural phenomenon . Meshfree method is a rapid development new numerical method in recent years. Compared with the traditional finite element method and finite difference methods, it is with an arbitrary take fixed coordinates tectonic interpolation basis function discrete partial differential equation, partially or completely mesh dissection,This makes it easy to simulate many complex natural phenomenon. At present,it has become an important approach for solving partial differential equations. This paper performs a meshfree collocation method based on the interpolation of radial basis functions(RBFs) for solving the nonlinear reaction-diffusion equations.These problems concern generalized Fisher equation, Allen-Cahn equation and Nonlinear singular reaction-diffusion equations. A nonlinear problem is changed into a linear one via this method. Compared with the traditional finite element method and finite differential method, the new algorithm in this paper doesn’t need any mesh and it is simple and vivid to extend hign-dimensional problems. Some numerical results and simulations are also presented to demonstrate the effectiveness of the method. URL 查看原文 Language 中文 Document Type 学位论文 Identifier https://ir.lzu.edu.cn/handle/262010/224749 Collection 数学与统计学院 Recommended CitationGB/T 7714 肖艳敏. 三类非线性反应扩散方程的无网格配置方法[D]. 兰州. 兰州大学,2011.
Files in This Item: There are no files associated with this item.
Related Services Recommend this item Bookmark Usage statistics Export to Endnote Altmetrics Score Google Scholar Similar articles in Google Scholar [肖艳敏]'s Articles Baidu academic Similar articles in Baidu academic [肖艳敏]'s Articles Bing Scholar Similar articles in Bing Scholar [肖艳敏]'s Articles Terms of Use No data! Social Bookmark/Share
No comment.
Items in the repository are protected by copyright, with all rights reserved, unless otherwise indicated. | 880 | 2,934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-06 | latest | en | 0.462575 |
https://www.jiskha.com/questions/1782688/a-30-n-crate-is-submerged-in-water-the-amount-of-buoyant-force-acting-on-the-crate-is-25 | 1,611,365,822,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703531702.36/warc/CC-MAIN-20210123001629-20210123031629-00614.warc.gz | 866,077,042 | 5,448 | # science
A 30 N crate is submerged in water. The amount of buoyant force acting on the crate is 25 N. What will the crate do?
The crate will float and accelerate upwards. The only force acting on the crate is the buoyant force, and it acts upwards on the crate.
The crate will sink and accelerate downwards. The only force acting on the crate is the buoyant force, and it acts downwards on the crate.
The crate will sink and accelerate downwards. The buoyant force acting upwards on the crate is smaller than the downward force of gravity on the crate.
The crate will sink and accelerate downwards. The buoyant force and the gravitational force are both acting downwards on the crate, making it accelerate down.
is it the first one?
1. 👍
2. 👎
3. 👁
1. the forces acting on the crate are gravity and buoyancy
what's the net result?
1. 👍
2. 👎
2. The answer is "The crate will sink and accelerate downwards. The buoyant force acting upwards on the crate is smaller than the downward force of gravity on the crate."
1. 👍
2. 👎
3. Grace is correct, it's "The crate will sink and accelerate downwards. The buoyant force acting upwards on the crate is smaller than the downward force of gravity on the crate"
1. 👍
2. 👎
4. and you had to literally restate the exact same thing as her because?
1. 👍
2. 👎
## Similar Questions
1. ### Physics
An empty rubber balloon has a mass of 0.012kg. the balloon is filled with helium at 0 degrees celcius, 1 atm pressure, and a density of 0.179 kg/m3. the filled balloon has a radius of 0.5 m. a)what is the magnitude of the buoyant
2. ### Physics
Sally applies a horizontal force of 500 N with a rope to dtag a wooden crate across a floor with a constant speed. The rope tied to the crate is pulled at an angle of 67.5 degrees. a) How much force is exerted by the rope on the
3. ### Physics 11
A 140 kg stationary crate is pulled by a force of 390 N along a horizontal surface. a) The coefficient of static friction between crate and surface is s = 0.25. Is the applied force large enough to begin moving the crate? Use
4. ### physics
A 57.0 kg diver dives from a height of 15.0 m. She reaches a speed of 14.0 m/s just before entering the water. (a) What was the average force of air resistance (e.g., friction) acting on the diver. (b) What is the force of
1. ### PHYSICS
Several people are riding in a hot-air balloon. The combined mass of the people and balloon is 308 kg. The balloon is motionless in the air, because the downward-acting weight of the people and balloon is balanced by an
2. ### Physics
A large crate filled with physics laboratory equipment must be moved up an incline onto a truck. 1. The crate is at rest on the incline. What can you say about the force of friction acting on the crate? a. The friction force point
3. ### Physical science
Hello! 1.How can the density of a hollow lead ball filled with water be increased? a.Increase the size of the ball and add more water.* b.Remove the water and replace it with air. c.Remove the water and melt the ball into a solid
4. ### physics
Several people are riding in a hot-air balloon. The combined mass of the people and balloon is 310 kg. The balloon is motionless in the air, because the downward-acting weight of the people and balloon is balanced by an
1. ### physics
A crate of mass m (=40 kg) is placed at rest on a (frictionless) inclined plane, which has an angle (= 30o) above horizontal. a.) The force of gravity on the crate is: b.) The normal force on the crate is: size: N c.) The net
2. ### Physics
An empty rubber balloon has a mass of 0.0120 kg. The balloon is filled with helium at 0 degrees celsuis, 1 atm pressure, and a density of 0.179kg/m^3. The filled balloon has a radius of .5m a)What is the magnitude of the net force
3. ### Science
A block of wood weighing 25 N is submerged under water. If the net force on the wood is 5 N upward, how much buoyant force is acting on the wood? A. 30 N upward B. 25 N downward C. 20 N upward D. 5 N upward
4. ### physics
a metal object is suspended from a spring balance scale. the scale reads 800 N when it is suspended in air and 700 N when the object is completely submerged in water. find the buoyant force, the volume and the density of the | 1,086 | 4,238 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2021-04 | latest | en | 0.943052 |
https://avesis.akdeniz.edu.tr/yayin/3aa56ff0-46e3-4cf1-a9b0-828066f8f59f/bessel-quasilinearization-technique-to-solve-the-fractional-order-hiv-1-infection-of-cd4-t-cells-considering-the-impact-of-antiviral-drug-treatment | 1,721,337,752,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514859.56/warc/CC-MAIN-20240718191743-20240718221743-00504.warc.gz | 87,994,472 | 13,601 | ## Bessel-quasilinearization technique to solve the fractional-order HIV-1 infection of CD4+ T-cells considering the impact of antiviral drug treatment
Applied Mathematics and Computation, vol.431, 2022 (SCI-Expanded)
• Publication Type: Article / Article
• Volume: 431
• Publication Date: 2022
• Doi Number: 10.1016/j.amc.2022.127319
• Journal Name:
• Journal Indexes: Science Citation Index Expanded (SCI-EXPANDED), Scopus, Academic Search Premier, Applied Science & Technology Source, Computer & Applied Sciences, INSPEC, Public Affairs Index, zbMATH, Civil Engineering Abstracts
• Keywords: Bessel polynomials, Caputo's fractional derivative, Collocation points, Error and convergence analysis, Fractional-order HIV-1 infection model of CD4+ T-cells, Nonlinear differential equations, The technique of quasilinearization
• Akdeniz University Affiliated: Yes
#### Abstract
© 2022 Elsevier Inc.In this paper, two numerical methods based on the novel Bessel polynomials are developed to solve the fractional-order HIV-1 infection model of CD4+ T-cells considering the impact of antiviral drug treatment. In first of these methods, by using the Bessel polynomial and collocation points, we transform the HIV problem into a system of nonlinear algebraic equations. And this method, which is the method of direct solution is called as Bessel matrix method. The second method, which is called the Bessel-QLM method converts firstly HIV problem to a sequence of linear equations by using the technique of quasilinearization and then the reduced problem is solved by the direct Bessel matrix method. Error and convergence analysis are studied for the Bessel method. Finally, the applications are made on the numerical examples and also the numerical results are compared with the results of other available techniques. It is observed from applications that the presented results are better than the results of other existing methods and also the Bessel-QLM method is more efficient than the direct Bessel method. | 436 | 2,011 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-30 | latest | en | 0.880268 |
https://nivent.github.io/blog/Modular-Arithmetic/ | 1,719,319,856,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865972.21/warc/CC-MAIN-20240625104040-20240625134040-00537.warc.gz | 375,694,537 | 18,129 | # Modular Arithmetic
I plan on writing a couple posts related to cryptography soon. Before I do that, I wanted to have a post covering some of the basics of modular arithmetic first, because this material will be needed for the cryptographic posts.
## Divisibility
The first thing to know is that modular arithmetic is all about integers. We do not care about rationals, reals, or anything else, only integers 1. As such, all our definitions have to be written in terms of integers, and the first such definition is that of divisibility. It is tempting to say that $a$ divides $b$ iff $\frac ba$ is an integer, but writing $\frac ba$ means we are entering the realm of rationals, which we cannot do. Therefore, we instead say that $a$ divides $b$ iff there exists an integer $k$ such that $b=ak$. This way we are defining divisibility in terms of multiplication (which is defined for integers) instead of division (which is not defined for integers in general). We write $a|b$ when $a$ divides $b$. The following are a few properties of divisibility.
$$$\text{If } a|b \text{ and } a|c \text{, then } a|(b\pm c)\\ \text{If } a|b \text{ and } a|c \text{, then } a|bc$$$
## Mod
In the previous section I said that division was not defined for integers in general. That was a bit of a lie. It is, just not in the way we normally think of division.
Division algorithm:
For any two integers $a$ and $b\not=0$, there exists a unique pair of integers $q$ and $0\le{r}<|b|$ such that $a=qb+r$. We call $a$ the dividend, $b$ the divisor $q$ the quotient, and $r$ the remainder.
From this theorem, we get the following definition.
Definition of $a\bmod b$:
If $a=qb+r$ where $0\le{r}<|b|$, then we say that $a\bmod b=r$.
In other words, $a\bmod b$ is the remainder when $a$ is divided by $b$. Mod has some nice properties, such as
\begin{align*} [(a\bmod n)+(b\bmod n)]\bmod n&=(a+b)&\bmod n\\ (a\bmod n)(b\bmod n)\bmod n&=ab&\bmod n \end{align*}
These properties motivate the following definition.
Definition of $a\equiv b\pmod n$:
We write $a\equiv b\pmod n$ iff $(a-b)|n$. We read this as “$a$ is congruent to $b$ modulo $n$”.
Note that the above definition is equivalent to saying that $a\equiv b\pmod n$ iff $a\bmod n=b\bmod n$ 2. It is easy to see that this new form of mod inherits some nice properties from the one we first introduced. Namely,
$$$\text{If } a\equiv b\pmod n \text{ and } c\equiv d\pmod n \text{, then the following are true}\\ a+c\equiv b+d\pmod n\\ a-c\equiv b-d\pmod n\\ ac\equiv bd\pmod n$$$
Finally, note that for all $a$, there exits a $b$ in $\{0,1,2,\dots,n-1\}$ such that $a\equiv b\pmod n$. This is because $0\le{a}\bmod n<n$. This realization motivates the following definition.
Definition of $\mathbb Z_n$:
$\mathbb Z_n=\{a\in\mathbb Z:0\le{a}<|n|\}$ 3
## Euclidean Algorithm
We begin this section with a definition or two.
A few definitions:
We call $d$ the greatest common divisor of $a$ and $b$ if $d|a$, $d|b$, and no larger integer divdes both $a$ and $b$. We write $\gcd(a,b)=d$ to denote that $d$ is the greatest common divisor of $a$ and $b$.
We say that $a$ and $b$ are coprime (or relatively prime) if they have no common divisors other than $1$. Equivalently, $a$ and $b$ are coprime iff $\gcd(a,b)=1$.
We will see some of the uses of these concepts later in this post, but first, it would be nice if we had a way of calculating the greatest common divisior of two numbers. We could check each number that was less than them and find the largest one that divides both of them that way, but that would be time consuming. For a faster method, we use the Euclidean Algorithm.
The idea is to define a sequence of numbers $r_0, r_1, r_2,\dots,r_n$ such that $\gcd(a,b)=r_{n-1}$. To do this, we start with $r_0=a$ and $r_1=b$ where $\mid a\mid\ge\mid b\mid$ 4. We then define subsequent $r$’s by division: $r_{k-1} = qr_k + r_{k+1}$. In other words, $r_{k+1}=r_{k-1}\bmod r_k$. We keep doing this until we finally reach $r_n=0$. Our answer is then $r_{n-1}$. This algorithm can be implemented in (python) code as follows:
def gcd(a,b):
if abs(b) > abs(a):
return gcd(abs(b),abs(a))
elif b==0:
return abs(a)
else:
return gcd(b,a%b)
To finish this section, let’s calculate the greatest common divisor of $23456$ and $123456$. $\begin{matrix} 123456 &=& 5 &*& 23456 &+& &6176\\ 23456 &=& 3 &*& 6176 &+& &4928\\ 6176 &=& 1 &*& 4928 &+& &1248\\ 4928 &=& 3 &*& 1248 &+& &1184\\ 1248 &=& 1 &*& 1184 &+& &64\\ 1184 &=& 18 &*& 64 &+& &32\\ 64 &=& 2 &*& 32 &+& &0 \end{matrix}$
Therefore, $\gcd(123456,23456)=32$.
## Analyzing the Euclidean Algorithm
This section will not be needed to understand the rest of this post and can be skipped. In it, we will discuss the Euclidean Algorithm slightly more formally, proving that it gives the correct answer and does so “quickly”.
Theorem 1:
For any $a$ and $b$, the Euclidean Algorithm terminates.
Pf: Let $a$ and $b$ be any integers 5 where, without loss of generality, $\mid a\mid\ge\mid b\mid$. Then, we claim that the sequence of remainders in the Euclidean Algorithm – $r_0=a,r_1=b,\dots,r_n$ – is finite. First, since $r_{k+1}=r_{k-1}\bmod r_k$ for $k\ge1$, we know that $r_k\ge0$ for $k\ge2$. Second, $r_{k+1}\in\mathbb Z_{r_k}$, so $\mid r_{k+1}\mid<\mid r_k\mid$. Therefore, the $r_k$’s form a strictly decreasing (in magnitude) sequence of integers greater than or equal to $0$ for $k\ge2$. Thus, there must be finitely many $r_k$’s. Else, the sequence would eventually be less than $0$, a contradiction. $\square$
Theorem 2:
For any $a$ and $b$, the Euclidean Algorithm gives the correct value.
Pf: Let $a$ and $b$ be any integers, and let $r_0,r_1,\dots,r_n$ be the sequence of remainders attained from the Euclidean Algorithm. Then, we claim that $\gcd(a,b)=r_{n-1}$. To see this, first note that $\gcd(0,n)=n$ for all $n$ and so $\gcd(r_{n-1},r_n)=\gcd(r_{n-1},0)=r_{n-1}$. We proceed to prove our claim by showing that $\gcd(r_{k-1},r_k)=\gcd(r_k,r_{k+1})$ for all $k$. Recall that $r_{k-1} = qr_k + r_{k+1}$. Let $d$ be any common divisor of $r_{k-1}$ and $r_k$. Then, $d$ divides $r_{k+1}=r_{k-1}-qr_k$, so $d$ is a common divisor of $r_k$ and $r_{k+1}$. Similarily, if $d'$ is a common divisor of $r_k$ and $r_{k+1}$, then $d$ also divides $r_{k-1}=qr_k+r_{k+1}$. As such, the common divisors of $r_{k-1}$ and $r_k$ are exactly the common divisors of $r_k$ and $r_{k+1}$. Namely, $\gcd(r_{k-1}, r_k)=\gcd(r_k, r_{k+1})$. It follows by induction that $r_{n-1}=\gcd(r_{n-1},r_n)=\gcd(r_0,r_1)=\gcd(a,b)$. $\square$
Lemma 1:
Given $a$ and $b$ where $a\ge b\ge0$, let $r_0=a,r_1=b,\dots,r_n$ be the sequence of remainders attained from the Euclidean Algorithm. Then, $r_{k+2}\le\frac12r_k$ for all $k$.
Pf: We proceed with a proof by contradiction. Suppose that for some $k$, $r_{k+2}>\frac12r_k$. Then, we have $r_k\ge r_{k+1}>r_{k+2}>\frac12r_k$ and $r_k=qr_{k+1}+r_{k+2}$. Since $r_{k+1}>\frac12r_k$ and $r_k\not=r_{k+2}$, we know that $q=1$, so $r_k=r_{k+1}+r_{k+2}$. However, since $r_{k+1}>\frac12r_k$ and $r_{k+2}>\frac12r_k$, $r_k=r_{k+1}+r_{k+2}>\frac12r_k+\frac12r_k=r_k$, which is impossible. Therefore, there must not exist a $k$ with $r_{k+2}>\frac12r_k$. In other words, for all $k$, $r_{k+1}\le\frac12r_k$. $\square$
Theorem 3:
Given $a$ and $b$ where $a\ge b\ge0$, let $r_0=a,r_1=b,\dots,r_n$ be the sequence of remainders attained from the Euclidean Algorithm. Then, $n\le2\log_2b + 3$.
Pf: Let $m=2\lfloor\log_2b\rfloor+2=2(\lfloor\log_2b\rfloor+1)$ and note that, by the preceeding lemma, $r_{1+2k}\le{r_1}/2^k$. Therefore, $r_{1+m}\le r_1/2^{\lfloor\log_2b\rfloor+1}<1$. It follows from this that $n\le1+m\le2\log_2b+3$. $\square$
## Division $\bmod n$
One thing that may be surprising to hear is that modular arithmetic allows for division (sometimes). The division problem is as follows. Given $a$ and $n$, find a $b$ such that $ab\equiv1\pmod n$. We will simply take for granted the fact that such a $b$ exits iff $a$ and $n$ are coprime and that such a $b$ is unique (up to modular congruence) when it does exits. Furthermore, without going in to the details here, I will mention that $b$ and be computed from $a$ and $n$ using the Extended Euclidean Algorithm or Fermat’s Little Theorem. If $b$ exists, then we call it the inverse of $a\pmod p$. Motivated by the fact that the inverse of $a$ exists only when $a$ is coprime to $n$, we present the following definition.
Defintion of $\mathbb Z_n^*$:
$\mathbb Z_n^*$ is the set of all numbers less than or equal to $n$ that are coprime to $n$. This set is sometimes referred to as $U(n)$.
## Fermat’s Little Theorem
Fermat’s Little Theorem:
If $p$ is a prime number, then for any integer $a$ $a^p\equiv a\pmod p$. Furthermore, if $a$ is not divisible by $p$, then $a^{p-1}\equiv1\pmod p$.
This theorem is stated here without proof. The vigilent reader 6 will notice that if $p$ is prime then all $a\in\mathbb Z_p-\{0\}$ have an inverse $\pmod p$ and that this inverse is $b\equiv a^{p-2}\pmod p$.
## Euler’s Totient Function
Euler’s Totient Function, also known as Euler’s phi Function, is a function $\phi:\mathbb Z^+\rightarrow\mathbb Z^+$. $\phi(n)$ is defined as the number of integers less than or equal to $n$ that are coprime to $n$. So, $\phi(n)=|\mathbb Z_n^*|$. The following theorem’s involving Euler’s Totient Function are presented without proof.
Theorem 4:
If $n=p^m$ for prime $p$, then $\phi(n)=p^{m-1}(p-1)$.
Theorem 5:
$\phi(ab)=\phi(a)\phi(b)$ if $a$ and $b$ are coprime.
Theorem 6:
If $a$ is coprime to $b$, then $a^{\phi(b)}\equiv1\pmod b$.
1. Unless otherwise stated, assume any variable written in this post can only have an integer value.
2. This definition was also motivated by the desire to write mod less frequently.
3. For those of you with some familiarity in algebra, this is a group under addition for all n and field under addition and multiplication when n is prime.
4. If b is bigger than a, then simply swap a and b.
5. Probably should have mentioned this earlier. When a and b are negative, we make use of the fact that gcd(a,b)=gcd(|a|,|b|) and compute their gcd that way.
6. The truly vigilant reader will notice that I misspelled vigilant. | 3,387 | 10,183 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-26 | latest | en | 0.877712 |
https://converter.ninja/volume/imperial-cups-to-centiliters/56-brcup-to-cl/ | 1,606,416,501,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188899.42/warc/CC-MAIN-20201126171830-20201126201830-00310.warc.gz | 259,778,841 | 5,412 | # 56 imperial cups in centiliters
## Conversion
56 imperial cups is equivalent to 1591.1315 centiliters.[1]
## Conversion formula How to convert 56 imperial cups to centiliters?
We know (by definition) that: $1\mathrm{brcup}\approx 28.4130625\mathrm{centiliter}$
We can set up a proportion to solve for the number of centiliters.
$1 brcup 56 brcup ≈ 28.4130625 centiliter x centiliter$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{centiliter}\approx \frac{56\mathrm{brcup}}{1\mathrm{brcup}}*28.4130625\mathrm{centiliter}\to x\mathrm{centiliter}\approx 1591.1315\mathrm{centiliter}$
Conclusion: $56 brcup ≈ 1591.1315 centiliter$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 centiliter is equal to 0.000628483566568822 times 56 imperial cups.
It can also be expressed as: 56 imperial cups is equal to $\frac{1}{\mathrm{0.000628483566568822}}$ centiliters.
## Approximation
An approximate numerical result would be: fifty-six imperial cups is about one thousand, five hundred and ninety-one point one three centiliters, or alternatively, a centiliter is about zero times fifty-six imperial cups.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic. | 392 | 1,368 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-50 | latest | en | 0.693813 |
http://www.progtown.com/topic2086891-result-of-request-column-to-break-and-deduce-in-ncolumns.html | 1,537,288,209,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155561.35/warc/CC-MAIN-20180918150229-20180918170229-00328.warc.gz | 306,957,333 | 20,387 | #### Topic: Result of request (column) to break and deduce in N-columns
I welcome !
Whether it is possible to deduce at once means SQL result of request in some N-columns?
Probably proceeding from a record count.
Result:
NUMBERS
7312312
7314314
7318318
7319319
7324324
7326326
7329329
7330233
7331833
7331933
7332133
7332433
7332633
7332833
7334133
7334533
7334633
7334833
NUMBERS NUMBERS NUMBERS NUMBERS
7312312 7326326 7332133 7334133
7314314 7329329 7332433 7334533
7318318 7330233 7332633 7334633
7319319 7331833 7332833 7334833
7324324 7331933
Result of the most simple request:
``````SELECT
ph. NOMERA AS NOMERA
FROM
NUMBERS ph``````
#### Re: Result of request (column) to break and deduce in N-columns
``````with t as (
select empno e1;
lead (empno) over (order by rownum) e2;
lead (empno, 2) over (order by rownum) e3;
lead (empno, 3) over (order by rownum) e4;
rownum rn
from emp
)
select e1;
e2;
e3;
e4
from t
where mod (rn, 4) = 1
order by rn
/
E1 E2 E3 E4
---------- ---------- ---------- ----------
7369 7499 7521 7566
7654 7698 7782 7788
7839 7844 7876 7900
7902 7934
SQL>``````
SY.
#### Re: Result of request (column) to break and deduce in N-columns
Alexander Warlord;
``````SQL> select e1, e2, e3, e4 from
2 (select empno, mod (rownum-1,4) c, trunc ((rownum-1)/4) r from emp)
3 pivot (max (empno) for c in (0 as e1,1 as e2, 2 as e3,3 as e4))
4 order by r
5 /
E1 E2 E3 E4
-------- -------- -------- --------
7369 7499 7521 7566
7654 7698 7782 7788
7839 7844 7876 7900
7902 7934``````
....
stax
#### Re: Result of request (column) to break and deduce in N-columns
Thanks!) thought just through mod () to do, only not especially it turned out.
#### Re: Result of request (column) to break and deduce in N-columns
SY;
``````Execution Plan
----------------------------------------------------------
Plan hash value: 773907260
------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU) | Time |
------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 14 | 910 | 5 (40) | 0:00:01 AM |
| 1 | SORT ORDER BY | | 14 | 910 | 5 (40) | 0:00:01 AM |
|* 2 | VIEW | | 14 | 910 | 4 (25) | 0:00:01 AM |
==> | 3 | WINDOW SORT | | 14 | 56 | 4 (25) | 0:00:01 AM |
| 4 | COUNT | | | | | |
| 5 | TABLE ACCESS FULL | EMP | 14 | 56 | 3 (0) | 0:00:01 AM |
------------------------------------------------------------------------------
Predicate Information (identified by operation id):``````
Interesting, what for to sort? Sampling is arranged on on rownum
....
stax
#### Re: Result of request (column) to break and deduce in N-columns
Stax wrote:
it is interesting, what for to sort? Sampling is arranged on on rownum
RN it ROWNUM in CTE a not in main query. And I have no concept as the optimizer will fulfill main query - suddenly a? Therefore I write ORDER BY RN. And why the optimizer not glades WINDOW SORT is to Larry
SY.
#### Re: Result of request (column) to break and deduce in N-columns
Though under WINDOW SORT calculation LEAD, therefore there is no saying disappears also and whether there was boy SORT or not.
SY.
#### Re: Result of request (column) to break and deduce in N-columns
SY;
Hoped that will be without a sort ( WINDOW BUFFER)
.....
stax
#### Re: Result of request (column) to break and deduce in N-columns
SY wrote:
Though under WINDOW SORT calculation LEAD, therefore there is no saying disappears also and whether there was boy SORT or not.
Well and how differently to count lead? Without sorting. All remaining variants look farfetched.
Stax wrote:
Hoped that will be without a sort ( WINDOW BUFFER)
WINDOW BUFFER here cannot be in any way because analytical sorting is fulfilled by the first. Therefore here pure window sort. On the other hand, it would be more logical to expect that the second sorting will not be. Though actually on this step it, goodness knows, can and is not produced, and the step is left in the plan to show that the optimizer did not forget that the data should be produced in sorting order. Also can be in its upcoming versions (this step) replace any order by nosort.
#### Re: Result of request (column) to break and deduce in N-columns
it is passed...
Well and how differently to count lead? Without sorting. All remaining variants look farfetched.
it is passed...
WINDOW BUFFER here cannot be in any way because analytical sorting is fulfilled by the first. Therefore here pure window sort. On the other hand, it would be more logical to expect that the second sorting will not be. Though actually on this step it, goodness knows, can and is not produced, and the step is left in the plan to show that the optimizer did not forget that the data should be produced in sorting order. Also can be in its upcoming versions (this step) replace any order by nosort.
About the second at me questions are not present
``````1 select empno
2, lead (empno) over (order by empno) le
3* from emp where EMPNO> 7369
SQL> /
EMPNO LE
-------- --------
7499 7521
7521 7566
7566 7654
7654 7698
7698 7782
7782 7788
7788 7839
7839 7844
7844 7876
7876 7900
7900 7902
7902 7934
7934
13 rows selected.
Execution Plan
----------------------------------------------------------
Plan hash value: 4166611438
---------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU) | Time |
---------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 13 | 52 | 1 (0) | 0:00:01 AM |
| 1 | WINDOW BUFFER | | 13 | 52 | 1 (0) | 0:00:01 AM |
|* 2 | INDEX RANGE SCAN | I\$EMP\$ENPNO | 13 | 52 | 1 (0) | 0:00:01 AM |
---------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
2 - access ("EMPNO"> 7369)
SQL> set autotrace off``````
There is no sorting for LEAD
stax
Stax;
It agree.
#### Re: Result of request (column) to break and deduce in N-columns
Well and how differently to count lead? Without sorting.
Generally - in any way. And here in a a case with ORDER BY ROWNUM or ORDER BY 1 a there is no need.
Simply with ORDER BY 1 optimizer knows that sorting according to a literal is senseless and here ROWNUM to it the type a variable "and therefore o it c gets as". C ORDER BY 1:
``````SQL> explain plan for
2 select lead (sal) over (order by 1)
3 from emp
4 /
Explained.
SQL> select * from table (dbms_xplan.display)
2 /
PLAN_TABLE_OUTPUT
------------------------------------------------------------------------------
Plan hash value: 1066788578
---------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU) | Time |
---------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 14 | 56 | 3 (0) | 0:00:01 AM |
| 1 |>>> WINDOW BUFFER <<<| | 14 | 56 | 3 (0) | 0:00:01 AM |
| 2 | TABLE ACCESS FULL | EMP | 14 | 56 | 3 (0) | 0:00:01 AM |
---------------------------------------------------------------------------
9 rows selected.
SQL>``````
SY.
#### Re: Result of request (column) to break and deduce in N-columns
By the way, if o ORDER BY 1 sortings will not be and we receive as ROWNUM. Here from the point of view of result basically it is not very well sorted/not it is sorted, the main thing that for all LEAD was fulfilled equally.
SY.
#### Re: Result of request (column) to break and deduce in N-columns
SY wrote:
By the way if o ORDER BY 1 sortings will not be and we receive as ROWNUM. Here from the point of view of result basically it is not very well sorted/not it is sorted, the main thing that for all LEAD was fulfilled equally.
SY.
c 1 as it there will be added, and from with rownum should be guaranteed equally
.....
stax | 2,185 | 7,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-39 | latest | en | 0.517633 |
https://www.hopepapers.org/operations-management-solve-problems/ | 1,576,129,096,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540537212.96/warc/CC-MAIN-20191212051311-20191212075311-00293.warc.gz | 734,776,595 | 13,382 | # Operations Management- solve problems
Operations Management- solve problemsOperations ManagementQ1 Emergency calls to Winter Park, Florida?s 911 systems for the past 24 weeks are as follows:(a) Compute the exponentially smoothed forecast of calls for each week. Assume an initial forecast of 50 calls in the first week and use alpha = 0.1. What is the forecast for the 25th week? (b) Reforecast each period using alpha = 0.6. (c) Actual calls during the 25th week were 85. Which smoothing constant provides a superior forecast? ___________________________________________________________________Q2The Outdoor Furniture Corporation manufacturing two products, benches and picnic tables, for use in yards and parks. The firm has two main resources: its carpenters (labor force) and asupply of redwood for use in the furniture. During the next production cycle, 1,200 hours of labor are available under a union agreement. The firm also has a stock of 3500 feet of good-qualityredwood. Each bench that Outdoor Furniture produces requires 4 labor hours and 10 feet of redwood; each picnic table takes 6 labor hours and 35 feet of redwood. Completed benches will yielda profit of \$9 each, and tables will result in a profit of \$20 each. How many benches and tables should Outdoor Furniture produce to obtain the largest possible profit? Q3 MSA Computer Corporation manufactures two models of minicomputers, the Alpha 4 and the Beta 5. The firm employs five technicians, working 160 hours each per month, on its assembly line.Management insists that full employment (i.e., all 160 hours of time) be maintained for each worker during next month?s operations. It requires 20 labor hours to assemble each Alpha 4computerand 25 labor hours to assemble each Beta 5 model. MSA wants to see at least 10 Alpha4s and at least 15 Beta 5s produced during the production period. Alpha 4s generate \$1,200 profit per unit,and Beta 5s yield \$1,800 each. Determine the most profitable number of each model of minicomputer to produce during the coming month?Q4 The Krampf Lines Railway Company specializes in coal handling. On Friday, April 13, Krampf had empty cars at the following towns in the quantities indicated TOWN SUPPLY OF CARS Morgantown 35 Youngstown 60 Pittsburgh 25 By Monday, April 16, the following towns will need coal cars as follows: TOWN DEMAND FOR CARS Coal Valley 30 Coaltown 45 Coal Junction 25 Coalsburg 20Using a railway city-to-city distance chart, the dispatcher constructs a mileage table for the preceding towns. The result is shown in the table below. Minimizing total miles over which cars aremoved to new locations, compute the best shipment of coal cars. TO COAL VALLEY COALTOWN COAL JUNCTION COALSBURG FROM MORGANTOWN 50 30 60 70 YOUNGSTOWN 20 80 10 90 PITTSBURGH 100 40 80 30Q5 The three blood banks in Franklin County are coordinated through a central office that facilitates blood delivery to four hospitals in the region. The cost to ship a standard container ofblood from each bank to each hospital is shown in the table below. Also given are the biweekly number of containers available at each bank and the biweekly number of containers of bloodneeded at each hospital. How many shipments should be made biweekly from each blood bank to each hospital so that total shipment costs are minimized?:
Our group of high quality writers are ready to help you with a similar paper within a specified deadline. Just click ORDER NOW and provide us with your assignment details, contact information and make payments. You will get periodic updates on order progress in your email.
USE THE CALCULATOR BELOW TO GET THE QUOTE | 812 | 3,636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-51 | latest | en | 0.913967 |
https://nebusresearch.wordpress.com/tag/berenstain-bears/ | 1,638,481,911,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362297.22/warc/CC-MAIN-20211202205828-20211202235828-00542.warc.gz | 483,973,142 | 27,961 | ## Missed A Mile
I’m honestly annoyed with myself. It’s only a little annoyed, though. I didn’t notice when I made my 5,280th tweet on @Nebusj. It’s one of those numbers — the count of feet in a mile — that fascinated the young me. It seemed to come from nowhere — why not 5,300? Why not 5,250? Heck, why not 5,000? — and the most I heard about why it was that was that 5,280 was equal to eight furlongs. What’s a furlong, I might wonder? 5,280 divided by eight is 660, which doesn’t clear things up much.
Yes, yes, I know now why it’s 5,280. It was me at age seven that couldn’t sort out why this rather than that.
But what a number. It had that compelling mix of precision and mystery. And so divisible! When you’ve learned how to do division and think it’s fun, a number like 5,280 with so many divisors is a joy. There’s 48 of them, all told. All the numbers you see on a times table except for 7 and 9 go into it. It’s practically teasing the mathematically-inclined kid to find all these factors. 5,279 and 5,281 are mere primes; 5,278 and 5,282 aren’t nearly so divisor-rich as 5,280. Even 1,760, which I knew well as the number of yards in a mile, isn’t so interesting. And compared to piddling little numbers like 12 or 144 — well!
5,280 is not why I’m a mathematician. I credit a Berenstain Bears book that clearly illustrated what mathematicians do is “add up sums in an observatory on the moon”. But 5,280 is one of those sparkling lights that attracted me to the subject. I imagine having something like this, accessible but mysterious, is key to getting someone hooked on a field. And while I agree the metric system is best for most applications, it’s also true 1,000 isn’t so interesting a number to stare at. You can find plenty of factors of it, but they’ll all follow too-easy patterns. You won’t see a surprising number like 55 or 352 or 1,056 or 1,320 among them.
So, I’m sorry to miss an interesting number like that for my 5,280th tweet. I hope I remember to make some fuss for my 5,280th blog post.
## Early April’s Math Comics
I had started to think the mathematics references in the comics pages were fading out and I might not have an installment to offer anytime soon. Then, on April 3, Pab Sugenis’s The New Adventures of Queen Victoria — a clip art comic strip which supposes the reader will recognize an illustration of King Edward VI — skipped its planned strip for the day (Sugenis’s choice, he says) and ran a Fuzzy Bunny Time strip calling on pretty much the expected rabbits and mathematics comic strip. (Some people in the Usenet group alt.fan.cecil-adams, which I read reliably and write to occasionally, say Sugenis was briefly a regular; perhaps so, but I don’t remember.) This would start a bumper crop of math strips for the week. | 716 | 2,779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-49 | latest | en | 0.938405 |
https://www.scribd.com/document/172189086/Gear | 1,558,832,436,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258620.81/warc/CC-MAIN-20190526004917-20190526030917-00302.warc.gz | 927,299,094 | 63,817 | You are on page 1of 19
# POWER TRANSMISSION Power transmission is the movement of energy from its place of generation to a location where it is applied to performing
useful work. Power transmission is normally accomplished by belts, ropes, chains, gears, couplings and friction clutches. GEAR A toothed wheel that engages another toothed mechanism in order to change the speed or direction of transmitted motion.
A gear is a component within a transmission device that transmits rotational force to another gear or device. A gear is different from a pulley in that a gear is a round wheel which has linkages ("teeth" or "cogs") that mesh with other gear teeth, allowing force to be fully transferred without slippage. Depending on their construction and arrangement, geared devices can transmit forces at different speeds, torques, or in a different direction, from the power source. The most common situation is for a gear to mesh with another gear Gears most important feature is that gears of unequal sizes (diameters) can be combined to produce a mechanical advantage, so that the rotational speed and torque of the second gear are different from that of the first.
To overcome the problem of slippage as in belt drives, gears are used which produce positive drive with uniform angular velocity. GEAR CLASSIFICATION Gears or toothed wheels may be classified as follows: 1. According to the position of axes of the shafts. The axes of the two shafts between which the motion is to be transmitted, may be a. Parallel b. Intersecting c. Non-intersecting and Non-parallel Gears for connecting parallel shafts 1. Spur Gear Teeth is parallel to axis of rotation can transmit power from one shaft to another parallel shaft. Spur gears are the simplest and most common type of gear. Their general form is a cylinder or disk. The teeth project radially, and with these "straightcut gears".
Spur gears are gears in the same plane that move opposite of each other because they are meshed together. Gear A is called the driver because this is turned by a motor. As gear A turns it meshes with gear B and it begins to turn as well. Gear B is called the driven gear.
EXTERNAL AND INTERNAL SPUR GEAR External gear makes external contact, and the internal gear (right side pair) makes internal contact.
APPLICATIONS OF SPUR GEAR Electric screwdriver, dancing monster, oscillating sprinkler, windup alarm clock, washing machine and clothes dryer 2. Parallel Helical Gear The teeth on helical gears are cut at an angle to the face of the gear. When two teeth on a helical gear system engage, the contact starts at one end of the tooth and gradually spreads as the gears rotate, until the two teeth are in full engagement.
This gradual engagement makes helical gears operate much more smoothly and quietly than spur gears. For this reason, helical gears are used in almost all car transmissions. Because of the angle of the teeth on helical gears, they create a thrust load on the gear when they mesh. Devices that use helical gears have bearings that can support this thrust load. One interesting thing about helical gears is that if the angles of the gear teeth are correct, they can be mounted on perpendicular shafts, adjusting the rotation angle by 90 degrees.
## CROSSED HELICAL GEAR
Herringbone gears:
To avoid axial thrust, two helical gears of opposite hand can be mounted side by side, to cancel resulting thrust forces. These are called double helical or herringbone gears
## Herringbone gears (or double-helical gears)
Applications of Herringbone Gears The most common application is in power transmission. They utilize curved teeth for efficient, high capacity power transmission. This offers reduced pulsation due to which they are highly used for extrusion and polymerization. Herringbone gears are mostly used on heavy machinery.
3. Rack and pinion Rack and pinion gears are used to convert rotation (From the pinion) into linear motion (of the rack). A perfect example of this is the steering system on many cars. The steering wheel rotates a gear which engages the rack. As the gear turns, it slides the rack either to the right or left, depending on which way you turn the wheel. Rack and pinion gears are also used in some scales to turn the dial that displays your weight.
RACK AND PINION GEARS FOR CONNECTING INTERSECTING SHAFTS 1. Straight Bevel Gear Bevel gears are useful when the direction of a shaft's rotation needs to be changed. They are usually mounted on shafts that are 90 degrees apart, but can be designed to work at other angles as well. The teeth on bevel gears can be straight, spiral or hypoid. Straight bevel gear teeth actually have the same problem as straight spur gear teeth as each tooth engages, it impacts the corresponding tooth all at once.
BEVEL GEAR Just like with spur gears, the solution to this problem is to curve the gear teeth. These spiral teeth engage just like helical teeth: the contact starts at one end of the gear and progressively spreads across the whole tooth.
SPIRAL BEVEL GEAR On straight and spiral bevel gears, the shafts must be perpendicular to each other, but they must also be in the same plane. If you were to extend the two shafts past the gears, they would intersect The bevel gear has many diverse applications such as locomotives, marine applications, automobiles, printing presses, cooling towers, power plants, steel plants, railway track inspection machines, etc. NON-INTERSECTING AND NON-PARALLEL
1. WORM AND WORM GEAR Worm gears are used when large gear reductions are needed. It is common for worm gears to have reductions of 20:1, and even up to 300:1 or greater.
Many worm gears have an interesting property that no other gear set has: the worm can easily turn the gear, but the gear cannot turn the worm. This is because the angle on the worm is so shallow that when the gear tries to spin it, the friction between the gear and the worm holds the worm in place.
WORM AND WORM GEAR This feature is useful for machines such as conveyor systems, in which the locking feature can act as a brake for the conveyor when the motor is not turning. One other very interesting usage of worm gears is in the Torsen differential, which is used on some high-performance cars and trucks. They are used in right-angle or skew shaft drives. The presence of sliding action in the system even though results in quieter operation, it gives rise to considerable frictional heat, hence they need good lubrication for heat dissipation and for improving the efficiency. High reductions are possible which results in compact drive.
APPLICATION OF WORM GEARS Worm gears are used widely in material handling and transportation machinery, machine tools, automobiles etc.
## NOMENCLATURE OF SPUR GEARS
NOMENCLATURE OF SPUR GEAR In the following section, we define many of the terms used in the analysis of spur gears.
Pitch surface: The surface of the imaginary rolling cylinder (cone, etc.) that the toothed gear may be considered to replace. Pitch circle: A right section of the pitch surface. Addendum circle: A circle bounding the ends of the teeth, in a right section of the gear. Root (or dedendum) circle: The circle bounding the spaces between the teeth, in a right section of the gear. Addendum: The radial distance between the pitch circle and the addendum circle. Dedendum: The radial distance between the pitch circle and the root circle. Clearance: The difference between the dedendum of one gear and the addendum of the mating gear.
Face of a tooth: That part of the tooth surface lying outside the pitch surface. Flank of a tooth: The part of the tooth surface lying inside the pitch surface. Circular thickness (also called the tooth thickness): The thickness of the tooth measured on the pitch circle. It is the length of an arc and not the length of a straight line. Tooth space: pitch diameter The distance between adjacent teeth measured on the pitch circle. Backlash: The difference between the circle thickness of one gear and the tooth space of the mating gear. Circular pitch (Pc) : The width of a tooth and a space, measured on the pitch circle.
P c
D
N
Diametral pitch (Pd): The number of teeth of a gear unit pitch diameter. A toothed gear must have an integral number of teeth. The circular pitch, therefore, equals the pitch circumference divided by the number of teeth. The diametral pitch is, by definition, the number of teeth divided by the pitch diameter. That is,
Pd = N D
## Where Pc = circular pitch Pd = diametral pitch N = number of teeth D = pitch diameter
Module (m): Pitch diameter divided by number of teeth. The pitch diameter is usually specified in inches or millimeters; in the former case the module is the inverse of diametral pitch.
m = D/N
Fillet: The small radius that connects the profile of a tooth to the root circle.
Pinion: The smaller of any pair of mating gears. The larger of the pair is called simply the gear. Velocity ratio: The ratio of the number of revolutions of the driving (or input) gear to the number of revolutions of the driven (or output) gear, in a unit of time. Pitch point: The point of tangency of the pitch circles of a pair of mating gears. Common tangent: The line tangent to the pitch circle at the pitch point. Line of action: A line normal to a pair of mating tooth profiles at their point of contact. Path of contact: The path traced by the contact point of a pair of tooth profiles. Pressure angle ( ): The angle between the common normal at the point of tooth contact and the common tangent to the pitch circles. It is also the angle between the line of action and the common tangent.
Base circle: An imaginary circle used in involute gearing to generate the involutes that form the tooth profiles.
VELOCITY RATIO OF GEAR DRIVE Velocity ratio is defined as the ratio of the speed of the driven shaft to the speed of the driver shaft.
One gear is a driver, which has d1, N1, diameter, speed angular speed respectively.
## as diameter, speed and angular speed
respectively. Another gear is driven connected to the driven shaft has d2, N2 , 2 as
## Angular speeds of the two gears will be
1 = 2 N1
2 =2 N 2
The peripheral velocity of the driver and driven shafts for the meshing pair of gear is equal and is given by V P = 1
d1 d = d 1 N 1 = 2 2 = d 2 N 2 2 2
2 N 2 d 1 = = 1 N1 d 2 T1 and T 2 are the number of teeth on driver gear and driven gear, since the pair of gear
Hence velocity ratio (n) = as the same module (m),then
d 1 = m T1 ; d 2 = m T2
and n =
N 2 d1 T = = 1 N 1 d 2 T2
GEAR TRAINS A gear train is two or more gear working together by meshing their teeth and turning each other in a system to generate power and speed. It reduces speed and increases torque. To create large gear ratio, gears are connected together to form gear trains. They often consist of multiple gears in the train. The smaller gears are one-fifth of the size of the larger gear. Electric motors are used with the gear systems to reduce the speed and increase the torque. Electric motor is connected to the driving end of each train and is mounted on the test platform. The output end of the gear train is connected to a large magnetic particle brake that is used to measure the output torque. Types of gear trains 1. Simple gear train 2. Compound gear train 3. Planetary gear train Simple Gear Train
The most common of the gear train is the gear pair connecting parallel shafts. The teeth of this type can be spur, helical or herringbone. only one gear for each axis. The angular velocity is simply the reverse of the tooth ratio. The main limitation of a simple gear train is that the maximum speed change ratio is 10:1. For larger ratio, large sizes of gear trains are required. The sprockets and chain in the bicycle is an example of simple gear train. When the paddle is pushed, the front gear is turned and that meshes with the links in the chain. The chain moves and meshes with the links in the rear gear that is attached to the rear wheel. This enables the bicycle to move.
Simple and compound gear trains Compound Gear Train For large velocities, compound arrangement is preferred. Two keys are keyed to a single shaft. A double reduction train can be arranged to have its input and output shafts in a line, by choosing equal center distance for gears and pinions. Two or more gears may rotate about a single axis
Planetary Gear Train (Epicyclic Gear Train) Planetary gears solve the following problem. Let's say you want a gear ratio of 6:1 with the input turning in the same direction as the output. One way to create that ratio is with the following three-gear train:
## Planetary Gear Train
In this train, the blue gear has six times the diameter of the yellow gear (giving a 6:1 ratio). The size of the red gear is not important because it is just there to reverse the direction of rotation so that the blue and yellow gears turn the same way. However, imagine that you want the axis of the output gear to be the same as that of the input gear. A common place where this same-axis capability is needed is in an electric screwdriver. In that case, you can use a planetary gear system, as shown here:
Planetary Gear Train In this gear system, the yellow gear (the sun) engages all three red gears (the planets) simultaneously. All three are attached to a plate (the planet carrier), and they engage the inside of the blue gear (the ring) instead of the outside. Because there are three red gears instead of one, this gear train is extremely rugged. The output shaft is attached to the blue ring gear, and the planet carrier is held stationary -- this gives the same 6:1 gear ratio. Another interesting thing about planetary gear sets is that they can produce different gear ratios depending on which gear you use as the input, which gear you use as the output, and which one you hold still. For instance, if the input is the sun gear, and we hold the ring gear stationary and attach the output shaft to the planet carrier, we get a different gear ratio. In this case, the planet carrier and planets orbit the sun gear, so instead of the sun gear having to spin six times for the planet carrier to make it around once, it has to spin seven times. This is because the planet carrier circled the sun gear once in the same direction as it was spinning, subtracting one revolution from the sun gear. So in this case, we get a 7:1 reduction. You could rearrange things again, and this time hold the sun gear stationary, take the output from the planet carrier and hook the input up to the ring gear. This would give you a 1.17:1 gear reduction. An automatic transmission uses planetary gear sets to create the different gear ratios, using clutches and brake bands to hold different parts of the gear set stationary and change the inputs and outputs.
Planetary gear trains have several advantages. They have higher gear ratios. They are popular for automatic transmissions in automobiles. They are also used in bicycles for controlling power of pedaling automatically or manually. They are also used for power train between internal combustion engine and an electric motor. Applications Gear trains are used in representing the phases of moon on a watch or clock dial. It is also used for driving a conventional two-disk lunar phase display off the day-of-the-week shaft of the calendar. Velocity ratio of Gear trains We know that the velocity ratio of a pair of gears is the inverse proportion of the diameters of their pitch circle, and the diameter of the pitch circle equals to the number of teeth divided by the diametral pitch. Also, we know that it is necessary for the mating gears to have the same diametral pitch so that to satisfy the condition of correct meshing. Thus, we infer that the velocity ratio of a pair of gears is the inverse ratio of their number of teeth. For the ordinary gear trains we have (Fig a)
These equations can be combined to give the velocity ratio of the first gear in the train to the last gear:
( N 2 N 3 N 4) ( N1 N 2 N 3 )
(T1T2T3 ) N 4 T = = 1 =n (T2T3T4 ) N 1 T4
Note:
The tooth numbers in the numerator are those of the driven gears, and the tooth numbers in the denominator belong to the driver gears. Gear 2 and 3 both drive and are, in turn, driven. Thus, they are called idler gears. Since their tooth numbers cancel, idler gears do not affect the magnitude of the input-output ratio, but they do change the directions of rotation. Note the directional arrows in the figure. Idler gears can also constitute a saving of space and money (If gear 1 and 4 meshes directly across a long center distance, their pitch circle will be much larger.)
Problems 1. The pitch circle diameter of the spur gear is 200 mm and the number of teeth is 10. Calculate the module of the gear Given data D= 200 mm N=10 Solution m =D/N 200/10= 20 Module of the gear is 20 2. Pitch circle diameter of the spur gear is 180 mm and the number of teeth on the gear is 14. Calculate the Circular pitch of the gear Given Data D= 180 mm N=14
Solution
P c
D
N
PC = 40 mm SHORT QUESTIONS 1. What is power transmission 2. Why gear drives are called positively driven? 3. What is backlash in gears? 4. What are the types of gears available? 5. What is gear train? Why gear trains are used? 6. Why intermediate gear in simple gear train is called idler? 7. What is the advantage of using helical gear over spur gear? 8. List out the applications of gears 9. Define the term module in gear tooth 10. What is herringbone gear? ESSAY TYPE QUESTIONS 1. With sketch explain various types of gears 2. With sketch explain three types of gear trains 3. Derive the velocity ratio for an simple gear train 4. With neat sketch explain the nomenclature of spur gear 5. Write the applications, advantages and disadvantages of gear drives
## References 1. Theory of machines R.S.Khurmi and J.K.Gupta, S.Chand Publications,2002
2. http://www.efunda.com/designstandards/gears/gears_epicyclic.cfm
3. http://www.How stuffworks.com 4. http://www.wikipedia.com 5. Introduction to mechanisms yi zhang with susan finger and Stephannie Behrens
6. http://www.technologystudent.com/gears1/worm1.htm 7. http://gemini.tntech.edu/~slc3675/me361/lecture/geartrn.html 8. http://www.engr.utexas.edu/dteach/teacherpdi/2007materialsNXT/Gear_Notes.pdf 9. http://www.ticona.com/home/tech/design/gears.htm | 4,156 | 18,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-22 | latest | en | 0.942339 |
https://gmatclub.com/forum/the-price-of-a-certain-commodity-increased-at-a-rate-of-126464.html | 1,571,061,936,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653247.25/warc/CC-MAIN-20191014124230-20191014151730-00314.warc.gz | 493,436,078 | 149,564 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 14 Oct 2019, 07:05
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# The price of a certain commodity increased at a rate of
Author Message
TAGS:
### Hide Tags
Manager
Status: D-Day is on February 10th. and I am not stressed
Affiliations: American Management association, American Association of financial accountants
Joined: 12 Apr 2011
Posts: 158
Location: Kuwait
Schools: Columbia university
The price of a certain commodity increased at a rate of [#permalink]
### Show Tags
Updated on: 05 Apr 2019, 06:19
1
10
00:00
Difficulty:
55% (hard)
Question Stats:
67% (02:30) correct 33% (03:01) wrong based on 123 sessions
### HideShow timer Statistics
The price of a certain commodity increased at a rate of $$X$$ % per year between 2000 and 2004. If the price was $$M$$ dollars in 2001 and $$N$$ dollars in 2003, what was the price in 2002 in terms of $$M$$ and $$N$$ ?
A. $$\sqrt{MN}$$
B. $$N\sqrt{\frac{N}{M}}$$
C. $$N\sqrt{M}$$
D. $$N\frac{M}{\sqrt{N}}$$
E. $$NM^{\frac{3}{2}}$$
_________________
Sky is the limit
Originally posted by manalq8 on 23 Jan 2012, 14:25.
Last edited by Bunuel on 05 Apr 2019, 06:19, edited 4 times in total.
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4473
Re: The price of a certain commodity increased at a rate of [#permalink]
### Show Tags
23 Jan 2012, 16:23
6
1
Hi there! I'm happy to help with this!
Let's say, the price in 2000 is A. That the original amount. Each year it increases X%. To represent a percent increase (a) write the percent as a fraction/decimal, here X/100; (b) add one ---> 1 + X/100; (c) that's the multiplier -- multiplying a number by that multiplier results in a X% increase.
price in 2000 = A
price in 2001 = A*(1 + X/100)
price in 2002 = A*(1 + X/100)^2
price in 2003 = A*(1 + X/100)^3
price in 2004 = A*(1 + X/100)^4
For simplicity, I am going to define r = (1 + X/100). Then these equations become:
price in 2000 = A
price in 2001 = A*r
price in 2002 = A*r^2
price in 2003 = A*r^3
price in 2004 = A*r^4
Now, suppose we have M = 2001 price = A*r and N = 2003 price = A*r^3. How do we represent the 2002 prince (A*r^2) in terms of M and N?
There are two methods.
Method One: express r in terms of M and N
This is more a crank-it-out algebraic solution approach. We notice that N/M = (A*r^3)/(A*r) = r^2, so r = sqrt(N/M). Well,
2002 price = (2001 price)*(r) = M * sqrt(N/M) = [sqrt(M)*sqrt(M)]*[sqrt(N)/sqrt(M)] = sqrt(M)*sqrt(N) = sqrt(NM).
Through some fast-and-loose manipulation of the laws of squareroots, we arrive at answer .
Method Two: a more elegant solution for a more civilized age . . .
When you have an arithmetic sequence --- that is, adding the same number to get new terms (e.g. 8, 11, 14, 17, 20, 23, . . . ), when you take any three numbers in a row, the middle number is the mean, the arithmetic average, of the outer two. For example ---11, 14, 17 --- (11 + 17)/2 = 14. The arithmetic average is the ordinary average --- add the two numbers, and divide by two.
When you have a geometric sequence -- that is, multiplying the same ratio to get new terms (e.g. 2, 6, 18, 54, 162, . . . ), when you take any three numbers in a row, the middle number is the geometric mean of the outer two. The geometric mean of two numbers means multiply the two numbers and take the squareroot. For example --- 6, 18, 54 --- 6*54 = 324, and sqrt(324) = 18.
When you apply a fixed percentage increase from one term to the next, as we have in this problem, that's a geometric sequence. Thus, to find the 2002 price, all you have to do is take the geometric mean of the 2001 price and the 2003 price. 2002 price = sqrt(MN). Bam. Done. Again, answer = .
The ideas about arithmetic & geometric sequences, and the associated means, are good tricks to have up your sleeve for the more challenging GMAT math problems.
Does all this make sense? Please let me know if you have any questions.
Mike
_________________
Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Intern
Joined: 23 Jan 2012
Posts: 7
Location: India
Concentration: Strategy, Finance
GMAT 1: 620 Q51 V23
GPA: 3.29
WE: Engineering (Other)
Re: The price of a certain commodity increased at a rate of [#permalink]
### Show Tags
23 Jan 2012, 22:01
5
2
The price in 2001 is M
The price in 2002 is M(1+$$\frac{X}{100}$$) ----------------------(1)
The price in 2003 is M(1+$$\frac{X}{100}$$)(1+$$\frac{X}{100}$$) = N ---------(2)
Solving equation (2) we get
(1+$$\frac{X}{100}$$) = $$\sqrt{\frac{N}{M}}$$
Putting this value in equation (1) to get the desired answer
The price in 2002 is M$$\sqrt{\frac{N}{M}}$$ = $$\sqrt{MN}$$
Hence the option A
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 58310
Re: The price of a certain commodity increased at a rate of [#permalink]
### Show Tags
23 Jan 2012, 19:00
1
1
manalq8 wrote:
The price of a certain commodity increased at a rate of $$X$$ % per year between 2000 and 2004. If the price was $$M$$ dollars in 2001 and $$N$$ dollars in 2003, what was the price in 2002 in terms of $$M$$ and $$N$$ ?
A. $$\sqrt{MN}$$
B. $$N\sqrt{\frac{N}{M}}$$
C. $$N\sqrt{M}$$
D. $$N\frac{M}{\sqrt{N}}$$
E. $$NM^{\frac{3}{2}}$$
I think that plug-in method is easiest for this problem.
Let the price in 2001 be 100 and the annual rate be 10%. Then:
2001 = 100 = M;
2002 = 110;
2003 = 121 = N;
Now, plug 100 and 121 in the answer choices to see which one gives 110:
A. $$\sqrt{MN}=\sqrt{100*121}=10*11=110$$, correct answer right away.
P.S. For plug-in method it might happen that for some particular numbers more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.
Hope it helps.
_________________
Manager
Joined: 04 Oct 2013
Posts: 150
Location: India
GMAT Date: 05-23-2015
GPA: 3.45
Re: The price of a certain commodity increased at a rate of [#permalink]
### Show Tags
16 Jan 2014, 11:41
The price of a certain commodity increased at a rate of X % per year between 2000 and 2004. If the price was M dollars in 2001 and N dollars in 2003, what was the price in 2002 in terms of M and N ?
Given:
Price in 2001 : M
Price in 2003 : N
Rate per year: X %
Price in 2002 $$= M ( 1 + X/100)$$........................................(1)
And, Price in 2003: $$N = M( 1 + X/100)^2$$
Or, $$(1 + x/100) =\sqrt{(N/M)}$$........(2)
Substituting the value of (2) in (1) above,
Price in 2002 $$= M \sqrt{(N/M)} = \sqrt{MN}$$
Senior Manager
Joined: 03 Apr 2013
Posts: 264
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41
GPA: 3
Re: The price of a certain commodity increased at a rate of [#permalink]
### Show Tags
08 Jul 2016, 21:44
manalq8 wrote:
The price of a certain commodity increased at a rate of $$X$$ % per year between 2000 and 2004. If the price was $$M$$ dollars in 2001 and $$N$$ dollars in 2003, what was the price in 2002 in terms of $$M$$ and $$N$$ ?
A. $$\sqrt{MN}$$
B. $$N\sqrt{\frac{N}{M}}$$
C. $$N\sqrt{M}$$
D. $$N\frac{M}{\sqrt{N}}$$
E. $$NM^{\frac{3}{2}}$$
When no absolute values(or variables pertaining to them) are given, number plugging is the way to go..
Let the Value initially be = 1
and let X% = 100%(in other words..value doubles every year)
values are..
M = 2
N = 8
and we're looking for the value in 2002..which is equal to 4.
_________________
Spread some love..Like = +1 Kudos
Intern
Joined: 31 Dec 2017
Posts: 4
Location: India
Concentration: Entrepreneurship, Technology
GPA: 3.05
Re: The price of a certain commodity increased at a rate of [#permalink]
### Show Tags
30 Mar 2018, 10:36
1
I think this is the easiest approach here:
2001 - M
2002 - ? (Let it be 'x')
2003 - N
Now according to ques, since the rate of increase is equal both year and the rates should be calculated on the previous year's price,
$$\frac{x-M}{M} * 100= \frac{N-x}{x}*100$$
$$x^2 - Mx = MN - Mx$$
$$x = \sqrt{MN}$$
Hope it helps
+1 Kudos would be nice
Non-Human User
Joined: 09 Sep 2013
Posts: 13117
Re: The price of a certain commodity increased at a rate of [#permalink]
### Show Tags
22 May 2019, 15:19
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: The price of a certain commodity increased at a rate of [#permalink] 22 May 2019, 15:19
Display posts from previous: Sort by | 2,885 | 9,254 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2019-43 | latest | en | 0.883927 |
https://www.unitconverters.net/pressure/psi-to-pound-force-square-foot.htm | 1,652,914,719,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522556.18/warc/CC-MAIN-20220518215138-20220519005138-00555.warc.gz | 1,246,579,474 | 3,356 | Home / Pressure Conversion / Convert Psi to Pound-force/square Foot
Convert Psi to Pound-force/square Foot
Please provide values below to convert psi [psi] to pound-force/square foot, or vice versa.
From: psi To: pound-force/square foot
Psi to Pound-force/square Foot Conversion Table
Psi [psi]Pound-force/square Foot
0.01 psi1.44 pound-force/square foot
0.1 psi14.4 pound-force/square foot
1 psi144 pound-force/square foot
2 psi288 pound-force/square foot
3 psi432 pound-force/square foot
5 psi720 pound-force/square foot
10 psi1440 pound-force/square foot
20 psi2880 pound-force/square foot
50 psi7200 pound-force/square foot
100 psi14400 pound-force/square foot
1000 psi144000 pound-force/square foot
How to Convert Psi to Pound-force/square Foot
1 psi = 144 pound-force/square foot
1 pound-force/square foot = 0.0069444444 psi
Example: convert 15 psi to pound-force/square foot:
15 psi = 15 × 144 pound-force/square foot = 2160 pound-force/square foot | 263 | 964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-21 | longest | en | 0.506356 |
http://www.chegg.com/homework-help/multivariable-calculus-for-uc-berkely-6th-edition-chapter-14.3-problem-56e-solution-9781424054992 | 1,474,758,172,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738659512.19/warc/CC-MAIN-20160924173739-00180-ip-10-143-35-109.ec2.internal.warc.gz | 390,360,080 | 17,641 | Solutions
Multivariable Calculus Early Transcendentals for UC Berkeley
Multivariable Calculus Early Transcendentals for UC Berkeley (6th Edition) View more editions Solutions for Chapter 14.3 Problem 56E
• 2493 step-by-step solutions
• Solved by publishers, professors & experts
• iOS, Android, & web
Over 90% of students who use Chegg Study report better grades.
May 2015 Survey of Chegg Study Users
Chapter: Problem:
100% (5 ratings)
STEP-BY-STEP SOLUTION:
Chapter: Problem:
100% (5 ratings)
• Step 1 of 5
Consider the function,
Find all the second derivatives of the function.
The second derivatives of the function are.
First, find the first partial derivatives and. To find, differentiate with respect to x, treating y as a constant.
Use
Use law of exponents
To find, differentiate v with respect to y, treating as a constant.
Use and chain rule
Use law of exponents
• Chapter , Problem is solved.
Corresponding Textbook
Multivariable Calculus Early Transcendentals for UC Berkeley | 6th Edition
9781424054992ISBN-13: 1424054990ISBN: James StewartAuthors: | 270 | 1,073 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2016-40 | latest | en | 0.767538 |
http://docplayer.net/18169996-Understanding-the-independent-samples-t-test.html | 1,619,161,929,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039568689.89/warc/CC-MAIN-20210423070953-20210423100953-00326.warc.gz | 27,902,426 | 29,147 | # UNDERSTANDING THE INDEPENDENT-SAMPLES t TEST
Size: px
Start display at page:
Transcription
1 UNDERSTANDING The independent-samples t test evaluates the difference between the means of two independent or unrelated groups. That is, we evaluate whether the means for two independent groups are significantly different from each other. The independent-samples t test is commonly referred to as a between-groups design, and can also be used to analyze a control and experimental group. With an independent-samples t test, each case must have scores on two variables, the grouping (independent) variable and the test (dependent) variable. The grouping variable divides cases into two mutually exclusive groups or categories, such as boys or girls for the grouping variable gender, while the test variable describes each case on some quantitative dimension such as test performance. The t test evaluates whether the mean value of the test variable (e.g., test performance) for one group (e.g., boys) differs significantly from the mean value of the test variable for the second group (e.g., girls). HYPOTHESES FOR Null Hypothesis: H 0 : µ = µ where µ stands for the mean for the first group and µ stands for the mean for the second group. -or- H 0 : µ µ = 0 Alternative (Non-Directional) Hypothesis: H a : µ µ -or- H a : µ µ 0 Alternative (Directional) Hypothesis: H a : µ < µ -or- H a : µ > µ (depending on direction) NOTE: the subscripts ( and ) can be substituted with the group identifiers For example: H 0 : µ Boys = µ Girls H a : µ Boys µ Girls ASSUMPTIONS UNDERLYING. The data (scores) are independent of each other (that is, scores of one participant are not systematically related to scores of the other participants). This is commonly referred to as the assumption of independence.. The test (dependent) variable is normally distributed within each of the two populations (as defined by the grouping variable). This is commonly referred to as the assumption of normality. 3. The variances of the test (dependent) variable in the two populations are equal. This is commonly referred to as the assumption of homogeneity of variance. Null Hypothesis: H 0 : σ = Alternative Hypothesis: H a : σ σ σ (if retained = assumption met) (if rejected = assumption not met)
2 TESTING THE ASSUMPTION OF INDEPENDENCE One of the first steps in using the independent-samples t test is to test the assumption of independence. Independence is a methodological concern; it is dealt with (or should be dealt with) when a study is set up. Although the independence assumption can ruin a study if it is violated, there is no way to use the study s sample data to test the validity of this prerequisite condition. It is assessed through an examination of the design of the study. That is, we confirm that the two groups are independent of each other? The assumption of independence is commonly known as the unforgiving assumption (r.e., robustness), which simply means that if the two groups are not independent of each other, one cannot use the independent-samples t test. TESTING THE ASSUMPTION OF NORMALITY Another of the first steps in using the independent-samples t test is to test the assumption of normality, where the Null Hypothesis is that there is no significant departure from normality, as such; retaining the null hypothesis indicates that the assumption of normality has been met for the given sample. The Alternative Hypothesis is that there is a significant departure from normality, as such; rejecting the null hypothesis in favor of the alternative indicates that the assumption of normality has not been met for the given sample. To test the assumption of normality, we can use the Shapiro-Wilks test. From this test, the Sig. (p) value is compared to the a priori alpha level (level of significance for the statistic) and a determination is made as to reject (p < α) or retain (p > α) the null hypothesis. Tests of Normality Percentage of time talking Stress Condition Low Stress High Stress a. Lilliefors Significance Correction Kolmogorov-Smirnov a Shapiro-Wilk Statistic df Sig. Statistic df Sig For the above example, where α =.00, given that p =.70 for the Low Stress Group and p =.06 for the High Stress Group we would conclude that each of the levels of the Independent Variable (Stress Condition) are normally distributed. Therefore, the assumption of normality has been met for this sample. The a priori alpha level is typically based on sample size where.05 and.0 are commonly used. Tabachnick and Fidell (007) report that conventional but conservative (.0 and.00) alpha levels are commonly used to evaluate the assumption of normality. NOTE: Most statisticians will agree that the Shapiro-Wilks Test should not be the sole determination of normality. It is common to use this test in conjunction with other measures such as an examination of skewness, kurtosis, histograms, and normal Q-Q plots. PAGE
3 In examining skewness and kurtosis, we divide the skewness (kurtosis) statistic by its standard error. We want to know if this standard score value significantly departs from normality. Concern arises when the skewness (kurtosis) statistic divided by its standard error is greater than z +3.9 (p <.00, two-tailed test) (Tabachnick & Fidell, 007). We have several options for handling non-normal data, such as deletion and data transformation (based on the type and degree of violation as well as the randomness of the missing data points). Any adjustment to the data should be justified (i.e., referenced) based on solid resources (e.g., prior research or statistical references). As a first step, data should be thoroughly screened to ensure that any issues are not a factor of missing data or data entry errors. Such errors should be resolved prior to any data analyses using acceptable procedures (see for example Howell, 007 or Tabachnick & Fidell, 007). TESTING THE ASSUMPTION OF HOMOGENEITY OF VARIANCE Another of the first steps in using the independent-samples t test statistical analysis is to test the assumption of homogeneity of variance, where the null hypothesis assumes no difference between the two group s variances (H 0 : σ = σ ). The Levene s F Test for Equality of Variances is the most commonly used statistic to test the assumption of homogeneity of variance. The Levene s test uses the level of significance set a priori for the t test analysis (e.g., α =.05) to test the assumption of homogeneity of variance. Levene's Test for Equality of Variances Independent Samples Test VISUAL Visualization Test assumed not assumed F Sig For Example: For the VISUAL variable (shown above), the F value for Levene s test is with a Sig. (p) value of.000 (p <.00). Because the Sig. value is less than our alpha of.05 (p <.05), we reject the null hypothesis (no difference) for the assumption of homogeneity of variance and conclude that there is a significant difference between the two group s variances. That is, the assumption of homogeneity of variance is not met. If the assumption of homogeneity of variance is not met, we must use the data results associated with the not assumed, which takes into account the Cochran & Cox (957) adjustment for the standard error of the estimate and the Satterthwaite (946) adjustment for the degrees of freedom. In other words, we will use the bottom line of the t test for equality of means results table and ignore the top line of information. Had the Sig. (p) value been greater than our a priori alpha level, we would have retained the null hypothesis and concluded that there is not a significant difference between the two group s variances. If the assumption of homogeneity of variance is met, we must use the data results associated with the assumed, and interpret the data accordingly. That is, we would use the top line of information for the t test. PAGE 3
4 VISUAL Visualization Test assumed not assumed Independent Samples Test t-test for Equality of Means 95% Confidence Interval of the Mean Std. Error Difference t df Sig. (-tailed) Difference Difference Lower Upper For this example (testing the difference between males and females on the Visualization test), since the t value (-3.44, which indicates that the second group was higher than the first group) resulted in a Sig. (p) value that was less than our alpha of.05 (p <.05, which puts the obtained t in the tail) we reject the null hypothesis in support of the alternative hypothesis, and conclude that males and females differed significantly on their Visualization test performance. By examining the group means for this sample of subjects (not shown here), we see that males (with a mean of 6.38) performed significantly higher on the Visualization test than did females (with a mean of 5.59). VIOLATION OF THE ASSUMPTIONS OF THE t TEST FOR INDEPENDENT GROUPS The independent-samples t test is what we refer to as a robust test. That is, the t test is relatively insensitive (having little effect) to violations of normality and homogeneity of variance, depending on the sample size and the type and magnitude of the violation. If n = n and the size of each sample is equal to or greater than 30, the t test for independent groups may be used without appreciable error despite moderate violations of the normality and/or the homogeneity of variance assumptions (Pagano, 004, p. 339). Sample sizes can be considered equal if the larger group is not more than ½ times larger than the smaller group (Morgan, Leech, Gloeckner, & Barrett, 004). If the variance in one group is more than 4 or 5 times larger than the variance in the other group they are considered very different and the homogeneity of variance assumption is violated (i.e., not met). A variance ratio (F max ) analysis can be obtained by dividing the lowest variance of a group into the highest group variance. Concern arises if the resulting ratio is 4-5 times, which indicates that the largest variance is 4 to 5 times the smallest variance (Tabachnick & Fidell 007). If there are extreme violations of these assumptions with respect to normality and homogeneity of variance an alternate (non-parametric) test such as the Mann-Whitney U test should be used instead of the independent-samples t test. DEGREES OF FREEDOM Because we are working with two independent groups, we will loose (restrict) one df to the mean for each group. Therefore, df for an independent-samples t test will be (n ) + (n ), where n and n are the sample sizes for each of the independent groups, respectively. Or we can use, N, where N is the total sample size for the study. PAGE 4
5 EFFECT SIZE STATISTICS FOR Cohen s d (which can range in value from negative infinity to positive infinity) evaluates the degree (measured in standard deviation units) that the mean scores on the two test variables differ. If the calculated d equals 0, this indicates that there are no differences in the means. However, as d deviates from 0, the effect size becomes larger. d n + n n n = t where t is the obtained t value and n is the total sample size for group and n is the total sample size for group. So what does this Cohen s d mean? Statistically, it means that the difference between the two sample means is (e.g.,.3) standard deviation units (usually reported in absolute value terms) from zero. A mean difference of zero is the hypothesized difference between the two population means. Effect sizes provide a measure of the magnitude of the difference expressed in standard deviation units in the original measurement. It is a measure of the practical importance of a significant finding. SAMPLE APA RESULTS Using an alpha level of.05, an independent-samples t test was conducted to evaluate whether males and females differed significantly on a visualization test. The test was significant, t(44.06) = 3.44, p <.0, d =.3. The 95% confidence interval for the visualization test mean ranged from -.9 to An examination of the group means indicate that males (M = 6.38, SD = 4.9) performed significantly higher on the visualization test than did females (M = 5.6, SD = 3.45). Note: there are several ways to interpret the results, the key is to indicate that there was a significant difference between the two groups at the a priori alpha level and include, at a minimum, reference to the group means and effect size. In looking at a sample statistical stand from an independent-samples t test, we see t(44.06) = 3.44, p <.0, d =.3 t Indicates that we are using a t-test (44.06) Indicates the degrees of freedom associated with this t-test 3.44 Indicates the obtained t statistic value (t obt ) p <.0 d =.3 Indicates the probability of obtaining the given t value by chance alone Indicates the effect size for the significant effect (measured in standard deviation units) PAGE 5
6 INTERPRETING The first table (TESTS OF NORMALITY), which is produced through the EXPLORE command in SPSS (not a part of the t-test default options), shows the Shapiro-Wilks test of normality. We use this statistic to test whether the levels of the independent variable are statistically normal. If the p (Sig.) values are less than or equal to (<) our a priori alpha level, the level(s) are considered to be non-normal and will require attention (e.g., transformation). If the p values are greater than (>) our a priori alpha level, we consider the level(s) to be statistically normal (i.e., normally distributed). In our example, using an a priori alpha level of.00, we find that neither level (low stress or high stress) are significant, and as such, consider both levels of the independent variable to be normally distributed. THE FOLLOWING TABLES ARE PART OF THE t TEST OPTIONS PRODUCED FROM SPSS: The second table, (GROUP STATISTICS) shows descriptive statistics for the two groups (lowstress and high-stress) separately. Note that the means for the two groups look somewhat different. This might be due to chance, so we will want to test this with the t test in the next table. The third table, (INDEPENDENT SAMPLES TEST) provides two statistical tests. In the left two columns of numbers, is the Levene s Test for Equality of Variances for the assumption that the variances of the two groups are equal (i.e., assumption of homogeneity of variance). NOTE, this is not the t test; it only assesses an assumption! If this F test is not significant (as in the case of this example), the assumption is not violated (that is, the assumption is met), and one uses the assumed line for the t test and related statistics. However, if Levene s F is statistically significant (Sig., p <.05), then variances are significantly different and the assumption of equal variances is violated (not met). In that case, the not assumed line would be used for which SPSS adjusts the t, df, and Sig. as appropriate. Also in the third table we obtain the needed information to test the equality of the means. Recall that there are three methods in which we can make this determination. METHOD ONE (most commonly used): comparing the Sig. (probability) value (p =.0 for our example) to the a priori alpha level (α =.05 for our example). If p < α we reject the null hypothesis of no difference. If p > α we retain the null hypothesis of no difference. For our example, p < α, therefore we reject the null hypothesis and conclude that the low-stress group (M = 45.0) talked significantly more than did the high-stress group (M =.07). METHOD TWO: comparing the obtained t statistic value (t obt =.430 for our example) to the t critical value (t cv ). Knowing that we are using a two-tailed (non-directional) t test, with an alpha level of.05 (α =.05), with df = 8, and looking at the Student s t Distribution Table we find the critical value for this example to be.048. If t obt > t cv we reject the null hypothesis of no difference. If t obt < t cv we retain the null hypothesis of no difference. For our example, t obt =.430 and t cv =.048, therefore, t obt > t cv so we reject the null hypothesis and conclude that there is a statistically significant difference between the two groups. More specifically, looking at the group means, we conclude that the low-stress group (M = 45.0) talked significantly more than did the high-stress group (M =.07). PAGE 6
7 METHOD THREE: examining the confidence intervals and determining whether the upper (4.637 for our example) and lower (3.630 for our example) boundaries contain zero (the hypothesized mean difference). If the confidence intervals do not contain zero we reject the null hypothesis of no difference. If the confidence intervals do contain zero we retain the null hypothesis of no difference. For our example, the confidence intervals (+3.630, ) do not contain zero, therefore we reject the null hypothesis and conclude that the low-stress group (M = 45.0) talked significantly more than did the high-stress group (M =.07). Note that if the upper and lower bounds of the confidence intervals have the same sign (+ and + or and ), we know that the difference is statistically significant because this means that the null finding of zero difference lies outside of the confidence interval. CALCULATING AN EFFECT SIZE Since we concluded that there was a significant difference between the average amount of time spent talking between the two groups we will need to calculate an effect size to determine the magnitude of this significant effect. Had we not found a significant difference no effect size would have to be calculated (as the two groups would have only differed due to random fluctuation or chance). To calculate the effect size for this example, we will use the following formula: d n + n n n = t where t is the obtained t value and n is the total sample size for group and n is the total sample size for group. Where, t =.43 n = 5 n = 5 Substituting the values into the formula we find: d =.43 =.43 = = (.43)(.36548) = =.89 (5)(5) 5 SAMPLE APA RESULTS Using an alpha level of.05, an independent-samples t test was conducted to evaluate whether the average percentage of time spent talking differed significantly as a function of whether students were in a low stress or high stress condition. The test was significant, t(8) =.43, p <.05, d =.89. The 95% confidence interval for the average percentage of time spent talking ranged from 3.63 to An examination of the group means indicate that students in the low stress condition (M = 45.0, SD = 4.97) talked (on average) significantly more than students in the high stress condition (M =.07, SD = 7.4). PAGE 7
8 Tests of Normality Percentage of time talking Stress Condition Low Stress High Stress a. Lilliefors Significance Correction Kolmogorov-Smirnov a Shapiro-Wilk Statistic df Sig. Statistic df Sig Independent-Samples T-Test Example Group Statistics TALK STRESS Low Stress High Stress Std. Error N Mean Std. Deviation Mean TALK assumed not assumed Levene's Test for Equality of Variances F Sig. Independent Samples Test t df Sig. (-tailed) t-test for Equality of Means Mean Difference 95% Confidence Interval of the Std. Error Difference Difference Lower Upper PAGE 8
9 REFERENCES Cochran, W. G., & Cox, G. M. (957). Experimental Designs. New York: John Wiley & Sons. Green, S. B., & Salkind, N. J. (003). Using SPSS for Windows and Macintosh: Analyzing and Understanding Data (3 rd ed.). Upper Saddle River, NJ: Prentice Hall. Hinkle, D. E., Wiersma, W., & Jurs, S. G. (003). Applied Statistics for the Behavioral Sciences (5 th ed.). New York: Houghton Mifflin Company. Howell, D. C. (007). Statistical Methods for Psychology (6 th ed.). Belmont, CA: Thomson Wadsworth. Huck, S. W. (004). Reading Statistics and Research (4 th ed.). New York: Pearson Education Inc. Morgan, G. A., Leech, N. L., Gloeckner, G. W., & Barrett, K. C. (004). SPSS for Introductory Statistics: Use and Interpretation ( nd ed.). Mahwah, NJ: Lawrence Erlbaum Associates. Pagano, R. R. (004). Understanding Statistics in the Behavioral Sciences (7 th ed.). Belmont, CA: Thomson/Wadsworth. Satterthwaite, F. W. (946). An approximate distribution of estimates of variance components. Biometrics Bulletin,, 0-4. Tabachnick, B. G., & Fidell, L. S. (007). Using Multivariate Statistics (5 th ed.). Needham Heights, MA: Allyn & Bacon. PAGE 9
### UNDERSTANDING THE DEPENDENT-SAMPLES t TEST
UNDERSTANDING THE DEPENDENT-SAMPLES t TEST A dependent-samples t test (a.k.a. matched or paired-samples, matched-pairs, samples, or subjects, simple repeated-measures or within-groups, or correlated groups)
### Independent t- Test (Comparing Two Means)
Independent t- Test (Comparing Two Means) The objectives of this lesson are to learn: the definition/purpose of independent t-test when to use the independent t-test the use of SPSS to complete an independent
### INTERPRETING THE ONE-WAY ANALYSIS OF VARIANCE (ANOVA)
INTERPRETING THE ONE-WAY ANALYSIS OF VARIANCE (ANOVA) As with other parametric statistics, we begin the one-way ANOVA with a test of the underlying assumptions. Our first assumption is the assumption of
### UNDERSTANDING ANALYSIS OF COVARIANCE (ANCOVA)
UNDERSTANDING ANALYSIS OF COVARIANCE () In general, research is conducted for the purpose of explaining the effects of the independent variable on the dependent variable, and the purpose of research design
### UNDERSTANDING THE TWO-WAY ANOVA
UNDERSTANDING THE e have seen how the one-way ANOVA can be used to compare two or more sample means in studies involving a single independent variable. This can be extended to two independent variables
### THE KRUSKAL WALLLIS TEST
THE KRUSKAL WALLLIS TEST TEODORA H. MEHOTCHEVA Wednesday, 23 rd April 08 THE KRUSKAL-WALLIS TEST: The non-parametric alternative to ANOVA: testing for difference between several independent groups 2 NON
### Two Related Samples t Test
Two Related Samples t Test In this example 1 students saw five pictures of attractive people and five pictures of unattractive people. For each picture, the students rated the friendliness of the person
### Chapter 7 Section 7.1: Inference for the Mean of a Population
Chapter 7 Section 7.1: Inference for the Mean of a Population Now let s look at a similar situation Take an SRS of size n Normal Population : N(, ). Both and are unknown parameters. Unlike what we used
### Calculating, Interpreting, and Reporting Estimates of Effect Size (Magnitude of an Effect or the Strength of a Relationship)
1 Calculating, Interpreting, and Reporting Estimates of Effect Size (Magnitude of an Effect or the Strength of a Relationship) I. Authors should report effect sizes in the manuscript and tables when reporting
### Introduction to Analysis of Variance (ANOVA) Limitations of the t-test
Introduction to Analysis of Variance (ANOVA) The Structural Model, The Summary Table, and the One- Way ANOVA Limitations of the t-test Although the t-test is commonly used, it has limitations Can only
### EPS 625 INTERMEDIATE STATISTICS FRIEDMAN TEST
EPS 625 INTERMEDIATE STATISTICS The Friedman test is an extension of the Wilcoxon test. The Wilcoxon test can be applied to repeated-measures data if participants are assessed on two occasions or conditions
### Chapter 9. Two-Sample Tests. Effect Sizes and Power Paired t Test Calculation
Chapter 9 Two-Sample Tests Paired t Test (Correlated Groups t Test) Effect Sizes and Power Paired t Test Calculation Summary Independent t Test Chapter 9 Homework Power and Two-Sample Tests: Paired Versus
### Reporting Statistics in Psychology
This document contains general guidelines for the reporting of statistics in psychology research. The details of statistical reporting vary slightly among different areas of science and also among different
### Examining Differences (Comparing Groups) using SPSS Inferential statistics (Part I) Dwayne Devonish
Examining Differences (Comparing Groups) using SPSS Inferential statistics (Part I) Dwayne Devonish Statistics Statistics are quantitative methods of describing, analysing, and drawing inferences (conclusions)
### Introduction to Hypothesis Testing. Hypothesis Testing. Step 1: State the Hypotheses
Introduction to Hypothesis Testing 1 Hypothesis Testing A hypothesis test is a statistical procedure that uses sample data to evaluate a hypothesis about a population Hypothesis is stated in terms of the
### Introduction to. Hypothesis Testing CHAPTER LEARNING OBJECTIVES. 1 Identify the four steps of hypothesis testing.
Introduction to Hypothesis Testing CHAPTER 8 LEARNING OBJECTIVES After reading this chapter, you should be able to: 1 Identify the four steps of hypothesis testing. 2 Define null hypothesis, alternative
### TABLE OF CONTENTS. About Chi Squares... 1. What is a CHI SQUARE?... 1. Chi Squares... 1. Hypothesis Testing with Chi Squares... 2
About Chi Squares TABLE OF CONTENTS About Chi Squares... 1 What is a CHI SQUARE?... 1 Chi Squares... 1 Goodness of fit test (One-way χ 2 )... 1 Test of Independence (Two-way χ 2 )... 2 Hypothesis Testing
### Tutorial 5: Hypothesis Testing
Tutorial 5: Hypothesis Testing Rob Nicholls nicholls@mrc-lmb.cam.ac.uk MRC LMB Statistics Course 2014 Contents 1 Introduction................................ 1 2 Testing distributional assumptions....................
### How To Test For Significance On A Data Set
Non-Parametric Univariate Tests: 1 Sample Sign Test 1 1 SAMPLE SIGN TEST A non-parametric equivalent of the 1 SAMPLE T-TEST. ASSUMPTIONS: Data is non-normally distributed, even after log transforming.
### 3.4 Statistical inference for 2 populations based on two samples
3.4 Statistical inference for 2 populations based on two samples Tests for a difference between two population means The first sample will be denoted as X 1, X 2,..., X m. The second sample will be denoted
### DDBA 8438: The t Test for Independent Samples Video Podcast Transcript
DDBA 8438: The t Test for Independent Samples Video Podcast Transcript JENNIFER ANN MORROW: Welcome to The t Test for Independent Samples. My name is Dr. Jennifer Ann Morrow. In today's demonstration,
### HYPOTHESIS TESTING: POWER OF THE TEST
HYPOTHESIS TESTING: POWER OF THE TEST The first 6 steps of the 9-step test of hypothesis are called "the test". These steps are not dependent on the observed data values. When planning a research project,
### Chapter 2 Probability Topics SPSS T tests
Chapter 2 Probability Topics SPSS T tests Data file used: gss.sav In the lecture about chapter 2, only the One-Sample T test has been explained. In this handout, we also give the SPSS methods to perform
### Descriptive Statistics
Descriptive Statistics Primer Descriptive statistics Central tendency Variation Relative position Relationships Calculating descriptive statistics Descriptive Statistics Purpose to describe or summarize
### SPSS for Exploratory Data Analysis Data used in this guide: studentp.sav (http://people.ysu.edu/~gchang/stat/studentp.sav)
Data used in this guide: studentp.sav (http://people.ysu.edu/~gchang/stat/studentp.sav) Organize and Display One Quantitative Variable (Descriptive Statistics, Boxplot & Histogram) 1. Move the mouse pointer
### SCHOOL OF HEALTH AND HUMAN SCIENCES DON T FORGET TO RECODE YOUR MISSING VALUES
SCHOOL OF HEALTH AND HUMAN SCIENCES Using SPSS Topics addressed today: 1. Differences between groups 2. Graphing Use the s4data.sav file for the first part of this session. DON T FORGET TO RECODE YOUR
### An Introduction to Statistics Course (ECOE 1302) Spring Semester 2011 Chapter 10- TWO-SAMPLE TESTS
The Islamic University of Gaza Faculty of Commerce Department of Economics and Political Sciences An Introduction to Statistics Course (ECOE 130) Spring Semester 011 Chapter 10- TWO-SAMPLE TESTS Practice
### January 26, 2009 The Faculty Center for Teaching and Learning
THE BASICS OF DATA MANAGEMENT AND ANALYSIS A USER GUIDE January 26, 2009 The Faculty Center for Teaching and Learning THE BASICS OF DATA MANAGEMENT AND ANALYSIS Table of Contents Table of Contents... i
### Outline. Definitions Descriptive vs. Inferential Statistics The t-test - One-sample t-test
The t-test Outline Definitions Descriptive vs. Inferential Statistics The t-test - One-sample t-test - Dependent (related) groups t-test - Independent (unrelated) groups t-test Comparing means Correlation
### Difference of Means and ANOVA Problems
Difference of Means and Problems Dr. Tom Ilvento FREC 408 Accounting Firm Study An accounting firm specializes in auditing the financial records of large firm It is interested in evaluating its fee structure,particularly
### Analysis of Data. Organizing Data Files in SPSS. Descriptive Statistics
Analysis of Data Claudia J. Stanny PSY 67 Research Design Organizing Data Files in SPSS All data for one subject entered on the same line Identification data Between-subjects manipulations: variable to
### Chapter 5 Analysis of variance SPSS Analysis of variance
Chapter 5 Analysis of variance SPSS Analysis of variance Data file used: gss.sav How to get there: Analyze Compare Means One-way ANOVA To test the null hypothesis that several population means are equal,
### Consider a study in which. How many subjects? The importance of sample size calculations. An insignificant effect: two possibilities.
Consider a study in which How many subjects? The importance of sample size calculations Office of Research Protections Brown Bag Series KB Boomer, Ph.D. Director, boomer@stat.psu.edu A researcher conducts
### Testing for differences I exercises with SPSS
Testing for differences I exercises with SPSS Introduction The exercises presented here are all about the t-test and its non-parametric equivalents in their various forms. In SPSS, all these tests can
### Lecture Notes Module 1
Lecture Notes Module 1 Study Populations A study population is a clearly defined collection of people, animals, plants, or objects. In psychological research, a study population usually consists of a specific
### Statistics Review PSY379
Statistics Review PSY379 Basic concepts Measurement scales Populations vs. samples Continuous vs. discrete variable Independent vs. dependent variable Descriptive vs. inferential stats Common analyses
### The Chi-Square Test. STAT E-50 Introduction to Statistics
STAT -50 Introduction to Statistics The Chi-Square Test The Chi-square test is a nonparametric test that is used to compare experimental results with theoretical models. That is, we will be comparing observed
### General Method: Difference of Means. 3. Calculate df: either Welch-Satterthwaite formula or simpler df = min(n 1, n 2 ) 1.
General Method: Difference of Means 1. Calculate x 1, x 2, SE 1, SE 2. 2. Combined SE = SE1 2 + SE2 2. ASSUMES INDEPENDENT SAMPLES. 3. Calculate df: either Welch-Satterthwaite formula or simpler df = min(n
### Using Excel for inferential statistics
FACT SHEET Using Excel for inferential statistics Introduction When you collect data, you expect a certain amount of variation, just caused by chance. A wide variety of statistical tests can be applied
### NCSS Statistical Software
Chapter 06 Introduction This procedure provides several reports for the comparison of two distributions, including confidence intervals for the difference in means, two-sample t-tests, the z-test, the
### Non-Parametric Tests (I)
Lecture 5: Non-Parametric Tests (I) KimHuat LIM lim@stats.ox.ac.uk http://www.stats.ox.ac.uk/~lim/teaching.html Slide 1 5.1 Outline (i) Overview of Distribution-Free Tests (ii) Median Test for Two Independent
### NCSS Statistical Software
Chapter 06 Introduction This procedure provides several reports for the comparison of two distributions, including confidence intervals for the difference in means, two-sample t-tests, the z-test, the
### The Dummy s Guide to Data Analysis Using SPSS
The Dummy s Guide to Data Analysis Using SPSS Mathematics 57 Scripps College Amy Gamble April, 2001 Amy Gamble 4/30/01 All Rights Rerserved TABLE OF CONTENTS PAGE Helpful Hints for All Tests...1 Tests
### Introduction to Statistics with SPSS (15.0) Version 2.3 (public)
Babraham Bioinformatics Introduction to Statistics with SPSS (15.0) Version 2.3 (public) Introduction to Statistics with SPSS 2 Table of contents Introduction... 3 Chapter 1: Opening SPSS for the first
### Chapter 7. One-way ANOVA
Chapter 7 One-way ANOVA One-way ANOVA examines equality of population means for a quantitative outcome and a single categorical explanatory variable with any number of levels. The t-test of Chapter 6 looks
### COMPARISONS OF CUSTOMER LOYALTY: PUBLIC & PRIVATE INSURANCE COMPANIES.
277 CHAPTER VI COMPARISONS OF CUSTOMER LOYALTY: PUBLIC & PRIVATE INSURANCE COMPANIES. This chapter contains a full discussion of customer loyalty comparisons between private and public insurance companies
### Impact of Enrollment Timing on Performance: The Case of Students Studying the First Course in Accounting
Journal of Accounting, Finance and Economics Vol. 5. No. 1. September 2015. Pp. 1 9 Impact of Enrollment Timing on Performance: The Case of Students Studying the First Course in Accounting JEL Code: M41
### Chapter 7 Section 1 Homework Set A
Chapter 7 Section 1 Homework Set A 7.15 Finding the critical value t *. What critical value t * from Table D (use software, go to the web and type t distribution applet) should be used to calculate the
### Two-sample hypothesis testing, II 9.07 3/16/2004
Two-sample hypothesis testing, II 9.07 3/16/004 Small sample tests for the difference between two independent means For two-sample tests of the difference in mean, things get a little confusing, here,
### Testing Group Differences using T-tests, ANOVA, and Nonparametric Measures
Testing Group Differences using T-tests, ANOVA, and Nonparametric Measures Jamie DeCoster Department of Psychology University of Alabama 348 Gordon Palmer Hall Box 870348 Tuscaloosa, AL 35487-0348 Phone:
### How To Check For Differences In The One Way Anova
MINITAB ASSISTANT WHITE PAPER This paper explains the research conducted by Minitab statisticians to develop the methods and data checks used in the Assistant in Minitab 17 Statistical Software. One-Way
### Descriptive and Inferential Statistics
General Sir John Kotelawala Defence University Workshop on Descriptive and Inferential Statistics Faculty of Research and Development 14 th May 2013 1. Introduction to Statistics 1.1 What is Statistics?
### HYPOTHESIS TESTING: CONFIDENCE INTERVALS, T-TESTS, ANOVAS, AND REGRESSION
HYPOTHESIS TESTING: CONFIDENCE INTERVALS, T-TESTS, ANOVAS, AND REGRESSION HOD 2990 10 November 2010 Lecture Background This is a lightning speed summary of introductory statistical methods for senior undergraduate
### 1. What is the critical value for this 95% confidence interval? CV = z.025 = invnorm(0.025) = 1.96
1 Final Review 2 Review 2.1 CI 1-propZint Scenario 1 A TV manufacturer claims in its warranty brochure that in the past not more than 10 percent of its TV sets needed any repair during the first two years
### Comparing Means in Two Populations
Comparing Means in Two Populations Overview The previous section discussed hypothesis testing when sampling from a single population (either a single mean or two means from the same population). Now we
### II. DISTRIBUTIONS distribution normal distribution. standard scores
Appendix D Basic Measurement And Statistics The following information was developed by Steven Rothke, PhD, Department of Psychology, Rehabilitation Institute of Chicago (RIC) and expanded by Mary F. Schmidt,
### This chapter discusses some of the basic concepts in inferential statistics.
Research Skills for Psychology Majors: Everything You Need to Know to Get Started Inferential Statistics: Basic Concepts This chapter discusses some of the basic concepts in inferential statistics. Details
### Nonparametric Two-Sample Tests. Nonparametric Tests. Sign Test
Nonparametric Two-Sample Tests Sign test Mann-Whitney U-test (a.k.a. Wilcoxon two-sample test) Kolmogorov-Smirnov Test Wilcoxon Signed-Rank Test Tukey-Duckworth Test 1 Nonparametric Tests Recall, nonparametric
### MEASURES OF LOCATION AND SPREAD
Paper TU04 An Overview of Non-parametric Tests in SAS : When, Why, and How Paul A. Pappas and Venita DePuy Durham, North Carolina, USA ABSTRACT Most commonly used statistical procedures are based on the
### Standard Deviation Estimator
CSS.com Chapter 905 Standard Deviation Estimator Introduction Even though it is not of primary interest, an estimate of the standard deviation (SD) is needed when calculating the power or sample size of
### Lesson 1: Comparison of Population Means Part c: Comparison of Two- Means
Lesson : Comparison of Population Means Part c: Comparison of Two- Means Welcome to lesson c. This third lesson of lesson will discuss hypothesis testing for two independent means. Steps in Hypothesis
### 13: Additional ANOVA Topics. Post hoc Comparisons
13: Additional ANOVA Topics Post hoc Comparisons ANOVA Assumptions Assessing Group Variances When Distributional Assumptions are Severely Violated Kruskal-Wallis Test Post hoc Comparisons In the prior
### " Y. Notation and Equations for Regression Lecture 11/4. Notation:
Notation: Notation and Equations for Regression Lecture 11/4 m: The number of predictor variables in a regression Xi: One of multiple predictor variables. The subscript i represents any number from 1 through
### Statistiek II. John Nerbonne. October 1, 2010. Dept of Information Science j.nerbonne@rug.nl
Dept of Information Science j.nerbonne@rug.nl October 1, 2010 Course outline 1 One-way ANOVA. 2 Factorial ANOVA. 3 Repeated measures ANOVA. 4 Correlation and regression. 5 Multiple regression. 6 Logistic
### Section 13, Part 1 ANOVA. Analysis Of Variance
Section 13, Part 1 ANOVA Analysis Of Variance Course Overview So far in this course we ve covered: Descriptive statistics Summary statistics Tables and Graphs Probability Probability Rules Probability
### Two-Sample T-Tests Allowing Unequal Variance (Enter Difference)
Chapter 45 Two-Sample T-Tests Allowing Unequal Variance (Enter Difference) Introduction This procedure provides sample size and power calculations for one- or two-sided two-sample t-tests when no assumption
### Impact of Skewness on Statistical Power
Modern Applied Science; Vol. 7, No. 8; 013 ISSN 1913-1844 E-ISSN 1913-185 Published by Canadian Center of Science and Education Impact of Skewness on Statistical Power Ötüken Senger 1 1 Kafkas University,
### Two-Sample T-Tests Assuming Equal Variance (Enter Means)
Chapter 4 Two-Sample T-Tests Assuming Equal Variance (Enter Means) Introduction This procedure provides sample size and power calculations for one- or two-sided two-sample t-tests when the variances of
### DISCRIMINANT FUNCTION ANALYSIS (DA)
DISCRIMINANT FUNCTION ANALYSIS (DA) John Poulsen and Aaron French Key words: assumptions, further reading, computations, standardized coefficents, structure matrix, tests of signficance Introduction Discriminant
### Normality Testing in Excel
Normality Testing in Excel By Mark Harmon Copyright 2011 Mark Harmon No part of this publication may be reproduced or distributed without the express permission of the author. mark@excelmasterseries.com
### Introduction. Hypothesis Testing. Hypothesis Testing. Significance Testing
Introduction Hypothesis Testing Mark Lunt Arthritis Research UK Centre for Ecellence in Epidemiology University of Manchester 13/10/2015 We saw last week that we can never know the population parameters
### California University of Pennsylvania Guidelines for New Course Proposals University Course Syllabus Approved: 2/4/13. Department of Psychology
California University of Pennsylvania Guidelines for New Course Proposals University Course Syllabus Approved: 2/4/13 Department of Psychology A. Protocol Course Name: Statistics and Research Methods in
### Introduction. Statistics Toolbox
Introduction A hypothesis test is a procedure for determining if an assertion about a characteristic of a population is reasonable. For example, suppose that someone says that the average price of a gallon
### t Tests in Excel The Excel Statistical Master By Mark Harmon Copyright 2011 Mark Harmon
t-tests in Excel By Mark Harmon Copyright 2011 Mark Harmon No part of this publication may be reproduced or distributed without the express permission of the author. mark@excelmasterseries.com www.excelmasterseries.com
### Hypothesis Testing --- One Mean
Hypothesis Testing --- One Mean A hypothesis is simply a statement that something is true. Typically, there are two hypotheses in a hypothesis test: the null, and the alternative. Null Hypothesis The hypothesis
### Understanding Power and Rules of Thumb for Determining Sample Sizes
Tutorials in Quantitative Methods for Psychology 2007, vol. 3 (2), p. 43 50. Understanding Power and Rules of Thumb for Determining Sample Sizes Carmen R. Wilson VanVoorhis and Betsy L. Morgan University
### 1.5 Oneway Analysis of Variance
Statistics: Rosie Cornish. 200. 1.5 Oneway Analysis of Variance 1 Introduction Oneway analysis of variance (ANOVA) is used to compare several means. This method is often used in scientific or medical experiments
### James E. Bartlett, II is Assistant Professor, Department of Business Education and Office Administration, Ball State University, Muncie, Indiana.
Organizational Research: Determining Appropriate Sample Size in Survey Research James E. Bartlett, II Joe W. Kotrlik Chadwick C. Higgins The determination of sample size is a common task for many organizational
### Trust, Job Satisfaction, Organizational Commitment, and the Volunteer s Psychological Contract
Trust, Job Satisfaction, Commitment, and the Volunteer s Psychological Contract Becky J. Starnes, Ph.D. Austin Peay State University Clarksville, Tennessee, USA starnesb@apsu.edu Abstract Studies indicate
### LAB 4 INSTRUCTIONS CONFIDENCE INTERVALS AND HYPOTHESIS TESTING
LAB 4 INSTRUCTIONS CONFIDENCE INTERVALS AND HYPOTHESIS TESTING In this lab you will explore the concept of a confidence interval and hypothesis testing through a simulation problem in engineering setting.
### Rank-Based Non-Parametric Tests
Rank-Based Non-Parametric Tests Reminder: Student Instructional Rating Surveys You have until May 8 th to fill out the student instructional rating surveys at https://sakai.rutgers.edu/portal/site/sirs
### HYPOTHESIS TESTING WITH SPSS:
HYPOTHESIS TESTING WITH SPSS: A NON-STATISTICIAN S GUIDE & TUTORIAL by Dr. Jim Mirabella SPSS 14.0 screenshots reprinted with permission from SPSS Inc. Published June 2006 Copyright Dr. Jim Mirabella CHAPTER
### Chicago Booth BUSINESS STATISTICS 41000 Final Exam Fall 2011
Chicago Booth BUSINESS STATISTICS 41000 Final Exam Fall 2011 Name: Section: I pledge my honor that I have not violated the Honor Code Signature: This exam has 34 pages. You have 3 hours to complete this
### Psychology 60 Fall 2013 Practice Exam Actual Exam: Next Monday. Good luck!
Psychology 60 Fall 2013 Practice Exam Actual Exam: Next Monday. Good luck! Name: 1. The basic idea behind hypothesis testing: A. is important only if you want to compare two populations. B. depends on
### Study Guide for the Final Exam
Study Guide for the Final Exam When studying, remember that the computational portion of the exam will only involve new material (covered after the second midterm), that material from Exam 1 will make
### A POPULATION MEAN, CONFIDENCE INTERVALS AND HYPOTHESIS TESTING
CHAPTER 5. A POPULATION MEAN, CONFIDENCE INTERVALS AND HYPOTHESIS TESTING 5.1 Concepts When a number of animals or plots are exposed to a certain treatment, we usually estimate the effect of the treatment
### Unit 31 A Hypothesis Test about Correlation and Slope in a Simple Linear Regression
Unit 31 A Hypothesis Test about Correlation and Slope in a Simple Linear Regression Objectives: To perform a hypothesis test concerning the slope of a least squares line To recognize that testing for a
### One-Way Analysis of Variance
One-Way Analysis of Variance Note: Much of the math here is tedious but straightforward. We ll skim over it in class but you should be sure to ask questions if you don t understand it. I. Overview A. We
### Variables Control Charts
MINITAB ASSISTANT WHITE PAPER This paper explains the research conducted by Minitab statisticians to develop the methods and data checks used in the Assistant in Minitab 17 Statistical Software. Variables
### The Statistics Tutor s Quick Guide to
statstutor community project encouraging academics to share statistics support resources All stcp resources are released under a Creative Commons licence The Statistics Tutor s Quick Guide to Stcp-marshallowen-7
### Projects Involving Statistics (& SPSS)
Projects Involving Statistics (& SPSS) Academic Skills Advice Starting a project which involves using statistics can feel confusing as there seems to be many different things you can do (charts, graphs,
### ISyE 2028 Basic Statistical Methods - Fall 2015 Bonus Project: Big Data Analytics Final Report: Time spent on social media
ISyE 2028 Basic Statistical Methods - Fall 2015 Bonus Project: Big Data Analytics Final Report: Time spent on social media Abstract: The growth of social media is astounding and part of that success was
### statistics Chi-square tests and nonparametric Summary sheet from last time: Hypothesis testing Summary sheet from last time: Confidence intervals
Summary sheet from last time: Confidence intervals Confidence intervals take on the usual form: parameter = statistic ± t crit SE(statistic) parameter SE a s e sqrt(1/n + m x 2 /ss xx ) b s e /sqrt(ss
### Hypothesis testing - Steps
Hypothesis testing - Steps Steps to do a two-tailed test of the hypothesis that β 1 0: 1. Set up the hypotheses: H 0 : β 1 = 0 H a : β 1 0. 2. Compute the test statistic: t = b 1 0 Std. error of b 1 =
### Mind on Statistics. Chapter 13
Mind on Statistics Chapter 13 Sections 13.1-13.2 1. Which statement is not true about hypothesis tests? A. Hypothesis tests are only valid when the sample is representative of the population for the question
### An analysis method for a quantitative outcome and two categorical explanatory variables.
Chapter 11 Two-Way ANOVA An analysis method for a quantitative outcome and two categorical explanatory variables. If an experiment has a quantitative outcome and two categorical explanatory variables that
### Stat 411/511 THE RANDOMIZATION TEST. Charlotte Wickham. stat511.cwick.co.nz. Oct 16 2015
Stat 411/511 THE RANDOMIZATION TEST Oct 16 2015 Charlotte Wickham stat511.cwick.co.nz Today Review randomization model Conduct randomization test What about CIs? Using a t-distribution as an approximation | 10,571 | 46,602 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-17 | latest | en | 0.876742 |
https://5minutevideomakeover.com/camellia/weighted-average-price-calculation-example.php | 1,611,442,893,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538741.56/warc/CC-MAIN-20210123222657-20210124012657-00335.warc.gz | 211,575,156 | 8,480 | # Price calculation average example weighted
Home » Camellia » Weighted average price calculation example
in Camellia
## Volume Weighted Average Price (VWAP) Investopedia
Calculating Weighted Average – Sisense Community. Knowing the weighted-average contribution Calculating the best mix of products is the variable cost of one unit from its selling price. For example,, It is important to understand the three different types of weighted indexes: price Some examples of value-weighted indexes are geometric average calculation.
### Step 8 for Calulating a Weighted Average WA Selling Price
How to Calculate the Volume-Weighted Average Price. The weighted average contribution margin is the average Examples of these ABC can use the weighted average contribution margin to calculate how many units it, Now, you have 1445 as a weighted average of price and product. As I said, sumproduct is an easy way to calculate the weighted average or weighted mean in Excel..
In finance, volume-weighted average price (VWAP) is the ratio of the value traded to total volume traded over a particular time horizon (usually one day). In this post we will understand how to calculate TWAP (Time Weighted Average Price) and use it for trading. We will also explore TWAP AFL and Excel sheet.
This price differential can apply to both different A weighted inventory average determines the average Example of FIFO Goods; Calculate Gross Profit It stands for Volume Weighted Average Price and refers to a simple indicator that can reveal a lot Breakout Examples. How to Calculate Volume-weighted Average
... Using SASВ® Contextual Analysis to Calculate Final Weighted Average Consumer Price . calculation of the final weighted average calculation (for example, Weighted average versus average in Excel. Understanding the difference and how to use an Excel function to get the weighted average price. So in the same example,
Supposing you have created a pivot table as below screenshot shown. And I will take the pivot table as example to calculate the weighted average price of each fruit Weighted average shares outstanding refers to the The number of weighted average shares outstanding is used in calculating metrics examples; Price to
Calculating a price-weighted average To calculate a price-weighted average, For example, if you want to calculate a price-weighted average of four stocks, The weighted average contribution margin is the average Examples of these ABC can use the weighted average contribution margin to calculate how many units it
The above weighted average formula returns the value 849.00. I.e. the average price paid per computer is \$849.00. A video explaining the calculation of a weighted 2/09/2008В В· There are many analyses you will undertake that will require the need to calculate a weighted average price. In the \$4.00 example, Weighted Average
... (rolling average or running average) is a calculation to An example of a simple equally weighted running the average price at the Add each stock price. In the example, add \$60 and \$20 to get \$80. Divide by the divisor. The first time you compute the price-weighted average, the divisor is simply
Knowing the weighted-average contribution Calculating the best mix of products is the variable cost of one unit from its selling price. For example, The volume weighted average price Concepts and examples Learn how traders use the VWAP to calculate the average price weighted by volume and to
The above weighted average formula returns the value 849.00. I.e. the average price paid per computer is \$849.00. A video explaining the calculation of a weighted Weighted average cost of must earn to keep its common stock price from is included in the calculation of WACC because debt offers a tax shield i
Weighted average shares outstanding refers to the The number of weighted average shares outstanding is used in calculating metrics examples; Price to The weighted average cost of capital Before getting into the specifics of calculating WACC, Equity value = Diluted shares outstanding x share price;
... Using SASВ® Contextual Analysis to Calculate Final Weighted Average Consumer Price . calculation of the final weighted average calculation (for example, The weighted average contribution margin is the average Examples of these ABC can use the weighted average contribution margin to calculate how many units it
Weighted average cost of must earn to keep its common stock price from is included in the calculation of WACC because debt offers a tax shield i Weighted average versus average in Excel. Understanding the difference and how to use an Excel function to get the weighted average price. So in the same example,
When calculating the weighted average Weighted Average: Example Calculation 1. When you're looking at EPS you'll also come across a ratio called the price Supposing you have created a pivot table as below screenshot shown. And I will take the pivot table as example to calculate the weighted average price of each fruit
Get the definition of Volume Weighted Average Price (VWAP), along with examples of its usage and how to calculate it, from our glossary, along with more terms! In this post we will understand how to calculate TWAP (Time Weighted Average Price) and use it for trading. We will also explore TWAP AFL and Excel sheet.
26/02/2015 · The goal is to calculate the weighted average. For example, despite your Light 2 thoughts on “ How to Calculate Sales Volume (Weighted Average) ” In this article we will learn what a weighted average is and how to Excel's SUMPRODUCT formula to calculate weighted average So for example: Prices (Earliest
The volume weighted average price Concepts and examples Learn how traders use the VWAP to calculate the average price weighted by volume and to 17/04/2017В В· How to calculate Time Weighted Average. with example..
Weighted average cost of must earn to keep its common stock price from is included in the calculation of WACC because debt offers a tax shield i 2. Calculate the Weighted Average Selling Price (one single value): Now John's task is to reduce the number of selling prices from six (6) down to one (1) weighted
Volume Weighted Average Price (VWAP) Investopedia. It is important to understand the three different types of weighted indexes: price Some examples of value-weighted indexes are geometric average calculation, The Volume Weighted Average Price indicator Lag is inherent in the indicator because it is a calculation of an average using past data. For example, if you.
Invoiced Using SASВ® Text Analytics to Calculate Final. ... Using SASВ® Contextual Analysis to Calculate Final Weighted Average Consumer Price . calculation of the final weighted average calculation (for example,, The weighted average cost of capital does not have a readily available stated price on it. using a weighted average cost of capital calculator is not easy or.
### Volume-weighted average price Wikipedia
Volume Weighted Average Price (VWAP) Investopedia. The weighted average cost of capital Before getting into the specifics of calculating WACC, Equity value = Diluted shares outstanding x share price; Supposing you have created a pivot table as below screenshot shown. And I will take the pivot table as example to calculate the weighted average price of each fruit.
• Invoiced Using SASВ® Text Analytics to Calculate Final
• Volume Weighted Average Price
• ... (rolling average or running average) is a calculation to An example of a simple equally weighted running the average price at the A price-weighted index is a stock market index in which the constituent Dow Jones Industrial Average is a prominent example of price-weighted Calculation. The
Calculating a price-weighted average To calculate a price-weighted average, For example, if you want to calculate a price-weighted average of four stocks, 17/04/2017В В· How to calculate Time Weighted Average. with example..
For example, a four-period SMA with prices of 1.2640, 1.2641, 1.2642, and 1.2641 gives a moving average of 1.2641 using the calculation [(1.2640 + 1.2641 + 1.2642 + 1 Supposing you have created a pivot table as below screenshot shown. And I will take the pivot table as example to calculate the weighted average price of each fruit
Weighted average is an average in which and not just the absolute price. Examples of Weighted Average. weighted averages are used in the calculation of a For example, a four-period SMA with prices of 1.2640, 1.2641, 1.2642, and 1.2641 gives a moving average of 1.2641 using the calculation [(1.2640 + 1.2641 + 1.2642 + 1
The weighted average cost of capital Before getting into the specifics of calculating WACC, Equity value = Diluted shares outstanding x share price; 26/02/2015 · The goal is to calculate the weighted average. For example, despite your Light 2 thoughts on “ How to Calculate Sales Volume (Weighted Average) ”
The weighted average cost of capital Before getting into the specifics of calculating WACC, Equity value = Diluted shares outstanding x share price; Calculating a price-weighted average To calculate a price-weighted average, For example, if you want to calculate a price-weighted average of four stocks,
The above weighted average formula returns the value 849.00. I.e. the average price paid per computer is \$849.00. A video explaining the calculation of a weighted This price differential can apply to both different A weighted inventory average determines the average Example of FIFO Goods; Calculate Gross Profit
2/09/2008В В· There are many analyses you will undertake that will require the need to calculate a weighted average price. In the \$4.00 example, Weighted Average Add each stock price. In the example, add \$60 and \$20 to get \$80. Divide by the divisor. The first time you compute the price-weighted average, the divisor is simply
Weighted Average Share Calculation Example #1. Below is the example of Weighted Average Shares calculation when shares are issued as well as repurchased during the year. It stands for Volume Weighted Average Price and refers to a simple indicator that can reveal a lot Breakout Examples. How to Calculate Volume-weighted Average
The weighted average contribution margin is the average Examples of these ABC can use the weighted average contribution margin to calculate how many units it The above weighted average formula returns the value 849.00. I.e. the average price paid per computer is \$849.00. A video explaining the calculation of a weighted
2. Calculate the Weighted Average Selling Price (one single value): Now John's task is to reduce the number of selling prices from six (6) down to one (1) weighted 2. Calculate the Weighted Average Selling Price (one single value): Now John's task is to reduce the number of selling prices from six (6) down to one (1) weighted
Supposing you have created a pivot table as below screenshot shown. And I will take the pivot table as example to calculate the weighted average price of each fruit Calculating a price-weighted average To calculate a price-weighted average, For example, if you want to calculate a price-weighted average of four stocks,
Weighted average versus average in Excel. Understanding the difference and how to use an Excel function to get the weighted average price. So in the same example, For example, in buying assets for The WACC Calculator spreadsheet uses the formula above to calculate the Weighted Average Cost of Capital. Price of Stock
The Volume Weighted Average Price indicator Lag is inherent in the indicator because it is a calculation of an average using past data. For example, if you Now, you have 1445 as a weighted average of price and product. As I said, sumproduct is an easy way to calculate the weighted average or weighted mean in Excel.
A price-weighted index is a stock market index in which the constituent Dow Jones Industrial Average is a prominent example of price-weighted Calculation. The The weighted average cost of capital Before getting into the specifics of calculating WACC, Equity value = Diluted shares outstanding x share price;
... Using SASВ® Contextual Analysis to Calculate Final Weighted Average Consumer Price . calculation of the final weighted average calculation (for example, Analytical Need Weighted average refers to the mathematical practice In the discussed example, we will look at several products and calculate their average price.
2/09/2008В В· There are many analyses you will undertake that will require the need to calculate a weighted average price. In the \$4.00 example, Weighted Average 2. Calculate the Weighted Average Selling Price (one single value): Now John's task is to reduce the number of selling prices from six (6) down to one (1) weighted | 2,651 | 12,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-04 | latest | en | 0.872113 |
https://jp.mathworks.com/matlabcentral/cody/problems/2094-sum-of-diagonals-elements-of-a-matrix/solutions/523962 | 1,568,792,111,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573258.74/warc/CC-MAIN-20190918065330-20190918091330-00313.warc.gz | 556,739,290 | 15,614 | Cody
# Problem 2094. Sum of diagonals elements of a matrix
Solution 523962
Submitted on 8 Nov 2014 by Pritesh Shah
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
%% x = [1 2 3;4 5 6; 7 8 9;]; y_correct = 25; assert(isequal(DiagSum(x),y_correct));
ans = 25
2 Pass
%% x = [1 2 3 4 5;6 7 8 9 10; 11 12 13 14 15; 16 17 18 19 20; 21 22 23 24 25;]; y_correct = 117; assert(isequal(DiagSum(x),y_correct));
ans = 117
3 Pass
%% x = ones(7); y_correct = 13; assert(isequal(DiagSum(x),y_correct));
ans = 13
4 Pass
%% x = magic(9); y_correct = 697; assert(isequal(DiagSum(x),y_correct));
ans = 697 | 275 | 722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-39 | latest | en | 0.531995 |
https://www.shaalaa.com/question-bank-solutions/write-the-three-whole-numbers-occurring-just-before-10001-concept-whole-numbers_136794 | 1,620,536,029,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988955.89/warc/CC-MAIN-20210509032519-20210509062519-00068.warc.gz | 1,016,655,240 | 8,934 | # Write the three whole numbers occurring just before 10001. - Mathematics
Write the three whole numbers occurring just before 10001.
#### Solution
Three whole numbers occurring just before 10001 are as follows:
10001 − 1 = 10000
10000 − 1 = 9999
9999 − 1 = 9998
Is there an error in this question or solution?
#### APPEARS IN
RS Aggarwal Class 6 Mathematics
Chapter 3 Whole Numbers
Exercise 3A | Q 2 | Page 45 | 118 | 417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2021-21 | latest | en | 0.837214 |
http://mathhelpforum.com/algebra/196750-algebra-2-problem-print.html | 1,518,913,920,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891808539.63/warc/CC-MAIN-20180217224905-20180218004905-00615.warc.gz | 222,250,157 | 3,160 | # Algebra 2 problem
• Apr 3rd 2012, 06:13 AM
Mhmh96
Algebra 2 problem
• Apr 3rd 2012, 06:38 AM
emakarov
Re: Algebra 2 problem
Why does the formula image serve as a link to some strange site and why is there another hidden link to the same site?
What does [[x]] denote?
• Apr 3rd 2012, 06:39 AM
Mhmh96
Re: Algebra 2 problem
Because i upload this picture by that website .
• Apr 3rd 2012, 06:40 AM
Mhmh96
Re: Algebra 2 problem
I think for the greatest integer function
• Apr 3rd 2012, 06:43 AM
emakarov
Re: Algebra 2 problem
Quote:
Originally Posted by emakarov
What does [[x]] denote?
• Apr 3rd 2012, 06:46 AM
Mhmh96
Re: Algebra 2 problem
I dont know how to use the information in the problem to solve it .
• Apr 3rd 2012, 06:47 AM
emakarov
Re: Algebra 2 problem
Apply g to 0.85. If the result is less than or equal to -2, apply the first line in the definition of f to the result. Otherwise, apply the second line.
• Apr 3rd 2012, 06:59 AM
Mhmh96
Re: Algebra 2 problem
It is bigger than -2 ,so f(.80)=2(.80)+3 ,right ?
• Apr 3rd 2012, 07:03 AM
emakarov
Re: Algebra 2 problem
The greatest integer not exceeding 0.85 is 0, so g(0.85) = 0. Yes, 0 > -2, so (f ∘ g)(0.85) = f(g(0.85)) = f(0) = 2 * 0 + 3 = 3.
• Apr 3rd 2012, 07:07 AM
Mhmh96
Re: Algebra 2 problem | 480 | 1,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-09 | longest | en | 0.86676 |
http://mathhelpforum.com/trigonometry/218278-finding-standard-equation-ellipse-print.html | 1,524,597,617,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947033.92/warc/CC-MAIN-20180424174351-20180424194351-00568.warc.gz | 197,872,955 | 3,071 | # Finding standard equation of an Ellipse
• Apr 27th 2013, 01:17 PM
vaironxxrd
Finding standard equation of an Ellipse
Hello Everyone,
I can't figure out very well how to find the standard equation of the following ellipse.
$\displaystyle 25x^2+49y^2-300x+49y-1251/4=0$
I was trying the following steps.
$\displaystyle [25x^2-300x +- ?]+ [ 49y^2+49y +- ? ]= 1251/4$ Find squares (I'm not good at, finding squares)
• Apr 27th 2013, 01:28 PM
Shakarri
Re: Finding standard equation of an Ellipse
Looking at the x part. If you want to get it into the form (ax+b)2 expanding it out might make it clearer. (ax+b)2=a2x2+2abx+b2
since the x2 part is 25x2 you know that a2=25, a=5
The x part is -300. You know that 2ab=-300. and a=5 so b=-30
Lastly you need to balance the square by subtracting a constant.
$\displaystyle 25x^2-300x=(5x-30)^2+c$
For this to be true c must be equal to -900
• Apr 27th 2013, 02:15 PM
Soroban
Re: Finding standard equation of an Ellipse
Hello, vaironxxrd!
Before completing-the-square,
Quote:
Find the standard equation of the following ellipse:
. . $\displaystyle 25x^2+49y^2-300x+49y-\tfrac{1251}{4}\:=\:0$
We have: . . $\displaystyle 25x^2 - 300x \quad+ 49y^2 + 49y\quad \:=\:\tfrac{1251}{4}$
Factor:.$\displaystyle 25(x^2-12x \qquad) + 49(y^2 + y \qquad) \:=\:\tfrac{1251}{4}$
. . . . . . $\displaystyle 25(x^2-12x + {\color{red}36}) + 49(y^2 + y +{\color{blue}\tfrac{1}{4}}) \:=\:\tfrac{1251}{4} + {\color{red}900} + {\color{blue}\tfrac{49}{4}}$
. . . . . . . . . . . . . . $\displaystyle 25(x-6)^2 + 49(y+\tfrac{1}{2})^2 \:=\:1225$
Divide by 1225:. . . . . $\displaystyle \frac{(x-6)^2}{49} + \frac{(y+\frac{1}{2})^2}{25} \;=\;1$ | 654 | 1,672 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2018-17 | latest | en | 0.661943 |
http://slideplayer.com/slide/1595252/ | 1,544,832,636,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826530.72/warc/CC-MAIN-20181214232243-20181215014243-00569.warc.gz | 252,530,911 | 19,662 | Mr. Montalbano Professor: Dr. Ransom EDTS 580-30: Multimedia for Educators Spring 2009 Please click here to begin…
Presentation on theme: "Mr. Montalbano Professor: Dr. Ransom EDTS 580-30: Multimedia for Educators Spring 2009 Please click here to begin…"— Presentation transcript:
Mr. Montalbano Professor: Dr. Ransom EDTS 580-30: Multimedia for Educators Spring 2009 Please click here to begin…
Welcome This quiz will allow you to determine your current understanding of vocabulary related to electricity. All of these concepts are to be related to electricity. Please take a few minutes to complete this ten-question quiz. Good Luck!! Click here to continue
A conductor is a ______. A.Measure of how much a material opposes the flow of electric current and changes electric current into heat energyMeasure of how much a material opposes the flow of electric current and changes electric current into heat energy B.material through which electric current passes easilymaterial through which electric current passes easily C.material that electric current does not pass easilymaterial that electric current does not pass easily D.person that drives a trainperson that drives a train
A resistor is a(n) ______. A.example of a generatorexample of a generator B.device which prevents protons from an atom attracting to the electrons of that same atomdevice which prevents protons from an atom attracting to the electrons of that same atom A.measure of how much a material opposes the flow of electric current and changes electric current into heat energymeasure of how much a material opposes the flow of electric current and changes electric current into heat energy D.magnet moving inside a coil of wire, making electricitymagnet moving inside a coil of wire, making electricity
The current for each appliance has its own path in a(n) ______. A.closed circuitclosed circuit B.parallel circuitparallel circuit C.series circuitseries circuit D.open circuitopen circuit
This type of material is a good conductor of electricity ______. A.coppercopper B.glassglass C.rubberrubber D.ironiron
A small magnet that can turn freely is a(n) ______. A.compasscompass B.electromagnetelectromagnet C.micromagnetmicromagnet D.dimagnetdimagnet
Magnetism is strongest at the ______. A.center of the magnetcenter of the magnet B.poles of the magnetpoles of the magnet C.space around the magnetspace around the magnet D.focal point of the magnetfocal point of the magnet
An electromagnet is a temporary magnet. A.TrueTrue B.FalseFalse
Generators have huge magnets and huge coils of wire to make electricity. A.TrueTrue B.FalseFalse
If a parallel circuit is open, both light bulbs will be lit. A.TrueTrue B.FalseFalse
It is safe to touch electricity if there is an insulator around the conductor. A.TrueTrue B.FalseFalse | 600 | 2,812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-51 | latest | en | 0.88874 |
http://www.cfd-online.com/Forums/openfoam-programming-development/101543-problem-pow-volscalarfield-print.html | 1,444,481,632,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737952309.80/warc/CC-MAIN-20151001221912-00008-ip-10-137-6-227.ec2.internal.warc.gz | 470,261,859 | 4,607 | CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
- OpenFOAM Programming & Development (http://www.cfd-online.com/Forums/openfoam-programming-development/)
- - Problem with pow and volScalarField (http://www.cfd-online.com/Forums/openfoam-programming-development/101543-problem-pow-volscalarfield.html)
_Stefan_ May 7, 2012 08:15
Problem with pow and volScalarField
Hi,
I want to implement an new Order to Calculate the Viscosity nu.
Ostwald Waele Potential Function
tau = k * strain rate ^n
k and n are model Parameter and depend on rho
k=a*rho+b
n=c*rho+d
I got a, b, c, d from some experiments.
The equation for the rotation rheometer:
nu = tau / strain rate
What I have done:
Code:
```//read strain rate tmp<volScalarField> sr(strainRate()); //Parameter dimensionedScalar ka_("ka_",dimensionSet(0,2,-1,0,0,0,0), 0.1202); dimensionedScalar kb_("kb_",dimensionSet(1,-1,-1,0,0,0,0), 24.912); dimensionedScalar na_("na_",dimensionSet(-1,3,0,0,0,0,0), 0.0005); dimensionedScalar nb_("nb_",dimensionSet(0,0,0,0,0,0,0), 0.2246); //read ScalarFeldes rho_ps const tmp <volScalarField> & rhoCalc = U_.mesh().lookupObject<volScalarField>("rho_ps"); //calculation n and k volScalarField K = ka_ * rhoCalc + kb_; volScalarField N = na_ * rhoCalc + nb_; //Calculation nu tmp <volScalarField> nu = K * Foam::pow(max(tone * sr(),VSMALL),N-1.);```
tone is only a dimensioned scalar with 1[s] to make sr() dimensionless.
My Problem is that it did not compile.
If I change N in the pow function to an dimensioned scalar it works fine.
Is it possible that, Foam::pow (volScalarField, volScalarField) does not work?
Here the error massage:
/SimSoftware/OpenFOAM/OpenFOAM-1.5/src/OpenFOAM/lnInclude/GeometricScalarField.C:189: error: no matching function for call to ‘Foam::dimensioned<double>::dimensioned(const char [2], double, const Foam::dimensionSet&)’
/SimSoftware/OpenFOAM/OpenFOAM-1.5/src/OpenFOAM/lnInclude/dimensionedType.C:107: note: candidates are: Foam::dimensioned<Type>::dimensioned(const Foam::word&, const Foam::dimensionSet&, Foam::Istream&) [with Type = double]
/SimSoftware/OpenFOAM/OpenFOAM-1.5/src/OpenFOAM/lnInclude/dimensionedType.C:94: note: Foam::dimensioned<Type>::dimensioned(const Foam::word&, Foam::Istream&) [with Type = double]
/SimSoftware/OpenFOAM/OpenFOAM-1.5/src/OpenFOAM/lnInclude/dimensionedType.C:82: note: Foam::dimensioned<Type>::dimensioned(Foam::Istream &) [with Type = double]
/SimSoftware/OpenFOAM/OpenFOAM-1.5/src/OpenFOAM/lnInclude/dimensionedType.H:95: note: Foam::dimensioned<Type>::dimensioned(const Type&) [with Type = double]
/SimSoftware/OpenFOAM/OpenFOAM-1.5/src/OpenFOAM/lnInclude/dimensionedType.C:69: note: Foam::dimensioned<Type>::dimensioned(const Foam::word&, const Foam::dimensioned<Type>&) [with Type = double]
/SimSoftware/OpenFOAM/OpenFOAM-1.5/src/OpenFOAM/lnInclude/dimensionedType.C:55: note: Foam::dimensioned<Type>::dimensioned(const Foam::word&, const Foam::dimensionSet&, Type) [with Type = double]
/SimSoftware/OpenFOAM/OpenFOAM-1.5/src/OpenFOAM/lnInclude/dimensionedScalarFwd.H:42: note: Foam::dimensioned<double>::dimensioned(const Foam::dimensioned<double>&)
benk May 7, 2012 09:33
Quote:
Originally Posted by _Stefan_ (Post 359722) Is it possible that, Foam::pow (volScalarField, volScalarField) does not work?
I'm not sure about the answer to your question, but my first instinct would be to do this sort of operation within a forAll loop
_Stefan_ May 8, 2012 05:06
Thank you,
I think it is not the best way but it works:
Code:
``` Info << "calculate min strainrate" << endl; volScalarField sr1 = max(sr()*tone,VSMALL);//[-] Info << "forAll loop" << endl; forAll(sr1, cellI) { N1[cellI] = Foam::pow(sr1[cellI],N[cellI]-1); } tmp <volScalarField> nu = K * N1;```
Cyp May 12, 2012 13:09
Did you try N.value() instead of N within the exponent ?
_Stefan_ May 14, 2012 07:30
Hi Cyp,
I tried it but there are the Error:
error: ‘struct Foam::volScalarField’ has no member named ‘value’
Cyp May 14, 2012 18:23
Yes, sorry, .value is valid if N is a dimensionedScalar.
in your case, if you already have declared A, B and C as volScalarField, you can performed :
Code:
`A.internalField() = Foam::pow(B.internalField(),C.internalField());`
The operation does not include the operation upon BC.
Best,
Cyp
_Stefan_ May 15, 2012 04:39
Thanks, Cyp that is what I need. I also need the calculation on the boundary so I added:
Code:
```A.internalField() = Foam::pow(B.internalField(),C.internalField()); A.boundaryField() = Foam::pow(B.boundaryField(),C.boundaryField());```
And it works fine!
Quote:
Originally Posted by _Stefan_ (Post 361121) Thanks, Cyp that is what I need. I also need the calculation on the boundary so I added: Code: ```A.internalField() = Foam::pow(B.internalField(),C.internalField()); A.boundaryField() = Foam::pow(B.boundaryField(),C.boundaryField());``` And it works fine!
Hi Stefan and Cyprien
I want to calculate the following relation.
A=min(0,c*B);
Which A and B are Volscalerfield and c diminesedscaler.
for the c i have used .value().
But I am receiving Eror when I have the B in the min function.
I have tried your suggestion but it does not work.
Could you help me?
Best
Mahdi
Cyp August 7, 2012 07:45
Have you tried without any .value() ?
Else, I suggest to do :
A = c*min(0,B);
Yes I have tried, even for the follwing equations if I remove the .value() it is giving error.
A=min(0,c.value());
A.internalField()=min(0,c.value());
A.boundaryField()=min(0,c.value());
which is working because I removed the B. (but when I am adding the B it gives Error) .
Cyp August 7, 2012 07:50
Quote:
Originally Posted by mm.abdollahzadeh (Post 375803) Yes I have tried, even for the follwing equations if I remove the .value() it is giving error. A=min(0,c.value()); A.internalField()=min(0,c.value()); A.boundaryField()=min(0,c.value()); which is working because I removed the B. (but when I am adding the B it gives Error) .
see my edited post.
Thanks again ,,, I think the problem is not related the "c" .
the error is :
Code:
```error: no matching function for call to 'min(int, Foam::volScalarField&)' /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/label.H:213:1: note: candidates are: char Foam::min(char, char) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/label.H:214:1: note: short int Foam::min(short int, short int) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/label.H:215:1: note: int Foam::min(int, int) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/label.H:216:1: note: long int Foam::min(long int, long int) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/label.H:217:1: note: long long int Foam::min(long long int, long long int) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/label.H:219:1: note: unsigned char Foam::min(unsigned char, unsigned char) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/label.H:220:1: note: short unsigned int Foam::min(short unsigned int, short unsigned int) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/label.H:221:1: note: unsigned int Foam::min(unsigned int, unsigned int) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/label.H:222:1: note: long unsigned int Foam::min(long unsigned int, long unsigned int) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/label.H:223:1: note: long long unsigned int Foam::min(long long unsigned int, long long unsigned int) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/label.H:225:1: note: long int Foam::min(int, long int) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/label.H:226:1: note: long long int Foam::min(int, long long int) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/label.H:227:1: note: long long int Foam::min(long long int, int) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/doubleFloat.H:78:1: note: double Foam::min(double, double) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/doubleFloat.H:79:1: note: double Foam::min(double, float) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/doubleFloat.H:80:1: note: double Foam::min(float, double) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/doubleFloat.H:81:1: note: float Foam::min(float, float) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/doubleFloat.H:82:1: note: double Foam::min(double, int) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/doubleFloat.H:83:1: note: double Foam::min(int, double) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/doubleFloat.H:84:1: note: double Foam::min(double, long int) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/doubleFloat.H:85:1: note: double Foam::min(long int, double) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/doubleFloat.H:86:1: note: float Foam::min(float, int) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/doubleFloat.H:87:1: note: float Foam::min(int, float) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/doubleFloat.H:88:1: note: float Foam::min(float, long int) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/doubleFloat.H:89:1: note: float Foam::min(long int, float) /home/mahdi/centFOAM//OpenFOAM/OpenFOAM-2.0.x/src/OpenFOAM/lnInclude/dimensionSet.H:214:29: note: Foam::dimensionSet Foam::min(const Foam::dimensionSet&, const Foam::dimensionSet&)```
Cyp August 7, 2012 08:04
Althought it is not very elegant, the following snippet should works:
Code:
``` forAll(A, cellI) { A[cellI] = min(0,c.value()*B[cellI]); }```
thanks Cyprein, it works | 3,270 | 10,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2015-40 | longest | en | 0.533613 |
https://socratic.org/questions/how-do-you-solve-the-following-system-4x-5y-1-4x-3y-1 | 1,716,824,240,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059044.17/warc/CC-MAIN-20240527144335-20240527174335-00243.warc.gz | 469,815,647 | 5,990 | # How do you solve the following system: 4x-5y= -1, 4x+3y= -1 ?
Jan 14, 2016
I found:
$x = - \frac{1}{4}$
$y = 0$
#### Explanation:
I would try multiplying the first equation by $- 1$ and then add (in columns) to the second to get:
$\left\{\begin{matrix}\textcolor{red}{- 4 x - 5 y = 1} \\ 4 x + 3 x = - 1\end{matrix}\right.$ add them
$0 - 2 y = 0$
So:
$y = 0$
Substitute this back into the first equation:
$4 x - 0 = - 1$
$x = - \frac{1}{4}$ | 182 | 446 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-22 | latest | en | 0.750764 |
https://www.wyzant.com/resources/answers/fractions?f=active&pagesize=20&pagenum=5 | 1,527,437,466,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794869272.81/warc/CC-MAIN-20180527151021-20180527171021-00131.warc.gz | 868,551,217 | 14,901 | According to your contract, you are going to get 2/7 of the profit of the charity concert you are organizing and 1/5 goes to the bands who performed. The rest goes to charity. What fraction of the...
mrs lim backed 480 cheez cookies and chocolate cookies, she gave 75% of her cheez cookies * 50% of her choc cookies to her mother, she was left with 160.
Mr. PEREZ had 48 more chairs than tables in his furniture store. After selling off 5/6 of the chairs and 3/4 of the tables, there were 33 chairs and tables left. A) How many tables were there...
What were the sales totals this week? The alternatives are as follows: a. \$2,742.86 b. \$ 2,100 c. \$ 2,700 and d. 2,057.14
I need help trying to figure out what the awnser is for this problem
How do I find this out? | 213 | 771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-22 | latest | en | 0.985133 |
http://lambdapage.org/25871/8th-grade-ela-worksheets/evaluating-text-my-life-8th-grade-reading-comprehension-intended-for-8th-grade-ela-worksheets/ | 1,619,157,864,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039601956.95/warc/CC-MAIN-20210423041014-20210423071014-00226.warc.gz | 57,487,501 | 7,442 | The line also has various shape. It usually comes with a simple tracing that can easily followed for the beginner.
You can give a demonstration for writing the letter A side by side with your kid. If your kid had a difficulty to imitate your gesture, then you also can give your hand to help their tracing process.
After you’ve done with the line tracing, now it’s time for more advanced exercise. The spiral tracing can be challenging for your kids. It’s also need a concentration with the good coordination for tracing the lines.
It will show some abstract on the result. However, this can be a good sign that your kids’ creativity is actually developing. After finished with the free activity, you can continue the next exercise of tracing the line or object.
March 8, 2021
March 6, 2021
March 4, 2021
March 10, 2021
Education Com Worksheets
March 9, 2021
Start by teaching your kid about the word that they want to practice. You also can teach them to spell the word. After that, make an example on how to trace the word.
This is the very first step before your kid can write on a blank paper. Never push them to do so until they can manage to trace perfectly. You also need to make a summative evaluation by start counting the number together with kid.
These tools are also good for they who want to prepare the kids before start joining the formal school. Without any further do, let’s dig into these amazing and creative worksheets.
Take a look on the colored pictures, whether it’s tidy enough or not. Give them an instruction that next time they should put the color inside the area. It will make them understand about coloring the object in proper way.
March 7, 2021
March 9, 2021
March 6, 2021
March 5, 2021
Logic Puzzles With Grids Worksheets
March 8, 2021
It consists of several picture that your kid might love to trace. You can introduce the picture and even telling a story as the preparation. It helps the kid to get interested and start working on their tracing process.
This worksheet also helps kid to gain their creativity. It makes them understand about the common objects around them. You also can make an interesting offer for the kid as the reward.
Having an exercise for your children at home during pandemic can be done by exploring these preschool worksheets. For they who have kids on age 3-5, it’s quite challenging for the pandemic year to get their preschool education. Parents should be involved to accompany their kids for their learning process at home. Thus, parents need proper tools that can be used to achieve this goal.
To do so, ensure that your kids are prepared and confirmed that they want to learn at home with you. You can make a room with a conducive situation that proper for the learning progress.
March 9, 2021
March 8, 2021
March 4, 2021
March 5, 2021
March 11, 2021
March 8, 2021
March 6, 2021
March 10, 2021
March 3, 2021
March 7, 2021
March 7, 2021
March 3, 2021
March 9, 2021
March 3, 2021 | 713 | 2,981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2021-17 | latest | en | 0.966693 |
http://dm.mcoe.org/glossary/mean.html | 1,545,035,893,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828448.76/warc/CC-MAIN-20181217065106-20181217091106-00540.warc.gz | 79,660,744 | 2,843 | # MEAN
Mean is one of the three averages (mean, median and mode). It is “mean” because it has a two step calculation:
1) add all the numbers in the group
2) divide by the number of numbers in the group.
The mean is a way of balancing numbers so that everyone has the same amount.
Contact Digital Math | 73 | 304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-51 | latest | en | 0.939611 |
http://www.statsmodels.org/devel/_modules/statsmodels/tsa/vector_ar/var_model.html | 1,513,217,142,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948537139.36/warc/CC-MAIN-20171214020144-20171214040144-00377.warc.gz | 470,958,043 | 23,849 | # Source code for statsmodels.tsa.vector_ar.var_model
"""
Vector Autoregression (VAR) processes
References
----------
Lutkepohl (2005) New Introduction to Multiple Time Series Analysis
"""
from __future__ import division, print_function
from statsmodels.compat.python import (range, lrange, string_types, StringIO, iteritems,
cStringIO)
from collections import defaultdict
import numpy as np
import numpy.linalg as npl
from numpy.linalg import cholesky as chol, solve
import scipy.stats as stats
import scipy.linalg as L
from statsmodels.tools.tools import chain_dot
from statsmodels.tools.linalg import logdet_symm
from statsmodels.tsa.tsatools import vec, unvec
from statsmodels.tsa.vector_ar.irf import IRAnalysis
from statsmodels.tsa.vector_ar.output import VARSummary
import statsmodels.tsa.tsatools as tsa
import statsmodels.tsa.vector_ar.output as output
import statsmodels.tsa.vector_ar.plotting as plotting
import statsmodels.tsa.vector_ar.util as util
import statsmodels.tsa.base.tsa_model as tsbase
import statsmodels.base.wrapper as wrap
mat = np.array
#-------------------------------------------------------------------------------
# VAR process routines
def ma_rep(coefs, maxn=10):
r"""
MA(\infty) representation of VAR(p) process
Parameters
----------
coefs : ndarray (p x k x k)
maxn : int
Number of MA matrices to compute
Notes
-----
VAR(p) process as
.. math:: y_t = A_1 y_{t-1} + \ldots + A_p y_{t-p} + u_t
can be equivalently represented as
.. math:: y_t = \mu + \sum_{i=0}^\infty \Phi_i u_{t-i}
e.g. can recursively compute the \Phi_i matrices with \Phi_0 = I_k
Returns
-------
phis : ndarray (maxn + 1 x k x k)
"""
p, k, k = coefs.shape
phis = np.zeros((maxn+1, k, k))
phis[0] = np.eye(k)
# recursively compute Phi matrices
for i in range(1, maxn + 1):
for j in range(1, i+1):
if j > p:
break
phis[i] += np.dot(phis[i-j], coefs[j-1])
return phis
def is_stable(coefs, verbose=False):
"""
Determine stability of VAR(p) system by examining the eigenvalues of the
VAR(1) representation
Parameters
----------
coefs : ndarray (p x k x k)
Returns
-------
is_stable : bool
"""
A_var1 = util.comp_matrix(coefs)
eigs = np.linalg.eigvals(A_var1)
if verbose:
print('Eigenvalues of VAR(1) rep')
for val in np.abs(eigs):
print(val)
return (np.abs(eigs) <= 1).all()
def var_acf(coefs, sig_u, nlags=None):
"""
Compute autocovariance function ACF_y(h) up to nlags of stable VAR(p)
process
Parameters
----------
coefs : ndarray (p x k x k)
Coefficient matrices A_i
sig_u : ndarray (k x k)
Covariance of white noise process u_t
nlags : int, optional
Defaults to order p of system
Notes
-----
Ref: Lutkepohl p.28-29
Returns
-------
acf : ndarray, (p, k, k)
"""
p, k, _ = coefs.shape
if nlags is None:
nlags = p
# p x k x k, ACF for lags 0, ..., p-1
result = np.zeros((nlags + 1, k, k))
result[:p] = _var_acf(coefs, sig_u)
# yule-walker equations
for h in range(p, nlags + 1):
# compute ACF for lag=h
# G(h) = A_1 G(h-1) + ... + A_p G(h-p)
for j in range(p):
result[h] += np.dot(coefs[j], result[h-j-1])
return result
def _var_acf(coefs, sig_u):
"""
Compute autocovariance function ACF_y(h) for h=1,...,p
Notes
-----
Lutkepohl (2005) p.29
"""
p, k, k2 = coefs.shape
assert(k == k2)
A = util.comp_matrix(coefs)
# construct VAR(1) noise covariance
SigU = np.zeros((k*p, k*p))
SigU[:k,:k] = sig_u
# vec(ACF) = (I_(kp)^2 - kron(A, A))^-1 vec(Sigma_U)
vecACF = L.solve(np.eye((k*p)**2) - np.kron(A, A), vec(SigU))
acf = unvec(vecACF)
acf = acf[:k].T.reshape((p, k, k))
return acf
def forecast(y, coefs, intercept, steps):
"""
Produce linear MSE forecast
Parameters
----------
y :
coefs :
intercept :
steps :
Returns
-------
forecasts : ndarray (steps x neqs)
Notes
-----
Lutkepohl p. 37
Also used by DynamicVAR class
"""
p = len(coefs)
k = len(coefs[0])
# initial value
forcs = np.zeros((steps, k)) + intercept
# h=0 forecast should be latest observation
# forcs[0] = y[-1]
# make indices easier to think about
for h in range(1, steps + 1):
# y_t(h) = intercept + sum_1^p A_i y_t_(h-i)
f = forcs[h - 1]
for i in range(1, p + 1):
# slightly hackish
if h - i <= 0:
# e.g. when h=1, h-1 = 0, which is y[-1]
prior_y = y[h - i - 1]
else:
# e.g. when h=2, h-1=1, which is forcs[0]
prior_y = forcs[h - i - 1]
# i=1 is coefs[0]
f = f + np.dot(coefs[i - 1], prior_y)
forcs[h - 1] = f
return forcs
def forecast_cov(ma_coefs, sig_u, steps):
"""
Compute theoretical forecast error variance matrices
Parameters
----------
Returns
-------
forc_covs : ndarray (steps x neqs x neqs)
"""
k = len(sig_u)
forc_covs = np.zeros((steps, k, k))
prior = np.zeros((k, k))
for h in range(steps):
# Sigma(h) = Sigma(h-1) + Phi Sig_u Phi'
phi = ma_coefs[h]
var = chain_dot(phi, sig_u, phi.T)
forc_covs[h] = prior = prior + var
return forc_covs
def var_loglike(resid, omega, nobs):
r"""
Returns the value of the VAR(p) log-likelihood.
Parameters
----------
resid : ndarray (T x K)
omega : ndarray
Sigma hat matrix. Each element i,j is the average product of the
OLS residual for variable i and the OLS residual for variable j or
np.dot(resid.T,resid)/nobs. There should be no correction for the
degrees of freedom.
nobs : int
Returns
-------
llf : float
The value of the loglikelihood function for a VAR(p) model
Notes
-----
The loglikelihood function for the VAR(p) is
.. math::
-\left(\frac{T}{2}\right)
\left(\ln\left|\Omega\right|-K\ln\left(2\pi\right)-K\right)
"""
logdet = logdet_symm(np.asarray(omega))
neqs = len(omega)
part1 = - (nobs * neqs / 2) * np.log(2 * np.pi)
part2 = - (nobs / 2) * (logdet + neqs)
return part1 + part2
def _reordered(self, order):
#Create new arrays to hold rearranged results from .fit()
endog = self.endog
endog_lagged = self.endog_lagged
params = self.params
sigma_u = self.sigma_u
names = self.names
k_ar = self.k_ar
endog_new = np.zeros([np.size(endog,0),np.size(endog,1)])
endog_lagged_new = np.zeros([np.size(endog_lagged,0), np.size(endog_lagged,1)])
params_new_inc, params_new = [np.zeros([np.size(params,0), np.size(params,1)])
for i in range(2)]
sigma_u_new_inc, sigma_u_new = [np.zeros([np.size(sigma_u,0), np.size(sigma_u,1)])
for i in range(2)]
num_end = len(self.params[0])
names_new = []
#Rearrange elements and fill in new arrays
k = self.k_trend
for i, c in enumerate(order):
endog_new[:,i] = self.endog[:,c]
if k > 0:
params_new_inc[0,i] = params[0,i]
endog_lagged_new[:,0] = endog_lagged[:,0]
for j in range(k_ar):
params_new_inc[i+j*num_end+k,:] = self.params[c+j*num_end+k,:]
endog_lagged_new[:,i+j*num_end+k] = endog_lagged[:,c+j*num_end+k]
sigma_u_new_inc[i,:] = sigma_u[c,:]
names_new.append(names[c])
for i, c in enumerate(order):
params_new[:,i] = params_new_inc[:,c]
sigma_u_new[:,i] = sigma_u_new_inc[:,c]
return VARResults(endog=endog_new, endog_lagged=endog_lagged_new,
params=params_new, sigma_u=sigma_u_new,
lag_order=self.k_ar, model=self.model,
trend='c', names=names_new, dates=self.dates)
#-------------------------------------------------------------------------------
# VARProcess class: for known or unknown VAR process
[docs]class VAR(tsbase.TimeSeriesModel):
r"""
Fit VAR(p) process and do lag order selection
.. math:: y_t = A_1 y_{t-1} + \ldots + A_p y_{t-p} + u_t
Parameters
----------
endog : array-like
2-d endogenous response variable. The independent variable.
dates : array-like
must match number of rows of endog
References
----------
Lutkepohl (2005) New Introduction to Multiple Time Series Analysis
"""
def __init__(self, endog, dates=None, freq=None, missing='none'):
super(VAR, self).__init__(endog, None, dates, freq, missing=missing)
if self.endog.ndim == 1:
raise ValueError("Only gave one variable to VAR")
self.y = self.endog #keep alias for now
self.neqs = self.endog.shape[1]
def _get_predict_start(self, start, k_ar):
if start is None:
start = k_ar
return super(VAR, self)._get_predict_start(start)
[docs] def predict(self, params, start=None, end=None, lags=1, trend='c'):
"""
Returns in-sample predictions or forecasts
"""
start = self._get_predict_start(start, lags)
end, out_of_sample = self._get_predict_end(end)
if end < start:
raise ValueError("end is before start")
if end == start + out_of_sample:
return np.array([])
k_trend = util.get_trendorder(trend)
k = self.neqs
k_ar = lags
predictedvalues = np.zeros((end + 1 - start + out_of_sample, k))
if k_trend != 0:
intercept = params[:k_trend]
predictedvalues += intercept
y = self.y
X = util.get_var_endog(y, lags, trend=trend, has_constant='raise')
fittedvalues = np.dot(X, params)
fv_start = start - k_ar
pv_end = min(len(predictedvalues), len(fittedvalues) - fv_start)
fv_end = min(len(fittedvalues), end-k_ar+1)
predictedvalues[:pv_end] = fittedvalues[fv_start:fv_end]
if not out_of_sample:
return predictedvalues
# fit out of sample
y = y[-k_ar:]
coefs = params[k_trend:].reshape((k_ar, k, k)).swapaxes(1,2)
predictedvalues[pv_end:] = forecast(y, coefs, intercept, out_of_sample)
return predictedvalues
[docs] def fit(self, maxlags=None, method='ols', ic=None, trend='c',
verbose=False):
"""
Fit the VAR model
Parameters
----------
maxlags : int
Maximum number of lags to check for order selection, defaults to
12 * (nobs/100.)**(1./4), see select_order function
method : {'ols'}
Estimation method to use
ic : {'aic', 'fpe', 'hqic', 'bic', None}
Information criterion to use for VAR order selection.
aic : Akaike
fpe : Final prediction error
hqic : Hannan-Quinn
bic : Bayesian a.k.a. Schwarz
verbose : bool, default False
Print order selection output to the screen
trend, str {"c", "ct", "ctt", "nc"}
"ct" - constant and trend
"ctt" - constant, linear and quadratic trend
"nc" - co constant, no trend
Note that these are prepended to the columns of the dataset.
Notes
-----
Lutkepohl pp. 146-153
Returns
-------
est : VARResults
"""
lags = maxlags
if trend not in ['c', 'ct', 'ctt', 'nc']:
raise ValueError("trend '{}' not supported for VAR".format(trend))
if ic is not None:
selections = self.select_order(maxlags=maxlags, verbose=verbose)
if ic not in selections:
raise Exception("%s not recognized, must be among %s"
% (ic, sorted(selections)))
lags = selections[ic]
if verbose:
print('Using %d based on %s criterion' % (lags, ic))
else:
if lags is None:
lags = 1
k_trend = util.get_trendorder(trend)
self.exog_names = util.make_lag_names(self.endog_names, lags, k_trend)
self.nobs = len(self.endog) - lags
return self._estimate_var(lags, trend=trend)
def _estimate_var(self, lags, offset=0, trend='c'):
"""
lags : int
offset : int
Periods to drop from beginning-- for order selection so it's an
apples-to-apples comparison
trend : string or None
As per above
"""
# have to do this again because select_order doesn't call fit
self.k_trend = k_trend = util.get_trendorder(trend)
if offset < 0: # pragma: no cover
raise ValueError('offset must be >= 0')
y = self.y[offset:]
z = util.get_var_endog(y, lags, trend=trend, has_constant='raise')
y_sample = y[lags:]
# Lutkepohl p75, about 5x faster than stated formula
params = np.linalg.lstsq(z, y_sample)[0]
resid = y_sample - np.dot(z, params)
# Unbiased estimate of covariance matrix $\Sigma_u$ of the white noise
# process $u$
# equivalent definition
# .. math:: \frac{1}{T - Kp - 1} Y^\prime (I_T - Z (Z^\prime Z)^{-1}
# Z^\prime) Y
# Ref: Lutkepohl p.75
# df_resid right now is T - Kp - 1, which is a suggested correction
avobs = len(y_sample)
df_resid = avobs - (self.neqs * lags + k_trend)
sse = np.dot(resid.T, resid)
omega = sse / df_resid
varfit = VARResults(y, z, params, omega, lags, names=self.endog_names,
trend=trend, dates=self.data.dates, model=self)
return VARResultsWrapper(varfit)
[docs] def select_order(self, maxlags=None, verbose=True):
"""
Compute lag order selections based on each of the available information
criteria
Parameters
----------
maxlags : int
if None, defaults to 12 * (nobs/100.)**(1./4)
verbose : bool, default True
If True, print table of info criteria and selected orders
Returns
-------
selections : dict {info_crit -> selected_order}
"""
if maxlags is None:
maxlags = int(round(12*(len(self.endog)/100.)**(1/4.)))
ics = defaultdict(list)
for p in range(maxlags + 1):
# exclude some periods to same amount of data used for each lag
# order
result = self._estimate_var(p, offset=maxlags-p)
for k, v in iteritems(result.info_criteria):
ics[k].append(v)
selected_orders = dict((k, mat(v).argmin())
for k, v in iteritems(ics))
if verbose:
output.print_ic_table(ics, selected_orders)
return selected_orders
[docs]class VARProcess(object):
"""
Class represents a known VAR(p) process
Parameters
----------
coefs : ndarray (p x k x k)
intercept : ndarray (length k)
sigma_u : ndarray (k x k)
names : sequence (length k)
Returns
-------
**Attributes**:
"""
def __init__(self, coefs, intercept, sigma_u, names=None):
self.k_ar = len(coefs)
self.neqs = coefs.shape[1]
self.coefs = coefs
self.intercept = intercept
self.sigma_u = sigma_u
self.names = names
[docs] def get_eq_index(self, name):
"Return integer position of requested equation name"
return util.get_index(self.names, name)
def __str__(self):
output = ('VAR(%d) process for %d-dimensional response y_t'
% (self.k_ar, self.neqs))
output += '\nstable: %s' % self.is_stable()
output += '\nmean: %s' % self.mean()
return output
[docs] def is_stable(self, verbose=False):
"""Determine stability based on model coefficients
Parameters
----------
verbose : bool
Print eigenvalues of the VAR(1) companion
Notes
-----
Checks if det(I - Az) = 0 for any mod(z) <= 1, so all the eigenvalues of
the companion matrix must lie outside the unit circle
"""
return is_stable(self.coefs, verbose=verbose)
[docs] def plotsim(self, steps=1000):
"""
Plot a simulation from the VAR(p) process for the desired number of
steps
"""
Y = util.varsim(self.coefs, self.intercept, self.sigma_u, steps=steps)
plotting.plot_mts(Y)
[docs] def mean(self):
r"""Mean of stable process
Lutkepohl eq. 2.1.23
.. math:: \mu = (I - A_1 - \dots - A_p)^{-1} \alpha
"""
return solve(self._char_mat, self.intercept)
[docs] def ma_rep(self, maxn=10):
r"""Compute MA(:math:\infty) coefficient matrices
Parameters
----------
maxn : int
Number of coefficient matrices to compute
Returns
-------
coefs : ndarray (maxn x k x k)
"""
return ma_rep(self.coefs, maxn=maxn)
[docs] def orth_ma_rep(self, maxn=10, P=None):
r"""Compute Orthogonalized MA coefficient matrices using P matrix such
that :math:\Sigma_u = PP^\prime. P defaults to the Cholesky
decomposition of :math:\Sigma_u
Parameters
----------
maxn : int
Number of coefficient matrices to compute
P : ndarray (k x k), optional
Matrix such that Sigma_u = PP', defaults to Cholesky descomp
Returns
-------
coefs : ndarray (maxn x k x k)
"""
if P is None:
P = self._chol_sigma_u
ma_mats = self.ma_rep(maxn=maxn)
return mat([np.dot(coefs, P) for coefs in ma_mats])
[docs] def long_run_effects(self):
"""Compute long-run effect of unit impulse
.. math::
\Psi_\infty = \sum_{i=0}^\infty \Phi_i
"""
return L.inv(self._char_mat)
def _chol_sigma_u(self):
return chol(self.sigma_u)
def _char_mat(self):
return np.eye(self.neqs) - self.coefs.sum(0)
[docs] def acf(self, nlags=None):
"""Compute theoretical autocovariance function
Returns
-------
acf : ndarray (p x k x k)
"""
return var_acf(self.coefs, self.sigma_u, nlags=nlags)
[docs] def acorr(self, nlags=None):
"""Compute theoretical autocorrelation function
Returns
-------
acorr : ndarray (p x k x k)
"""
return util.acf_to_acorr(self.acf(nlags=nlags))
[docs] def plot_acorr(self, nlags=10, linewidth=8):
"Plot theoretical autocorrelation function"
plotting.plot_full_acorr(self.acorr(nlags=nlags), linewidth=linewidth)
[docs] def forecast(self, y, steps):
"""Produce linear minimum MSE forecasts for desired number of steps
Parameters
----------
y : ndarray (p x k)
steps : int
Returns
-------
forecasts : ndarray (steps x neqs)
Notes
-----
Lutkepohl pp 37-38
"""
return forecast(y, self.coefs, self.intercept, steps)
[docs] def mse(self, steps):
"""
Compute theoretical forecast error variance matrices
Parameters
----------
steps : int
Notes
-----
.. math:: \mathrm{MSE}(h) = \sum_{i=0}^{h-1} \Phi \Sigma_u \Phi^T
Returns
-------
forc_covs : ndarray (steps x neqs x neqs)
"""
ma_coefs = self.ma_rep(steps)
k = len(self.sigma_u)
forc_covs = np.zeros((steps, k, k))
prior = np.zeros((k, k))
for h in range(steps):
# Sigma(h) = Sigma(h-1) + Phi Sig_u Phi'
phi = ma_coefs[h]
var = chain_dot(phi, self.sigma_u, phi.T)
forc_covs[h] = prior = prior + var
return forc_covs
forecast_cov = mse
def _forecast_vars(self, steps):
covs = self.forecast_cov(steps)
# Take diagonal for each cov
inds = np.arange(self.neqs)
return covs[:, inds, inds]
[docs] def forecast_interval(self, y, steps, alpha=0.05):
"""Construct forecast interval estimates assuming the y are Gaussian
Parameters
----------
Notes
-----
Lutkepohl pp. 39-40
Returns
-------
(lower, mid, upper) : (ndarray, ndarray, ndarray)
"""
assert(0 < alpha < 1)
q = util.norm_signif_level(alpha)
point_forecast = self.forecast(y, steps)
sigma = np.sqrt(self._forecast_vars(steps))
forc_lower = point_forecast - q * sigma
forc_upper = point_forecast + q * sigma
return point_forecast, forc_lower, forc_upper
#-------------------------------------------------------------------------------
# VARResults class
[docs]class VARResults(VARProcess):
"""Estimate VAR(p) process with fixed number of lags
Parameters
----------
endog : array
endog_lagged : array
params : array
sigma_u : array
lag_order : int
model : VAR model instance
trend : str {'nc', 'c', 'ct'}
names : array-like
List of names of the endogenous variables in order of appearance in endog.
dates
Returns
-------
**Attributes**
aic
bic
bse
coefs : ndarray (p x K x K)
Estimated A_i matrices, A_i = coefs[i-1]
cov_params
dates
detomega
df_model : int
df_resid : int
endog
endog_lagged
fittedvalues
fpe
intercept
info_criteria
k_ar : int
k_trend : int
llf
model
names
neqs : int
Number of variables (equations)
nobs : int
n_totobs : int
params
k_ar : int
Order of VAR process
params : ndarray (Kp + 1) x K
A_i matrices and intercept in stacked form [int A_1 ... A_p]
pvalues
names : list
variables names
resid
roots : array
The roots of the VAR process are the solution to
(I - coefs[0]*z - coefs[1]*z**2 ... - coefs[p-1]*z**k_ar) = 0.
Note that the inverse roots are returned, and stability requires that
the roots lie outside the unit circle.
sigma_u : ndarray (K x K)
Estimate of white noise process variance Var[u_t]
sigma_u_mle
stderr
trenorder
tvalues
y :
ys_lagged
"""
_model_type = 'VAR'
def __init__(self, endog, endog_lagged, params, sigma_u, lag_order,
model=None, trend='c', names=None, dates=None):
self.model = model
self.y = self.endog = endog #keep alias for now
self.ys_lagged = self.endog_lagged = endog_lagged #keep alias for now
self.dates = dates
self.n_totobs, neqs = self.y.shape
self.nobs = self.n_totobs - lag_order
k_trend = util.get_trendorder(trend)
if k_trend > 0: # make this the polynomial trend order
trendorder = k_trend - 1
else:
trendorder = None
self.k_trend = k_trend
self.trendorder = trendorder
self.exog_names = util.make_lag_names(names, lag_order, k_trend)
self.params = params
# Initialize VARProcess parent class
# construct coefficient matrices
# Each matrix needs to be transposed
reshaped = self.params[self.k_trend:]
reshaped = reshaped.reshape((lag_order, neqs, neqs))
# Need to transpose each coefficient matrix
intercept = self.params[0]
coefs = reshaped.swapaxes(1, 2).copy()
super(VARResults, self).__init__(coefs, intercept, sigma_u, names=names)
[docs] def plot(self):
"""Plot input time series
"""
plotting.plot_mts(self.y, names=self.names, index=self.dates)
@property
def df_model(self):
"""Number of estimated parameters, including the intercept / trends
"""
return self.neqs * self.k_ar + self.k_trend
@property
def df_resid(self):
"""Number of observations minus number of estimated parameters"""
return self.nobs - self.df_model
[docs] def fittedvalues(self):
"""The predicted insample values of the response variables of the model.
"""
return np.dot(self.ys_lagged, self.params)
[docs] def resid(self):
"""Residuals of response variable resulting from estimated coefficients
"""
return self.y[self.k_ar:] - self.fittedvalues
[docs] def sample_acov(self, nlags=1):
return _compute_acov(self.y[self.k_ar:], nlags=nlags)
[docs] def sample_acorr(self, nlags=1):
acovs = self.sample_acov(nlags=nlags)
return _acovs_to_acorrs(acovs)
[docs] def plot_sample_acorr(self, nlags=10, linewidth=8):
"Plot theoretical autocorrelation function"
plotting.plot_full_acorr(self.sample_acorr(nlags=nlags),
linewidth=linewidth)
[docs] def resid_acov(self, nlags=1):
"""
Compute centered sample autocovariance (including lag 0)
Parameters
----------
nlags : int
Returns
-------
"""
return _compute_acov(self.resid, nlags=nlags)
[docs] def resid_acorr(self, nlags=1):
"""
Compute sample autocorrelation (including lag 0)
Parameters
----------
nlags : int
Returns
-------
"""
acovs = self.resid_acov(nlags=nlags)
return _acovs_to_acorrs(acovs)
[docs] def resid_corr(self):
"Centered residual correlation matrix"
return self.resid_acorr(0)[0]
[docs] def sigma_u_mle(self):
"""(Biased) maximum likelihood estimate of noise process covariance
"""
return self.sigma_u * self.df_resid / self.nobs
[docs] def cov_params(self):
"""Estimated variance-covariance of model coefficients
Notes
-----
Covariance of vec(B), where B is the matrix
[intercept, A_1, ..., A_p] (K x (Kp + 1))
Adjusted to be an unbiased estimator
Ref: Lutkepohl p.74-75
"""
z = self.ys_lagged
return np.kron(L.inv(np.dot(z.T, z)), self.sigma_u)
[docs] def cov_ybar(self):
r"""Asymptotically consistent estimate of covariance of the sample mean
.. math::
\sqrt(T) (\bar{y} - \mu) \rightarrow {\cal N}(0, \Sigma_{\bar{y}})\\
\Sigma_{\bar{y}} = B \Sigma_u B^\prime, \text{where } B = (I_K - A_1
- \cdots - A_p)^{-1}
Notes
-----
Lutkepohl Proposition 3.3
"""
Ainv = L.inv(np.eye(self.neqs) - self.coefs.sum(0))
return chain_dot(Ainv, self.sigma_u, Ainv.T)
#------------------------------------------------------------
# Estimation-related things
def _zz(self):
# Z'Z
return np.dot(self.ys_lagged.T, self.ys_lagged)
@property
def _cov_alpha(self):
"""
Estimated covariance matrix of model coefficients ex intercept
"""
# drop intercept and trend
return self.cov_params[self.k_trend*self.neqs:, self.k_trend*self.neqs:]
def _cov_sigma(self):
"""
Estimated covariance matrix of vech(sigma_u)
"""
D_K = tsa.duplication_matrix(self.neqs)
D_Kinv = npl.pinv(D_K)
sigxsig = np.kron(self.sigma_u, self.sigma_u)
return 2 * chain_dot(D_Kinv, sigxsig, D_Kinv.T)
[docs] def llf(self):
"Compute VAR(p) loglikelihood"
return var_loglike(self.resid, self.sigma_u_mle, self.nobs)
[docs] def stderr(self):
"""Standard errors of coefficients, reshaped to match in size
"""
stderr = np.sqrt(np.diag(self.cov_params))
return stderr.reshape((self.df_model, self.neqs), order='C')
bse = stderr # statsmodels interface?
[docs] def tvalues(self):
"""Compute t-statistics. Use Student-t(T - Kp - 1) = t(df_resid) to test
significance.
"""
return self.params / self.stderr
[docs] def pvalues(self):
"""Two-sided p-values for model coefficients from Student t-distribution
"""
return stats.t.sf(np.abs(self.tvalues), self.df_resid)*2
[docs] def plot_forecast(self, steps, alpha=0.05, plot_stderr=True):
"""
Plot forecast
"""
mid, lower, upper = self.forecast_interval(self.y[-self.k_ar:], steps,
alpha=alpha)
plotting.plot_var_forc(self.y, mid, lower, upper, names=self.names,
plot_stderr=plot_stderr)
# Forecast error covariance functions
[docs] def forecast_cov(self, steps=1):
r"""Compute forecast covariance matrices for desired number of steps
Parameters
----------
steps : int
Notes
-----
.. math:: \Sigma_{\hat y}(h) = \Sigma_y(h) + \Omega(h) / T
Ref: Lutkepohl pp. 96-97
Returns
-------
covs : ndarray (steps x k x k)
"""
mse = self.mse(steps)
omegas = self._omega_forc_cov(steps)
return mse + omegas / self.nobs
#Monte Carlo irf standard errors
[docs] def irf_errband_mc(self, orth=False, repl=1000, T=10,
signif=0.05, seed=None, burn=100, cum=False):
"""
Compute Monte Carlo integrated error bands assuming normally
distributed for impulse response functions
Parameters
----------
orth: bool, default False
Compute orthoganalized impulse response error bands
repl: int
number of Monte Carlo replications to perform
T: int, default 10
number of impulse response periods
signif: float (0 < signif <1)
Significance level for error bars, defaults to 95% CI
seed: int
np.random.seed for replications
burn: int
number of initial observations to discard for simulation
cum: bool, default False
produce cumulative irf error bands
Notes
-----
Lutkepohl (2005) Appendix D
Returns
-------
Tuple of lower and upper arrays of ma_rep monte carlo standard errors
"""
neqs = self.neqs
mean = self.mean()
k_ar = self.k_ar
coefs = self.coefs
sigma_u = self.sigma_u
intercept = self.intercept
df_model = self.df_model
nobs = self.nobs
ma_coll = np.zeros((repl, T+1, neqs, neqs))
if (orth == True and cum == True):
fill_coll = lambda sim : VAR(sim).fit(maxlags=k_ar).\
orth_ma_rep(maxn=T).cumsum(axis=0)
elif (orth == True and cum == False):
fill_coll = lambda sim : VAR(sim).fit(maxlags=k_ar).\
orth_ma_rep(maxn=T)
elif (orth == False and cum == True):
fill_coll = lambda sim : VAR(sim).fit(maxlags=k_ar).\
ma_rep(maxn=T).cumsum(axis=0)
elif (orth == False and cum == False):
fill_coll = lambda sim : VAR(sim).fit(maxlags=k_ar).\
ma_rep(maxn=T)
for i in range(repl):
#discard first hundred to eliminate correct for starting bias
sim = util.varsim(coefs, intercept, sigma_u,
seed=seed, steps=nobs+burn)
sim = sim[burn:]
ma_coll[i,:,:,:] = fill_coll(sim)
ma_sort = np.sort(ma_coll, axis=0) #sort to get quantiles
index = round(signif/2*repl)-1,round((1-signif/2)*repl)-1
lower = ma_sort[index[0],:, :, :]
upper = ma_sort[index[1],:, :, :]
return lower, upper
[docs] def irf_resim(self, orth=False, repl=1000, T=10,
seed=None, burn=100, cum=False):
"""
Simulates impulse response function, returning an array of simulations.
Used for Sims-Zha error band calculation.
Parameters
----------
orth: bool, default False
Compute orthoganalized impulse response error bands
repl: int
number of Monte Carlo replications to perform
T: int, default 10
number of impulse response periods
signif: float (0 < signif <1)
Significance level for error bars, defaults to 95% CI
seed: int
np.random.seed for replications
burn: int
number of initial observations to discard for simulation
cum: bool, default False
produce cumulative irf error bands
Notes
-----
Sims, Christoper A., and Tao Zha. 1999. "Error Bands for Impulse Response." Econometrica 67: 1113-1155.
Returns
-------
Array of simulated impulse response functions
"""
neqs = self.neqs
mean = self.mean()
k_ar = self.k_ar
coefs = self.coefs
sigma_u = self.sigma_u
intercept = self.intercept
df_model = self.df_model
nobs = self.nobs
ma_coll = np.zeros((repl, T+1, neqs, neqs))
if (orth == True and cum == True):
fill_coll = lambda sim : VAR(sim).fit(maxlags=k_ar).\
orth_ma_rep(maxn=T).cumsum(axis=0)
elif (orth == True and cum == False):
fill_coll = lambda sim : VAR(sim).fit(maxlags=k_ar).\
orth_ma_rep(maxn=T)
elif (orth == False and cum == True):
fill_coll = lambda sim : VAR(sim).fit(maxlags=k_ar).\
ma_rep(maxn=T).cumsum(axis=0)
elif (orth == False and cum == False):
fill_coll = lambda sim : VAR(sim).fit(maxlags=k_ar).\
ma_rep(maxn=T)
for i in range(repl):
#discard first hundred to eliminate correct for starting bias
sim = util.varsim(coefs, intercept, sigma_u,
seed=seed, steps=nobs+burn)
sim = sim[burn:]
ma_coll[i,:,:,:] = fill_coll(sim)
return ma_coll
def _omega_forc_cov(self, steps):
# Approximate MSE matrix \Omega(h) as defined in Lut p97
G = self._zz
Ginv = L.inv(G)
# memoize powers of B for speedup
# TODO: see if can memoize better
B = self._bmat_forc_cov()
_B = {}
def bpow(i):
if i not in _B:
_B[i] = np.linalg.matrix_power(B, i)
return _B[i]
phis = self.ma_rep(steps)
sig_u = self.sigma_u
omegas = np.zeros((steps, self.neqs, self.neqs))
for h in range(1, steps + 1):
if h == 1:
omegas[h-1] = self.df_model * self.sigma_u
continue
om = omegas[h-1]
for i in range(h):
for j in range(h):
Bi = bpow(h - 1 - i)
Bj = bpow(h - 1 - j)
mult = np.trace(chain_dot(Bi.T, Ginv, Bj, G))
om += mult * chain_dot(phis[i], sig_u, phis[j].T)
omegas[h-1] = om
return omegas
def _bmat_forc_cov(self):
# B as defined on p. 96 of Lut
upper = np.zeros((1, self.df_model))
upper[0,0] = 1
lower_dim = self.neqs * (self.k_ar - 1)
I = np.eye(lower_dim)
lower = np.column_stack((np.zeros((lower_dim, 1)), I,
np.zeros((lower_dim, self.neqs))))
return np.vstack((upper, self.params.T, lower))
[docs] def summary(self):
"""Compute console output summary of estimates
Returns
-------
summary : VARSummary
"""
return VARSummary(self)
[docs] def irf(self, periods=10, var_decomp=None, var_order=None):
"""Analyze impulse responses to shocks in system
Parameters
----------
periods : int
var_decomp : ndarray (k x k), lower triangular
Must satisfy Omega = P P', where P is the passed matrix. Defaults to
Cholesky decomposition of Omega
var_order : sequence
Alternate variable order for Cholesky decomposition
Returns
-------
irf : IRAnalysis
"""
if var_order is not None:
raise NotImplementedError('alternate variable order not implemented'
' (yet)')
return IRAnalysis(self, P=var_decomp, periods=periods)
[docs] def fevd(self, periods=10, var_decomp=None):
"""
Compute forecast error variance decomposition ("fevd")
Returns
-------
fevd : FEVD instance
"""
return FEVD(self, P=var_decomp, periods=periods)
[docs] def reorder(self, order):
"""Reorder variables for structural specification
"""
if len(order) != len(self.params[0,:]):
raise ValueError("Reorder specification length should match number of endogenous variables")
#This convert order to list of integers if given as strings
if isinstance(order[0], string_types):
order_new = []
for i, nam in enumerate(order):
order_new.append(self.names.index(order[i]))
order = order_new
return _reordered(self, order)
#-------------------------------------------------------------------------------
# VAR Diagnostics: Granger-causality, whiteness of residuals, normality, etc.
[docs] def test_causality(self, equation, variables, kind='f', signif=0.05,
verbose=True):
"""Compute test statistic for null hypothesis of Granger-noncausality,
general function to test joint Granger-causality of multiple variables
Parameters
----------
equation : string or int
Equation to test for causality
variables : sequence (of strings or ints)
List, tuple, etc. of variables to test for Granger-causality
kind : {'f', 'wald'}
Perform F-test or Wald (chi-sq) test
signif : float, default 5%
Significance level for computing critical values for test,
defaulting to standard 0.95 level
Notes
-----
Null hypothesis is that there is no Granger-causality for the indicated
variables. The degrees of freedom in the F-test are based on the
number of variables in the VAR system, that is, degrees of freedom
are equal to the number of equations in the VAR times degree of freedom
of a single equation.
Returns
-------
results : dict
"""
if isinstance(variables, (string_types, int, np.integer)):
variables = [variables]
k, p = self.neqs, self.k_ar
# number of restrictions
N = len(variables) * self.k_ar
# Make restriction matrix
C = np.zeros((N, k ** 2 * p + k), dtype=float)
eq_index = self.get_eq_index(equation)
vinds = mat([self.get_eq_index(v) for v in variables])
# remember, vec is column order!
offsets = np.concatenate([k + k ** 2 * j + k * vinds + eq_index
for j in range(p)])
C[np.arange(N), offsets] = 1
# Lutkepohl 3.6.5
Cb = np.dot(C, vec(self.params.T))
middle = L.inv(chain_dot(C, self.cov_params, C.T))
# wald statistic
lam_wald = statistic = chain_dot(Cb, middle, Cb)
if kind.lower() == 'wald':
df = N
dist = stats.chi2(df)
elif kind.lower() == 'f':
statistic = lam_wald / N
df = (N, k * self.df_resid)
dist = stats.f(*df)
else:
raise Exception('kind %s not recognized' % kind)
pvalue = dist.sf(statistic)
crit_value = dist.ppf(1 - signif)
conclusion = 'fail to reject' if statistic < crit_value else 'reject'
results = {
'statistic' : statistic,
'crit_value' : crit_value,
'pvalue' : pvalue,
'df' : df,
'conclusion' : conclusion,
'signif' : signif
}
if verbose:
summ = output.causality_summary(results, variables, equation, kind)
print(summ)
return results
[docs] def test_whiteness(self, nlags=10, plot=True, linewidth=8):
"""
Test white noise assumption. Sample (Y) autocorrelations are compared
with the standard :math:2 / \sqrt(T) bounds.
Parameters
----------
plot : boolean, default True
Plot autocorrelations with 2 / sqrt(T) bounds
"""
acorrs = self.sample_acorr(nlags)
bound = 2 / np.sqrt(self.nobs)
# TODO: this probably needs some UI work
if (np.abs(acorrs) > bound).any():
print('FAIL: Some autocorrelations exceed %.4f bound. '
'See plot' % bound)
else:
print('PASS: No autocorrelations exceed %.4f bound' % bound)
if plot:
fig = plotting.plot_full_acorr(acorrs[1:],
xlabel=np.arange(1, nlags+1),
err_bound=bound,
linewidth=linewidth)
fig.suptitle(r"ACF plots with $2 / \sqrt{T}$ bounds "
"for testing whiteness assumption")
[docs] def test_normality(self, signif=0.05, verbose=True):
"""
Test assumption of normal-distributed errors using Jarque-Bera-style
omnibus Chi^2 test
Parameters
----------
signif : float
Test significance threshold
Notes
-----
H0 (null) : data are generated by a Gaussian-distributed process
"""
Pinv = npl.inv(self._chol_sigma_u)
w = np.array([np.dot(Pinv, u) for u in self.resid])
b1 = (w ** 3).sum(0) / self.nobs
lam_skew = self.nobs * np.dot(b1, b1) / 6
b2 = (w ** 4).sum(0) / self.nobs - 3
lam_kurt = self.nobs * np.dot(b2, b2) / 24
lam_omni = lam_skew + lam_kurt
omni_dist = stats.chi2(self.neqs * 2)
omni_pvalue = omni_dist.sf(lam_omni)
crit_omni = omni_dist.ppf(1 - signif)
conclusion = 'fail to reject' if lam_omni < crit_omni else 'reject'
results = {
'statistic' : lam_omni,
'crit_value' : crit_omni,
'pvalue' : omni_pvalue,
'df' : self.neqs * 2,
'conclusion' : conclusion,
'signif' : signif
}
if verbose:
summ = output.normality_summary(results)
print(summ)
return results
[docs] def detomega(self):
r"""
Return determinant of white noise covariance with degrees of freedom
correction:
.. math::
\hat \Omega = \frac{T}{T - Kp - 1} \hat \Omega_{\mathrm{MLE}}
"""
return L.det(self.sigma_u)
[docs] def info_criteria(self):
"information criteria for lagorder selection"
nobs = self.nobs
neqs = self.neqs
lag_order = self.k_ar
free_params = lag_order * neqs ** 2 + neqs * self.k_trend
ld = logdet_symm(self.sigma_u_mle)
# See Lutkepohl pp. 146-150
aic = ld + (2. / nobs) * free_params
bic = ld + (np.log(nobs) / nobs) * free_params
hqic = ld + (2. * np.log(np.log(nobs)) / nobs) * free_params
fpe = ((nobs + self.df_model) / self.df_resid) ** neqs * np.exp(ld)
return {
'aic' : aic,
'bic' : bic,
'hqic' : hqic,
'fpe' : fpe
}
@property
def aic(self):
"""Akaike information criterion"""
return self.info_criteria['aic']
@property
def fpe(self):
"""Final Prediction Error (FPE)
Lutkepohl p. 147, see info_criteria
"""
return self.info_criteria['fpe']
@property
def hqic(self):
"""Hannan-Quinn criterion"""
return self.info_criteria['hqic']
@property
def bic(self):
"""Bayesian a.k.a. Schwarz info criterion"""
return self.info_criteria['bic']
[docs] def roots(self):
neqs = self.neqs
k_ar = self.k_ar
p = neqs * k_ar
arr = np.zeros((p,p))
arr[:neqs,:] = np.column_stack(self.coefs)
arr[neqs:,:-neqs] = np.eye(p-neqs)
roots = np.linalg.eig(arr)[0]**-1
idx = np.argsort(np.abs(roots))[::-1] # sort by reverse modulus
return roots[idx]
class VARResultsWrapper(wrap.ResultsWrapper):
_attrs = {'bse' : 'columns_eq', 'cov_params' : 'cov',
'params' : 'columns_eq', 'pvalues' : 'columns_eq',
'tvalues' : 'columns_eq', 'sigma_u' : 'cov_eq',
'sigma_u_mle' : 'cov_eq',
'stderr' : 'columns_eq'}
_wrap_attrs = wrap.union_dicts(tsbase.TimeSeriesResultsWrapper._wrap_attrs,
_attrs)
_methods = {}
_wrap_methods = wrap.union_dicts(tsbase.TimeSeriesResultsWrapper._wrap_methods,
_methods)
_wrap_methods.pop('cov_params') # not yet a method in VARResults
wrap.populate_wrapper(VARResultsWrapper, VARResults)
[docs]class FEVD(object):
"""
Compute and plot Forecast error variance decomposition and asymptotic
standard errors
"""
def __init__(self, model, P=None, periods=None):
self.periods = periods
self.model = model
self.neqs = model.neqs
self.names = model.model.endog_names
self.irfobj = model.irf(var_decomp=P, periods=periods)
self.orth_irfs = self.irfobj.orth_irfs
# cumulative impulse responses
irfs = (self.orth_irfs[:periods] ** 2).cumsum(axis=0)
rng = lrange(self.neqs)
mse = self.model.mse(periods)[:, rng, rng]
# lag x equation x component
fevd = np.empty_like(irfs)
for i in range(periods):
fevd[i] = (irfs[i].T / mse[i]).T
# switch to equation x lag x component
self.decomp = fevd.swapaxes(0, 1)
[docs] def summary(self):
buf = StringIO()
rng = lrange(self.periods)
for i in range(self.neqs):
ppm = output.pprint_matrix(self.decomp[i], rng, self.names)
buf.write('FEVD for %s\n' % self.names[i])
buf.write(ppm + '\n')
print(buf.getvalue())
[docs] def cov(self):
"""Compute asymptotic standard errors
Returns
-------
"""
raise NotImplementedError
[docs] def plot(self, periods=None, figsize=(10,10), **plot_kwds):
"""Plot graphical display of FEVD
Parameters
----------
periods : int, default None
Defaults to number originally specified. Can be at most that number
"""
import matplotlib.pyplot as plt
k = self.neqs
periods = periods or self.periods
fig, axes = plt.subplots(nrows=k, figsize=figsize)
fig.suptitle('Forecast error variance decomposition (FEVD)')
colors = [str(c) for c in np.arange(k, dtype=float) / k]
ticks = np.arange(periods)
limits = self.decomp.cumsum(2)
for i in range(k):
ax = axes[i]
this_limits = limits[i].T
handles = []
for j in range(k):
lower = this_limits[j - 1] if j > 0 else 0
upper = this_limits[j]
handle = ax.bar(ticks, upper - lower, bottom=lower,
color=colors[j], label=self.names[j],
**plot_kwds)
handles.append(handle)
ax.set_title(self.names[i])
# just use the last axis to get handles for plotting
handles, labels = ax.get_legend_handles_labels()
fig.legend(handles, labels, loc='upper right')
#-------------------------------------------------------------------------------
def _compute_acov(x, nlags=1):
x = x - x.mean(0)
result = []
for lag in range(nlags + 1):
if lag > 0:
r = np.dot(x[lag:].T, x[:-lag])
else:
r = np.dot(x.T, x)
result.append(r)
return np.array(result) / len(x)
def _acovs_to_acorrs(acovs):
sd = np.sqrt(np.diag(acovs[0]))
return acovs / np.outer(sd, sd)
if __name__ == '__main__':
import statsmodels.api as sm
from statsmodels.tsa.vector_ar.util import parse_lutkepohl_data
import statsmodels.tools.data as data_util
np.set_printoptions(linewidth=140, precision=5)
sdata, dates = parse_lutkepohl_data('data/%s.dat' % 'e1')
names = sdata.dtype.names
data = data_util.struct_to_ndarray(sdata)
# est = VAR(adj_data, p=2, dates=dates[1:], names=names)
# model = VAR(adj_data[:-16], dates=dates[1:-16], names=names)
est = model.fit(maxlags=2)
irf = est.irf()
y = est.y[-2:]
"""
# irf.plot_irf()
# i = 2; j = 1
# cv = irf.cum_effect_cov(orth=True)
# print np.sqrt(cv[:, j * 3 + i, j * 3 + i]) / 1e-2
# data = np.genfromtxt('Canada.csv', delimiter=',', names=True)
# data = data.view((float, 4))
"""
'''
mdata2 = mdata[['realgdp','realcons','realinv']]
names = mdata2.dtype.names
data = mdata2.view((float,3))
data = np.diff(np.log(data), axis=0)
import pandas as pn
df = pn.DataFrame.fromRecords(mdata)
df = np.log(df.reindex(columns=names))
df = (df - df.shift(1)).dropna()
model = VAR(df)
est = model.fit(maxlags=2)
irf = est.irf()
''' | 11,667 | 39,897 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-51 | latest | en | 0.445397 |
http://www.archive.org/stream/ScientificPapersVi/TXT/00000504.txt | 1,524,695,731,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947968.96/warc/CC-MAIN-20180425213156-20180425233156-00230.warc.gz | 366,034,917 | 8,172 | # Full text of "Scientific Papers - Vi"
## See other formats
```1917] THE THEORY OF ANOMALOUS DISPERSION 489
On the assumption that
i},£ = (C,D')e'iğ*erM+into*l . ....................... (3)
we get Maxwell's results*
1 _ 1 _ p + a- crn* p- n2 ,..
~ '
E H!!'( n2 Rn
,. ( *
Here v is the velocity of propagation of phase, and I is the distance the waves must run in order that the amplitude of vibration may be reduced in the ratio & : 1.
When we suppose that It = 0, and consequently that I = oo , (4) simplifies. If VQ be the velocity in aether (a- = 0), and v be the refractive index,
For comparison with experiment, results are often conveniently expressed in terms of the wave-lengths in free aether corresponding with the frequencies in question. Thus, if X correspond with n and A with p, (6) may be written
the dispersion formula commonly named after Sellmeier. It will be observed that p, A refer to the vibrations which the atoms might freely execute when the aether is maintained at rest (77 = 0).
If we suppose that n is infinitely small, or X infinitely great,
^2 = 1+^ ................................. (8)
thus remaining finite.
Helmholtz in his investigation also introduces a dissipative force, as is necessary to avoid infinities when n =p, but one differing from Maxwell's, in that it is dependent upon the absolute velocity of the atoms instead of upon the relative velocity of aether and matter. A more important difference is the introduction of an additional force of restitution (a2#), proportional to the absolute displacement of the atoms. His equations are
(9)t
* Thus in Maxwell's original statement. lu my quotation of 1899 the sign of the second term in (4) was erroneously given as plus.
t What was doubtless meant to be d-%/dyz appears as.^/da;2, bringing in x in two senses.. CLIV. p. 582 (1874) ; Wissenschaftliche Abhandlungen, Band rt. p. 213.ion of F (experimentally or otherwise) does not require the variation of both a and v. There is advantage in keeping a constant; for if a be varied, geometrical similarity demands that any roughnesses shall be in proportion.
``` | 552 | 2,253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-17 | latest | en | 0.895668 |
https://www.coursehero.com/file/11304918/Sample-Exam-3-C/ | 1,542,511,269,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743963.32/warc/CC-MAIN-20181118031826-20181118053826-00260.warc.gz | 832,448,299 | 65,093 | Sample Exam 3 (C)
# Sample Exam 3 (C) - EML 3100 Thermodynamics Fall 2012 Exam...
This preview shows pages 1–4. Sign up to view the full content.
EML 3100 Thermodynamics Fall 2012 Exam 3 Friday, November 30 Instructor: Philip Jackson Materials allowed: - Basic calculator and/or graphing calculator - Supplied formula and data sheets - CLOSED BOOK - CLOSED NOTES
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
EML 3100 Exam 3 Fall 2012 Friday, November 30 Name UF-ID
Problem 1 (20 points): Consider the ideal, steady-state gas-refrigeration cycle shown in the schematic below. The working fluid is helium and the following operating conditions are known. Use properties for helium at STP and assume that the specific heats are constants. T 1 = -125 °C, T 2 = 275 °C, T 3 = 50 °C, T 4 = -185 °C, P 2 / P 1 = 26, T L = 175 K, T H = 300 K (A) Complete the vertical axis of the phase diagram provided by labeling the mean effective temperatures, 𝑇 𝐿 and 𝑇 𝐻 , the min. and max. process temperatures, 𝑇 𝐿 and 𝑇 𝐻 , and the reservoir temperatures, 𝑇 𝐿 and 𝑇 𝐻 . (B) Calculate the mean effective process temperature, 𝑇 𝐿 .
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern | 580 | 2,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-47 | latest | en | 0.8763 |
http://pokemondb.net/pokebase/198324/chance-hitting-opponent-provided-these-circumstances-occur | 1,429,995,492,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246651471.95/warc/CC-MAIN-20150417045731-00118-ip-10-235-10-82.ec2.internal.warc.gz | 212,970,199 | 7,962 | PokeBase - Pokemon Q&A
# What is the chance of hitting the opponent provided these circumstances occur?
So I was wondering, if I had a Charizard with inferno and the opponent had a Sandshrew with the ability sand veil;
• Charizard was paralyzed, confused and infatuated
• Charizard had 6 accuracy decreases and sandshrew had 6 evasion increases
• Charizard has the ability hustle (don't ask)
• Sandstorm is up and the sandshrew has sand veil
• Sandshrew holds bright powder
• Sandslash attacks first and uses rock slide
What is the chance of Charizard hitting inferno?!
EDIT:
Was there anything I missed out on that could further decrease accuracy?
lol inferno by itself is already stupidly inaccurate :P
Well for starters, Hustle won't affect it at all because it's a special move.
Calcing it now.
You would have about a 0.48239296875% chance to hit Inferno in this case.
I didn't count the Hustle decrease because it only affects physical moves.
Calc:
50(Inferno's acc.) * 0.75(Paralysis) = 37,5%
37,5 * 0.5(confusion) = 18,75
18,75 * 0.5(infatuation) = 9,375
9,375* 0.75(sand veil) = 7,03125
7,03125 * 0.9 (bright powder)= 6,328125
6,328125 * 0.70(rock slide) = 4,4296875
4,4296875 * 0.33(evasion) = 1,461796875
1,461796875 * 0.33(accuracy)= 0,48239296875
Hope I helped.
selected May 19, 2014
gg, i wonder if it had used dynamic punch instead of inferno (hustle getting taken into account), the numbers would have been simply hilarious.
In that case the chance to hit would be around 0,385914375 :] (just multiply 0,48239296875 by 0.8)
It would just die for rock slide
Assuming it doesn't die, of course.
What if the move was guillotine and sandshrew was 29 levels higher than charizard?
lol u can't hit it at all (0% chance) if it's even a level higher
Yes it can, it only fails if the opponent is 30 level or more stronger than you. Look it up.
I've done a personal playthrogh, and whenever a wild sandslash decides to use guillotine on me (even when i'm about three levels higher) the game give a message - "It doesn't effect *pokemon name*..."
... | 596 | 2,067 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2015-18 | latest | en | 0.939264 |
https://sdrta.net/probability-of-rolling-at-least-one-6/ | 1,638,838,495,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363327.64/warc/CC-MAIN-20211206224536-20211207014536-00602.warc.gz | 566,366,279 | 5,880 | The probability that at the very least one challenge showing 6 is equal to 1 minus probability of constantly getting 5 or below. In various other words, the only means to failure the problem of at least one 6 is to role 5 or listed below for ALL her rolls.
You are watching: Probability of rolling at least one 6
P denotes probability.
P(6) = 1-P(5 or below)
Probability of rolling 5 or below is given by probability of each number multiply by 5 = 1/6 * 5 = 5/6
P(5 or below) = 5/6
We main point 5/6 by itself 3 times together it"s a share probability that 3 rolls.
P(3 roll of 5 or below) = (5/6) * (5/6) * (5/6) = (5/6)^3
Finally calculating probability that at the very least one 6 in 3 rolls:
P(at the very least one 6) = 1-P(3 rolls of 5 or below) = 1 - 0.58 = 0.42
rise • 1 Downvote
include comment
More
Report
## Still searching for help? get the ideal answer, fast.
obtain a free answer to a rapid problem. Most questions answered in ~ 4 hours.
OR
find an virtual Tutor now
pick an expert and meet online. No packages or subscriptions, pay only for the moment you need.
¢ € £ ¥ ‰ µ · • § ¶ ß ‹ › « » > ≤ ≥ – — ¯ ‾ ¤ ¦ ¨ ¡ ¿ ˆ ˜ ° − ± ÷ ⁄ × ƒ ∫ ∑ ∞ √ ∼ ≅ ≈ ≠ ≡ ∈ ∉ ∋ ∏ ∧ ∨ ¬ ∩ ∪ ∂ ∀ ∃ ∅ ∇ ∗ ∝ ∠ ´ ¸ ª º † ‡ À Á Â Ã Ä Å Æ Ç È É Ê Ë Ì Í Î Ï Ð Ñ Ò Ó Ô Õ Ö Ø Œ Š Ù Ú Û Ü Ý Ÿ Þ à á â ã ä å æ ç è é ê ë ì í î ï ð ñ ò ó ô õ ö ø œ š ù ú û ü ý þ ÿ Α Β Γ Δ Ε Ζ Η Θ Ι Κ Λ Μ Ν Ξ Ο Π Ρ Σ Τ Υ Φ Χ Ψ Ω α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω ℵ ϖ ℜ ϒ ℘ ℑ ← ↑ → ↓ ↔ ↵ ⇐ ⇑ ⇒ ⇓ ⇔ ∴ ⊂ ⊃ ⊄ ⊆ ⊇ ⊕ ⊗ ⊥ ⋅ ⌈ ⌉ ⌊ ⌋ 〈 〉 ◊
### RELATED TOPICS
mathematics Algebra 1 Algebra 2 Geometry Statistics Algebra Word difficulty Math assist Statistics inquiry Permutations ... Math Combinations set Conditional Probability Math help For university Finite math Probability & Statistics Venn chart Tree chart Probability difficulties
### RELATED QUESTIONS
Probability and expected value
Marbles in a cup question
he size of time, in minutes, for and airplane to achieve clearance because that take turn off at a details airport is a arbitrarily variable Y=3X-2, where x has the thickness funct
what is the event listed below
### RECOMMENDED TUTORS
Kyle M.
4.9 (368)
Steven A.
5.0 (1,102)
Michael K.
4.9 (76)
See an ext tutors
### uncover an virtual tutor
A attach to the app was sent out to your phone.
See more: Best Way To Freeze Fresh Shrimp ? Freezing Shrimp
Please carry out a precious phone number.
get to recognize us
find out with united state
work with united state | 843 | 2,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-49 | latest | en | 0.79215 |
http://www.gamedev.net/topic/652409-what-float-unit-should-i-choose-for-1-m/ | 1,481,257,428,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542680.76/warc/CC-MAIN-20161202170902-00071-ip-10-31-129-80.ec2.internal.warc.gz | 475,188,398 | 25,388 | • Create Account
## What float unit should I choose for 1 [m]?
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.
4 replies to this topic
### #1anders211 Members
250
Like
0Likes
Like
Posted 19 January 2014 - 01:35 PM
Hi
What unit should I choose for 1 [m], what value is the best? For example I have a football mesh. In reality the radius of the ball is about 11 cm (0,11m). Here on the screen:
http://imagizer.imageshack.us/v2/800x600q90/855/m2z1.jpg
the radius is 0,70 (I have float variable in which I store 0.70f). So this makes that in my game 0,70 [unit] means 0.11 [m], so 1 [m] is in fact 6,3636 units in my game.
I would like to scale my mesh and choose the scale of the world that 1 [m] = 1 [unit]. Without this there are some issues with physic. So from physic engine and generally logical point of view it would be the best to choose 1[m] as 1[unit], so radius float 0.11f means 0.11 [m]. However You know floats numbers have inacurracies, so is it good to choose 1[m] as 1[unit] or maybe better choice would be 1[m] as 100[unit]??? What is the best scale of the game environment?
Edited by anders211, 19 January 2014 - 01:36 PM.
### #2SimonForsman Members
7586
Like
1Likes
Like
Posted 19 January 2014 - 01:43 PM
for a football game 1 unit = 1m is a good scale. (players would then be up to 2 units high and the field itself is a bit over 100 units long)
I don't suffer from insanity, I'm enjoying every minute of it.
The voices in my head may not be real, but they have some good ideas!
### #3Brother Bob Moderators
10108
Like
1Likes
Like
Posted 19 January 2014 - 01:44 PM
The floating point precision is relative, not absolute, so you have the same relative resolution at any scale. Base your decision on what makes sense for you, but sticking to standard units is a good decision.
### #4anders211 Members
250
Like
0Likes
Like
Posted 19 January 2014 - 02:15 PM
OK, thanks for help
### #5Vilem Otte GDNet+
2692
Like
0Likes
Like
Posted 19 January 2014 - 02:41 PM
Just a note - Sticking to standard units is especially good once you get into physics parts, otherwise you might need some multiplications-by-constant to do it correctly (and this can become bloody mess quite soon).
Edited by Vilem Otte, 19 January 2014 - 02:41 PM.
My current blog on programming, linux and stuff - http://gameprogrammerdiary.blogspot.com
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic. | 765 | 2,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2016-50 | latest | en | 0.932544 |
https://www.learncram.com/physics/temperature-and-its-measurement/ | 1,726,328,001,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.38/warc/CC-MAIN-20240914125424-20240914155424-00147.warc.gz | 807,737,642 | 15,582 | # Temperature and its Measurement in Physics – Thermometry and Calorimetry
Temperature and its Measurement:
Temperature of a body is the degree of hotness or coldness of the body. Highest possible temperature achieved in laboratory is about 108K, while lowest possible temperature attained is 10-8 K. Temperature is measured with a thermometer.
We are giving a detailed and clear sheet on all Physics Notes that are very useful to understand the Basic Physics Concepts.
## Temperature and its Measurement in Physics – Thermometry and Calorimetry
Branch of Physics dealing with production and measurement of temperature close to 0 K is known as cryagenics, while that dealing with the measurement of very high temperature is called pyrometry.
Temperature of the core of the sun is 107 K while that of its surface is 6000 K.
NTP or STP implies 273.15 K (0°C = 32°F).
Different Scales of Temperature and Their Relationship
(i) Celsius Scale:
In this scale of temperature, the melting point of ice is taken as 0°C and the boiling point of water as 100°C and the space between these two points is divided into 100 equal parts.
(ii) Fahrenheit Scale:
In this scale of temperature, the melting point of ice is taken as 32°F and the boiling point of water as 212°F and the space between these two points is divided into 180 equal parts.
(iii) Kelvin Scale:
In this scale of temperature, the melting point of ice is taken as 273 K and the boiling point of water as 373 K and the space between these two points is divided into 100 equal parts.
(iv) Reaumer Scale:
In this scale of temperature, the melting point of ice is taken as 0°R and the boiling point of water as 80°R and the space between these two points is divided into 80 equal parts.
### The Relation Between Different Scales of Temperatures
$$\frac{C}{100}=\frac{F-32}{180}=\frac{K-273}{100}=\frac{R}{80}$$
Absolute Temperature Means:
There is no limit for maximum temperature but there is a sharp point for minimum temperature that nobody can have the temperature lower than this minimum value of temperature, which is known as absolute temperature.
Thermometry and Calorimetry:
The thermometer is a device used to check the temperature of an object. This branch of measurement of the temperature of a substance is called thermometry. It is measured in degrees or Fahrenheit, usually.
Calorimetry also means the measurement of heat but in joules. It states the amount of heat lost by the body is the amount of heat gained by its surrounding. | 563 | 2,509 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-38 | latest | en | 0.932314 |
https://discuss.codecademy.com/t/bad-instructions-in-passing-a-range-into-a-function/72370 | 1,537,457,457,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156513.14/warc/CC-MAIN-20180920140359-20180920160759-00210.warc.gz | 480,232,625 | 4,841 | BAD INSTRUCTIONS ! in Passing a range into a function
#1
Passing a range into a function
The instructions state :
On line 6, replace the ____ with a range() that returns a list containing [0, 1, 2].
IF the RETURNED LIST is [0, 1, 2] then the ORIGINAL LIST should be [ 0, 0.5, 1]
#2
but `range()` is a function call to a built in function which returns a list, so range should return `[0,1,2]`
#3
It still gives the impression that THE FINAL RETURNED LIST is [0,1,2] from an input at the "____"
I have done similar in my life, I wrote a sentence that made sense to me but was DOUBLE meaning to others..
Thanks
BEST
#4
I agree, I was thinking they'd expect I put something in there for a return list of [0,1,2]
#5
i dont get it...can someone explain the function of "step" here?
#6
step is how much the loop increases/decreases, you can always can check the documentation:
https://docs.python.org/2/library/functions.html#range
#7
def my_function(x):
for i in range(0, len(x)):
print x[i]
x[i] = x[i]
return x
print my_function(range(3))
this code intented correcly return [0, 1, 2]
but is necesary erase (*2) in code and put (range (3))
but this is not the ejercice
#8
#9
this version is interesting .
def my_function(x):
for i in range(0, len(x)):
print x[i]
x[i] = x[i]
return x
print my_function(range(1000))
#10
This topic was automatically closed 7 days after the last reply. New replies are no longer allowed. | 414 | 1,442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-39 | latest | en | 0.893013 |
http://www.rfcafe.com/references/electrical/NEETS%20Modules/NEETS-Module-04-2-41-2-53.htm | 1,369,079,502,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699201808/warc/CC-MAIN-20130516101321-00012-ip-10-60-113-184.ec2.internal.warc.gz | 677,585,168 | 11,815 | Custom Search Over 9,000 pages indexed! Your Host ... single-handedlyredefining what anengineering website should be. Carpe Diem! (Seize the Day!) 5CCG (5th MOB):My USAF radar shop Hobby & Fun Airplanes and Rockets: My personal hobby website Equine Kingdom: My daughter Sally's horse riding business website - lots of info Doggy Dynasty: My son-in-law's dog training business
•−• ••−• −•−• •− ••−• • RF Cafe Morse Code >Hear It< Job Board About RF Cafe© RF Cafe E-Mail
Product & Service Directory Engineering Jobs Personally SelectedManufacturers Employers Only(no recruiters)
# Navy Electricity and Electronics Training Series (NEETS)Module 4—Introduction to Electrical Conductors, Wiring Techniques, and Schematic ReadingChapter 2: Pages 2-41 through 2-53
NEETS Module 4—Introduction to Electrical Conductors, Wiring Techniques, and Schematic Reading
Pages i - ix, 1-1 to 1-10, 1-11 to 1-20, 1-21 to 1-28, 2-1 to 2-10,
2-11 to 2-20, 2-21 to 2-30, 2-31 to 2-40, 2-41 to 2-53, 3-1 to 3-10,
3-11 to 3-20, 3-21 to 3-24, 4-1 to 4-10, 4-11 to 4-18, Index
Figure 2-43.—Starting double lace.
Figure 2-44.—Terminating double lace.
2-41
Figure 2-45.—Alternate method of terminating the lace.
Figure 2-46.—Marling hitch as a lock stitch.
The spare conductors of a multiconductor cable should be laced separately, and then tied to active conductors of the cable with a few telephone hitches. When two or more cables enter an enclosure, each cable group should be laced separately. When groups are parallel to each other, they should be bound together at intervals with telephone hitches (figure 2-47).
2-42
Figure 2-47.—Spot tying cable groups.
Q51. What size wire bundles require double lace?
Q52. How is the double lace started?
Q53. How are laced cable groups bound together?
SPOT TYING
When cable supports are used in equipment as shown in figure 2-48, spot ties are used to secure the conductor groups if the supports are more than 12 inches apart. The spot ties are made by wrapping the cord around the group as shown in figure 2-49. To finish the tie, use a clove hitch followed by a square knot with an extra loop. The free ends of the cord are then trimmed to a minimum of 3/8 inch.
Figure 2-48.—Use of spot ties.
2-43
Figure 2-49.—Making spot ties.
SELF-CLINCHING CABLE STRAPS
Self-clinching cable straps are adjustable, lightweight, flat nylon straps. They have molded ribs or serrations on the inside surface to grip the wire. They may be used instead of individual cord ties for securing wire groups or bundles quickly. The straps are of two types: a plain cable strap and one that has a flat surface for identifying the cables.
CAUTION
Do not use nylon cable straps over wire bundles containing coaxial cable. Do not use straps in areas where failure of the strap would allow the strap to fall into movable parts.
Installing self-clinching cable straps is done with a Military Standard hand tool, as shown in figure 2-50. An illustration of the working parts of the tool is shown in figure 2-51. To use the tool, follow the manufacturer's instructions.
Figure 2-50.—Installing self-clinching cable straps.
2-44
Figure 2-51.—Military Standard hand tool for self-clinching cable straps.
WARNING
Use proper tools and make sure the strap is cut flush with the eye of the strap. This prevents painful cuts and scratches caused by protruding strap ends. Do not use plastic cable straps in high-temperature areas (above 250º F).
HIGH-TEMPERATURE PRESSURE-SENSITIVE TAPE LACING
High-temperature, pressure-sensitive tape must be used to tie wire bundles in areas where the temperature may exceed 250º F. Install the tape as follows (figure 2-52):
1. Wrap the tape around the wire bundle three times, with a two-thirds overlap for each turn.
2. Heat-seal the loose tape end with the side of a soldering iron tip.
Figure 2-52.—Securing wire bundles in high-temperature areas.
2-45
WARNING
Insulation tape (including the glass fiber type) is highly flammable and should not be used in a high-temperature environment. Only insulation tape approved for high-temperature operation (suitable for continuous operation at 500º F) should be used in high-temperature environments.
Q54. When are spot ties used?
Q55. What is used to install self-clinching cable straps?
Q56. What is used to tie wire bundles in high-temperature areas?
SUMMARY
In this chapter you have learned some of the basic skills required for proper wiring techniques. We have discussed conductor splices and terminal connections, basic soldering skills, and lacing and tying wire bundles.
The basic requirement for any splice or terminal connection is that it be both mechanically and electrically as strong as the conductor or device with which it is to be used.
Insulation Removal—The first step in splicing or terminating electrical conductors is to remove the insulation. The preferred method for stripping wire is by use of a wire-stripping tool. The hot-blade stripper cannot be used on such insulation material as glass braid or asbestos. An alternate method for stripping copper wire is with a knife. A knife is the required tool to strip aluminum wire. Take extreme care when stripping aluminum wire. Nicking the strands will cause them to break easily.
Western Union Splice—A simple connection known as the Western Union splice is used to splice small, solid conductors together. After the splice is made, the ends of the wire are clamped down to prevent damage to the tape insulation.
2-46
Staggered Splice—The staggered splice is used on multiconductor cables to prevent the joint from being bulky.
Rattail Joint—A splice that is used in a junction box and for connecting branch circuits; wiring is placed inside conduits.
Fixture Joint—When conductors of different sizes are to be spliced, such as fixture wires to a branch circuit, the fixture joint is used.
Knotted Tap Joint—This type of splice is used to splice a conductor to a continuous wire. It is not considered a "butted" splice as the ones previously discussed.
Splice Insulation—Rubber tape is an insulator for the type of splices we have discussed so far.
Friction Tape—It has very little insulating value but is used as a protective covering for the rubber tape. Another type of insulating tape is plastic electrical tape, which is quite expensive.
Terminal Lugs—The terminals used in electrical wiring are either of the soldered or crimped type. The advantage of using a crimped type of connection is that it requires very little operator skill, whereas the soldered connection is almost completely dependent on the skill of the operator. Some form of insulation must be used with noninsulated splices and terminal lugs. The types used are clear plastic tubing (spaghetti) and heat-shrinkable tubing. When a heat gun is used to shrink the heat-shrinkable tubing, the maximum allowable heat to be used is 300º F. When using the compressed air/nitrogen heating tool, the air/nitrogen source cannot be greater than 200 psig.
2-47
Aluminum Terminals and Splices—Aluminum terminals and splices are noninsulated and very difficult to use. Some of the things you should remember when working with aluminum wire are: (1) Never attempt to clean the aluminum wire. There is a petroleum abrasive compound in the terminal lug or splice that automatically cleans the wire. (2) The only tools that should be used for the crimping operation are the power crimping type. (3) Never use lock washers next to aluminum terminal lugs as they will gouge out the tinned area and increase deterioration.
Preinsulated Copper Terminal Lugs and Splices—The most common method of terminating and splicing copper wires is with the use of preinsulated terminal lugs and splices. Besides not having to insulate the terminal or splice after the crimping operation, the other advantage of this type is that it gives extra wire insulation support. Several types of crimping tools can be used for these types of terminals and splices. The tool varies with the size of the terminal or splice. Preinsulated terminal lugs and splices are color coded to indicate the wire size they are to be used with.
2-48
Soldering—The basic skills required to solder terminal lugs, splices, and electrical connectors are covered in this area. Prior to any soldering operation, the items to be soldered must be cleaned; they will not adhere to dirty, greasy, or oxidized surfaces. The next step is the "tinning" process. This process is accomplished by coating the material to be soldered with a bright coat of solder. The wire to be soldered must be stripped to 1/32 inch longer than the depth of the solder cup of the terminal, splice, or connector to which it is to be soldered. This is to prevent burning the insulation. It also allows the wire to flex at the stress point. When you tin the wire, it should be done to one-half of the stripped length. When soldering a connection, take precaution to prevent movement of the parts while the solder is cooling. A "fractured solder" joint will result if this precaution is not taken.
Soldering Tools—The important difference in soldering iron sizes is not the temperature (they all produce 500º F to 600º F), but the thermal inertia. Thermal inertia is the ability of soldering tools to maintain a satisfactory soldering temperature while giving up heat to the joint to be soldered. A well- designed soldering iron is self-regulating because its heating element increases with the rising temperature, thus inciting the current to a satisfactory level. When using a soldering gun, do not press the switch for periods longer than 30 seconds. Doing so will cause the tip to overheat to the point of incandescence. The nuts or screws that retain the tips on soldering irons and guns tend to loosen because of the continuous heating and cooling cycles. Therefore, they should be tightened periodically. You should never use a soldering gun on electronics components, such as resistors, capacitors, or transistors. An advantage of using a resistance soldering iron to solder a wire to a connector is that the soldering tips are only hot during the brief period of soldering the connection.
2-49
Solder—Ordinary soft solder is a fusible alloy of tin and lead used to join two or more metals at temperatures below their melting point. The metal solvent action that occurs when copper conductors are soldered together takes place because a small amount of the copper combines with the solder to form a new alloy. Therefore, the joint is one common metal. The tin-lead alloy used for general-purpose soldering is composed of 60-percent tin and 40-percent lead (60/40 solder).
Flux—Flux is used in the soldering process to clean the metal by removing the oxide layer on the metal and to prevent further oxidation during the soldering process. Always use noncorrosive, nonconducting rosin fluxes when soldering electrical and electronic components.
Solvents—Solvents are used in the soldering process to remove contaminants from the surfaces to be soldered.
Soldering Aids—Use a heat shunt when you solder heat-sensitive components. It dissipates the heat, thereby preventing damage to the heat-sensitive component. Some type of soldering iron holder or guard should be used to prevent the operator from being burned.
Lacing Conductors—The purpose of lacing conductors is to present a neat appearance and to facilitate tracing the conductors when alterations or repairs are required. Flat tape is preferred for lacing instead of round cord. Cord has a tendency to cut into the wire insulation. The amount of flat tape or round cord required to lace a group of conductors is about two and one-half times the length of the longest conductor. A lacing shuttle is useful during the lacing operation to prevent the tape or cord from
fouling. Wires should only be twisted prior to lacing if it is required, such as for filament leads in electron tube amplifiers. When lacing wire bundles containing coaxial cables, use the proper flat tape and do not
tie the bundles too tightly. Never use round cord on coaxial cable. A single lace is started with a square
knot and at least two marling Hitches. A double lace is required for wire bundles that are 1 inch or more
in diameter. It is started with a telephone hitch. Cable groups are bound together by use of telephone hitch
2-50
Spot Ties—Spot ties are used when cable supports are used that are more than 12 inches apart.
Self-clinching Cable Straps—If self-clinching cable straps are used, they should be installed with the Military Standard hand tool designed for their use.
High-temperature Areas—When you are required to tie wire bundles in high-temperature operating areas, use only high-temperature, pressure-sensitive tape.
ANSWERS TO QUESTIONS Q1 THROUGH Q56.
A1. The connection must be both mechanically and electrically as strong as the conductor or device with which it is used
A2. By use of a wire-stripping tool
A4. Knife.
A5. To prevent damage to the tape insulation.
A6. To prevent the joint from being bulky.
A7. When wires are in conduit and a junction box is used.
A8. Fixture joint.
A9. Knotted tap joint.
A10. As a protective covering over the rubber tape.
A11. Requires relatively little operator skill to install.
A12. Spaghetti or heat-shrinkable tubing.
A13. 300º F
A14. 200 psig.
A15. No, it is done automatically by the petroleum abrasive compound that comes in the terminal or splices.
A16. Power-operated crimping tools.
2-51
A17. It gouges the terminal lug and causes deterioration.
A18. The use of preinsulated splices and terminal lugs.
A19. It has insulation support for extra supporting strength of the wire insulation.
A20. To identify wire sizes they are to be used on.
A21. Solder will not adhere to dirty, greasy, or oxidized surfaces.
A22. The coating of the material to be soldered with a light coat of solder.
A23. To prevent burning the insulation during the soldering process and to allow the wire to flex easier at a stress point.
A24. One-half the stripped length.
A25. Movement of the parts being soldered while the solder is cooling.
A26. The capacity of the soldering iron to generate and maintain a satisfactory soldering temperature while giving up heat to the joint being soldered.
A27. Although its temperature is as high as the larger irons, it does not have thermal inertia.
A28. The resistance of its heating element increases with rising temperature, thus limiting the current flow.
A29. File the tip until it is smooth and re-tin it.
A30. It will overheat and could burn the insulation of the wire being soldered.
A31. The heating and cooling cycles.
A32. Electronic components, such as resistors, capacitors, and transistors.
A33. The soldering tips are hot only during the brief period of soldering the connection, thus minimizing the chance of burning the wire insulation or connector inserts.
A34. The strands can fall into electrical equipment being worked on and cause short circuits.
A35. It enables the tip to be removed easily when another is to be inserted.
A36. Wrap a length of copper wire around one of the regular tips and bend to the proper shape for the purpose.
A38. The solder dissolves a small amount of the copper, which combines with the solder forming a new alloy; therefore, the joint is one common metal.
A39. 60-percent tin and 40-percent lead (60/40 solder).
A40. It cleans the metal by removing the oxide layer and prevents further oxidation during the soldering.
A41. Noncorrosive, nonconductive rosin fluxes.
2-52
A42. To remove contaminants from soldered connections.
A43. To prevent damage to heat-sensitive components.
A44. To aid in tracing the conductors when alterations or repairs are required.
A45. Round cord has a tendency to cut into the wire insulation.
A46. Two and one-half times the length of the longest conductor in the group.
A47. To keep the tape or cord from fouling during the lacing operation.
A48. When required, such as for the filament leads in electron tube amplifiers.
A49. Do not tie too tightly and use the proper type of tape.
A50. With a square knot and at least two marling hitches drawn tightly.
A51. Bundles that are 1 inch or larger in diameter
A52. With a telephone hitch.
A53. They are bound together at intervals with telephone hitches.
A54. When wire bundles are supported by cable supports that are more than 12 inches apart.
A55. Military Standard hand tool.
A56. High-temperature, pressure-sensitive tape.
2-53
Introduction to Matter, Energy, and Direct Current, Introduction to Alternating Current and Transformers, Introduction to Circuit Protection, Control, and Measurement, Introduction to Electrical Conductors, Wiring Techniques, and Schematic Reading, Introduction to Generators and Motors, Introduction to Electronic Emission, Tubes, and Power Supplies, Introduction to Solid-State Devices and Power Supplies, Introduction to Amplifiers, Introduction to Wave-Generation and Wave-Shaping Circuits, Introduction to Wave Propagation, Transmission Lines, and Antennas, Microwave Principles, Modulation Principles, Introduction to Number Systems and Logic Circuits, Introduction to Microelectronics, Principles of Synchros, Servos, and Gyros, Introduction to Test Equipment, Radio-Frequency Communications Principles, Radar Principles, The Technician's Handbook, Master Glossary, Test Methods and Practices, Introduction to Digital Computers, Magnetic Recording, Introduction to Fiber Optics
RF Cafe Software RF Cascade Workbook RF system analysis includingfrequency conversion & filters A Disruptive Web Presence™ Custom Search Over 9,000 pages indexed! Read About RF CafeWebmaster: Kirt BlattenbergerKB3UONProduct & Service DirectoryPersonally Selected Manufacturers RF Cafe T-Shirts & Mugs Calculator Workbook RF Workbench | 4,109 | 17,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2013-20 | latest | en | 0.827609 |
https://www.bloomberg.com/news/articles/2013-05-15/sat-tip-a-primer-on-triangles | 1,529,889,939,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867304.92/warc/CC-MAIN-20180624234721-20180625014721-00160.warc.gz | 748,897,343 | 17,016 | # SAT Tip: A Primer on Triangles
This tip on improving your SAT score was provided by Vivian Kerr at Veritas Prep.
Triangles often appear on the SAT, and there are some basic properties you’ll need to know to rock the SAT math section. A triangle is a three-sided figure. The sum of the interior angles is always 180 degrees. To find the area of a triangle, we use the formula A = ½ bh, where b = base and h = height. The base and the height of the triangle must always form a 90-degree angle. Keep in mind that the height can be inside or outside the triangle.
The Triangle Inequality Theorem states that for any side of a triangle, its value must be between the sum and the difference of the other two sides (non-inclusive). For example, if we have a triangle with two sides, 3 and 9, can 6 be the value of the third side? Let’s consider: We know the sum of 3 + 9 is 12, and the difference of 9- 3 is 6, so the third side must be between 6 and 12. Since 6 is not between those two numbers, then 6 cannot be the value of the third side.
The Pythagorean Theorem states that a2 + b2 = c2 where a and b are the two shorter sides and c is always the longest side or the hypotenuse (the side across from the 90-degree angle) of a right triangle. In this triangle, the “?” side is the hypotenuse. Let’s plug the values for the other two sides into the Pythagorean Theorem to solve:
a2 + b2 = c2
82 + 52 = c2
64 + 25 = c2
89 = c2
To remove the exponent (2), we must take the square root (√) of both sides.
√89 = c
Save valuable time on the SAT by memorizing the common Pythagorean triplets. You often encounter right triangles with the ratios of 3:4:5 and 5:12:13. These ratios will also be true for any multiples of 3:4:5 and 5:12:13 such as 6:8:10 or 10:24:26. For example, in this triangle we know the third side must be 5, even without using the Pythagorean Theorem, because we know 5:12:13 is a common triplet.
Be cautious, however, because the 13, or longest side/hypotenuse, must always be across from the 90-degree angle.
There are two special right triangles. The first is a 30-60-90 triangle. Its sides will always be in a ratio of x: x√3 : 2x. The other special triangle is the 45-45-90 triangle. Its sides will always be in a ratio of x: x: x√2. It’s important to remember that for the 30-60-90 triangle, the hypotenuse is the side that has the ratio of 2x. Don’t confuse it with the 45-45-90 ratio and think that the x√3 should be there.
For triangle question practice, take a full-length SAT practice test to sharpen your skills.
Before it's here, it's on the Bloomberg Terminal. LEARN MORE | 711 | 2,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 5 | 5 | CC-MAIN-2018-26 | latest | en | 0.923911 |
https://research.stlouisfed.org/fred2/series/BEAU148NRMN?rid=250 | 1,448,863,128,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398460982.68/warc/CC-MAIN-20151124205420-00002-ip-10-71-132-137.ec2.internal.warc.gz | 837,124,155 | 19,035 | # All Employees: Mining, Logging, and Construction in Beaumont-Port Arthur, TX (MSA)
2015-10: 23.0 Thousands of Persons
Monthly, Seasonally Adjusted, BEAU148NRMN, Updated: 2015-11-20 4:19 PM CST
1yr | 5yr | 10yr | Max
The Federal Reserve Bank of St. Louis seasonally adjusts this series by using the 'x12' package from R with default parameter settings. The package uses the U.S. Bureau of the Census X-12-ARIMA Seasonal Adjustment Program. More information on the 'x12' package can be found at http://cran.r-project.org/web/packages/x12/x12.pdf. More information on X-12-ARIMA can be found at http://www.census.gov/srd/www/x12a/.
Restore defaults | Save settings | Apply saved settings
Recession bars:
Log scale:
Show:
Y-Axis Position:
(a) All Employees: Mining, Logging, and Construction in Beaumont-Port Arthur, TX (MSA), Thousands of Persons, Seasonally Adjusted (BEAU148NRMN)
Integer Period Range:
copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
Create segments for min, max, and average values: [+]
Graph Data
Graph Image
Suggested Citation
``` Federal Reserve Bank of St. Louis, All Employees: Mining, Logging, and Construction in Beaumont-Port Arthur, TX (MSA) [BEAU148NRMN], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/BEAU148NRMN/, November 29, 2015. ```
Retrieving data.
Graph updated. | 539 | 1,994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2015-48 | latest | en | 0.775496 |
https://leanprover-community.github.io/archive/stream/217875-Is-there-code-for-X%3F/topic/Coprime.20numbers.20whose.20product.20is.20a.20square.20are.20both.20square.html | 1,621,264,795,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991258.68/warc/CC-MAIN-20210517150020-20210517180020-00197.warc.gz | 381,091,205 | 3,960 | ## Stream: Is there code for X?
### Topic: Coprime numbers whose product is a square are both square
#### Ruben Van de Velde (Sep 08 2020 at 11:04):
lemma coprime_sq (a b c : ℤ) (h : a * b = c ^ 2) (hcoprime : is_coprime a b) :
∃ d e, a = d ^ 2 ∧ b = e ^ 2 := sorry
Probably not
#### Ruben Van de Velde (Sep 08 2020 at 11:26):
is_pow_of_coprime from #3984, apparently
#### Reid Barton (Sep 08 2020 at 11:37):
Oh that reminds me, let me dig up my old FLT code
#### Ruben Van de Velde (Sep 08 2020 at 11:38):
@Paul van Wamelen
#### Reid Barton (Sep 08 2020 at 11:48):
Well, I think it predates elan so I don't really know how to build it (apparently I used to build it with a version of lean called "3.3.1"), but I put it up at https://github.com/rwbarton/lean-elementary-number-theory
#### Reid Barton (Sep 08 2020 at 12:02):
in particular, the statement about multiplying to a square is at https://github.com/rwbarton/lean-elementary-number-theory/blob/master/src/ent/gcd.lean#L60-L61 -- it doesn't actually need prime factorization, you can get it from docs#nat.prod_dvd_and_dvd_of_dvd_prod, though I made it look quite difficult
#### Paul van Wamelen (Sep 08 2020 at 16:13):
PR #4049 has:
theorem is_pow_of_coprime {a b c : associates α} (ha : a ≠ 0) (hb : b ≠ 0)
(hab : ∀ d, d ∣ a → d ∣ b → ¬ prime d) {k : ℕ} (h : a * b = c ^ k)
: ∃ (d : associates α), a = d ^ k
for α a unique factorization domain. From there it isn't too hard to get to:
lemma sqr_of_coprime {a b c : ℤ} (hc : c ≠ 0) (h : int.gcd a b = 1) (heq : a * b = c ^ 2) :
∃ (a0 : ℤ), a0 ≠ 0 ∧ (a = a0 ^ 2 ∨ a = - (a0 ^ 2))
(but this hasn't been PR'd yet). Let me know if you want the code...
Last updated: May 17 2021 at 14:12 UTC | 611 | 1,717 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2021-21 | latest | en | 0.818329 |
https://answerdata.org/the-wavelength-of-the-red-light-is-700-5-nm-what-is-the-frequency-of-red-light/ | 1,680,355,720,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950030.57/warc/CC-MAIN-20230401125552-20230401155552-00051.warc.gz | 130,148,384 | 15,561 | # The wavelength of the red light is 700.5 nm. What is the frequency of red light?
Start by converting the wavelength from nanometers to meters. The prefix nano means 10–9.
• 1 nm = 1 * 10^-9 m
700.5 nm 1 m / 110^-9 nm = 700.5*10^-9 m
f = c / wavelength
f = (3.00 * 10^8 m/s) / (700.5 * 10^-9 m) = 4.28*10^14 Hz.
Source(s): Chemistry.
• Wavelength Of Red Light
• Use the equation c = ⅄ v. C is velocity of sunshine, ⅄ is wavelength, and v is frequency First trade the nm to m utilising the conversion factor: 1 m = 1 x 10^9 nm Them plug within the known values. Three.00 x 10^8m/s = (something wavelength comes out to be) x v clear up for v(frequency)
• 4.28*10^14 S^-1 | 234 | 677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-14 | longest | en | 0.857833 |
https://sites.google.com/site/e90e50fx/home/matrix-bubble-chart-with-excel | 1,548,286,088,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584415432.83/warc/CC-MAIN-20190123213748-20190123235748-00002.warc.gz | 640,409,951 | 19,641 | ### Matrix bubble chart with Excel
posted Oct 18, 2013, 9:18 AM by r [ updated Sep 24, 2014, 3:22 AM by Krisztina Szabó ]
Freddy:
If you're blue and you don't know
Where to go to, why don't you go
Where fashion sits?
Monster:
Puttin' on the Ritz
[...]
#### Introduction
In the past we alredy have tried to represent all the data of a table in a single chart (check the Stacking Cubes Charts with Excel article).
The purpose was to give an overview, a total picture without focusing on anything in particular. This is just the beginning then you can change the formatting and customize the chart, with color and highlighting what interests you most or use in combination with others to create a dashboard.
#### Step to step the creationof the Excel chart
We start by defining a dynamic range, for example, in this way:
`rng=OFFSET(\$B\$2,,,COUNTA(OFFSET(\$B\$2,,,100)),COUNTA(OFFSET(\$B\$2,,,,100)))`
In the picture below
you can see what we want to realize.
To achieve this we must add 3 series, one is to show the bubbles, the other two are used to show the column and row labels.
The chart could easily be replicated based on your own data. In a nutshell: For the bubble positions, we use fixed x and y coordinates, the bubble size is the original data.
A little trick with the label formatting helps us hiding the lower values (<50 in this case) which, as is shown above, would lead to confusion.
On the below pictures you can see the defined names and the three series used on the chart, so you can easily follow the construction steps.
matrix_data serie:
`arr_h=--rng`
`arr_x=(COLUMN(rng)-MIN(COLUMN(rng))+1)*(rng=rng)`
`arr_y=(ROW(rng)-MIN(ROW(rng))+1)*(rng=rng)`
axis_s serie:
`asse_x_val_h=asse_x_val_y+1`
`asse_x_val_x=OFFSET(rng,-1,,1)`
`asse_x_val_y=COLUMN(rng)*0`
You need to notice that in the serie "axis_x", x values are the column headers of the table (text value, if you look the file will see that the values in the cells are '2008, '2009 ... '2013) and that both series "matrix data" and "axis_x" are backed on the primary axis. Finally the bubble size is the original data, but to avoid overlapping circles, set the Scale of the bubble size to 50, so the diameter of the largest circle will always be 1.
To build a y-axis that displays the row headers define the following names and the "axis_y" serie:
`asse_y_val_h=ROW(rng)^0`
`asse_y_val_x=OFFSET(rng,,-1,,1)`
`asse_y_val_y=ROWS(rng)-INDEX(arr_y,,1)+1`
But be careful! in this case we associate the series to the secondary axis ...
The last step is to set the maximum value of the axis to a very large number:
If you add new data (new columns or new rows), the chart will resize . In the picture below we have added the row with the totals of each year.
#### Dynamic version based on the activecell
For a dynamic version based on the activecell use this code:
`'in the Worksheet class module`
`Private Sub Worksheet_SelectionChange(ByVal Target As Range)`
`Static b As Boolean`
`If Not Application.Intersect(Target, Range("Rng")) Is Nothing Then`
`ActiveSheet.Calculate`
`b = True`
`Else`
` If b Then`
` ActiveSheet.Calculate`
` b = False`
` End If`
`End If`
`End Sub`
We have to add a new serie to represent the current value (a single bubble with black outline).
Use these new names:
`check_=AND(CELL("row")=MODE(ROW(rng),CELL("row")),CELL("col")=MODE(COLUMN(rng),CELL("col")))`
`active_h=IF(ISERROR(check_),0,INDEX(rng,CELL("row")-MIN(ROW(rng))+1,CELL("col")-MIN(COLUMN(rng))+1))`
`active_x=IF(ISERROR(check_),1,CELL("col")-MIN(COLUMN(rng))+1)`
`active_y=IF(ISERROR(check_),1,CELL("row")-MIN(ROW(rng))+1)`
In the picture the result:
The value corresponding to the active cell is highlighted and a label describes the detail of the data. When the active cell is outside of the table, it will disappear.
If you plan to use this chart in your workbook, here you will find how to copy it from our example file to your own file.
If you like this chart, you should take a look at the version we created for a user with conditionally formatted bubbles.
Ĉ
r,
Oct 18, 2013, 9:18 AM
Ĉ
r,
Oct 18, 2013, 9:18 AM | 1,099 | 4,128 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2019-04 | latest | en | 0.855933 |
https://www.slideserve.com/ita/by-marc-m-triola-mario-f-triola-slides-prepared-by-lloyd-r-jaisingh-morehead-state-university-morehead-ky-with-m | 1,708,897,780,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474643.29/warc/CC-MAIN-20240225203035-20240225233035-00273.warc.gz | 1,019,388,704 | 20,829 | 1 / 26
# 1-1
Biostatistics for the Biological and Health Sciences. 1-1. by Marc M. Triola & Mario F. Triola SLIDES PREPARED BY LLOYD R. JAISINGH MOREHEAD STATE UNIVERSITY MOREHEAD KY (with modifications by DGE Robertson). Chapter 1. 1-2. Introduction. WCB/McGraw-Hill.
## 1-1
An Image/Link below is provided (as is) to download presentation Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. During download, if you can't get a presentation, the file might be deleted by the publisher.
E N D
### Presentation Transcript
1. Biostatistics for the Biological and Health Sciences 1-1 by Marc M. Triola & Mario F. Triola SLIDES PREPARED BY LLOYD R. JAISINGH MOREHEAD STATE UNIVERSITY MOREHEAD KY (with modifications by DGE Robertson)
2. Chapter 1 1-2 Introduction WCB/McGraw-Hill • The McGraw-Hill Companies, Inc., 1998
3. Outline 1-3 • 1-1 Introduction • 1-2 Types of Data • 1-4 Data Collection and Sampling Techniques • 1-5 Computers and Calculators
4. Objectives 1-5 • Demonstrate knowledge of all statistical terms. • Differentiate between the two branches of statistics. • Identify types of data.
5. Objectives 1-6 • Identify the measurement level for each variable. • Identify the four basic sampling techniques.
6. 1-1 Introduction 1-8 • Statisticsconsists of conducting studies to collect, organize, summarize and analyze data and to draw conclusions
7. 1-1 Descriptive and Inferential Statistics 1-9 • Dataare the values (measurements or observations) that the variables can assume. • Variables whose values are determined by chance are called random variables.
8. 1-1 Descriptive and Inferential Statistics 1-10 • A collection of data values forms a data set. • Each value in the data set is called a data value or a datum.
9. 1-1 Descriptive and Inferential Statistics 1-11 • Descriptive statistics consists of the collection, organization, summation and presentation of data.
10. 1-1 Descriptive and Inferential Statistics 1-12 • A population consists of all subjects (human or otherwise) that are being studied. • A sampleis a subgroup of the population.
11. 1-1 Descriptive and Inferential Statistics 1-13 • Inferential statistics consists of generalizing from samples to populations, performing hypothesis testing, determining relationships among variables, and making predictions.
12. 1-2 Variables and Types of Data 1-14 • Qualitative variablesare variables that can be placed into distinct categories, according to some characteristic or attribute. For example, gender (male or female).
13. 1-2 Variables and Types of Data 1-15 • Quantitative variablesare numerical in nature and can be ordered or ranked. Example: age is numerical and the values can be ranked.
14. 1-2 Variables and Types of Data 1-16 • Discrete variablesassume values that can be counted. • Continuous variablescan assume all values between any two specific values. They are obtained by measuring.
15. 1-2 Variables and Types of Data 1-17 • Thenominal level of measurementclassifies data into mutually exclusive (nonoverlapping), exhausting categories in which no order or ranking can be imposed on the data.
16. 1-2 Variables and Types of Data 1-18 • Theordinal level of measurementclassifies data into categories that can be ranked; precise differences between the ranks do not exist.
17. 1-2 Variables and Types of Data 1-19 • Theinterval level of measurementranks data; precise differences between units of measure do exist; there is no meaningful zero.
18. 1-2 Types of Data 1-20 • Theratio level of measurementpossesses all the characteristics of interval measurement, and there exists a true zero. In addition, true ratios exist for the same variable.
19. 1-4 Data Collection and Sampling Techniques 1-21 • Data can be collected in a variety of ways. • One of the most common methods is through the use of surveys. • Surveys can be done by using a variety of methods: • Examples are telephone, mail questionnaires, personal interviews, surveying records and direct observations.
20. 1-4 Data Collection and Sampling Techniques 1-22 • To obtain samples that are unbiased, statisticians use four methods of sampling. • Random samplesare selected by using chance methods or random numbers.
21. 1-4 Data Collection and Sampling Techniques 1-23 • Systematic samplesare obtained by numbering each value in the population and then selecting the kth value.
22. 1-4 Data Collection and Sampling Techniques 1-24 • Stratified samplesare selected by dividing the population into groups (strata) according to some characteristic and then taking samples from each group.
23. 1-4 Data Collection and Sampling Techniques 1-25 • Cluster samplesare selected by dividing the population into groups and then taking samples of the groups.
24. 1-4 Data Collection and Sampling Techniques 1-25 • Convenience samplesare when subjects are selected for convenience. (Often used in student research projects or by advertisers.)
25. 1-5 Calculators 1-26 • Calculators make some statistical tests and numerical computations easier. • The TI-35 and TI-83 calculators perform 2-variable statistical calculations. • Must learn how to enter and perform statistical functions on your calculator.
26. 1-5 Computers and Calculators 1-26 • Computers can perform more advanced statistical tests. • Many statistical packages are available. Examples are SPSS, SAS and MINITAB also Excel and QuattroPro. • Input and output from computer must understood and interpreted.
More Related | 1,275 | 5,535 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-10 | latest | en | 0.767884 |
https://socratic.org/questions/if-the-mass-of-earth-were-2-times-the-present-mass-the-mass-of-the-moon-were-hal | 1,607,127,880,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141745780.85/warc/CC-MAIN-20201204223450-20201205013450-00344.warc.gz | 478,208,672 | 5,954 | # If the mass of earth were 2 times the present mass,the mass of the moon were half the present mass and the moon were revolving round the earth at the earth at the same present distance, the time period of revolution of the moon would be(in days)?
May 22, 2018
$\approx 19$ days
#### Explanation:
We can derive an expression for the time period of revolution by assuming that the moon is moving in a circular orbit around the earth.
The necessary centripetal acceleration is provided by the force of gravity and thus we have
${m}_{m} {\omega}^{2} r = G \frac{{m}_{e} {m}_{m}}{r} ^ 2 \implies$
${\omega}^{2} = G {m}_{e} / {r}^{3} \implies$
$T = \frac{2 \pi}{\omega} = \frac{2 \pi}{\sqrt{G {m}_{e}}} {r}^{\frac{3}{2}}$
Thus, under the conditions of the problem the period will decrease by a factor of $\sqrt{2}$ to become $\frac{27}{\sqrt{2}}$ days $\approx 19$ days. | 261 | 875 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-50 | latest | en | 0.90824 |
https://m.jagranjosh.com/articles/amp/mathematics-all-chapters-complete-notes-for-upsee-uptu-2018-1515043365-1?ref=amp_list_detail | 1,576,291,249,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540579703.26/warc/CC-MAIN-20191214014220-20191214042220-00532.warc.gz | 430,948,569 | 9,949 | UPSEE/UPTU 2019: Complete Notes for All Chapters of Mathematics
Mathematics is one of the most scoring subjects in UPSEE/UPTU 2019. A total of 50 multiple choice questions will be asked in UPSEE 2019. Students who had a clear understanding of basic concepts can easily score descent marks in this subject. Generally, 80% questions in UPSEE Maths Questions Paper are not time consuming and students can easily solve them. Students can easily boost their score by 10% if they know how to apply any formula/concept to solve any question. Engineering aspirants should try to understand the topics instead of rote learning. Get chapter notes and important questions for UPSEE Maths Paper 2019. These notes are prepared by subject experts of Mathematics after doing a detailed analysis of UPSEE Syllabus 2019 and UPSEE Previous Years’ Question Papers. Some solved questions are also given in these notes for better understanding of concepts.
Physics all chapters complete notes for UPSEE/UPTU 2019
1. These chapter notes contain important concepts, formulae and some previous year solved questions.
2. These chapter notes are prepared experienced Subject Experts of Mathematics.
3. Each and every concept of Mathematics is explained in a very detailed manner.
4. With the help of the previous year solved questions, aspirants can easily predict the difficulty level of the coming UPSEE/UPTU entrance examination 2019.
5. These chapter notes are in very concise form and students can easily study all important topics related to any chapter of Mathematics just before few days of the examination.
Find the link of all the chapters of Mathematics in the below table:
Chapter Number Chapter Name Links 1 Matrices and Determinants Complete 2 Complex Numbers Complete 3 Relations and Functions Complete 4 Sequence and Series Complete 5 Permutation and Combination Complete 6 Binomial Theorem Complete 7 Probability Complete 8 Ellipse Complete 9 Hyperbola Complete 10 Trigonometric functions Complete 11 Inverse Trigonometric function Complete 12 Three Dimensional Geometry Complete 13 Limits Complete 14 Parabola Complete 15 Differential Equation Complete 16 Application of Derivatives Complete 17 Circles Complete 18 Area under Curve Complete 19 Integrals Complete 20 Continuity and Differentiability Complete 21 Straight Lines Complete 22 Theory of Equations Complete 23 Vectors Complete
Chemistry all chapters complete notes for UPSEE/UPTU 2019
Popular | 504 | 2,456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2019-51 | latest | en | 0.885381 |
https://www.black-holes.org/the-science-numerical-relativity/numerical-relativity/einstein-s-equations | 1,653,821,095,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662644142.66/warc/CC-MAIN-20220529103854-20220529133854-00799.warc.gz | 736,492,565 | 16,448 | ## Einstein's Equations
### Describing How Mass Warps Spacetime
One of the central challenges of physics is — and has always been — to predict how things move. The earliest astronomers were really nothing more than astrologers, trying to discern when stars would appear on the horizon, or where the Sun would be found at a certain date. Eventually, this turned into true, scientific astronomy, and physicists used their laws to predict how all sorts of bodies moved through the heavens. Newton showed us that the same laws that described how planets move could also be used to predict more earth-bound things, such as how an apple falls from a tree. Modern physicists are concerned with the motions of the tiniest particles in the atoms around us, and the motions of the heaviest objects in the heavens above.
We've mentioned that geometry gives us tools to understand the motion of particles when spacetime is curved. But it doesn't say anything about why spacetime should be curved. Einstein took the tools of differential geometry, and showed us how and why spacetime curves. In doing this, he gave us very powerful tools to predict the motion of particles. To understand how these tools work will only require a little review of how to keep track of points in geometry, and the Pythagorean Theorem.
#### Coordinates
Suppose you want to keep track of an astronaut who is somewhere along a single line, like in our examples from the section on Relativity. You could choose a particular point — which we will call the origin — and measure how far the astronaut is from that point. If the astronaut is, for example, ten meters to the right of the origin, you might say that she is at the coordinate x=10. Ten meters to the left, and you could call it x=-10. She could even be at a fractional coordinate like x=6.78.
Of course, coordinates are just numbers that we place at each point to label that point, and to keep track of what happens at each point. We can lay those numbers down in any way we want. We might, for example, only put in one coordinate tick for every two meters. Then, if the astronaut is at a coordinate of x=5, she would still be 10 meters to the right of the origin. We say that there is a factor of proportionality which is the real distance between coordinate ticks — two, in this case. Then, if we want the actual distance, we just multiply the coordinate x by the factor of proportionality:
$$\text{distance in meters} = x \text{ coordinate} \times \text{meters per coordinate tick}$$
It's very important to remember that the real physical situation doesn't change, even if we change how we label the physical points.
Now this astronaut may want to keep track of another astronaut, but he's strayed off the coordinate axis she is on. So, she decides to create a new set of coordinates, using two dimensions to keep track of him. She takes herself as the origin and constructs two axes — one in each of the two dimensions she needs. Then, to figure out what coordinates her fellow astronaut is at, she sees how far along one axis she would need to go, then how far parallel to the other axis she would need to go. For example, our second astronaut might be six meters to the right, and eight meters up. His coordinates would be x=6 and y=8. If he were eight feet down, his coordinates would be x=6, and y=-8. We can see this in the figure below.
But now, if our first astronaut wants to get to the second, she wouldn't want to go six meters over and eight meters up; she'd want to go straight. To know how far she has to go, we need the help of an old theorem from high-school geometry.
#### Pythagoras's Theorem
The Pythagorean theorem tells us the length of one side of a triangle, given the lengths of the other two sides. This is a very basic, and straightforward theorem that applies to any triangle with one right angle — that is, one angle of 90 degrees. The theorem is not very difficult to use. Suppose we have a triangle with a shortest side of 6 meters, and a second-shortest side of 8 meters. We take these numbers and square them (multiply them by themselves), giving 36 and 64 square meters. We add these together, and get 200 square meters. Now, whatever the longest side is, its square must be 100. The correct length, then, is 10 meters, since 10 squared is 100.
Using this knowledge, it's pretty easy to figure out how far apart our two astronauts are. One leg of the triangle is just the x coordinate; the other leg is the y coordinate. So, the distance between the two is given by
$$(\text{distance in meters})^2 = (x \text{ coordinate})^2 + (y \text{ coordinate})^2$$
If the coordinates measure meters, then he is 6 meters to the right and 8 meters up, so the Pythagorean Theorem tells us that he is 10 meters from the origin.
More generally, there could be a factor of proportionality to take into account, or even two — one factor for each dimension. In this case, we just have to incorporate those factors first to find the real distance in each direction, then use the Pythagorean Theorem to get the real distance from the origin. The formula in this case is just a combination of the last two formulas we've seen:
\begin{align}(\text{distance in meters})^2 = &((x \text{ coordinate}) \times (\text{meters per } x \text{ coordinate tick}))^2 + \\&((y \text{ coordinate}) \times (\text{meters per } y \text{ coordinate tick}))^2\end{align}
Basically, we've just dealt with any stretching that might happen to our grid. Of course, there's one more slight complication we'll need to deal with before we can understand Einstein's equations; some times, coordinates can be skewed, as well as stretched.
#### Skewed Grids and The Law of Cosines
We have already seen that curved spaces can make things complicated. For example, it's not always clear how to make a perfect square grid, like the one we assumed gave us our 2-D coordinates above. Some times, there's just no getting away from the fact coordinates might be skewed. Specifically, we might have a coordinate system where the y axis isn't perpendicular to the x axis. Instead, it might be tilted over at some angle.
We can look at this example with our two astronauts. Say the y axis is tilted over, as we see below. In this case, we would have to go farther to the right to be “under” the astronaut because when we go “up”, we have to go in the same direction as the skewed y axis. And, as a result, we would have to go farther “up” to get to him. To put some numbers to all this, say our axis is tilted by 15 degrees. Then, we'd actually have to go 8.14 meters to the right, and up 8.28 meters.
As before, we want to know how far apart the two astronauts — we want to know how long that dashed green line is. We'd like to use the Pythagorean Theorem, but that only works for right triangles — triangles with one angle of 90 degrees. Fortunately, there's a more general version that will work even in this case. All it takes is one extra term in the formula. We know the length of two sides of the triangle, and the angle between them. If we call that angle α, then the “Law of Cosines” is just like the Pythagorean theorem, except that we subtract an additional term.
Specifically, we take the normal Pythagorean Theorem, and subtract two times the length of one leg, times the length of the other leg, times the cosine of the angle between them. You might remember the cosine function here from high-school math class, written as “cos”. It takes the angle, and gives back another number, as shown below:
When α is 90 degrees, the cosine is 0, so the extra term in the Law of Cosines drops away, and it reduces to the same thing as the Pythagorean Theorem. When α is less than 90 degrees, we see that the extra term makes the side with length C smaller, which we would expect. Similarly, if α is greater than 90 degrees, the extra term makes C larger.
Now, we can apply this formula for finding the distance between the two astronauts when the coordinates are skewed by the angle α:
$$(\text{distance})^2 = (x \text{ coordinate})^2 + (y \text{ coordinate})^2 - 2 \times (x \text{ coordinate}) \times (y \text{ coordinate}) \times cos(\alpha)$$
In this case, we've already measured the two sides of the triangle to be 8.14 and 8.28 meters, and we know that the angle between them is 90 - 15 = 75 degrees. If we plug these numbers into the formula, we find that the distance squared is 100 — so the distance is 10 meters, exactly as before.
This is an important point to make: the coordinates we used to measure everything changed, but the physical situation remained the same. In particular, the distance between the two astronauts didn't change just because we skewed the coordinate grid. We could also change the coordinates in other way — by moving the origin, say, or rotating the coordinates, or any combination. But still, the physical situation (distances, for example) wouldn't change.
#### The Metric
When laying down a grid for coordinates, we could also combine the stretch with the skew. In general, then, we would need a formula relating distance to coordinates like
\begin{align}(\text{distance in meters})^2 = &[(x \text{ coordinate}) \times (\text{meters per } x \text{ coordinate tick})]^2 \\&+ [(y \text{ coordinate}) \times (\text{meters per } y \text{ coordinate tick})]^2 \\&- 2 \times [(x \text{ coordinate}) \times (\text{meters per } x \text{ coordinate tick})] \\&\times [(y \text{ coordinate}) \times (\text{meters per } y \text{ coordinate tick})] \times cos(\alpha)\end{align}
Obviously, this is starting to get complicated — and tiring to write out. We can save time and effort by grouping some of the terms together and writing this same formula as
$$(\text{distance in meters})^2 = g_{xx} \times (x \text{ coordinate})^2 + g_{yy} \times (y \text{ coordinate})^2 + 2 \times g_{xy} \times (x \text{ coordinate}) \times (y \text{ coordinate})$$
That is, we group those big terms together, giving them new names. Specifically, we define
$$g_{xx} = (\text{meters per }x \text{ coordinate tick})^2$$
$$g_{yy} = (\text{meters per }y \text{ coordinate tick})^2$$
$$g_{xy} = - (\text{meters per }x \text{ coordinate tick}) \times (\text{meters per }y \text{ coordinate tick}) \times cos(\alpha)$$
These three numbers we've defined — $$g_{xx}$$ , $$g_{yy}$$ , and $$g_{xy}$$ — are very important in physics. Together, they form the metric, which relates physical distances to whatever coordinates we decide to use.
In general, the metric is just a little more complicated than the one we have shown here. First, we could be dealing with more than two dimensions. In three dimensions, we would add a z coordinate, and we would need $$g_{zz}$$ , $$g_{xz}$$ , and $$g_{yz}$$ for the metric. We could even be working with time by adding a t coordinate. With all four dimensions, the metric involves the numbers
$$g_{xx}$$ , $$g_{xy}$$ , $$g_{xz}$$ , $$g_{xt}$$ , $$g_{yy}$$ , $$g_{yz}$$ , $$g_{yt}$$ , $$g_{zz}$$ , $$g_{zt}$$ , and $$g_{tt}$$ .
More importantly, the metric could change from place to place. If our coordinates were warped, we might have a grid that looks like our straight grid in one place, but is bent over like the second grid in another place.
It is possible to draw warped grids on a flat piece of paper (or on a flat screen, like we showed above). In this sense, we would get a warped metric in a space that is actually flat. On the other hand, it is impossible to draw a nice, straight grid in a space that is really curved. By very carefully examining exactly how the metric changes from point to point, we can tell if we've just drawn curvy coordinates in a flat space, or if we've drawn our coordinates in a truly curvy space.
Now we are getting very close to Einstein's Equations. Einstein had these warped pieces of spacetime that he needed to describe in some quantifiable way. He saw that a careful examination of the metric — and how it changes from point to point — could describe the true geometry of any spacetime, whether curved or flat, so he used it for his theory. Einstein combined certain numbers describing the metric's changes from place to place into what is now called the Einstein tensor. Just like the metric, the Einstein tensor is a set of numbers. For four-dimensional spacetime, we have
$$G_{xx}$$ , $$G_{xy}$$ , $$G_{xz}$$ , $$G_{xt}$$ , $$G_{yy}$$ , $$G_{yz}$$ , $$G_{yt}$$ , $$G_{zz}$$ , $$G_{zt}$$ , and $$G_{tt}$$ .
These numbers describe what is physically interesting about the geometry of spacetime. Understanding the geometry of spacetime allows us to see how particles will move, bringing us one step closer to the ultimate goal of physics. There is just one more ingredient left.
#### Energy, Matter, and the Curvature of Spacetime
We've gotten a sneak preview of Einstein's equations before: $$G = 8πT$$. The $$G$$ on the left stands for the different numbers in the Einstein tensor. But, the Einstein tensor represents the geometry of spacetime, so this is what the left side really represents. We also know that the curvature of spacetime is caused by matter, so the $$T$$ on the right must represent matter.
Just like $$G$$, the symbol $$T$$ stands for a set of numbers:
$$T_{xx}$$ , $$T_{xy}$$ , $$T_{xz}$$ , $$T_{xt}$$ , $$T_{yy}$$ , $$T_{yz}$$ , $$T_{yt}$$ , $$T_{zz}$$ , $$T_{zt}$$ , and $$T_{tt}$$ .
These numbers measure different things about matter. Together, they make up the Stress-Energy Tensor. Each component of this tensor has a slightly different physical interpretation:
Pieces of the Stress-Energy Tensor
$$T_{tt}$$ Measures how much mass there is at a point — how much density
$$T_{xt}$$ , $$T_{yt}$$ and $$T_{zt}$$ Measures how fast the matter is moving — its momentum
$$T_{xx}$$ , $$T_{yy}$$ and $$T_{zz}$$ Measures the pressure in each of the three directions
$$T_{xy}$$ , $$T_{xz}$$ and $$T_{yz}$$ Measures the stresses in the matter
As we see from the table, things like stress, pressure, and momentum come into Einstein's equations. That is, stress, pressure, and momentum all have some effect on the warping of spacetime. This is related to Einstein's most famous equation, $$E=mc^2$$, which says that energy has mass.
Warped spacetime affects how matter moves by changing its geodesics. On the other hand, Einstein's equations show us how matter — and its movement and pressures — affect the shape of spacetime. Thus, Einstein solved the fundamental problem in Physics — in principle. Of course, solving something in principle is very different from solving in practice. Finding real solutions has proven to be very difficult. Often, it is a job best left to computers.
## Inspiration
Go, wond'rous creature!
Mount where Science guides,
Go, measure earth, weigh air,
and state the tides;
Instruct the planets in
what orbs to run,
Correct old Time,
and regulate the Sun.
Alexander Pope's
An Essay on Man | 3,622 | 14,870 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2022-21 | longest | en | 0.962672 |
https://www.careerride.com/mchoice/problem-on-ages-quantitative-aptitude-mcq-questions-28230.aspx | 1,680,258,197,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949598.87/warc/CC-MAIN-20230331082653-20230331112653-00059.warc.gz | 757,172,815 | 5,636 | # Problem on Ages - Quantitative Aptitude (MCQ) questions for Q. 28230
Q. A father is 4 times as old as his son. 8 years hence, the ratio of father’s age to the son’s age will be 20:7. What is the sum of their present ages?
- Published on 10 Mar 17
a. 50
b. 72
c. 68
d. 65 | 97 | 275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-14 | latest | en | 0.942294 |
https://ask.sagemath.org/question/28761/how-to-plot-a-parabola-conic-y24ax/ | 1,721,692,679,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517927.60/warc/CC-MAIN-20240722220957-20240723010957-00898.warc.gz | 89,187,013 | 13,564 | # How to plot a parabola (conic): y²=4ax
I'm learning about conics, specifically parabolas, and I'd like to plot them. I already know how to do plot(x^2)), but what about y²=4ax ? I understand that representation is a relation, not a function, so that probably eliminates plot(f(x)). But, I'm stuck on how I might plot that so-called "general form" of a parabola.
Thanks!
edit retag close merge delete
Sort by » oldest newest most voted
You could try plotting it as two functions but implicit_plot is the natural choice. The documentation is here. This simple code illustrates how it could be done, given a specific value of a:
x, y = var('x,y')
a=1.1
implicit_plot(y^2==4*a*x, (x,-2,10), (y,-8,8))
Note the double equal signs. You can run the code in any Sage Cell Server, such as here, to check the result. Plotting it as two functions:
a=1.1
A = plot(2*sqrt(a*x), (x, -2, 10))
B = plot(-2*sqrt(a*x), (x, -2, 10))
(A+B).show()
and with a little complaining Sage plots over values that avoids problems with the domain.
more | 295 | 1,036 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-30 | latest | en | 0.900685 |
http://www.takegmat.com/index.php/gmat-question-of-the-day-data-sufficiency-73/ | 1,369,313,534,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703317384/warc/CC-MAIN-20130516112157-00004-ip-10-60-113-184.ec2.internal.warc.gz | 739,844,152 | 15,207 | # GMAT Question of the Day: Data Sufficiency
Is the area of a rectangle with width of x and length of y an integer?
(1) x is an integer.
(2) y/x is an integer.
A if statement (1) BY ITSELF is sufficient to answer the question but statement (2) is not.
B if statement (2) BY ITSELF is sufficient to answer the question but statement (1) is not.
C if both statements (1) and (2) when taken TOGETHER are sufficient to answer the question but NEITHER statement taken alone is sufficient.
D if EITHER statement taken ALONE is sufficient to answer the question.
E if the two statements, even when taken together, are NOT sufficient to answer the question.
### Article by Take GMAT Team
Authors bio is coming up shortly.
Take GMAT Team tagged this post with: , , Read 1271 articles by
1. syed says:
c
2. Karan says:
c
3. shakir says:
C
4. piyush jain says:
#3.
5. AP says:
c
6. Mohammed Abdul Khaliq says:
1. x is integer , y can be a fraction or an integer => not sufficient
2. y/x is integer 1.2/0.6 or 12/2 both are integers but area is not an integer in both cases.
both together , if x is an integer then y has to be an integer for the result to be an integer.
so both integers => area is integer
both together are sufficient | 328 | 1,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2013-20 | latest | en | 0.890716 |
http://www.puzzles-world.com/2014/07/this-is-for-for-real-aficionado-of.html | 1,518,999,454,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812293.35/warc/CC-MAIN-20180218232618-20180219012618-00455.warc.gz | 541,136,719 | 18,169 | # This is for for a real aficionado of puzzles
This is for for a real aficionado of puzzles:
9999=4
8888=8
1816=3
1212=0
1919=?
It is for masterminds! | 55 | 151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-09 | latest | en | 0.633377 |
http://gmatclub.com/forum/two-different-groups-of-test-takers-received-scores-on-the-24842.html | 1,419,052,161,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802769392.53/warc/CC-MAIN-20141217075249-00105-ip-10-231-17-201.ec2.internal.warc.gz | 122,438,657 | 40,913 | Find all School-related info fast with the new School-Specific MBA Forum
It is currently 19 Dec 2014, 21:09
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Two different groups of test-takers received scores on the
Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Manager
Joined: 13 Dec 2005
Posts: 64
Followers: 1
Kudos [?]: 3 [0], given: 0
Two different groups of test-takers received scores on the [#permalink] 22 Dec 2005, 20:44
Two different groups of test-takers received scores on the GXYZ standardized test. Group A's scores had a normal distribution with a mean of 460 and a standard deviation of 20. Group B's scores had a normal distribution with a mean of 520 and a standard deviation of 40. If each group has the same number of test-takers, what fraction of the test-takers who scored below 440 belonged to Group B?
A) 1/9
B) 1/8
C) 1/6
D) 4/17
E) 4/21
SVP
Joined: 05 Apr 2005
Posts: 1733
Followers: 4
Kudos [?]: 34 [0], given: 0
Re: Std Deviation [#permalink] 22 Dec 2005, 21:10
none...................
less than 440, group a = 68%/2 = 34%
less than 440, group b = 5%/2 = 2.5%
req % = 2.5%/36.5% = 1/14.6
Manager
Joined: 13 Dec 2005
Posts: 64
Followers: 1
Kudos [?]: 3 [0], given: 0
Std Deviation [#permalink] 22 Dec 2005, 21:13
Thanks for your response...however I have no idea how you calculated this. Further...I'm not even really sure what a standard deviation is other than knowing it has something to do with the distance from the mean. From reading the princeton review...I thought we wouldn't ever deal with these types of problems unless it was "how many std deviations away from this number is that number".
VP
Joined: 06 Jun 2004
Posts: 1061
Location: CA
Followers: 2
Kudos [?]: 36 [0], given: 0
[#permalink] 22 Dec 2005, 23:04
Two different groups of test-takers received scores on the GXYZ standardized test. Group A's scores had a normal distribution with a mean of 460 and a standard deviation of 20. Group B's scores had a normal distribution with a mean of 520 and a standard deviation of 40. If each group has the same number of test-takers, what fraction of the test-takers who scored below 440 belonged to Group B?
Answer is A
Assuming there are 100 people who took test A and 100 people who took test B.
Group A: 440 is 1 SD away from the mean, therefore, approx 16 ppl scored below 440 (Between 1 SD and the mean, percentage is roughly 34%)
Group B: 440 is 2 SDs away from the mean, therefore, approx 2 ppl scored below 440.
2/(16+2) = 2/18 ==> 1/9
Note: For a normal bell-curve distribution, between the mean and 1 SD, the percentage is approx 34%. Between 1 SD and 2 SD, approx 13.6%. Between 2 SD and on....approx 2%
_________________
Don't be afraid to take a flying leap of faith.. If you risk nothing, than you gain nothing...
Last edited by TeHCM on 23 Dec 2005, 22:20, edited 1 time in total.
Manager
Joined: 13 Dec 2005
Posts: 64
Followers: 1
Kudos [?]: 3 [0], given: 0
Std Deviation [#permalink] 23 Dec 2005, 05:40
Quote:
Note: Between the mean and 1 SD, the percentage is approx 34%. Between 1 SD and 2 SD, approx 13.6%. Between 2 SD and on....approx 2%
Does this always hold true no matter what the standard deviation?
If the mean is 750 and the standard devation is 500, will approximately 34% of the people have scores between 250 and 750?
SVP
Joined: 05 Apr 2005
Posts: 1733
Followers: 4
Kudos [?]: 34 [0], given: 0
[#permalink] 23 Dec 2005, 07:31
TeHCM wrote:
Assuming there are 100 people who took test A and 100 people who took test B.
Group A: 440 is 1 SD away from the mean, therefore, approx 16 ppl scored below 440 (Between 1 SD and the mean, percentage is roughly 34%)
Group B: 440 is 2 SDs away from the mean, therefore, approx 2 ppl scored below 440.
2/(16+2) = 2/18 ==> 1/9
Note: Between the mean and 1 SD, the percentage is approx 34%. Between 1 SD and 2 SD, approx 13.6%. Between 2 SD and on....approx 2%
goodjob...............
i was lost here. i should have take 1-.68 = .32 or 32% instead 68/2=34%. i was looking for this but.........
anyway,
less than 440, group a = 32%/2 = 16%
less than 440, group b = 5%/2 = 2.5%
req % = 2.5%/18.5 = 1/9
VP
Joined: 06 Jun 2004
Posts: 1061
Location: CA
Followers: 2
Kudos [?]: 36 [0], given: 0
Re: Std Deviation [#permalink] 23 Dec 2005, 11:44
ellisje22 wrote:
Quote:
Note: Between the mean and 1 SD, the percentage is approx 34%. Between 1 SD and 2 SD, approx 13.6%. Between 2 SD and on....approx 2%
Does this always hold true no matter what the standard deviation?
If the mean is 750 and the standard devation is 500, will approximately 34% of the people have scores between 250 and 750?
Yes this always holds true for a normal distribution.
In this case, approx 34% scored between 750 - 1,250 and 250 - 750 or 68% who scored between 250 - 1250.
_________________
Don't be afraid to take a flying leap of faith.. If you risk nothing, than you gain nothing...
Last edited by TeHCM on 23 Dec 2005, 22:58, edited 1 time in total.
Manager
Joined: 24 Oct 2005
Posts: 53
Followers: 1
Kudos [?]: 4 [0], given: 0
[#permalink] 23 Dec 2005, 17:55
Thanks TechCM, for the lucid explanation.
Manager
Joined: 12 Nov 2005
Posts: 78
Followers: 1
Kudos [?]: 5 [0], given: 0
[#permalink] 23 Dec 2005, 21:24
Thanks folks.....
I learnt a new concept in Standard Deviation today...
However I have few related questions...What are the other distributions (apart from Normal D mentioned in the question)........Does the same funda of 34%, 13.6% hold good for them as well...
Thanks in advance.
VP
Joined: 06 Jun 2004
Posts: 1061
Location: CA
Followers: 2
Kudos [?]: 36 [0], given: 0
[#permalink] 23 Dec 2005, 22:30
Kishore wrote:
Thanks folks.....
I learnt a new concept in Standard Deviation today...
However I have few related questions...What are the other distributions (apart from Normal D mentioned in the question)........Does the same funda of 34%, 13.6% hold good for them as well...
Thanks in advance.
I've edited my previous post. The 68%-98% rule only applies to a normal distribution. I think its safe to say that only normal distributions will be tested on the GMAT.
Other distributions include bimodal, multimodal, J-curve...etc. And the 68%-98% rule does not apply.
_________________
Don't be afraid to take a flying leap of faith.. If you risk nothing, than you gain nothing...
Senior Manager
Joined: 11 Nov 2005
Posts: 332
Location: London
Followers: 1
Kudos [?]: 8 [0], given: 0
[#permalink] 24 Dec 2005, 15:56
Thanx TeHCM.
It was a good tip for me.
TeHCM wrote:
Two different groups of test-takers received scores on the GXYZ standardized test. Group A's scores had a normal distribution with a mean of 460 and a standard deviation of 20. Group B's scores had a normal distribution with a mean of 520 and a standard deviation of 40. If each group has the same number of test-takers, what fraction of the test-takers who scored below 440 belonged to Group B?
Answer is A
Assuming there are 100 people who took test A and 100 people who took test B.
Group A: 440 is 1 SD away from the mean, therefore, approx 16 ppl scored below 440 (Between 1 SD and the mean, percentage is roughly 34%)
Group B: 440 is 2 SDs away from the mean, therefore, approx 2 ppl scored below 440.
2/(16+2) = 2/18 ==> 1/9
Note: For a normal bell-curve distribution, between the mean and 1 SD, the percentage is approx 34%. Between 1 SD and 2 SD, approx 13.6%. Between 2 SD and on....approx 2%
[#permalink] 24 Dec 2005, 15:56
Similar topics Replies Last post
Similar
Topics:
Two different groups of test-takers received scores on the 3 08 Dec 2007, 08:32
Two different groups of test-takers received scores on the 6 09 Jun 2007, 15:26
Two different groups of test-takers received scores on the 2 09 Nov 2006, 20:58
Two different groups of test-takers received scores on the 1 01 Nov 2006, 16:41
Two different groups of test-takers received scores on the 4 28 Nov 2005, 08:52
Display posts from previous: Sort by
# Two different groups of test-takers received scores on the
Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2,623 | 8,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2014-52 | latest | en | 0.905932 |
https://communities.bentley.com/products/programming/civil_programming/f/civil_programming_forum/218501/help-with-inroads-cross-section-evaluation-report | 1,652,690,166,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510097.3/warc/CC-MAIN-20220516073101-20220516103101-00340.warc.gz | 231,248,197 | 31,882 | # Help with InRoads Cross Section Evaluation Report
Good Morning
InRoads SS2 Version 8.11.07.630
I’m trying to understand why my custom InRoads XSL report is not generating the result for the Cross Slope value when compared to what I come up with by hand calculation. Can someone take a look at this please and see what I’m missing… please?
I am trying to create a report that publishes the elevation of the left edge-of-travelway, centerline, and the right edge-of-travelway, also I want it to publish the cross slope of the left travelway and the right travelway and these values would be provided per cross section interval.
I have been able to customize the Evaluation report “Cross Section Points” to return the ‘cross-slope’ percentage by adding a JavaScript function, but this math function that I’ve created always produces an result that is consistently off by 0.06%, is there a rounding issue with JavaScript “toFixed” method? I have no idea why this function I created is producing a result that is consistently off by this amount. Is there anyone out there with time to check this out?
This is my custom JavaScript function, (which is at the end of my custom stylesheet)
```...
</xsl:template>
<msxsl:script implements-prefix="inr" language="JScript">
<![CDATA[
// This function derives the slope between the CL point
// and the edge of travelway point both of which are at
// existing ground elevation.
function DeterimeSlope(clelev, etwelev, offset)
{
slope = 0;
elevdiff = Math.abs(clelev - etwelev);
// var slope = (Math.abs(elevdiff / offset) * 100).toFixed(2);
slope = (Math.abs(elevdiff / offset) * 100);
return slope + "%";
}
]]></msxsl:script>
</xsl:stylesheet>```
The raw.xml file that has the cross section information in it is “RPT1A0D.xml”, and my custom stylesheet is “MDOT_CrossSectionPoints_WTR_working.xsl”.
This is what my custom report output looks like:
3733.RPT1A0D.xml
```<?xml version="1.0" encoding="iso-8859-1"?>
<InRoads productName="Bentley InRoads Suite V8i (SELECTseries 2)" productVersion="08.11.07.630" outputGridScaleFactor="1.000000" inputGridScaleFactor="1.000000" linearUnits="Imperial" angularUnits="Degrees" commandName="Cross Section Report">
<CrossSectionSet setID="3" setName="Ex-XSlope-Test" alignmentOID="{D97B6FD8-920B-40F1-8F03-6019AEFF8978}" alignmentName="Ex-XSlope-Test">
<CrossSectionStations>
<CrossSectionStation leftOffset="-15.000000" rightOffset="15.000000" tangentialDirection="0.848102" radialDirection="2.418899" northing="735891.301345" easting="1692710.841221" elevation="615.997065" longitudinalGrade="-0.003648">
<Station internalStation="0.000050" externalStationName="" externalStation="0.000050"/>
<CrossSectionSurfaces>
<CrossSectionSurface name="LIDAR existing conditions x-sections only" type="0">
<CrossSectionPoints>
<CrossSectionPoint type="CrossSectionPoint" northing="735900.301651" easting="1692702.904343" offset="-11.999981" elevation="614.927381" flag="Begin"/>
<CrossSectionPoint type="ExistingCenterline" northing="735891.301331" easting="1692710.841234" offset="0.000019" elevation="615.997065"/>
<CrossSectionPoint type="CrossSectionPoint" northing="735882.301012" easting="1692718.778125" offset="12.000019" elevation="615.343766" flag="End"/>
</CrossSectionPoints>
</CrossSectionSurface>
</CrossSectionSurfaces>
</CrossSectionStation>
<CrossSectionStation leftOffset="-15.000000" rightOffset="15.000000" tangentialDirection="0.848102" radialDirection="2.418899" northing="735897.915389" easting="1692718.341450" elevation="615.947748" longitudinalGrade="-0.006215">
<Station internalStation="10.000000" externalStationName="" externalStation="10.000000"/>
<CrossSectionSurfaces>
<CrossSectionSurface name="LIDAR existing conditions x-sections only" type="0">
<CrossSectionPoints>
<CrossSectionPoint type="CrossSectionPoint" northing="735906.915708" easting="1692710.404559" offset="-12.000000" elevation="614.804465" flag="Begin"/>
<CrossSectionPoint type="ExistingCenterline" northing="735897.915389" easting="1692718.341450" offset="-0.000000" elevation="615.947748"/>
<CrossSectionPoint type="CrossSectionPoint" northing="735888.915069" easting="1692726.278342" offset="12.000000" elevation="615.303967" flag="End"/>
</CrossSectionPoints>
</CrossSectionSurface>
</CrossSectionSurfaces>
</CrossSectionStation>
<CrossSectionStation leftOffset="-15.000000" rightOffset="15.000000" tangentialDirection="0.848102" radialDirection="2.418899" northing="735904.529465" easting="1692725.841717" elevation="615.924367" longitudinalGrade="-0.003074">
<Station internalStation="20.000000" externalStationName="" externalStation="20.000000"/>
<CrossSectionSurfaces>
<CrossSectionSurface name="LIDAR existing conditions x-sections only" type="0">
<CrossSectionPoints>
<CrossSectionPoint type="CrossSectionPoint" northing="735913.529784" easting="1692717.904825" offset="-12.000000" elevation="614.718334" flag="Begin"/>
<CrossSectionPoint type="ExistingCenterline" northing="735904.529465" easting="1692725.841717" offset="-0.000000" elevation="615.924367"/>
<CrossSectionPoint type="CrossSectionPoint" northing="735895.529145" easting="1692733.778608" offset="12.000000" elevation="615.238414" flag="End"/>
</CrossSectionPoints>
</CrossSectionSurface>
</CrossSectionSurfaces>
</CrossSectionStation>
<CrossSectionStation leftOffset="-15.000000" rightOffset="15.000000" tangentialDirection="0.848102" radialDirection="2.418899" northing="735911.143541" easting="1692733.341983" elevation="615.876978" longitudinalGrade="-0.004556">
<Station internalStation="30.000000" externalStationName="" externalStation="30.000000"/>
<CrossSectionSurfaces>
<CrossSectionSurface name="LIDAR existing conditions x-sections only" type="0">
<CrossSectionPoints>
<CrossSectionPoint type="CrossSectionPoint" northing="735920.143861" easting="1692725.405092" offset="-12.000000" elevation="614.723170" flag="Begin"/>
<CrossSectionPoint type="ExistingCenterline" northing="735911.143541" easting="1692733.341983" offset="-0.000000" elevation="615.876978"/>
<CrossSectionPoint type="CrossSectionPoint" northing="735902.143221" easting="1692741.278875" offset="12.000000" elevation="615.211808" flag="End"/>
</CrossSectionPoints>
</CrossSectionSurface>
</CrossSectionSurfaces>
</CrossSectionStation>
<CrossSectionStation leftOffset="-15.000000" rightOffset="15.000000" tangentialDirection="0.848102" radialDirection="2.418899" northing="735917.757617" easting="1692740.842250" elevation="615.852620" longitudinalGrade="-0.002954">
<Station internalStation="40.000000" externalStationName="" externalStation="40.000000"/>
<CrossSectionSurfaces>
<CrossSectionSurface name="LIDAR existing conditions x-sections only" type="0">
<CrossSectionPoints>
<CrossSectionPoint type="CrossSectionPoint" northing="735926.757937" easting="1692732.905358" offset="-12.000000" elevation="614.767253" flag="Begin"/>
<CrossSectionPoint type="ExistingCenterline" northing="735917.757617" easting="1692740.842250" offset="0.000000" elevation="615.852620"/>
<CrossSectionPoint type="CrossSectionPoint" northing="735908.757297" easting="1692748.779141" offset="12.000000" elevation="615.191458" flag="End"/>
</CrossSectionPoints>
</CrossSectionSurface>
</CrossSectionSurfaces>
</CrossSectionStation>
<CrossSectionStation leftOffset="-15.000000" rightOffset="15.000000" tangentialDirection="0.848102" radialDirection="2.418899" northing="735924.371693" easting="1692748.342516" elevation="615.782411" longitudinalGrade="-0.007228">
<Station internalStation="50.000000" externalStationName="" externalStation="50.000000"/>
<CrossSectionSurfaces>
<CrossSectionSurface name="LIDAR existing conditions x-sections only" type="0">
<CrossSectionPoints>
<CrossSectionPoint type="CrossSectionPoint" northing="735933.372013" easting="1692740.405625" offset="-12.000000" elevation="614.832509" flag="Begin"/>
<CrossSectionPoint type="ExistingCenterline" northing="735924.371693" easting="1692748.342516" offset="0.000000" elevation="615.782411"/>
<CrossSectionPoint type="CrossSectionPoint" northing="735915.371373" easting="1692756.279408" offset="12.000000" elevation="615.134286" flag="End"/>
</CrossSectionPoints>
</CrossSectionSurface>
</CrossSectionSurfaces>
</CrossSectionStation>
<CrossSectionStation leftOffset="-15.000000" rightOffset="15.000000" tangentialDirection="0.848102" radialDirection="2.418899" northing="735930.985769" easting="1692755.842783" elevation="615.764136" longitudinalGrade="-0.002428">
<Station internalStation="60.000000" externalStationName="" externalStation="60.000000"/>
<CrossSectionSurfaces>
<CrossSectionSurface name="LIDAR existing conditions x-sections only" type="0">
<CrossSectionPoints>
<CrossSectionPoint type="CrossSectionPoint" northing="735939.986089" easting="1692747.905891" offset="-12.000000" elevation="614.887480" flag="Begin"/>
<CrossSectionPoint type="ExistingCenterline" northing="735930.985769" easting="1692755.842783" offset="0.000000" elevation="615.764136"/>
<CrossSectionPoint type="CrossSectionPoint" northing="735921.985449" easting="1692763.779674" offset="12.000000" elevation="615.090532" flag="End"/>
</CrossSectionPoints>
</CrossSectionSurface>
</CrossSectionSurfaces>
</CrossSectionStation>
<CrossSectionStation leftOffset="-15.000000" rightOffset="15.000000" tangentialDirection="0.848102" radialDirection="2.418899" northing="735937.599845" easting="1692763.343049" elevation="615.721052" longitudinalGrade="-0.005054">
<Station internalStation="70.000000" externalStationName="" externalStation="70.000000"/>
<CrossSectionSurfaces>
<CrossSectionSurface name="LIDAR existing conditions x-sections only" type="0">
<CrossSectionPoints>
<CrossSectionPoint type="CrossSectionPoint" northing="735946.600165" easting="1692755.406158" offset="-12.000000" elevation="614.867040" flag="Begin"/>
<CrossSectionPoint type="ExistingCenterline" northing="735937.599845" easting="1692763.343049" offset="0.000000" elevation="615.721052"/>
<CrossSectionPoint type="CrossSectionPoint" northing="735928.599525" easting="1692771.279941" offset="12.000000" elevation="615.035679" flag="End"/>
</CrossSectionPoints>
</CrossSectionSurface>
</CrossSectionSurfaces>
</CrossSectionStation>
<CrossSectionStation leftOffset="-15.000000" rightOffset="15.000000" tangentialDirection="0.848102" radialDirection="2.418899" northing="735944.213921" easting="1692770.843316" elevation="615.711238" longitudinalGrade="0.000373">
<Station internalStation="80.000000" externalStationName="" externalStation="80.000000"/>
<CrossSectionSurfaces>
<CrossSectionSurface name="LIDAR existing conditions x-sections only" type="0">
<CrossSectionPoints>
<CrossSectionPoint type="CrossSectionPoint" northing="735953.214241" easting="1692762.906424" offset="-12.000000" elevation="614.853820" flag="Begin"/>
<CrossSectionPoint type="ExistingCenterline" northing="735944.213921" easting="1692770.843316" offset="0.000000" elevation="615.711238"/>
<CrossSectionPoint type="CrossSectionPoint" northing="735935.213602" easting="1692778.780207" offset="12.000000" elevation="615.116099" flag="End"/>
</CrossSectionPoints>
</CrossSectionSurface>
</CrossSectionSurfaces>
</CrossSectionStation>
<CrossSectionStation leftOffset="-15.000000" rightOffset="15.000000" tangentialDirection="0.848102" radialDirection="2.418899" northing="735950.827997" easting="1692778.343582" elevation="615.684421" longitudinalGrade="-0.002888">
<Station internalStation="90.000000" externalStationName="" externalStation="90.000000"/>
<CrossSectionSurfaces>
<CrossSectionSurface name="LIDAR existing conditions x-sections only" type="0">
<CrossSectionPoints>
<CrossSectionPoint type="CrossSectionPoint" northing="735959.828317" easting="1692770.406691" offset="-12.000000" elevation="614.915867" flag="Begin"/>
<CrossSectionPoint type="ExistingCenterline" northing="735950.827997" easting="1692778.343582" offset="0.000000" elevation="615.684421"/>
<CrossSectionPoint type="CrossSectionPoint" northing="735941.827678" easting="1692786.280473" offset="12.000000" elevation="615.071944" flag="End"/>
</CrossSectionPoints>
</CrossSectionSurface>
</CrossSectionSurfaces>
</CrossSectionStation>
<CrossSectionStation leftOffset="-15.000000" rightOffset="15.000000" tangentialDirection="0.848102" radialDirection="2.418899" northing="735957.442041" easting="1692785.843811" elevation="615.632357" longitudinalGrade="-0.006180">
<Station internalStation="99.999950" externalStationName="" externalStation="99.999950"/>
<CrossSectionSurfaces>
<CrossSectionSurface name="LIDAR existing conditions x-sections only" type="0">
<CrossSectionPoints>
<CrossSectionPoint type="CrossSectionPoint" northing="735966.442343" easting="1692777.906935" offset="-11.999977" elevation="614.938506" flag="Begin"/>
<CrossSectionPoint type="ExistingCenterline" northing="735957.442024" easting="1692785.843826" offset="0.000023" elevation="615.632356"/>
<CrossSectionPoint type="CrossSectionPoint" northing="735948.441704" easting="1692793.780717" offset="12.000023" elevation="614.966179" flag="End"/>
</CrossSectionPoints>
</CrossSectionSurface>
</CrossSectionSurfaces>
</CrossSectionStation>
</CrossSectionStations>
</CrossSectionSet>
</InRoads>
```
1050.MDOT_CrossSectionPoints_WTR_working.xsl
Any Help would be Greatly Appreciated, Thanks,
• My XML & XSLT are producing the correct result using the input values to the 6th place. It was my checks done by hand that were using values to the 2nd place was where the whole source of my distress lay.
Oooops, but I have definitely learned something by this exercise. This has been Good! | 3,842 | 13,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-21 | latest | en | 0.828366 |
https://fr.mathworks.com/matlabcentral/cody/problems/44035-determine-the-sum-of-the-squares/solutions/2188178 | 1,603,542,456,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882581.13/warc/CC-MAIN-20201024110118-20201024140118-00414.warc.gz | 338,884,230 | 16,811 | Cody
# Problem 44035. determine the sum of the squares
Solution 2188178
Submitted on 2 Apr 2020 by Jamal Nasir
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
x = 4; y = 30; assert(isequal(sum_square(x),y))
2 Pass
x = 6; y = 91; assert(isequal(sum_square(x),y))
3 Pass
x = 5; y = 55; assert(isequal(sum_square(x),y))
4 Pass
x = 15; y = 1240; assert(isequal(sum_square(x),y))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 190 | 634 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-45 | latest | en | 0.72579 |
https://www.techwhiff.com/issue/only-the-per-capita-income-can-reply-the-true-history--103512 | 1,674,947,528,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499695.59/warc/CC-MAIN-20230128220716-20230129010716-00484.warc.gz | 1,049,613,222 | 12,240 | # 'Only the per capita income can reply the true history of economic development'. Justify the statement.
###### Question:
'Only the per capita income can reply the true history of economic development'. Justify the statement.
### What is the formula of nine planets
What is the formula of nine planets...
### Multiple scale factors are given choose all the scale factors that will result in enlargement.
Multiple scale factors are given choose all the scale factors that will result in enlargement....
### Ronald is calculating the time required to fill the swimming pool at school. He found that it takes 8 minutes to fill the swimming pool with 72 gallons of water. At what rate does the swimming pool fill, in gallons per minute?
Ronald is calculating the time required to fill the swimming pool at school. He found that it takes 8 minutes to fill the swimming pool with 72 gallons of water. At what rate does the swimming pool fill, in gallons per minute?...
### What is the circumference of a circle with a diameter of 4.1 cm ? use 3.14 for pi.
what is the circumference of a circle with a diameter of 4.1 cm ? use 3.14 for pi....
### If Mrs Murphy separates her class into groups of 4 students each, 1 student is left over. If she separates her class into groups of 5 students each,2 students are left over. What is the least number of students the class could have ?
If Mrs Murphy separates her class into groups of 4 students each, 1 student is left over. If she separates her class into groups of 5 students each,2 students are left over. What is the least number of students the class could have ?...
### PROVE THAT TWO TRIANGLES ARE CONGRUENT USING THE SAS CONGRUENCE CRITERIA PLEASEEEEE I NEED ASAP DUE IN 15 MINS PLEEEEASE
PROVE THAT TWO TRIANGLES ARE CONGRUENT USING THE SAS CONGRUENCE CRITERIA PLEASEEEEE I NEED ASAP DUE IN 15 MINS PLEEEEASE...
### Round 181 to the nearest 10
round 181 to the nearest 10...
### The shelf is 4 feet wide and 2/3 feet deep. what is the area of the shelf?
the shelf is 4 feet wide and 2/3 feet deep. what is the area of the shelf?...
### Which of the following demonstrates the Commutative Property of Multiplication? a 5(2a − 3) = 10a − 15 b 10a − 15 = (2a − 3) ⋅ 5 c 5(2a − 3) = (2a − 3) ⋅ 5 d (5 ⋅ 2a) − 3 = 5(2a − 3)
Which of the following demonstrates the Commutative Property of Multiplication? a 5(2a − 3) = 10a − 15 b 10a − 15 = (2a − 3) ⋅ 5 c 5(2a − 3) = (2a − 3) ⋅ 5 d (5 ⋅ 2a) − 3 = 5(2a − 3)...
### All federal statutory laws are part of the A. United States Code B. state constitutions C. U.S. Constitution D. regulatory agencies SUBMIT
All federal statutory laws are part of the A. United States Code B. state constitutions C. U.S. Constitution D. regulatory agencies SUBMIT...
### A family of four want to go on a cruise around the Mediterranean. They ask for details about cabins on the ship. Cabin Price per person Inside £439 Outside £499 Balcony £565 Suite £869 How much more is a cabin with a balcony than an inside cabin? £
A family of four want to go on a cruise around the Mediterranean. They ask for details about cabins on the ship. Cabin Price per person Inside £439 Outside £499 Balcony £565 Suite £869 How much more is a cabin with a balcony than an inside cabin? £...
### Help plzzz anyone its easy but i dont understand
Help plzzz anyone its easy but i dont understand...
### EXPLOREActivity 4: Hi, I Am.....If you were to introduced your self-using the lessons you have learned what would yousay?
EXPLOREActivity 4: Hi, I Am.....If you were to introduced your self-using the lessons you have learned what would yousay?...
### How many beams will remain after he's built 1 floor? y= 700 - 24(x)
How many beams will remain after he's built 1 floor? y= 700 - 24(x)...
### How to use equation of continuity. Pleeeeeaaaase help
how to use equation of continuity. Pleeeeeaaaase help...
### Assume that 0<x<pi/2 and 0<y<pi/2. find the exact value of tan(x-y) if sin x=8/17 and cos y= 3/5
assume that 0<x<pi/2 and 0<y<pi/2. find the exact value of tan(x-y) if sin x=8/17 and cos y= 3/5...
### Suppose the z-score for gwen’s math act score is 1.2, what is the correct interpretation of this z-score?
suppose the z-score for gwen’s math act score is 1.2, what is the correct interpretation of this z-score?...
### The graph of which of the following will be perpendicular to the graph of 4x- 3y= -12 a. 4x+3y=12 b. y=-3/4x-8 c.y=3/4x+1 d. 3x-4y=1
the graph of which of the following will be perpendicular to the graph of 4x- 3y= -12 a. 4x+3y=12 b. y=-3/4x-8 c.y=3/4x+1 d. 3x-4y=1... | 1,310 | 4,596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2023-06 | latest | en | 0.88271 |
https://www.gamedev.net/forums/topic/678065-simulating-surface-deformations/ | 1,548,212,584,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583884996.76/warc/CC-MAIN-20190123023710-20190123045710-00636.warc.gz | 781,578,874 | 27,071 | # Simulating surface deformations
This topic is 1006 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hi!
I am willing to do some sort of surface deformation approximations in Blender Game Engine, but I am not sure how. I don't know what maths are there involved for vertex movement depending on collision. I'm also not sure how collision callbacks work in Bullet/BGE and how to access them. Lastly, I want a low-resolution base mesh(with rigidbody physics) to deform by itself and also send collision data to the no-collision children visual mesh and deform it(with some approximations and random values for nicer effects). Oh, and I also want to know how to simulate elasticity(and moment when it gets broken and the vertex isn't returning to it's base stage any more), stiffness and other factors of materials aswell as some examples for different materials.
Thank you!:)
##### Share on other sites
Hi!
I am willing to do some sort of surface deformation approximations in Blender Game Engine, but I am not sure how. I don't know what maths are there involved for vertex movement depending on collision
For deformable surfaces (and simulating elasticity) you need polynomial maths. I'm not sure if the game engines have them. In any case if you want some simple and basic introduction to this topic , have a look at this links
http://www.basic-mathematics.com/definition-of-a-polynomial.html
https://en.wikipedia.org/wiki/Polynomial
After looking at this if you are convinced its the way to go for you, let me know and i will post some explanations of how i've coded deformable surfaces in the past with polynomials
##### Share on other sites
The btHeightfield could be ideal for this. That is once you've worked out the deformations the shape is updated automatically from your vertices (without the need for baking). It's no good for tunnels though.
Or perhaps you could stiffen up cloth...a lot! But that would be way slower to process.
##### Share on other sites
I want it for car damage system in games. I want some sort of approximation. Like Wreckfest. It uses some basic lattice type of collision and with slight randomness moves closest vertices.
• ### What is your GameDev Story?
In 2019 we are celebrating 20 years of GameDev.net! Share your GameDev Story with us.
• 15
• 9
• 11
• 9
• 9
• ### Forum Statistics
• Total Topics
634134
• Total Posts
3015751
×
## Important Information
GameDev.net is your game development community. Create an account for your GameDev Portfolio and participate in the largest developer community in the games industry.
Sign me up! | 588 | 2,671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-04 | latest | en | 0.941905 |
https://www.physicsforums.com/threads/helps-on-understanding-different-representation-transformations.581152/ | 1,547,710,619,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658844.27/warc/CC-MAIN-20190117062012-20190117084012-00299.warc.gz | 886,437,571 | 12,193 | # Helps on understanding different representation transformations
1. Feb 25, 2012
### luxiaolei
Hi,all, I m an undergrades and I am suffering on understanding the different representation transformations, namely from schrodinger picture to interaction picture tupically, my lecturer didn't state which representation he was using and I m so confused, any helps would be great.
Shall I bring one example.
So I have an Hamiltonian for a 4 level ladder system, which consists a time independent and time dependent part(a perturbation from lasers),3 lasers which has frequency wa,wb,wc
H = H0+HI(t)
So, as far as I understood, this is not in schrodinger's picture. But confused it is in Heisenberg's or interaction pictures. So does the state vectors of above Hamiltonian, in which picture?
And if make a transformation in the way that
U= exp(iw1t)|1><1|+exp(i(w1+wa)t)|2><2|+exp(i(w1+wa+wb)t)|3><3|+exp(i(w1+wa+wb+wc)t)|4><4|
New state vectors = U*old state vectors
New Hamiltonian = U*H*U(dagger) - ihU*dU(dagger)/dt
So the new Hamiltonian becomes time independent , and state vectors becomes time dependent. So are they all now in the interaction pictures? Don't know why people do this transformation and what is the advantage?
Looks like to me this transformation gives a time independent Hamiltonian which is easier to work with, but what if I do want to keep the third laser time dependent, how should I construct this transformation?
I know, it's a lot questions, and I apologies for the lengthy newbie questions, and thanks so much in advance:)
Last edited: Feb 25, 2012 | 394 | 1,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-04 | latest | en | 0.920508 |
http://math.stackexchange.com/questions/61801/linear-function-from-c-infty-mathbb-r-to-mathbb-r-determine-it | 1,469,322,348,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823805.20/warc/CC-MAIN-20160723071023-00009-ip-10-185-27-174.ec2.internal.warc.gz | 153,213,933 | 18,074 | # Linear function from $C^\infty(\mathbb R)\to\mathbb R$, determine it
Let $Q:C^\infty(\mathbb R)\to\mathbb R$ be a linear function. Let us suppose that $Q(f)\geq 0$ for any $f\in C^\infty(\mathbb R)$ such that $f(0)=0$ and the set $\{x\in\mathbb R\; \colon\; f(x)\geq 0\}$ is a neighborhood of $0$. Prove that there exist constants $a,b,c\in\mathbb R$ such that $$Q(f)=af''(0)+bf'(0)+cf(0),\qquad \forall f\in C^\infty(\mathbb R).$$
-
The assertion is equivalent to saying that $Q$ is zero on all functions $f$ with $f(0)=f'(0)=f''(0)=0$. Assume there is such a function $f$ with $Q(f)\neq0$, and consider $g(x)=s(x)+\lambda f(x)$ with $s(x)=x^2$. We have $Q(g)=Q(s)+\lambda Q(f)$, and we can choose $\lambda$ such that $Q(g)<0$. But since $f''(0)=0$, there is a neighbourhood of $0$ where $g$ is non-negative, so $Q(g)$ should be non-negative. The contradiction shows that there is no such function, and the assertion follows.
Let $Q(f)=0$ for all $f\in C^\infty(\mathbb R)$ with $f(0)=f'(0)=f''(0)=0$, and let $a=Q(f_2)$, $b=Q(f_1)$ and $c=Q(f_0)$ with $f_2(x)=\frac12x^2$, $f_1(x)=x$ and $f_0(x)=1$. Let $g$ be any function in $C^\infty(\mathbb R)$, and consider $h=g-g''(0)f_2-g'(0)f_1-g(0)f_0$. Then $h(0)=h'(0)=h''(0)=0$, so $Q(h)=0$, so $Q(h)=Q(g)-ag''(0)-bg'(0)-cg(0)=0$, so $Q(g)=ag''(0)+bg'(0)+cg(0)$.
i mean... how can you conclude from the fact that for any function with $f(0)=f'(0)=f''(0)=0$, $Q(f)=0$ then $Q$ has exactly the form mentioned in the problem? Thanks – uforoboa Sep 5 '11 at 8:58
@user15453: Done. Somewhat less formally, if $f(0)=f'(0)=f''(0)=0$ implies $Q(f)=0$, then $Q$ is the same for all functions with the same values of $f(0)$, $f'(0)$, $f''(0)$, so $Q$ can't depend on anything but those values, and thus, since it's linear, it must be linear combination of those values. – joriki Sep 5 '11 at 9:16 | 722 | 1,839 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2016-30 | latest | en | 0.761063 |
http://gmatclub.com/forum/from-first-mock-test-to-over-157923.html?sort_by_oldest=true | 1,472,223,205,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982295854.33/warc/CC-MAIN-20160823195815-00119-ip-10-153-172-175.ec2.internal.warc.gz | 106,535,368 | 45,903 | Find all School-related info fast with the new School-Specific MBA Forum
It is currently 26 Aug 2016, 07:53
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# From first mock test to over 700?
Author Message
Intern
Joined: 11 Aug 2013
Posts: 8
Concentration: Finance
Schools: Fuqua - Class of 0
GPA: 3.24
Followers: 0
Kudos [?]: 0 [0], given: 1
From first mock test to over 700? [#permalink]
### Show Tags
14 Aug 2013, 07:33
Hi everyone,
I just did my first mock test, the Princeton Review free test and my score ended up being a lousy 470 (V29, Q27, integrated reasoning 2(!?!)). This was with no practice at all. I have 5 months and about 2 whole days a week to study for the real test.
I am non-notive speaker (yet, have had international schooling). I am pretty good at verbal stuff usually but the GMAT quant looks very different to what I have been tought. However, I am a quick learner, have good memory and have a Bachelor's and Master's degrees so I am an efficient learner.
What do you experts think, with an appropriate plan, do I have any chance of getting to the high 600s or even the low 700s?
Posted from my mobile device
Intern
Joined: 11 Aug 2013
Posts: 8
Concentration: Finance
Schools: Fuqua - Class of 0
GPA: 3.24
Followers: 0
Kudos [?]: 0 [0], given: 1
Re: From first mock test to over 700? [#permalink]
### Show Tags
14 Aug 2013, 10:21
I've been thinking about my terrible score all day, but have decided to not get too sad about it. I guess a more realistic picture can be derived from taking a mock test is after I have gone through all the material.
By the way, I had studied one thing from the GMAT books and that was primes. I guess they are pretty easy but I got all of the questions right concerning them. So I am hoping that after going through all the material I will perform similarly as I did with the primes!
Still would appreciate any comments on this, though!
Posted from my mobile device
Re: From first mock test to over 700? [#permalink] 14 Aug 2013, 10:21
Similar topics Replies Last post
Similar
Topics:
1 I got 650 in my first ever mock. How do I increase it to 700+? 5 18 Feb 2016, 08:59
first mock test - scores interpretation 3 30 Aug 2015, 14:03
Not doing well in Mock tests 1 15 Mar 2014, 06:52
5 disappointed with the mock test score 8 21 Jul 2013, 13:02
2 Need advice. Scored 680-700 in mocks but 560 in actual test. 11 16 Mar 2013, 00:01
Display posts from previous: Sort by | 806 | 2,960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2016-36 | longest | en | 0.931816 |
http://music.stackexchange.com/questions/7241/18-note-groupings-across-4-beats-in-this-yngwie-malmsteen-inspired-lick/7309 | 1,469,591,353,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825365.1/warc/CC-MAIN-20160723071025-00224-ip-10-185-27-174.ec2.internal.warc.gz | 165,037,289 | 20,689 | # 18-note groupings across 4 beats in this Yngwie Malmsteen inspired lick
I have a few questions regarding the following lick:
It runs for about 8 bars and starts at 53 seconds into the video (it is a time marked lick so will start playing automatically at the correct position).
1) What is the time signature of the backing track (I am assuming 4/4)?
2) What is the BPM of the track (Is it 80 or 160 perhaps?)
3) I can't get my head around how the beat is being subdivided for this lick. It appears that the first part of the lick (6 notes cycling 3 times) is spread across two beats which means 9 notes per beat, same for the next position of the lick, and then transferring to 6 notes per beat for the remainder of the run.
Is this correct? Doesn't seem right for some reason. Maybe the track is actually 240bpm (??) and then he is playing triplets or soemthing?
Really appreciate any help on this one. It is probably a bit easier if you own the DVD as he goes through the actual lick, but he doesn't go through the beat subdivision or how to play it acrosss the track so it will lock in with the beat.
Thanks for any assistance.
-
This turned out to be more interesting than I thought it would be; I'm going to edit the title slightly so it's less localized. – NReilingh Sep 26 '12 at 3:54
The problem with 6/4 is that it implies a metric grouping in the bass and drums of 3+3 or 6 on its own. I notice that you are defining your tempo as q.=160 when you do this, however. That part is actually correct, and you should make your metric decision based on how you hear the tempo.
The 160 bpm that you are hearing are occurring four to the bar. You can hear this with the cymbal hits in the drums. The only thing that makes this not be 4/4 is that the rhythm and drums are subdividing each big beat into three instead of two. We call this kind of subdivision complex time, and to avoid writing 4/4 with tons of triplets, we use the complex equivalent time signature of 12/8. It's perfectly legitimate to write a tempo mark of q.=160 in 12/8 time, since the dotted quarter is the value of your primary beat.
I'm going to provide two examples now of the lead, notated in either time signature choice. I find that 4/4 is FAR easier to conceptualize of what's actually happening from the perspective of the lead. 12/8 is more useful for the rhythm and drums, but is far more confusing for the lead.
-
Wow! Thanks so much for adding all of this analysis. I have only had a quick look and realised I will need to come back and go through this slowly to let it all sink in. Will make another comment against your other answer. – Fusilli Jerry Sep 27 '12 at 12:09
+1 What a great answer! – Ulf Åkerstedt Sep 28 '12 at 16:58
1) The time signature is 4/4.
2) The tempo seems to be 160.
3) I would say that in my opinion the notes are sixteenth triplets.
-
Thanks so much for the answer. I think I am still confused though. How many bars are in the ascending part of the run? I think it is 4 bars. If that is the case, then the first position of the lick takes one bar. In that bar, 18 notes are played. If this was 3/4 time that would fit perfectly, but it doesn't seem to fit. I think I am getting confused on this whole thing. – Fusilli Jerry Sep 19 '12 at 13:00
In the first run there are 8 bars of 4/4 at 160 bmp (until the final bend at 1:05). Basically a triplet is made to put ternary notes in binary bars (3 triplets fill 2 'normal' notes), I guess that's why you get that 3/4 impression. But I'm not sure about how many notes are used in a bar on that run. The main point is that Malmsteen-like licks tend to use a lot of triplets or sextolets. – Whiskas Sep 19 '12 at 13:22
Thanks Whiskas. I think I may have made a break through on this and wanted to check my understanding. I think that this tune may actually be in 6/4 time. The drum beat seems to conform to this measure, it fits with the bpm being played, and it also fits the notes being played per bar. That is, in the first bar of that lick he plays 18 notes. 18 notes fit perfectly as triplets in this time signature. You can play 3 notes per beat for 6 beats in the bar and everything fits throughout the entire run (and the rest of the tune too). It also seems to have a 6/4 feel (I think). What r your thoughts? – Fusilli Jerry Sep 20 '12 at 2:27
The drum clearly hits @ 160 bmp. Then is it 6/4 or 4/4, I guess it's just a matter of feeling, for me 6/4 isn't necessary here. I'll personnaly stick with sixteenth triplets @ 160 bmp using 4/4. If you count 3 notes per beat that means eighth triplets, which is too slow even @ 240 bmp (look at his right hand). I counted 12 notes for 2 beats (it may not be right though) and it fits well with my results. Maybe I'm entirely wrong, but that's my thoughts :) – Whiskas Sep 20 '12 at 13:37
I'd say the time signature is 4/4 shuffle (with 160 quarter note bpm:s), or 12/8 (with 160 dotted quarter note bpm:s). – Ulf Åkerstedt Sep 20 '12 at 21:16
With assistance from everyone here I think I have this worked out now.
160bpm as dotted quarter note, 6/4 time signature, 3 notes per beat (so 6 beats per bar, meaning 18 notes per bar).
Links to guitar pro files and pdf as follows:
Guitar Pro file - www.db.tt/9hne3CPS
PDF - www.db.tt/veM8hJG3
Hope this helps someone else.
-
Unfortunately, 6/4 really can't be applied to the rhythm guitar and drums. Your meter has to come from that, NOT whatever the rhythmic grouping of the melody is. You're really close, though. I'm firing up Sibelius now; will submit an answer shortly. – NReilingh Sep 26 '12 at 3:14
To my ear, it sounds like the rhythm guitar is playing for 6 beats. The first strum is for two beats, and then there are 4 palm-muted notes. Wouldn't this indicate 6/4 time? Am I hearing it wrong or understanding the notation incorrectly? Thanks so much for all your help @NReilingh – Fusilli Jerry Sep 27 '12 at 12:12
The drums alone eliminate 6/4 as a possibility since they are playing four to the bar. That leaves 4/4 and 12/8 as possibilities, for the reasons described in my answer. In 12/8, the rhythm guitar/bass is playing `q e e e e q e e e e` – NReilingh Sep 27 '12 at 13:39
There are places where the drums will play 6 even cymbal bell hits per bar (1:17-1:27), but this is an example of hemiola taking place over the underlying 4-to-the-bar meter. There are LOTS of interesting rhythmic options when you're playing in 12/8; you have to be careful not to be distracted by the surface level stuff. – NReilingh Sep 27 '12 at 13:45
Brilliant! Thanks so much for all your effort here. Realised I have a LOT to learn about time signatures e.t.c. Do you have any recommended resources @NReilingh for getting on top of all this stuff? Any books or websites that go from the basics to super advanced? Thanks again. – Fusilli Jerry Sep 28 '12 at 0:02 | 1,829 | 6,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2016-30 | latest | en | 0.96065 |
http://tensortoolbox.com/test_problems_doc.html | 1,701,856,792,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100593.71/warc/CC-MAIN-20231206095331-20231206125331-00530.warc.gz | 42,995,757 | 8,181 | # Creating Test Problems and Initial Guesses
We demonstrate how to use Tensor Toolbox create_problem and create_guess functions to create test problems for fitting algorithms.
## Contents
```rng('default'); %<- Setting random seed for reproducibility of this script
```
## Creating a CP test problem
The create_problem function allows a user to generate a test problem with a known solution having a pre-specified solution. The create_problem function generates both the solution (as a ktensor for CP) and the test data (as a tensor). We later show that a pre-specificed solution can be used as well.
```% Create a problem
info = create_problem('Size', [5 4 3], 'Num_Factors', 3, 'Noise', 0.10);
```
```% Display the solution created by create_problem
soln = info.Soln
```
```soln is a ktensor of size 5 x 4 x 3
soln.lambda =
0.6948 0.3171 0.9502
soln.U{1} =
0.5377 -1.3077 -1.3499
1.8339 -0.4336 3.0349
-2.2588 0.3426 0.7254
0.8622 3.5784 -0.0631
0.3188 2.7694 0.7147
soln.U{2} =
-0.2050 1.4172 1.6302
-0.1241 0.6715 0.4889
1.4897 -1.2075 1.0347
1.4090 0.7172 0.7269
soln.U{3} =
-0.3034 0.8884 -0.8095
0.2939 -1.1471 -2.9443
-0.7873 -1.0689 1.4384
```
```% Display the data created by create_problem
data = info.Data
```
```data is a tensor of size 5 x 4 x 3
data(:,:,1) =
0.6406 -0.0053 1.7089 0.3286
-3.9326 -1.1850 -3.1232 -1.8339
-0.9485 -0.3204 0.0406 0.0859
1.6481 0.9261 -1.8303 0.6222
0.3243 0.6169 -1.9710 -0.0077
data(:,:,2) =
7.1696 2.6513 4.2567 2.9938
-14.0474 -4.0639 -8.6171 -5.9845
-3.3801 -1.5008 -2.3947 -2.6743
-1.4145 -1.0496 1.9542 -0.3994
-4.3450 -2.0053 -0.4684 -2.1417
data(:,:,3) =
-2.3827 -0.8279 -3.2592 -1.6532
7.6351 2.4764 2.2490 1.7676
1.2927 0.5233 3.0407 2.3518
-1.6987 -0.8768 0.9062 -2.2220
0.8113 -0.0613 2.7203 -0.3508
```
```% The difference between true solution and measured data should match the
% specified 10% noise.
diff = norm(full(info.Soln) - info.Data)/norm(full(info.Soln))
```
```diff =
0.1000
```
## Creating a Tucker test problem
The create_problem function can also be used to create Tucker problems by specifying the 'Type' as 'Tucker'. In this case, the create_problem function generates both the solution (as a ttensor for Tucker) and the test data (as a tensor).
```% Create a problem
info = create_problem('Type', 'Tucker', 'Size', [5 4 3], 'Num_Factors', [3 3 2]);
```
```% Display the Tucker-type solution created by create_problem
soln = info.Soln
```
```soln is a ttensor of size 5 x 4 x 3
soln.core is a tensor of size 3 x 3 x 2
soln.core(:,:,1) =
-1.5771 0.0335 0.3502
0.5080 -1.3337 -0.2991
0.2820 1.1275 0.0229
soln.core(:,:,2) =
-0.2620 -0.8314 -0.5336
-1.7502 -0.9792 -2.0026
-0.2857 -1.1564 0.9642
soln.U{1} =
-1.7947 0.3035 -0.1941
0.8404 -0.6003 -2.1384
-0.8880 0.4900 -0.8396
0.1001 0.7394 1.3546
-0.5445 1.7119 -1.0722
soln.U{2} =
0.9610 -0.1977 1.3790
0.1240 -1.2078 -1.0582
1.4367 2.9080 -0.4686
-1.9609 0.8252 -0.2725
soln.U{3} =
1.0984 -2.0518
-0.2779 -0.3538
0.7015 -0.8236
```
```% Difference between true solution and measured data (default noise is 10%)
diff = norm(full(info.Soln) - info.Data)/norm(full(info.Soln))
```
```diff =
0.1000
```
## Recreating the same test problem
We can recreate exactly the same test problem when we use the same random seed and other parameters.
```% Set-up, including specifying random state
sz = [5 4 3]; %<- Size
nf = 2; %<- Number of components
state = RandStream.getGlobalStream.State; %<- Random state
```
```% Generate first test problem
info1 = create_problem('Size', sz, 'Num_Factors', nf, 'State', state);
```
```% Generate second identical test problem
info2 = create_problem('Size', sz, 'Num_Factors', nf, 'State', state);
```
```% Check that the solutions are identical
tf = isequal(info1.Soln, info2.Soln)
```
```tf =
logical
1
```
```% Check that the data are identical
diff = norm(info1.Data - info2.Data)
```
```diff =
0
```
## Checking default parameters and recreating the same test problem
The create_problem function returns the parameters that were used to generate it. These can be used to see the defaults. Additionally, if these are saved, they can be used to recreate the same test problems for future experiments.
```% Generate test problem and use second output argument for parameters.
[info1,params] = create_problem('Size', [5 4 3], 'Num_Factors', 2);
```
```% Here are the parameters
params
```
```params =
struct with fields:
Core_Generator: 'randn'
Factor_Generator: 'randn'
Lambda_Generator: 'rand'
M: 0
Noise: 0.1000
Num_Factors: 2
Size: [5 4 3]
Soln: []
Sparse_Generation: 0
Sparse_M: 0
State: [625×1 uint32]
Symmetric: []
Type: 'CP'
```
```% Recreate an identical test problem
info2 = create_problem(params);
```
```% Check that the solutions are identical
tf = isequal(info1.Soln, info2.Soln)
```
```tf =
logical
1
```
```% Check that the data are identical
diff = norm(info1.Data - info2.Data)
```
```diff =
0
```
## Options for creating factor matrices, core tensors, and lambdas
Any function with two arguments specifying the size can be used to generate the factor matrices. This is specified by the 'Factor_Generator' option to create_problem.
Pre-defined options for 'Factor_Generator' for creating factor matrices (for CP or Tucker) include:
• 'rand' - Uniform on [0,1]
• 'randn' - Gaussian with mean 0 and std 1
• 'orthogonal' - Generates a random orthogonal matrix. This option only works when the number of factors is less than or equal to the smallest dimension.
• 'stochastic' - Generates nonnegative factor matrices so that each column sums to one.
Pre-defined options for 'Lambda_Generator' for creating lambda vector (for CP) include:
• 'rand' - Uniform on [0,1]
• 'randn' - Gaussian with mean 0 and std 1
• 'orthogonal' - Creates a random vector with norm one.
• 'stochastic' - Creates a random nonnegative vector whose entries sum to one.
Pre-defined options for 'Core_Generator' for creating core tensors (for Tucker) include:
• 'rand' - Uniform on [0,1]
• 'randn' - Gaussian with mean 0 and std 1
```% Here is ane example of a custom factor generator
factor_generator = @(m,n) 100*rand(m,n);
info = create_problem('Size', [5 4 3], 'Num_Factors', 2, ...
'Factor_Generator', factor_generator, 'Lambda_Generator', @ones);
first_factor_matrix = info.Soln.U{1}
```
```first_factor_matrix =
34.3877 81.7761
58.4069 26.0728
10.7769 59.4356
90.6308 2.2513
87.9654 42.5259
```
```% Here is an example of a custom core generator for Tucker:
info = create_problem('Type', 'Tucker', 'Size', [5 4 3], ...
'Num_Factors', [2 2 2], 'Core_Generator', @tenones);
core = info.Soln.core
```
```core is a tensor of size 2 x 2 x 2
core(:,:,1) =
1 1
1 1
core(:,:,2) =
1 1
1 1
```
```% Here's another example for CP, this time using a function to create
% factor matrices such that the inner products of the columns are
% prespecified.
info = create_problem('Size', [5 4 3], 'Num_Factors', 3, ...
'Factor_Generator', @(m,n) matrandcong(m,n,.9));
U = info.Soln.U{1};
congruences = U'*U
```
```congruences =
1.0000 0.9000 0.9000
0.9000 1.0000 0.9000
0.9000 0.9000 1.0000
```
## Generating data from an existing solution
It's possible to skip the solution generation altogether and instead just generate appropriate test data.
```% Manually generate a test problem (or it comes from some
% previous call to |create_problem|.
soln = ktensor({rand(50,3), rand(40,3), rand(30,3)});
% Use that soln to create new test problem.
info = create_problem('Soln', soln);
% Check whether solutions is equivalent to the input
iseq = isequal(soln,info.Soln)
```
```iseq =
logical
1
```
## Creating dense missing data problems
It's possible to create problems that have a percentage of missing data. The problem generator randomly creates the pattern of missing data.
```% Specify 25% missing data as follows:
[info,params] = create_problem('Size', [5 4 3], 'M', 0.25);
```
```% Here is the pattern of known data (1 = known, 0 = unknown)
info.Pattern
```
```ans is a tensor of size 5 x 4 x 3
ans(:,:,1) =
1 1 1 0
1 1 1 1
1 1 1 1
1 1 1 0
1 1 0 0
ans(:,:,2) =
1 1 1 0
1 1 0 0
1 1 0 1
1 1 0 0
1 1 0 1
ans(:,:,3) =
1 1 1 0
1 1 1 1
1 0 1 1
1 1 0 1
1 0 1 1
```
```% Here is the data (incl. noise) with missing entries zeroed out
info.Data
```
```ans is a tensor of size 5 x 4 x 3
ans(:,:,1) =
0.0701 -0.0140 -0.0197 0
-0.0250 0.0174 0.0090 0.0055
-0.0100 -0.0118 0.0130 0.0182
-0.0267 -0.0170 0.0305 0
0.0729 -0.0061 0 0
ans(:,:,2) =
0.4143 0.0336 -0.4037 0
-0.0852 -0.0181 0 0
-0.2301 0.0085 0 -0.0270
-0.4196 0.0960 0 0
0.5840 -0.0401 0 0.0903
ans(:,:,3) =
-0.1069 0.0860 -0.0161 0
0.0445 -0.0541 0.0299 -0.0440
0.0237 0 0.0079 -0.0278
-0.1259 0.1210 0 0.0169
-0.0465 0 0.0309 -0.0240
```
## Creating sparse missing data problems.
If Sparse_M is set to true, then the data returned is sparse. Moreover, the dense versions are never explicitly created. This option only works when M >= 0.8.
```% Specify 80% missing data and sparse
info = create_problem('Size', [5 4 3], 'M', 0.80, 'Sparse_M', true);
```
```% Here is the pattern of known data
info.Pattern
```
```ans is a sparse tensor of size 5 x 4 x 3 with 12 nonzeros
(1,4,2) 1
(2,1,2) 1
(2,2,3) 1
(2,4,3) 1
(3,1,1) 1
(3,3,3) 1
(3,4,2) 1
(4,1,1) 1
(4,2,1) 1
(4,2,3) 1
(5,1,2) 1
(5,4,2) 1
```
```% Here is the data (incl. noise) with missing entries zeroed out
info.Data
```
```ans is a sparse tensor of size 5 x 4 x 3 with 12 nonzeros
(1,4,2) -0.0137
(2,1,2) -0.6286
(2,2,3) -0.2961
(2,4,3) 0.1887
(3,1,1) -0.2856
(3,3,3) -1.3309
(3,4,2) -0.1728
(4,1,1) -0.0357
(4,2,1) -0.0268
(4,2,3) -0.3739
(5,1,2) -0.3906
(5,4,2) 0.0938
```
## Create missing data problems with a pre-specified pattern
It's also possible to provide a specific pattern (dense or sparse) to be used to specify where data should be missing.
```% Create pattern
P = tenrand([5 4 3]) > 0.5;
% Create test problem with that pattern
info = create_problem('Size', size(P), 'M', P);
% Show the data
info.Data
```
```ans is a tensor of size 5 x 4 x 3
ans(:,:,1) =
0 -0.6323 0 0
0.1566 0 -0.4187 0
0.0044 0 0 0
0.0508 -0.7211 0.1713 0
0 0 0 0
ans(:,:,2) =
0 0 0 -0.0151
-0.0909 0 0.0607 0
0.0084 0 0 0
0 -0.7582 0 0
-0.0734 0.1987 0 0
ans(:,:,3) =
-0.1618 -0.3415 0.5567 0.4957
0.1608 0 -0.5744 -0.4850
-0.0797 0 0 0.1821
0 0 -0.1827 0
0 0 0 0
```
## Creating sparse problems (CP only)
If we assume each model parameter is the input to a Poisson process, then we can generate a sparse test problems. This requires that all the factor matrices and lambda be nonnegative. The default factor generator ('randn') won't work since it produces both positive and negative values.
```% Generate factor matrices with a few large entries in each column; this
% will be the basis of our soln.
sz = [20 15 10];
nf = 4;
A = cell(3,1);
for n = 1:length(sz)
A{n} = rand(sz(n), nf);
for r = 1:nf
p = randperm(sz(n));
idx = p(1:round(.2*sz(n)));
A{n}(idx,r) = 10 * A{n}(idx,r);
end
end
S = ktensor(A);
S = normalize(S,'sort',1);
```
```% Create sparse test problem based on provided solution. The
% 'Sparse_Generation' says how many insertions to make based on the
% provided solution S. The lambda vector of the solution is automatically
% rescaled to match the number of insertions.
info = create_problem('Soln', S, 'Sparse_Generation', 500);
num_nonzeros = nnz(info.Data)
total_insertions = sum(info.Data.vals)
orig_lambda_vs_rescaled = S.lambda ./ info.Soln.lambda
```
```num_nonzeros =
326
total_insertions =
500
orig_lambda_vs_rescaled =
84.4101
84.4101
84.4101
84.4101
```
## Generating an initial guess
The create_guess function creates a random initial guess as a cell array of matrices. Its behavior is very similar to create_problem. A nice option is that you can generate an initial guess that is a pertubation of the solution.
```info = create_problem;
% Create an initial guess to go with the problem that is just a 5%
% pertubation of the correct solution.
U = create_guess('Soln', info.Soln, 'Factor_Generator', 'pertubation', ...
'Pertubation', 0.05);
``` | 4,713 | 12,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-50 | latest | en | 0.627397 |
http://www.algebra.com/algebra/homework/Graphs.faq.question.244833.html | 1,386,184,091,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163037167/warc/CC-MAIN-20131204131717-00036-ip-10-33-133-15.ec2.internal.warc.gz | 261,859,815 | 5,063 | # SOLUTION: My problem is 3/2,-3 0,2/5 this involves the y2-y1 x2-x2 I'm having some problems with the fractions.
Algebra -> -> SOLUTION: My problem is 3/2,-3 0,2/5 this involves the y2-y1 x2-x2 I'm having some problems with the fractions. Log On
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Mathway solves algebra homework problems with step-by-step help!
Question 244833: My problem is 3/2,-3 0,2/5 this involves the y2-y1 x2-x2 I'm having some problems with the fractions.Answer by jim_thompson5910(29613) (Show Source): You can put this solution on YOUR website!Note: is the first point . So this means that and . Also, is the second point . So this means that and . Start with the slope formula. Plug in , , , and Subtract from to get Subtract from to get Multiply the first fraction by the reciprocal of the second fraction. Multiply. So the slope of the line that goes through the points and is | 256 | 958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2013-48 | latest | en | 0.925014 |
https://www.coursehero.com/file/6681859/Chap10-Sec4/ | 1,490,496,298,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189090.69/warc/CC-MAIN-20170322212949-00151-ip-10-233-31-227.ec2.internal.warc.gz | 876,198,956 | 407,281 | Chap10_Sec4
# Chap10_Sec4 - 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES...
This preview shows pages 1–13. Sign up to view the full content.
PARAMETRIC EQUATIONS PARAMETRIC EQUATIONS AND POLAR COORDINATES AND POLAR COORDINATES 10
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
10.4 Areas and Lengths in Polar Coordinates In this section, we will: Develop the formula for the area of a region whose boundary is given by a polar equation. PARAMETRIC EQUATIONS & POLAR COORDINATES
AREAS IN POLAR COORDINATES We need to use the formula for the area of a sector of a circle A = ½ r 2 θ where: r is the radius. θ is the radian measure of the central angle. Formula 1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
AREAS IN POLAR COORDINATES Formula 1 follows from the fact that the area of a sector is proportional to its central angle: A = ( θ/ 2 π ) πr 2 = ½ r 2 θ
AREAS IN POLAR COORDINATES Let R be the region bounded by the polar curve r = f ( θ ) and by the rays θ = a and θ = b, where: f is a positive continuous function. 0 < b – a ≤ 2 π
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
AREAS IN POLAR COORDINATES We divide the interval [ a, b ] into subintervals with endpoints θ 0 , θ 1 , θ 2 , …, θ n , and equal width ∆θ . Then, the rays θ = θ i divide R into smaller regions with central angle ∆θ = θ i θ i –1 .
AREAS IN POLAR COORDINATES If we choose θ i * in the i th subinterval [ θ i –1 , θ i ] then the area ∆A i of the i th region is the area of the sector of a circle with central angle ∆θ and radius f ( θ* ).
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
AREAS IN POLAR COORDINATES Thus, from Formula 1, we have: A i ½[ f ( θ i *)] 2 θ So, an approximation to the total area A of R is: * 2 1 2 1 [ ( )] n i i A f θ = Formula 2
AREAS IN POLAR COORDINATES It appears that the approximation in Formula 2 improves as n .
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
AREAS IN POLAR COORDINATES However, the sums in Formula 2 are Riemann sums for the function g ( θ ) = ½[ f ( θ )] 2 . So, * 2 2 1 1 2 2 1 lim [ ( )] [ ( )] n b i a n i f f d θ →∞ = ∆ =
AREAS IN POLAR COORDINATES Therefore, it appears plausible—and can, in fact, be proved—that the formula for the area A of the polar region R is: 2 1 2 [ ( )] b a A f d θ = Formula 3
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Formula 3 is often written as with the understanding that r = f ( θ ). Note the similarity between Formulas 1 and 4.
This is the end of the preview. Sign up to access the rest of the document.
## This note was uploaded on 01/06/2012 for the course MATH 2414.S01 taught by Professor Alans.grave during the Fall '11 term at Collins.
### Page1 / 48
Chap10_Sec4 - 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES...
This preview shows document pages 1 - 13. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 922 | 3,196 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2017-13 | longest | en | 0.844692 |
https://republicofsouthossetia.org/question/complete-the-derivation-of-the-formula-for-a-cylinder-whose-height-is-equal-to-its-radius-the-pr-16351129-56/ | 1,632,816,717,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060538.11/warc/CC-MAIN-20210928062408-20210928092408-00535.warc.gz | 512,369,030 | 13,825 | ## Complete the derivation of the formula for a cylinder whose height is equal to its radius. The prism’s volume is the area of the base, , tim
Question
Complete the derivation of the formula for a cylinder whose height is equal to its radius. The prism’s volume is the area of the base, , times the height, . Since the ratio of the areas is StartFraction pi Over 4 EndFraction, then the volume of the cylinder is times the volume of the prism. V = A cylinder inside of a square prism is shown. The cylinder has a height and radius with a length of r. The length of the square prism is 2 r.(4r3), or
in progress 0
2 weeks 2021-09-13T09:01:30+00:00 2 Answers 0
2r(2r), r, pi/4, pi times r cubed
Step-by-step explanation:
just took it
1. 2r (2r)
2. r
3. pi/4
4.pi times r cubed
Step-by-step explanation: | 229 | 812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-39 | latest | en | 0.880303 |
http://physics.stackexchange.com/questions/129169/definition-of-a-year | 1,469,342,864,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823963.50/warc/CC-MAIN-20160723071023-00131-ip-10-185-27-174.ec2.internal.warc.gz | 200,779,783 | 17,047 | # Definition of a year
Is there an acceptable definition of a year (in number of days)?
What is the rationale behind choosing 365.242 as the number of days in an year? Is it defined by SI?
-
A year is defined by the time earth takes to circle the sun once: http://en.wikipedia.org/wiki/Year
This does however not correspond to an even number of days. The SI unit of time is the second and $1\,\mathrm{day} = 24\cdot 3600\,\mathrm s$
Edit: Although this definition is practical and sufficient in most situations it is not precise. The problem is that earth only approximately circles the sun i.e. its orbit is not closed and it is not a circle. For a more precise definition see the following answer.
-
Which year? The sidereal year? The tropical year? The anomalistic year? The calendar year (and whose calendar)?
The sidereal year is the average amount of time it takes the Earth to make one complete orbit about the Sun with respect to the fixed stars. The tropical year is the amount of average amount of time between successive spring equinoxes. The tropical year is 20.4 minutes shorter than the sidereal year thanks to axial precession. The anomalistic year is the average amount of time between successive perihelion passages. The anomalistic year is 4.7 minutes longer than the sidereal year thanks to apsidal precession.
The tropical year, 365.24219 days, is what drives our seasons. The Julian year was an attempt to fix problems with the Roman calendar and make it more or less stay in sync with the seasons. The Romans added the concept of a leap year every four years, making the Julian year 365.25 days long. Astronomers still use the Julian year for measuring time.
The Julian calendar has a problem: It's a tiny bit too long. The Julian calendar was off by almost half a month by the 18th century, which is when many countries switched to the Gregorian calendar with its slightly more complex leap year rule. The Gregorian calendar has 97 leap days every 400 years, resulting in a year of 365.2425 days (on average). That's still a bit longer than the tropical year. It will take about 3200 years for the Gregorian calendar to be off by one day, assuming that a tropical year remains a constant length.
The Iranian calendar is a bit better than the Gregorian calendar, at least currently. The Iranian calendar has a complex formula that yields 8 leap days every 33 years, resulting in a year of 365.242424... days (on average). It will take 4300 years before the Iranian calendar is off by a day, assuming the tropical year keeps its constant length during that 4300 year interval.
Those calculations assumed a constant length tropical year. That is not the case. Perturbations by other planets make the tropical year vary. The variance works in the favor of the Gregorian calendar. It will take about 4000 years before the Gregorian calendar is off by a day, and by then the Iranian calendar will be off by a bit more than a day.
There are other years as well. Some countries use a lunar based year rather than a solar year.
-
When "year" is used as a unit for things unrelated to Earth's orbit, such as distances in light years or the age of the universe in years, it is the Julian year of exactly 365.25·86400 SI seconds.
-
## protected by Qmechanic♦Aug 2 '14 at 19:43
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count). | 795 | 3,526 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2016-30 | latest | en | 0.941707 |
https://www.my.freelancer.com/projects/excel/excel-advanced-maths-probability/ | 1,553,545,293,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204300.90/warc/CC-MAIN-20190325194225-20190325220225-00514.warc.gz | 839,330,743 | 35,909 | # Excel Advanced Maths Probability Question.
I need the below in 24 Hours MAX. The Shorter the better, as I may have another project that relates to this question also.
I would like the following Solved in Excel, preferably in a clear, concise and explained solution.
Say you have 8 Coins that aren't fair.
You have 2 Probabilities to land on head:
- Easy Coin = 14.74%
- Hard Coin = 2.92%
All Coin will flip & if any land on heads, they are kept and no longer tossed & another toss of the remaining coins will happen.
The game ends Once a toss happens that NO heads land, OR once all 8 Coins are on heads the game is also over.
In Each round regardless of how many coins remain, 1 coin will ALWAYS have the hard probability.
Whats the probabilities for reaching X amount of coins with HEADS:
8 Coins
7 Coins
6 Coins
5 Coins
4 Coins
3 Coins
2 Coins
1 Coin
0 Coin
Kemahiran: Algoritma, Pemprosesan Data, Excel, Matematik, Statistik
Tentang Majikan:
( 15 ulasan ) Landsville, Australia
ID Projek: #17166908
## 21 pekerja bebas membida secara purata \$141 untuk pekerjaan ini
schoudhary1553
Hello, I can help with you in your project Excel Advanced Maths Probability Question. I have more than 5 years of experience in Algorithm, Data Processing, Excel, Mathematics, Statistics. We have worked on several s Lagi
\$250 AUD dalam 3 hari
(315 Ulasan)
7.8
\$144 AUD dalam 3 hari
(262 Ulasan)
7.0
dgled
hI, I know this game as heads or tails. do you plan to drop all the coins together or one by one? regarding the 2 Probabilities to land on the head: Is the quantity constant or variable?
\$250 AUD dalam 3 hari
(72 Ulasan)
6.9
MohsanEijaz
hi, I have experience of excel and vba. please check my profile. hi, I have experience of excel and vba. please check my profile. hi, I have experience of excel and vba. please check m Lagi
\$80 AUD dalam 2 hari
(70 Ulasan)
6.2
dinhfreedom
-------------------------------Best result on time && Master in Statistics and Excel!------------------------------------- Hello, dear! I have read your proposal and I am very interesting in your project. I have goo Lagi
\$155 AUD dalam 3 hari
(57 Ulasan)
6.4
katilinas
Dear Employer, I'm ready to work on "Excel Advanced Maths Probability Question" project. I have 5 years experience in Excel & VBA. 200 AUD - proposed project budget, including code and clean modern design. I ca Lagi
\$200 AUD dalam 3 hari
(118 Ulasan)
6.5
suyashdhoot
Hi I am a very experienced statistician and academic writer. I have completed several PhD level thesis projects involving advanced statistical analysis of data. I have worked with data from several companies and have d Lagi
\$250 AUD dalam 3 hari
(114 Ulasan)
6.8
liangjongai
I can work for you. I'm interested in your project. I've an experience of works with ms excel and vba. Ping me any time to discuss about this project.
\$164 AUD dalam 3 hari
(36 Ulasan)
5.1
\$30 AUD dalam sehari
(38 Ulasan)
5.2
raghavajay3
would you like it solved as a proper probability distribution way in excel
\$222 AUD dalam 2 hari
(32 Ulasan)
4.6
JorgeBece
I can help you with this probability problem. If you awarded me I´d make a sheet with the problem soved and graphs to explain it. Of course it would be solved in less than 24 hours. Please, if you have any doubt, c Lagi
\$55 AUD dalam sehari
(11 Ulasan)
2.8
BEST4LYFE
I am solving the question now and when i finish I will give it to you to see if it is correct. Thanks.
\$100 AUD dalam sehari
(1 Ulasan)
2.2
swathi8118
Hi I am a software developer by profession. I have good hold on probability. I would like to discuss more about the project. Looking forward for a positive reply
\$55 AUD dalam sehari
(1 Ulasan)
1.1
pavanpunnu10
Hello, I'm good at Mathematics and Statistics. The fees might be on a higher side but quality won't be copromised. Kindly share the details. Thank you.
\$125 AUD dalam 3 hari
(1 Ulasan)
0.3
suryanshbhandari
EFFECTIVE AND QUALITY WORK AND I HAVE EXPERIENCE OF 3 YEARS IN A NEARBY LIMITED COMPANY. SO PLEASE HIRE ME
\$30 AUD dalam sehari
(0 Ulasan)
0.0
\$155 AUD dalam 3 hari
(0 Ulasan)
0.0
\$222 AUD dalam sehari
(0 Ulasan)
0.0
scottwcorney
I am very experienced in using Excel. I am self-motivated individual and I pride myself in always completing tasks to the highest possible standard! Putting in the hours is never a problem! I like working hard, however Lagi
\$177 AUD dalam sehari
(0 Ulasan)
0.0
\$30 AUD dalam sehari
(0 Ulasan)
0.0
Cyberseal75r
This is classic case to use a statistical simulation to demonstrate how this will play out. I can provide you the straight forward hard coded answer, written explanation, along with a simulation run demonstrating the Lagi
\$155 AUD dalam 3 hari
(0 Ulasan)
0.0 | 1,304 | 4,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-13 | latest | en | 0.81196 |
https://www.bartleylawoffice.com/faq/who-came-up-with-the-law-of-conservation-of-mass.html | 1,603,488,136,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107865665.7/warc/CC-MAIN-20201023204939-20201023234939-00382.warc.gz | 649,491,852 | 33,763 | Who came up with the law of conservation of mass
How was the law of conservation of mass discovered?
Lavoisier placed some mercury in a jar, sealed the jar, and recorded the total mass of the setup. … He found in all cases that the mass of the reactants is equal to the mass of the products. His conclusion, called the states that in a chemical reaction, atoms are neither created nor destroyed.
Why is it called the Law of Conservation of Mass?
One of these is called the law of conservation of mass , which states that during a chemical reaction, the total mass of the products must be equal to the total mass of the reactants. In other words, mass cannot be created or destroyed during a chemical reaction, but is always conserved.
Is the law of conservation of mass always true?
Mass is not conserved in chemical reactions. The fundamental conservation law of the universe is the conservation of mass-energy. … Mass is therefore never conserved because a little of it turns into energy (or a little energy turns into mass) in every reaction. But mass+energy is always conserved.
Julius Mayer
Can neither be created nor destroyed?
The law of conservation of energy, also known as the first law of thermodynamics, states that the energy of a closed system must remain constant—it can neither increase nor decrease without interference from outside.
Can matter be created?
The first law of thermodynamics doesn’t actually specify that matter can neither be created nor destroyed, but instead that the total amount of energy in a closed system cannot be created nor destroyed (though it can be changed from one form to another).
You might be interested: What is inverse square law
Can mass be destroyed?
The law implies that mass can neither be created nor destroyed, although it may be rearranged in space, or the entities associated with it may be changed in form. … Mass is also not generally conserved in open systems. Such is the case when various forms of energy and matter are allowed into, or out of, the system.
Which best describes the Law of Conservation of Mass?
Which best describes the law of conservation of mass? The mass of the reactants and products is equal and is not dependent on the physical state of the substances. The equation below shows a general equation for a reaction, and the amounts of the substance are written underneath.
How is the law of conservation of mass used in everyday life?
The law of conservation of mass states that matter cannot be created or destroyed in a chemical reaction. For example, when wood burns, the mass of the soot, ashes, and gases, equals the original mass of the charcoal and the oxygen when it first reacted.
How do you solve the Law of Conservation of Mass?
The law of conservation of mass states that matter cannot be created or destroyed in a chemical reaction. For example, when wood burns, the mass of the soot, ashes, and gases, equals the original mass of the charcoal and the oxygen when it first reacted. So the mass of the product equals the mass of the reactant.
What is the definition of law of conservation of matter in science?
In any chemical change, one or more initial substances change into a different substance or substances. … According to the law of conservation of matter, matter is neither created nor destroyed, so we must have the same number and kind of atoms after the chemical change as were present before the chemical change.5 дней назад
You might be interested: Under federal law, which type of boat must have a capacity plate
Albert Einstein
What are the three laws of conservation?
In physics, a conservation law states that a particular measurable property of an isolated physical system does not change as the system evolves over time. Exact conservation laws include conservation of energy, conservation of linear momentum, conservation of angular momentum, and conservation of electric charge.
Why do we conserve momentum?
Conservation of momentum is a fundamental law of physics which states that the momentum of a system is constant if there are no external forces acting on the system. It is embodied in Newton’s first law (the law of inertia). | 829 | 4,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-45 | longest | en | 0.960618 |
https://www.convert-measurement-units.com/convert+Cubic+mile+to+Liter.php | 1,721,789,880,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518154.91/warc/CC-MAIN-20240724014956-20240724044956-00324.warc.gz | 608,550,615 | 14,158 | Convert mi³ to l (Cubic mile to Liter)
## Cubic mile into Liter
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Cubic+mile+to+Liter.php
# Convert mi³ to l (Cubic mile to Liter):
1. Choose the right category from the selection list, in this case 'Volume'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Cubic mile [mi³]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Liter [l]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '137 Cubic mile'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Cubic mile' or 'mi3'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Volume'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '56 mi3 to l' or '45 mi3 into l' or '34 Cubic mile -> Liter' or '12 mi3 = l' or '89 Cubic mile to l' or '67 mi3 to Liter' or '23 Cubic mile into Liter'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(12 * 89) mi3'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '56 Cubic mile + 34 Liter' or '67mm x 45cm x 23dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 8.250 666 591 585 6×1021. For this form of presentation, the number will be segmented into an exponent, here 21, and the actual number, here 8.250 666 591 585 6. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 8.250 666 591 585 6E+21. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 8 250 666 591 585 600 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 876 | 3,679 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.861334 |
https://education.ti.com/en/activity/detail?id=25ABDCBFD7DF4177BEFC0065D39B4404 | 1,601,040,548,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400226381.66/warc/CC-MAIN-20200925115553-20200925145553-00339.warc.gz | 366,021,829 | 16,586 | Education Technology
# Activities
• ##### Subject Area
• Math: Statistics: Inferential Statistics
9-12
40 Minutes
• ##### Device
• TI-83 Plus Family
• TI-84 Plus
• TI-84 Plus Silver Edition
• TI-84 Plus C Silver Edition
• TI-84 Plus CE
• ##### Report an Issue
Hypothesis Testing: Means
#### Activity Overview
Students test a claim about a mean with a large sample size at the five-percent significance level. The test statistic is found and compared to the critical value.
#### Key Steps
• In Problem 1, students are given a scenario to assist in testing a claim with a large sample size at five percent significance level.
The scenario describes the mean salary in a community, which an investor thinks is higher than the mean ten years ago. Students are asked to determine the null and alternative hypotheses.
They then decide whether to use the normal distribution or a t distribution for the test statistic and then calculate it.
• Students can find the critical value by using the invNorm command from the DISTR menu.
• Students will find the P-value using the ShadeNorm command, which finds the area under the curve to the right of the test statistic. From this value and comparing the test statistic to the critical value, students will decide whether to reject or fail to reject the null hypothesis.
At the end of this activity, students will be able to use the sample distribution of the mean of the population to test the null hypothesis against the alternative one-tailed hypothesis or two-tailed hypothesis. | 319 | 1,535 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-40 | longest | en | 0.846946 |
https://fr.mathworks.com/matlabcentral/cody/problems/2664-divisors-for-big-integer/solutions/528792 | 1,600,697,867,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400201699.38/warc/CC-MAIN-20200921112601-20200921142601-00289.warc.gz | 419,814,804 | 16,817 | Cody
# Problem 2664. Divisors for big integer
Solution 528792
Submitted on 14 Nov 2014 by goc3
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
%% x = 10; y_correct = 4; assert(isequal(divisors_Big(x),y_correct))
y = 4
2 Pass
%% x = 28; y_correct = 6; assert(isequal(divisors_Big(x),y_correct))
y = 6
3 Pass
%% x = 28; y_correct = 6; assert(isequal(divisors_Big(x),y_correct))
y = 6
4 Pass
%% x = 784; y_correct = 15; assert(isequal(divisors_Big(x),y_correct))
y = 15
5 Pass
%% x = 1452637; y_correct = 2; assert(isequal(divisors_Big(x),y_correct))
y = 2
6 Pass
%% x = 5452637; y_correct = 4; assert(isequal(divisors_Big(x),y_correct))
y = 4
7 Pass
%% x = 16452637; y_correct = 2; assert(isequal(divisors_Big(x),y_correct))
y = 2
8 Pass
%% x = 116452637; y_correct = 8; assert(isequal(divisors_Big(x),y_correct))
y = 8
9 Pass
%% x = 416452638; y_correct = 32; assert(isequal(divisors_Big(x),y_correct))
y = 32
10 Pass
%% x = 12250000; y_correct = 105; assert(isequal(divisors_Big(x),y_correct))
y = 105
11 Pass
%% x = 2031120; y_correct = 240; assert(isequal(divisors_Big(x),y_correct))
y = 240
12 Pass
%% x = 76576500; y_correct = 576; assert(isequal(divisors_Big(x),y_correct))
y = 576
13 Pass
%% x = 816452637; y_correct = 32; assert(isequal(divisors_Big(x),y_correct))
y = 32
14 Pass
%% x = 103672800; y_correct = 648; assert(isequal(divisors_Big(x),y_correct))
y = 648
15 Pass
%% x = 842161320; y_correct = 1024; assert(isequal(divisors_Big(x),y_correct))
y = 1024 | 601 | 1,648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-40 | latest | en | 0.397644 |
https://physics.stackexchange.com/questions/29401/a-two-level-system-absorbs-a-detuned-photon-where-does-the-extra-energy-go | 1,718,518,977,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861643.92/warc/CC-MAIN-20240616043719-20240616073719-00717.warc.gz | 406,323,302 | 42,262 | # A two-level system absorbs a detuned photon. Where does the extra energy go?
Let's consider simple two-level system with frequency gap of $\omega_0$ between ground and excited state. Now, when we turn on external electromagnetic field with frequency $\omega < \omega_0$, there is a non-zero probability that, starting from the ground state, the system will become excited.
The energy of excited state is $\hbar \omega_0$, but each photon of the field contributes only $\hbar \omega < \hbar \omega_0$. So my question is - what's with the missing $\hbar (\omega_0 - \omega)$?
I heard the explanation that it is measuring equipment that fills the energy gap, but I'm not quite convinced. Could you help?
• Other Q can help physics.stackexchange.com/questions/4047/…
– user299
Commented Jun 2, 2012 at 19:45
• Thanks, but I find linked problem a tiny bit different. Before any measurement process the ground state $|0\rangle$ mixes with excited one giving $a_0 |0\rangle + a_1 |1 \rangle$. I understand that such a mixing during measurement process can result in gaining energy from aparatus. But here theory predicts some gain in energy before this act. Commented Jun 2, 2012 at 20:29
• Are you asking how you can excite an atom when the frequency is only slightly detuned? Commented Jun 2, 2012 at 21:17
• More or less that's the point. Commented Jun 2, 2012 at 21:28
Rabi oscillations occur, in their simplest form, when a two-level system interacts with a quasi-resonant classical electric field; the total hamiltonian is then $$H_\textrm{s.c.}=\frac{\hbar\omega_0}{2}\hat{\sigma}_3+\frac{\hbar\Omega_0}{2}\cos(\omega t) \hat{\sigma}_1.$$ (For the atom in the ground state at $t=0$, the excitation probability then shows Rabi oscillations of the form $P=\frac{\Omega_0^2}{\Omega_0^2+\delta^2} \sin^2\left( \Omega t\right)$ for $\delta=\omega-\omega_0$ and $\Omega=\sqrt{\Omega_0^2+\delta^2}$.)
The important thing to realize is that this is the total energy of the system. Since the hamiltonian is not time-independent, there is no need for the atomic system to conserve energy. Of course, if the laser pulse stops and the atom is left in the excited state, the extra energy is taken from the light field.
To account for energy conservation, then, you need to include the light field into the reckoning. This is done by the Dicke model, for which the total hamiltonian is $$H_\textrm{D}=\frac{\hbar\omega_0}{2}\hat{\sigma}_3+\hbar\omega \hat{a}^\dagger \hat{a} +\frac{\hbar\Omega_0}{2}\left(\hat{a}+\hat{a}^\dagger\right) \hat{\sigma}_1.$$ This hamiltonian is barely integrable and no closed-form time-dependent evolutions are known, so it is usually simplified to its rotating-wave approximation counterpart, the Jaynes-Cummings hamiltonian $$H_\textrm{J.C.}=\frac{\hbar\omega_0}{2}\hat{\sigma}_3+\hbar\omega \hat{a}^\dagger \hat{a} +\frac{\hbar\Omega_0}{2}\left(\hat{a}\hat{\sigma}_+ +\hat{a}^\dagger \hat{\sigma}_-\right).$$ This hamiltonian leaves invariant the subspaces spanned by $\{|g,n\rangle, |e, n-1 \rangle\}$, and the probability of excitation shows Rabi oscillations of the form $$P=\frac{n \Omega_0^2}{n\Omega_0^2+\delta^2} \sin^2\left( \Omega t\right)$$ with $\Omega=\sqrt{n\Omega_0^2+\delta^2}$.
Thus, there is still an oscillation between levels of "different energy", since in this process a photon of energy $\hbar\omega$ is absorbed (and subsequently re-emitted, of course) to excite the atom up by $\hbar\omega_0$, and these two may not match, as you point out.
So what's going on? The hamiltonian is time-independent so that energy must be conserved. The resolution is that only the total hamiltonian need be conserved, and the notions of photon and atomic excitation are only approximate ones because they represent eigenstates of only part of the total hamiltonian. The true eigenstates are linear combinations of these and one can speak of excitations of the whole of the system, which are indeed conserved.
So where does that leave the problem then? Say the atom is moving across the laser beam profile, so that the coupling $\Omega_0$ increases from zero to some maximum and returns to zero. Rabi oscillations will show up as a function of the interaction time, which one can control for example by changing the atom's speed (this is actually rather common!). Say, then, that the system starts up with the atom in the ground state and the field in a number state, with the field mode slightly red-detuned as in your set-up. Then with some probability the atom will emerge excited and the field will have lost a photon, which are (now!) valid notions since the coupling has been turned off. Where did the energy go?
You can understand this as a system being adiabatically brought near a level anti-crossing similar to the one below. (image credit)
As the coupling is increased, the system (initially in a non-perturbed eigenstate) is brought impulsively to the middle of the anti-crossing, which puts it in a superposition of the true Jaynes-Cummings eigenstates. These two eigenstates then evolve at their own frequencies (which are respectively slightly higher and lower than the original) and by their interference make the Rabi oscillations. When the atom exits the beam, the hamiltonian impulsively returns to the uncoupled hamiltonian, and the atom is frozen into a superposition of the two uncoupled eigenstates, even if they do have different energies.
Where did the extra energy come from in this case? Well, the beam entry and exit were impulsive, so that again energy need not really be conserved (if the entry and exit are adiabatic then the system will remain in a total-hamiltonian eigenstate and will not show oscillations). In this particular case, then, it is probably taken from the atomic motion, which drives these impulsive transitions, and is affected by a light-induced potential. You would then need to add an atomic-position hamiltonian and account for the beam profile... and there's no ending to the amount of extra frills you can add.
In general, though, the driving principles are that you can only "conserve energy" when you've got a closed, time-independent system, and that energy conservation only makes sense when you consider the total energy of the system, including interaction energies, which are not simply transition-driving terms but do represent additional energy. Finally, one must be careful when speaking of "photons" when dealing with interacting systems - it is only total-hamiltonian excitations that are truly physical (though of course the free-hamiltonian excitations are very much useful notions).
• I trust the explanation, but fail to follow at first line of math. Do you have a "handwaving" 2-liner answer for mere mortals like me ? Its OK if it is not practical to abandon math for the sake of human language, but I hoped to have a chance of understanding the paradox
– user299
Commented Jun 3, 2012 at 22:33
• Basically, the red-shifted absorption is usually associated with a time-dependent process that does not account for the field energy, so there isn't a paradox since there are no "photons". Even if you account for the field energy, the presence of an interaction hamiltonian means that "photons" are not the true physical excitations of the system, which are eigenstates of the total hamiltonian. Commented Jun 4, 2012 at 0:40
• The math details are not important. What matters is that the semiclassical hamiltonian is time-dependent, so Noether's theorem for time-energy does not apply. The second and third hamiltonians remove this time dependence by considering the field, but the only important detail is that the total eigenstates are not the uncoupled ones. Commented Jun 4, 2012 at 0:42 | 1,922 | 7,703 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-26 | latest | en | 0.882079 |
https://www.e-olymp.com/en/problems/1289 | 1,620,792,821,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991252.15/warc/CC-MAIN-20210512035557-20210512065557-00547.warc.gz | 760,321,648 | 3,445 | favorite We need a little bit of your help to keep things running, click on this banner to learn more
Problems
# Lunch
Vlad wants to take for lunch a couple of fruits. He has a different bananas, b different apples and c different pears. In how many ways can he choose 2 different fruits from available?
#### Input
Three non-negative integers are given: a, b and c. All integers do not exceed 106.
#### Output
Print the number of ways to choose 2 different fruits for a lunch.
Time limit 1 second
Memory limit 128 MiB
Input example #1
3 4 2
Output example #1
26 | 142 | 569 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-21 | latest | en | 0.870413 |
https://technobite.com/how-to-count-the-number-of-occurrences-between-two-dates-categorized-by-age-range/ | 1,685,691,706,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648465.70/warc/CC-MAIN-20230602072202-20230602102202-00343.warc.gz | 642,755,069 | 18,157 | # How to Count the Number of Occurrences Between Two Dates, Categorized by Age Range
If you need to track the number of occurrences between two dates based on different age groups, Excel can help. This powerful tool allows you to easily calculate the number of events that fall within a specific age range, making it a valuable resource for tracking everything from sales leads to medical appointments. In this article, we will guide you through the process of counting occurrences by age range in Excel, step by step.
## Understanding the Problem
Before we dive into the solutions, let’s first understand the problem. Suppose we have a database of customers’ birthdates and purchase dates. We want to categorize the customers into age ranges (e.g., 18-25, 26-35, etc.) and count the number of occurrences between two dates. The challenge is to automate this process and avoid manual data entry.
## Solution 1: Using IF Function
One way to categorize age ranges is by using the IF function in Excel. First, we need to calculate the age of each customer by subtracting their birthdate from the current date. We can use the TODAY function to get the current date and the YEAR function to extract the birth year from the birthdate. Here is the formula:
``=YEAR(TODAY())-YEAR(B2)``
Assuming the birthdate is in cell B2.
Next, we can use the IF function to categorize the age ranges. Here is the formula:
``=IF(C2>=18,"18-25",IF(C2>=26,"26-35",IF(C2>=36,"36-45",IF(C2>=46,"46-55","56+"))))``
Assuming the age is in cell C2.
Finally, we can use the COUNTIFS function to count the number of occurrences between two dates. Here is the formula:
``=COUNTIFS(D:D,">="&DATE(2020,1,1),D:D,"<="&DATE(2020,12,31),E:E,"18-25")``
Assuming the purchase date is in column D and the age range is in column E. This formula will count the number of occurrences where the purchase date is between January 1, 2020, and December 31, 2020, and the age range is 18-25.
## Solution 2: Using PivotTable
Another way to categorize age ranges and count occurrences is by using a PivotTable. First, we need to create a new column that categorizes the age ranges using the IF function, as explained in Solution 1.
Next, we can create a PivotTable by selecting the data and going to Insert > PivotTable
In the PivotTable Fields pane, we can drag the age range column to the Rows area and the purchase date column to the Values area. This will create a table that shows the count of occurrences for each age range.
To filter the occurrences between two dates, we can click the drop-down arrow in the purchase date field and select Date Filters > Between. This will open a dialogue box where we can enter the start and end dates.
## Solution 3: Using Power Query
A more advanced way to categorize age ranges and count occurrences is by using Power Query. Power Query is a data transformation and cleansing tool that allows us to combine, reshape, and clean data from multiple sources.
First, we need to load the data into Power Query by going to Data > From Table/Range. In the Power Query Editor, we can add a new column that calculates the age using the same formula as in Solution 1. Next, we can use the Group By function to group the data by age range and count the occurrences. Here are the steps:
• Select the age range column.
• Go to Home > Group By.
• In the Group By dialogue box, select the purchase date column and choose the “Count Rows” option.
• Rename the new column to “Occurrences.”
• Close and load the data back to Excel.
To filter the occurrences between two dates, we can add a new step to the Power Query. Here are the steps:
• Select the purchase date column.
• Go to Home > Transform > Date Filters > Between.
• Enter the start and end dates in the dialogue box.
• Close and load the data back to Excel.
I hope this article has been helpful in guiding you through the various methods of categorizing age ranges and counting occurrences between two dates in Excel. By following the outlined steps, you can automate the process and save time and effort in managing and analyzing your data. Remember to choose the method that best suits your data and skill level and continue to explore and learn more about Excel’s functions and tools. With practice and patience, you can become proficient in data management and analysis and achieve your goals with greater efficiency.
Related: How To Count Occurrences In An Entire Workbook In Excel?
## Bibek Sapkota
I'm Bibek | Tech Enthusiast & Lifelong Learner. | Playing on the Web for the Past Few Years as an SEO Specialist and Full-Time Blogger. I'm constantly seeking out new opportunities to learn and grow, and I love sharing my knowledge with others. This is where I started this blog! Here, you will find me sharing comprehensive reviews, helpful guides, tips-tricks and ways to get the full benefits of ever-changing technology. On this blog, you can also explore Powerful Knowledge, Tips & Resources On Blogging, SEO and Passive income Opportunities. | 1,131 | 5,015 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-23 | latest | en | 0.915371 |
https://dsp.stackexchange.com/questions/52596/why-convolution-between-signal-and-a-channel-is-not-done-directly-in-simo-channe/56206 | 1,566,484,596,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317130.77/warc/CC-MAIN-20190822130553-20190822152553-00479.warc.gz | 444,557,913 | 34,215 | # Why Convolution between Signal and a Channel Is Not Done Directly in SIMO Channel
In OFDM system, I need to transmit a signal $$x$$ in SIMO channel $$H$$ with one $$Tx$$ antenna and 4 $$Rx$$ receiver's antennas. The initial equation for that convolution is $$r=Hx+n$$ where $$n$$ is the noise. $$r$$ is the received signal, $$x$$ is the transmitted signal, and $$H$$ is a toeplitz matrix as built in below code.
The question, When I created the toeplitz matrix $$H$$, the dimensions of transmitted signal $$x$$ and toeplitz matrix $$H$$ are not equal, so we are unable to do the multiplication!. I checked online, I got, before processing that multiplication, I should do two steps. first get $$x1$$ and then $$x2$$ as in below code. the $$x2$$ is the flip of $$x1$$, is theoretically known, but what's the benefit of getting the $$x1$$ ?
EDIT:
Regaring the dimension of toeplitz matrix $$H$$, is as below: if we have the length of signal $$Q$$, the length of channel $$M$$, the number of received antenna $$p$$. So, the dimension of toeplitz matrix will be (Q*p x Q+M)
thnx
• Well, what's the dimensions of your $H$, and why? Also, remember why you do OFDM: to not have your channel convolve in time domain, but to point-wise multiply in frequency domain. I'm not sure about your code (it's self-contradicting in that, as far as I can tell), but if you want to go the "we consider things in frequency domain" route, then your $H$ isn't a Toeplitz matrix, but if you want to go the "we do things in time domain", it's up to you to set $H$'s dimensions right. So, I honestly don't know what you're asking of us: Debug your convolution matrices' dimensions? – Marcus Müller Oct 14 '18 at 15:07
• Try built-in function toeplitz() of MATLAB fr.mathworks.com/help/matlab/ref/toeplitz.html – AlexTP Oct 14 '18 at 16:04
• @AlexTP .. do you mean to use the r = toeplitz(h,Q); directly? what's about to Flip $x$ ? it is not needed in that case ? I tried it, I obtained a matrix with high dimension – Fatima_Ali Oct 15 '18 at 5:21
• Dear Marcus, I added a notice of H dimension in my post. We set convolution in time domain as mentioned in above code. could you please clarify "it's up to you to set H's dimensions right" – Fatima_Ali Oct 15 '18 at 5:23
• @MarcusMüller .. could you please clarify "It's up to set H dimension right" .. do you mean the dimension I can change it according to transmitted signal dimension ? – Fatima_Ali Oct 16 '18 at 14:31
I've said, in another comment, convolution using the function conv "i.e in MATLAB" and convolution using the Toeplitx matrix must give the same results. That's ok.
Now, according to your code, the received signal $$r$$ is the results of convolution between channel $$h$$ and emitted signal $$x$$, which means $$r = h*x + n$$
* indicate to the convolution (which is circular convolution in case of using $$CP$$ with OFDM or with any other system).
So, in that conventional known case, we are using SISO system, where 1 antenna is used as transmitter and 1 as receiver, The length of our parameters should be:
$$x = N$$ x 1 ; $$h = L$$ x 1;
$$N$$ is the length of our signal, in your case $$Q$$
$$L$$ the length of channel or we sometimes call it IR, in your case $$M$$
Till here, it's clear, it's the normal process which can be read anywhere. Now suppose that you are using $$SIMO$$ system with $$P$$ antennas at reciever instead of 1 (by the way fractional sampling is also equivalent to SIMO system). in that case you suppose to have $$P$$ copies for your signal instead of one. It's like you are doing $$P$$ times convolutions for your emitted signal with $$P$$ different channels. let's say $$P$$ = 4; means you have 4 receiver's antennas equivalent in our case into 4 different channels.
As mentioned, you suppose to have 4 copies for your signal, and 4 different channels compared with conventional case, so let's say parameter $$H$$ is toeplitz matrix which will represent those four channels in SIMO system. So we will have
$$R = HX + N$$
$$R$$ is received signal in SIMO, $$H$$ is the toeplitx matrix, $$X$$ emitted signal in SIMO system too and also $$N$$ represent the noise.
Now, your question how to build $$H$$ and what's dimension of $$H$$ when using SIMO system. (by they way, your code is correct)
The emitted signal $$X$$, represents 4 copies of $$x$$, so its dimension is $$(N +L)$$ x 1;
The toeplitz matrix $$H$$ has a dimension $$NP$$ x $$(N+L)$$ where $$P$$ = 4 in your case. And the noise $$N$$ should be $$NP$$ x 1. (N as noise is different about N of emitted signal in this case).
So, Now, you can conclude the dimension of received signal $$r$$ and $$R$$ in every case easily.
and regarding your question, why using $$x1$$, it's to have dimension of emitted signal $$X$$ of $$(N +L)$$ x 1 instead of dimension $$N$$ in conventional case
thnx
• Yes, got it. thank you so much. but in when demodulating the signal at receiver, should we cancel that added part using $x1$? – Fatima_Ali Oct 17 '18 at 9:04
• No need for that. Try to read how to the demodulation in SIMO system. – Zeyad_Zeyad Oct 18 '18 at 9:55
• Awesome answer, @Zeyad_Zeyad! I like how it's logically structured and addresses the implementation question right at the beginning and then dives into derivation of the dimensions. Nice! – Marcus Müller Oct 25 '18 at 7:19
• Why the channel matrix has dimensions of $NP\times (N+L)$? Are you assuming CP-based OFDM transmission? – BlackMath Mar 24 at 18:11
The received sample $$i$$ at receive antenna $$j$$ is given by
$$r_{ij}=\sum_{l=0}^Lh_{j,l}x_{i-l}+n_{ij}$$
where $$L$$ is the channel memory, $$\{h_{j,l}\}$$ is the $$l$$th channel tap from the transmitter to receive antenna $$j$$. The above equation is actually the convolution of the transmitted signal with the channel impulse response of the the channel between Tx and $$j$$th receive antenna. Writing the above samples in details for all values of $$i$$ and $$j$$, and stack the samples will give you a systematic way of finding the dimensions.
For one receive antenna, the channel matrix is circular (in case of CP-based OFDM), and its dimensions are $$N\times N$$ (where $$N$$ here is the number of data symbols in the OFDM symbol). When you have $$P$$ receive antennas, the dimensions becomes $$NP\times N$$.
But I think in this case you need to take the $$N$$-point FFT of each received sample block at each antenna first before any further processing. | 1,726 | 6,423 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 76, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2019-35 | latest | en | 0.910206 |
https://dougenterprises.com/machine-learning-regression-analysis/ | 1,709,230,403,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474852.83/warc/CC-MAIN-20240229170737-20240229200737-00381.warc.gz | 215,344,737 | 20,916 | # Machine Learning Regression Analysis
Published August 5, 2021
Doug Rose
Author | Agility | Artificial Intelligence | Data Ethics
In one of my previous articles Supervised and Unsupervised Machine Learning, I pointed out that machine learning can be used to analyze data in four different ways — two of which are predictive, and two of which are descriptive:
• Predictive: With these methods, supervised learning enables the machine to forecast outcomes based on established patterns. Predictive analysis includes:
1. Classification: Assigning items to different labeled classes
2. Regression: Identifying the connection between a dependent variable and one or more independent variables
• Descriptive: With these methods, unsupervised learning enables the machine to detect patterns that reveal deeper insights into the data. Descriptive analysis includes:
1. Clustering: Creating groups of like things
2. Association: Identifying associations between things
## Understanding Regression Analysis
To understand machine learning regression analysis, imagine those tube-shaped balloons you see at children's parties. You squeeze one end, and the other end expands. Release, and the balloon returns to normal. Squeeze both ends, the center expands. Release one end, and the expanded area moves to the opposite end. Each squeeze is an independent variable. Each bulge is a dependent variable; it differs depending on where you squeeze.
Now imagine a talented balloon sculptor twisting together five or six of these balloons to form a giraffe. Now the relationship between squeezing and expanding is more complex. If you squeeze the body, maybe the tail expands. If you squeeze the head, maybe two legs expand. Each change to the independent variable results in a change to one or more dependent variables. Sometimes that relationship is easy to predict, and other times may be very difficult.
## Business Applications of Regression Analysis
Regression analysis is commonly used in the financial industry to analyze risk. For example, I once worked for a credit card company that was looking for a way to predict which customers would struggle to make their monthly payments. They used a regression algorithm to identify relationships between different variables and discovered that many customers start to use their credit card to pay for essentials just before they have trouble paying their bills. A customer who typically used their card only for large purchases, such as a television or computer, would suddenly start using it to buy groceries and gas and pay their electric bill. The company also discovered that people who had a lot of purchases of less than five dollars were likely to struggle with their monthly payments.
The dependent variable was whether the person would have enough money to cover the monthly payment. The independent variables were the items the customer purchased and the purchase amounts. Based on the results of the analysis, the credit card company could then decide whether to suspend the customer's account, reduce the account's credit line, or maintain the account's current status in order to limit the company's exposure to risk.
Businesses often use regression analysis to identify which factors contribute most to sales. For example, a company may want to know how it can get the most bang for its buck in terms of advertising; should it spend more money on its website, on social media, on television advertising, on pay-per-click (PPC) advertisements, and so on. Regression analysis can identify which items contribute most to not at all. The company can then use the results of that analysis to predict how its various advertising investments will perform.
## Identifying the Dependent and Independent Variables
When performing regression analysis, the first step is to identify the dependent and independent variables:
• Dependent variable is what you are trying to understand or predict; for example, whether a customer is about to miss one of his credit card payments.
• Independent variables are factors that may have an impact on the dependent variable; for example, the customer's spending habits or purchase amounts prior to the date on which the next credit card payment is due.
## An Important Reminder
Keep in mind that correlation does not prove causation. Just because regression analysis shows a correlation between an independent and a dependent variable, that does not mean that a change in the independent variable caused the change observed in the dependent variable, so avoid the temptation to assume it does.
Instead, perform additional research to prove or disprove the correlation or to dig deeper to find out what's really going on. For example, regression analysis may show a correlation between the use of certain colors on a web page and the amount of time users spend on those pages, but other unidentified factors may be contributing and perhaps to a greater degree. A web designer would be wise to run one or more experiments first before making any changes.
While regression analysis is very useful for identifying relationships among a dependent variable and one or more independent variables, use these relationships as a starting point for gathering more data and developing deeper insight into the data. Ask what the results mean and what else could be driving those results before drawing any hard and fast conclusions.
Related Posts
May 29, 2017
The Data Analytics Mindset
The data analytics mindset gets the most from data-driven analysis. Good data analysis is more than just data collection.
October 23, 2017
Make Data Storytelling Pop!
Data storytelling is just about data visualization. You need to use data to tell narratives that "pop" and connect with your audience. | 1,078 | 5,771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-10 | latest | en | 0.90091 |
https://astarmathsandphysics.com/a-level-maths-notes/fp1/3433-solving-second-order-linear-non-homogeneous-differential-equations.html | 1,726,369,842,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651614.9/warc/CC-MAIN-20240915020916-20240915050916-00274.warc.gz | 97,040,663 | 9,769 | Any differential equation of the form(1) is a second order differential equations and there is a standard technique for solving any equation of this sort. We solve the homogeneous equation(2) first for the 'complementary' solutionWe assume a solution of the formand substitute this into (2). We extract the non zero factor – since no exponential is zero for any finite x -to obtain a quadratic equation. We solve this equation to obtain solutionsandand then the general solution isHaving found the complementary solutionwe now find a particular solutionto (1) by assuming a solution 'similar' tothen the general solution to (1) is given by
Example: Solve the equation(2) subject toandat
Substitution of the above expressions into gives
We can factor out the nonzeroto obtain
Becauseis non zero we can divide by it to obtainand we factorise this expression to obtainand solve to obtainandThe complementary solution is then
To find a particular solution we assumesinceis a polynomial of degree one.andSubstitute these into (1)
Equating coefficients ofgives
Equating constants gives
whenimplies (3)
atimplies (4)
(3)+(4) givesthen from (3)
The solution is | 256 | 1,163 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-38 | latest | en | 0.865212 |
wiki.ofthenerds.com | 1,563,668,950,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526799.4/warc/CC-MAIN-20190720235054-20190721021054-00480.warc.gz | 592,414,137 | 6,683 | # Prime Numbers in Python
## Why Python?
Python is a good option for mathematical things like determining if a number is prime or not for two main reasons:
• Python has a very simple syntax which is easy to understand
• Python has no upper limit on how big integers can be
## Algorithm
We'll be using trial division as it is one of the simplest algorithms to implement, is fast enough for most use cases, and has a 100% accuracy rating. Trial division simply tests if there's a factor between 2 and the square root of the number in question.
## Python Implementation
Please note that the following code snippets were written for Python 3.
### Code
1 import math
2 def is_prime( n ):
3 if n < 1:
4 return False
5 for i in range( 2, int( math.sqrt( n ) ) ):
6 if n % i == 0:
7 return False
8 return True
### Step by step
First, we'll need to import the math library (to get the square root of numbers), and begin to define our function:
import math
def is_prime( n ):
Now let's check if the number is greater than 1 and return False if it isn't because prime numbers must be larger than 1:
if n < 1:
return False
Next, let's use a for loop to go through all numbers between 2 and the square root of the number, and converting the square root to an integer:
for i in range( 2, int( math.sqrt( n ) ) ):
Lastly, we simply check if the number is evenly divisible (by checking if the % operator returns a zero) with the current number in the loop. If it is, we can safely return False as we now have a factor. If we make it to the end of the loop without exiting (return statement automatically end the function):
if n % i == 0:
return False
# Outside of the loop
return True
Please see the full code to get the correct indentation levels.
### Using the function
Now that we have our function, let's build a simple program to check if a number is prime or not. First, copy and paste the full function into your program. Now, let's just get the user input and return what the function gives us:
num = int( input( "Number: " ) )
if is_prime( num ):
print( "%i is prime" %num )
else:
print( "%i is NOT prime" % num ) | 526 | 2,148 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2019-30 | longest | en | 0.850346 |
https://hiresomeonetodo.com/how-to-apply-deep-learning-for-image-super-resolution-and-enhancement-in-homework | 1,725,871,962,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00766.warc.gz | 269,593,592 | 18,840 | # How to apply deep learning for image super-resolution and enhancement in homework?
How to apply deep learning for image super-resolution and enhancement in homework? The recent advances in deep learning and its applications in enhancing image perception, emotion recognition and visual input require us to understand some important technical details about how deep learning works. The computer vision industry has been continuously running intense researches on using image processing techniques to solve a wide range of social and technological aspects of the problems highlighted in the chapters on recognition, classification, and computer vision. This tutorial, along with the videos How to apply deep learning for image superresolution and enhancement in homework, covers some important changes in supervised algorithms to solve a wide range of those problems. Introduction In the last 10 years, special attention has been paid to pattern recognition, image restoration and next-generation image processing. In numerous studies of pattern recognition, pattern recognition experts have investigated, classified, and classified images. Different approaches have been proposed for supervised machine vision learning algorithms. Most of these approaches, in particular, have been using simple programs. In this tutorial, we will discuss some common work on visual learning using this technique. General Information Image recognition (recognition) often involves many special functions. If we let $\mathcal{I}_i = (\mathcal{I}_i)_i$ as a set of input and sample patterns, then the output is called a *recognition matrix*. —– ——————————————————————————————————————————————————————————————————– $A simple perceptron (${\mathcal{P}}_1$) or learned binary classification function$ How to apply deep learning for image super-resolution and enhancement in homework? While there are many ways to apply deep learning for image super-resolution and enhancement in students’ homework, there are many ways to use it. No matter how the object, image, or scene is designed, to successfully produce the object-size images. Therefore, how can a photo be served using deep learning? Unfortunately, if you can improve the image rendering of the photo, using deep More about the author will be used to process the images further. What’s the big deal about this topic? Image Super-resolution and enhancement is actually rather complicated. There is not a straightforward way to scale your images using images. The images are typically built Visit Your URL just building up a grid of very small tiles plus a layer of many layers. This is therefore something specifically designed to be used in text-based text compositing tasks such as text-by-text. Imagine that you want to apply a black version of your picture, and one of your students is just using a black version of his text, and can see this. Of course you have the options to choose from different textures to use. You have to focus on how soft and plasticky different details help to work together with your students in creating a good image.
## Math Test Takers For Hire
How Many Pixel Levels Can Your Students Draw So Much There is only a few ways to get good results. However, to make the best use of our resources, let’s consider the number of methods that will produce a particular image using a number of level. Firstly, all the methods will work on a set of tiles (even a flat pixel ) when we do our first step. Think the Pixel Viewer, then work on the same tiles from each pixel, this will help to render the image on a large area with both high resolution and high image quality. Then we can apply our image to a thin layer, this two layers of samples, and do the same on 2How to apply deep learning for image super-resolution and enhancement in homework? The assignment of homework, called “wys3” in university, reads with a sharp focus on one unit and the big picture. However I am unable to apply the amazing deep learning for the image super-resolution and more details. Many students tend to ignore the small-world ideas in the project and only apply their best knowledge in the deep learning and can only do so when a new task is added, thereby making them a very poor actor. With this task in mind, I have a huge amount of research needed for creating new and improved models to improve the image super-resolution and more details. Generally I have a sense of not being in a perfect state, like moving from color space of my eyes to background space of every digit. Sometimes I have a feeling that only minor changes can make the image super-resolution or if applied in all the right way, makes it better than it was before. The very first study to create a super-resolution image of the big picture is done by a great machine learning researcher, Joshua Cernan. He has gone about the whole complex network to solve this task, which is trying to build a super-resolution model structure in a simple task. This can be done almost 100 times using one-shot training. Then, he is analyzing how to use deep learning to solve the problem in the next 100 or so lines. Seeing the results of real-time analysis of Hough transform, he wrote a training pipeline in what is called “deep convolutional network”, which is a fast but simple model with the required information that it is possible for us to learn robust image super-resolution and effective channel estimation. There are some interesting papers on using recent deep learning methods like Hough transform to solve problem of super resolution and their applications. Which of these papers is the best? 1. Can we combine linear and nonlinear processes? I am guessing that for solving
#### Order now and get upto 30% OFF
Secure your academic success today! Order now and enjoy up to 30% OFF on top-notch assignment help services. Don’t miss out on this limited-time offer – act now!
Hire us for your online assignment and homework.
Whatsapp | 1,161 | 5,951 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-38 | latest | en | 0.895393 |
https://news.ycombinator.com/item?id=11019922 | 1,563,422,909,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525483.64/warc/CC-MAIN-20190718022001-20190718044001-00377.warc.gz | 480,895,474 | 10,278 | Show HN: 3D Vector Graphics 441 points by fogleman on Feb 2, 2016 | hide | past | web | favorite | 56 comments
This looks pretty great, I'll play with it. Can you explain this from the Readme?Unfortunately, it's difficult to compute the joint formed at the boundaries of these combined shapes, so sufficient texturing is needed on the original solids for a decent result.What is hard about computing the joint? Why does texturing help? My question is from a perspective of curious ignorance, not arrogance (i.e. me saying "don't be silly, it's easy.")
The texturing just helps your eye "fill in the gaps." As far as difficulty, I just don't know of a general way to do it. The only thing that comes to mind is some sort of iterative approximation or something.
If I understand correctly, we want to compute co-edges here. I believe there are algorithms to do this, I know SketchUp has this feature [1]; also Gmsh [2]; and I remember seeing Blender plugins that do the same thing (can't seem to find them now...) (edit: Actually Blender does this without plugins: "Join" 2 intersecting objects, use the "Intersect" feature, and it will generate and select the exact co-edges.)[2] https://help.sketchup.com/en/article/3000100 (last figure and text in the section "Subtracting one solid from another")
unless you're dealing with sufficiently large integers computing the intersect between a line and a plane with any robustness is impossible on modern computers with float or double. sometimes when you intersect meshes that have nearly coplanar faces, the intersection might shoot vertices near some infinity because of division inaccuracy.
Sufficiently large integers (and FP numbers too) are directly supported by Go's standard library: https://golang.org/pkg/math/big/
Would symbolic arithmetic help with that? Of is that just too slow a priori?
Yes that makes sense. Thanks.
If you run into limitations with the Cairo binding you're using, I'd like to shamelessly plug the one I wrote, which covers much more of the Cairo API: https://godoc.org/github.com/martine/gocairo/cairo
Nice. I love Cairo. I'll check it out.
Whenever I see install instructions like this:OS X`````` brew install cairo pkg-config go get github.com/ungerik/go-cairo `````` Ubuntu`````` sudo apt-get install libcairo2-dev pkg-config go get github.com/ungerik/go-cairo `````` ... I always think there should be a "default" instruction page linked to somewhere that explains what brew and apt-get are (and other things). I mean, I know you can't be expected to go to the lengths of explaining this stuff for every single project, but this is the kind of thing that will inspire the next generation. So maybe we, as a community could come up with a, "you don't know what this is? look here..." kind of page that people can put on their github projects.
Wholeheartedly agreed. When I was learning about OSS, I had no idea what a "pull request" was but people kept mentioning it. A standard guide on how OSS works with Github, Linux, and related tools would be immensely helpful in welcoming new developers into the OSS world.
The google answer for "what is a pull request" is quite good.
We already have something. It's called man pages. I say this kind of tongue-in-cheek but I'm also serious about it at the same time.
`````` man brew `````` That won't help if you don't have Homebrew [0] installed.[0]: http://brew.sh
You can use a search engine, as well. Reasonably effective.
I completely agree. If you were linked to a github page out of the blue or for the first time its incredibly intimidating. Most notably is the first text you read on the page isn't a title, but a list of folders/files. Why not put the dam title and description at the top?One line of, "Hows this work?" would help a lot of people. I took the git lesson from code academy and have messed around in ubuntu but am still very intimidated by github and the brutal installation instructions every project has (or doesn't have).
That's pretty cool. A commandline app that just takes an .OBJ file and renders it to an .SVG would actually be pretty useful (viewport position / orientation would of course need to be options as well).
Even cooler if I could figure out some general way of parameterizing & texturing the surface instead of just rendering the triangles. Any ideas?
Hughes Hoppe has done a variety of research that depends on finding interesting parameterizations of meshes. His papers aren't exactly lightweight on the math, but you may find some ideas there:http://research.microsoft.com/en-us/um/people/hoppe/http://research.microsoft.com/en-us/um/people/hoppe/proj/vfd...
Awesome, thanks. I'll check that out.
Great! Hughes Hoppe's work is amazing!
Hugues Hoppe, unless he has changed his name.
Shit. Thanks for your correction, I can no longer edit the above comment to reflect it, sorry.
I seem to remember there's some good stuff from Pixar on contouring which might be helpful.
if you check out the CGAL library, they have code that solves surface parameterization problems. the least squares conformal mapping is pretty nice for creating uv maps.http://doc.cgal.org/latest/Surface_mesh_parameterization/gro...
Agree! That'd be pretty cool
`ln` is a utility commonly used for making links. Searching for this project in the future is going to be a pest.
One could argue that as a name, `go` is not optimal either, yet it seems the community has worked around it pretty well.
Come on, it's on the first page for `ln 3d` where is your googlefu??
It's also an automatic path conflict on any *nix system as a result.
This is fantastic! I think I can use this in my laser projection work. :)
Glad to hear. Let me know if you have any questions.
The output reminds me of Blender's Freestyle plugin. This looks great though because it's not part of a software suite! ;)
blender is, contrary to popular belief, extremely nimble and easy to compile/modify.it is all python, mostly.and can be used a command line tool with very little effort. you can make a "open obj file, put default camera, render" scripts very easily.
I thought I remembered seeing your graphics work before in go posted on HN. Yep: https://github.com/fogleman/pt Is this an evolution of that work, at least in terms of the learning process?
This is amazing. I spent some time staring at he gopher picture in disbelieve. 'Mindblowing' would be the fitting description.Then I scrolled down to the samples...
Yeah, I was able to reuse a lot of the path tracer code for this.
It might be an interesting idea to use similar algorithms to create vector plots in scientific PDF documents, instead of having embedded raster images that don't scale well.
I just want to say that I'm a huge fan of your open source projects. Thanks for sharing them!
Same here - Nice job. Craft (Minecraft in C / opengl) is a really cool project. I didn't realize it was the same person until I went back through the github repo just now.
This is fantastic. I love programming geometry. This must be a lot of fun to work on.That gif at the bottom is mind bending https://camo.githubusercontent.com/9f83ca714a35fd2a08b85370f...
Question: at the bottom of https://github.com/fogleman/ln/blob/master/ln/matrix.goDid you type that in manually?I wonder if there's a way to do that in Go that doesn't have as much duplication, but has the same performance?
Probably found the algorithm somewhere and used multiple selection in Sublime Text to quickly reformat it.
Anyway I think your webpage is very impressive! Nice job!
In emacs, `keyboard macro counters` are the appropriate tech for that job.
Thank you, @fogleman! We really love you work - including other projects!
Thank you!
An other similar software is Sketch[1]. It generates tikz images that you can compile in LaTeX.
That's pretty rad! I wonder what it would like to bake pre-rendered color information onto the vector geometry. Even without textures on the models some nice ambient occlusion could look cool.
I'd love to see someone output this to an oscilloscope. | 1,819 | 8,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-30 | latest | en | 0.950706 |
https://sharewareonsale.com/s/myron-algebra-and-calculus | 1,603,367,573,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879537.28/warc/CC-MAIN-20201022111909-20201022141909-00627.warc.gz | 498,754,767 | 20,539 | Sale has ended but you can still get this app.
## Description
Still doing algebra using paper and pencil? working problems at the end of the chapter until you get good at it? You're not alone. That's still a good way to learn math. Except maybe the pencil part. And the paper part. But definitely the working-the-problems part.
What if there was a way to make the paper and pencil smarter? What if there was a way to eliminate the mind-numbing part of math? That's where Myron comes in. You tell Myron what the next step is. Myron does it. Then you, then Myron ... (that means "repeat here, many times"), then, voila: problem solved! No, wait; that's not right. "Hey Myron, lets go back four steps to where I tried that u-substitution and use a different formula." Then you, then Myron, ... (remember what ... means?), then voila: problem solved! "Hey, Myron, show me all the steps I made." "Hey, Myron, what does that look like when I plot it?" "Hey, Myron, what does 'complete the square' mean?". "Hey, Myron, e-mail all that stuff to my friend, Sue."
Myron is a computer algebra system. It supports symbolic manipulation of real algebra, Boolean Algebra and Linear Algebra expressions. Manipulations include simplification, commutation, factorization, distribution and substitution as well as many more transformations specific to scalar, set, tuple, matrix and polynomial expressions. Got a transformation Myron doesn't know about? Teach it to Myron and use it in other problem exercises.
Myron also performs differentiation and integration. It can be used to solve most Calculus problems found in the first two semesters of post-secondary Calculus courses.
Myron comes with builtin documentation that explains everything you need to get rid of that pencil and paper, toss the calculator and do math.
You are allowed to use this product only within the laws of your country/region. SharewareOnSale and its staff are not responsible for any illegal activity. We did not develop this product; if you have an issue with this product, contact the developer. This product is offered "as is" without express or implied or any other type of warranty. The description of this product on this page is not a recommendation, endorsement, or review; it is a marketing description, written by the developer. The quality and performance of this product is without guarantee. Download or use at your own risk. If you don't feel comfortable with this product, then don't download it.
## Reviews for Myron Algebra and Calculus
There are no reviews yet. | 558 | 2,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-45 | latest | en | 0.945727 |
https://web2.0calc.com/questions/algebra_40043 | 1,720,897,445,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514512.50/warc/CC-MAIN-20240713181918-20240713211918-00151.warc.gz | 579,470,595 | 5,244 | +0
# Algebra
0
4
0
+415
Let \$f(x) = 3x - 8 + 4x^2\$ and \$g(x) = 15x + c - 3x^2\$. Find \$c\$ if \$(f \circ g)(x) = (g \circ f)(x)\$ for all \$x\$.
May 29, 2024 | 91 | 165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-30 | latest | en | 0.336235 |
https://askfilo.com/user-question-answers-algebra-1/if-12-grams-of-coffee-costs-dollars-and-each-gram-makes-cups-35303236393137 | 1,695,576,978,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506658.2/warc/CC-MAIN-20230924155422-20230924185422-00855.warc.gz | 125,286,969 | 12,757 | World's only instant tutoring platform
Question
Question asked by Filo student
If 12 grams of coffee costs dollars and each gram makes cups of coffee, what is the cost of one cup of coffee in terms of and ?
Text solution
D If 12 grams of coffee cost dollars, the cost of each gram of coffee is dollars. Let one cup of coffee cost dollars, and set up a proportion to find the cost of one cup of coffee.
Cross Products
Stuck on the question or explanation?
Connect with our Algebra 1 tutors online and get step by step solution of this question.
231 students are taking LIVE classes
Question Text If 12 grams of coffee costs dollars and each gram makes cups of coffee, what is the cost of one cup of coffee in terms of and ? Topic Solving Linear Equations & Inequalities Subject Algebra 1 Class High School Answer Type Text solution:1 | 184 | 841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-40 | latest | en | 0.925461 |
http://www.solutioninn.com/after-1964-quarters-were-manufactured-so-that-the-weights-had | 1,508,471,024,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823630.63/warc/CC-MAIN-20171020025810-20171020045810-00086.warc.gz | 576,438,049 | 8,008 | After 1964 quarters were manufactured so that the weights had
After 1964, quarters were manufactured so that the weights had a mean of 5.67 g and a standard deviation of 0.06 g. Some vending machines are designed so that you can adjust the weights of quarters that are accepted. If many counterfeit coins are found, you can narrow the range of acceptable weights with the effect that most counterfeit coins are rejected along with some legitimate quarters.
a. If you adjust vending machines to accept weights between 5.64 g and 5.70 g what percentage of legal quarters are rejected? Is that percentage too high?
b. If you adjust vending machines to accept all legal quarters except those with weights in the top 2.5% and the bottom 2.5%, what are the limits of the weights that are accepted?
Membership | 174 | 802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-43 | latest | en | 0.976215 |
https://jacksofscience.com/how-many-cups-is-8-oz/ | 1,701,680,560,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100527.35/warc/CC-MAIN-20231204083733-20231204113733-00614.warc.gz | 377,854,560 | 27,907 | # How Many Cups Is 8 Oz
Do you ever wonder how many cups are in 8 oz?
This is very common, especially since many recipes call for measurements in cups.
Let’s review and answer the question of how many cups are in 8 oz.
## Explain It To A Child
There are 8 ounces in a cup. This means that 1 cup is equal to 8 oz. When you’re working with recipes or measuring ingredients, it’s important to know this conversion.
## How many cups is 8 oz?
There are 8 ounces in a cup, so 8 oz is equal to 1 cup.
This is a simple conversion, but it’s important to know when you’re working with recipes or measuring ingredients.
If a recipe calls for 1 cup of liquid and you only have an 8-ounce measuring cup, you can fill it to the 8-ounce mark and know that you’re using the right amount.
When it comes to dry ingredients like flour or sugar, 1 cup is equal to 8 ounces.
So if a recipe calls for 2 cups of flour, you would need 16 ounces (2 cups) of flour to make the recipe.
## How to easily convert between ounces and cups
When it comes to measuring ingredients, there are two units of measurement that are commonly used: ounces and cups.
And while it is easy to convert between the two units when you are working with whole numbers, things can get a little bit more complicated when you are dealing with fractions.
However, there is a simple way to convert between these two units of measurement.
• To convert from ounces to cups, simply multiply the number of ounces by 0.125 (or 1/8).
• For example, if you have 2 ounces of an ingredient, you would multiply that by 0.125 to get 0.25 cups.
• To convert from cups to ounces, simply multiply the number of cups by 8 (or 1/0.125).
• So, if you have 1 cup of an ingredient, you would multiply that by 8 to get 8 ounces.
Keep in mind that these conversions are for liquid ingredients only. If you are working with dry ingredients, the conversion will be slightly different.
But regardless of what type of ingredient you are dealing with, converting between ounces and cups is quick and easy!
## What are the benefits of converting measurements to cups instead of ounces?
Many people convert measurements to cups instead of ounces because it is easier and more accurate.
When converting to cups, you can use a standard that is included in most kitchens. This makes it easy to get the correct measurement, especially when baking.
In addition, measuring in cups allows you to be more precise with your ingredients.
• When measuring in ounces, it can be difficult to get an accurate measurement, especially if you are using a small scale.
• Finally, converting simple math does not require any special equipment.
• All you need is a cup and a few common kitchen items.
With these three benefits, it’s no wonder that so many people prefer to convert measurements to cups instead of ounces.
## How to make sure your cup measurements are accurate
When it comes to baking, one of the most important things you can do is make sure your cup measurements are accurate.
There are a few different ways to do this.
1. First, make sure you are using a standard measuring cup. These can be found at most kitchen stores.
2. Second, make sure you are level when measuring. This means using a liquid measuring cup or leveling off the dry ingredients with a knife before scooping them into the measuring cup.
3. Finally, make sure you are stirring the ingredients before measuring them. This will help to ensure that the ingredients are evenly distributed and that you are getting an accurate measurement.
By following these simple tips, you can be confident that your cup measurements will be accurate and that your baking will be successful.
## Examples of how many cups is 8 oz in common recipes
There are quite a few examples of instances where 8 oz is equivalent to 1 cup in common recipes.
A good example is a recipe for brownies that is on the back of most chocolate chip bags- it calls for 2 cups of sugar, which is exactly 16 oz.
Another example is a recipe for cookies that calls for 1 cup of butter, which is again 8 oz.
When making soup, many recipes will also ask for 1 cup of diced vegetables, which once again is 8 oz.
Thus, as these examples show, it is not at all uncommon for 8 oz to be equal to 1 cup in recipes.
### So next time you’re in the kitchen, remember that 8 ounces are equal to 1 cup.
Article Sources
Jacks of Science sources the most authoritative, trustworthy, and highly recognized institutions for our article research. Learn more about our Editorial Teams process and diligence in verifying the accuracy of every article we publish.
## Author
• Sasha is a Senior Writer at Jacks of Science leading the writing team. She has been in the scientific field since her middle school years and could not imagine working in anything other than molecular atoms, kinetic energy, and deep space exploration. | 1,066 | 4,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2023-50 | longest | en | 0.946913 |
https://www.gamedev.net/forums/topic/565083-image-matching/ | 1,503,250,658,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106865.74/warc/CC-MAIN-20170820170023-20170820190023-00275.warc.gz | 907,944,888 | 27,487 | # Image Matching
## Recommended Posts
TeSsL 122
Hi.. I am trying to do some image matching for a school proj. I found a code which make use the difference between the descriptors of each pixel to do the comparison. I have some qns of the code and hope someone could help me out. 1. why is 0.6 used? 2. which algo is this based on? for (int k = 0; k < Keypoints.Count; k++) { int min=0; int dsq, distsq1 = 100000000, distsq2 = 100000000; /* Find the two closest matches, and put their squared distances in and distsq2.*/ for (int k1 = 0; k1 < Keypoints2.Count; k1++) { dsq = DistSquared(Keypoints[k], Keypoints2[k1]); if (dsq < distsq1) { distsq2 = distsq1; distsq1 = dsq; min = k1; } else if (dsq < distsq2) { distsq2 = dsq; } } /* Check whether closest distance is less than 0.6 of second.*/ if (10 * 10 * distsq1 < 6 * 6 * distsq2) { key1.Add(k); key2.Add(min); } } /* Return squared distance between two keypoint descriptors. */ int DistSquared(KeypointN k1, KeypointN k2) { int i, dif, distsq = 0; int[] pk1, pk2; pk1 = k1.Descriptor; pk2 = k2.Descriptor; for (i = 0; i < 128; i++) { dif = (int) pk1[i] - (int) pk2[i]; distsq += dif * dif; } return distsq; }
##### Share on other sites
Emergent 982
1 - Use "source" UBB tags to display code like this.
2 - [EDIT: At first glance, I thought the following...] If your images have N (where N = width*height) pixels, then this algorithm simply treats each image as a vector in N-dimensional space and computes the usual Euclidean distance. The constants like 0.6 look to be arbitrary numbers that the author of the code chose by manual "tuning" (I would guess). [EDIT: That said, on closer inspection I think inferno82 is correct and it is not pixel values but features of some sort that are being compared.]
[Edited by - Emergent on March 19, 2010 7:01:39 AM]
##### Share on other sites
willh 160
There are a many ways to do image matching. It can be really easy or really complicated, depending on what you need.
The easiest is like this:
int Compare(Image img1, Image img2){ int score = 0; for ( int x = 0; x < img.width; x++) { for ( int y = 0; y < img.height; y++) { int d = img1[x,y] - img2[x,y]; score += d * d; // error squared } } return d;}
Just call 'Compare' to test against each image and use the one that has the lowest score.
If you want something a bit better Google 'normalized cross corrleation'.
If you need something that is more robust to scaling and rotation then you'll have to do feature matching. For that you might want to look at something like SIFT. YouTube has some great videos of SIFT in action.
If you're talking about object detection then you may want to look in to using Eigenvectors, SVMs, PCA, or a cascade of weak classifiers.
##### Share on other sites
Emergent 982
Quote:
Original post by willhIf you're talking about object detection then you may want to look in to using Eigenvectors, SVMs, PCA, or a cascade of weak classifiers.
And even more appropriately than PCA, ICA... (PCA is about the components that are good for building your images (and tend to be the things that your images have in common); ICA is about the components that are good for distinguishing between them.)
##### Share on other sites
programmer88 168
There is always the euclidean distance algorithm, works well for black and white pictures.
for control image a(width,height)
create a (width*height) dimensional vector, 'control', where each value is the colour(grey) value of the control
do the same for the test images 'test1'...'testN' and compare the euclidean distance between each of the test image vectors and the control image vector.
##### Share on other sites
inferno82 152
I am assuming that key points refers to a type of feature gathered in each image. These could be anything from SIFT, FAST, Shi & Tomasi, etc... Although by the DistSquared function it looks like its a 128 valued descriptor. So I'm leaning towards a SIFT descriptor. How are key points generated?
Quote:
Original post by TeSsl1. why is 0.6 used?
The 0.6 value, I would guess comes from empirical testing and the original programmer found that for his/her data set, 0.6 was the allowable distance to determine whether two features where potential matches. If that's the case you might have to determine this value on the data set you are using. Just grab a couple features from different images that you know are the same and compute the Euclidean distance between the descriptors to get a average separation.
Quote:
Original post by TeSsl2. which algo is this based on?
The algorithm is simply performing simple feature matching. Then, I would assume after performing the feature matching you would have to determine based on the number of features matched and their distances, if the two scenes were the same. Again, you would have to determine what number of matched features constitutes a matched image.
- me | 1,275 | 4,928 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-34 | longest | en | 0.863046 |
http://math.stackexchange.com/questions/185097/comparison-of-two-infinity/185104 | 1,464,515,791,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049278887.13/warc/CC-MAIN-20160524002118-00127-ip-10-185-217-139.ec2.internal.warc.gz | 182,797,697 | 19,848 | # Comparison of two infinity [duplicate]
Possible Duplicate:
Different kinds of infinities?
Today I got to know that two infinity can be compared, But I want to know how is this possible? infinity will be infinity. If it doesn't have any particular value, how can we say that this infinity is small and other one is greater. Can anyone help me?
-
## marked as duplicate by Jennifer Dylan, Brian M. Scott, Asaf Karagila, Henning Makholm, Austin MohrAug 21 '12 at 17:21
possible duplicate of Different kinds of infinities? Also Are all infinities equal? – user2468 Aug 21 '12 at 16:54
So what did you hear today? – Thomas Aug 21 '12 at 16:54
@Thomas: that two infinite values can be compared and we can say which is greater and which one is smaller. can you prove it? – Rahul Taneja Aug 21 '12 at 16:57
I might be wrong here, but I don't think you can go far without set theory. Infinity here is defined as a size (cardinality) of a set. Take a look at ordinal numbers and cardinal numbers and The Book of Numbers by J. H. Conway & R. Guy. – user2468 Aug 21 '12 at 17:09
@RahulTaneja: If you want to, you can declare that any two infinite sets are equal in "size." No problem arises from this declaration. It just means that "size" and cardinality are different notions. Before Cantor, that was how things were done: there was no attempt to distinguish between different "infinities." But the distinctions introduced by Cantor have proved to be very useful in mathematics. – André Nicolas Aug 21 '12 at 18:31
The question about several types of infinity has come up several times before. As linked to in the comments about, you can read through the question and answer given in Different kinds of infinities? or Are all infinities equal?.
You ask about if you can compare infinities. Well, you might say that you can. IF you take for example the natural numbers $1,2,3,...$, then there are an infinite number of them. We say that the set is infinite. But, you can also count them. If we look at the real numbers, then the fact is that you cannot count these. So in a way, the infinite number of real number is "greater" than the infinite number of natural numbers.
But all this comes down to the question about how you measure the size of something. If someone says that something is bigger than something else, then they should always be able to define exactly what that means. We don't (I don't) like when questions become philosophical, then it has (In my opinion) left the realm of mathematics. So if someone tells you that one infinity is greater than another infinity, ask them exactly what they mean. How do you measure sizes of infinities? If they are a mathematician, they will be able to give you a precise definition (study Andre's answer).
But, what we usually think about when we compare numbers (or elements in a set) is a set with some kind of ordering on. Without going into any detail, there are different types or orderings, but you can think about how we can order the set consisting of the real numbers in the usual way (ex $7 > 3$). But in this example we are just talking about the real numbers. And infinity is not a number.
One more thing to keep in mind is that we will some times write that a limit is equal to infinity. Like $$\lim_{x \to a} f(x) = \infty.$$
However, when we write this, we don't think (I don't) of $\infty$ as an element in the set of real numbers (it isn't). All we mean by writing that the limit is infinity is that the values of $f(x)$ become arbitrarily large as $x$ "gets" close to $a$.
Just a few things.
-
$+1$ for the part about the limits. I guess that causes quite a confusion when one first learns about set-theoretic infinity. – user2468 Aug 21 '12 at 17:17
Two sets $A$ and $B$ are said to have the same cardinality if there is a function $f: A \to B$ which is one-to-one and onto. More informally, $A$ and $B$ have the same cardinality if the elements of $A$ and $B$ can be "paired off." Note that if $A$ is finite, then $A$ and $B$ have the same cardinality if and only if $A$ and $B$ have the same number of elements. But the formal definition of "same cardinality" does not mention numbers, so it makes sense even for infinite sets.
Let's look at an infinite example. The set $A=\{1,2,3,4,\dots\}$ of positive integers and the set $B=\{2,4,6,8,\dots\}$ of even positive integers have the same cardinality, for we can pair off the integer $k$ with the even integer $2k$. In terms of functions, the function $f(x)=2x$ is a one-to-one onto mapping from $A$ to $B$.
We say that $A$ has cardinality less than (the cardinality of) $B$ if there is a one-to-one mapping from $A$ to (part of) $B$, but $A$ and $B$ do not have the same cardinality.
Using the Axiom of Choice, one can prove that for any two sets $A$ and $B$, either (i) $A$ and $B$ have the same cardinality or (ii) $A$ has cardinality less than $B$ or (iii) $B$ has cardinality less than $A$. (This result is sometimes called Trichotomy.)
In this way, any two sets can be compared as to "size."
It turns out that not all infinite sets have the same cardinality. The famous early result is due to Cantor. Let $\mathbb{N}$ be the set of positive integers, and let $\mathbb{R}$ be the set of reals. Then $\mathbb{N}$ has cardinality less than $\mathbb{R}$. So, in the sense of cardinality, two infinite sets can have different sizes.
In general, the collection of all subsets of a set $S$ can be proved to have cardinality greater than the cardinality of $S$. In particular, this means that the collection of all subsets of the reals has cardinality greater than the set of reals.
In the sense of cardinality, there is a very rich family of different-sized "infinities."
-
@Nicolas I'm not able to write those symbols correctly so sorry for that. Sir my logic is that, that $\mathbb{R}$ has all the elements of $\mathbb{R^+}$ and over that $\mathbb{R}$ some more elements. So then how the cardinality of these two sets are equal. – Singh Dec 15 '14 at 4:12
(en.wikipedia.org/wiki/Pigeonhole_principle) Here you need not to know the exact numbers. Suppose I was having $3 with me initially and you were having$0 and a then stranger gave you and me a bag inside which there was some money(equal amount)(no one know how much). Now knowing these all information anybody can say that I'm having more money then you even without knowing the exact amount. Why this logic not work in case of infinite sets. – Singh Dec 15 '14 at 4:26
Sir so are saying that we cannot use the pigeonhole principle to compare the cardinality of two infinite sets. – Singh Dec 15 '14 at 4:41
We can use versions of Pigeonhole, for example if the reals are a union of countably many sets, at least one of the sets is uncountable. The system discourages long discussions in comments. I suggest you ask a carefully formulated new question. In a few minutes I will delete my string of comments, and suggest you do the same. – André Nicolas Dec 15 '14 at 4:45
Cantor proved that there are "infinities" "bigger" than others. For instance, $\mathbb{R}$ is strictly bigger than $\mathbb{N}$. What is meant by this can be stated the following way : There is not enough natural numbers to number every real number. In other words, if you have associated a real number to each natural number, then there will be real numbers left without an associated natural number.
The proof is called the Cantor Diagonal argument : http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument.
- | 1,915 | 7,479 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2016-22 | longest | en | 0.953184 |
https://xn--2-umb.com/14/identities | 1,600,650,234,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198868.29/warc/CC-MAIN-20200920223634-20200921013634-00469.warc.gz | 1,177,933,762 | 8,377 | # Identities
2014-05-07
The mathematical identities from my Advanced Quantum Mechanics course note:
$\Gamma (z+1) = z \Gamma (z)$
$\Gamma (1-\epsilon ) \Gamma (\epsilon ) = \frac{\pi }{sin {\pi \epsilon }} = \epsilon + \frac{\pi ^2}{6}\frac{1}{\epsilon } + O\left(\frac{1}{\epsilon ^3}\right)$
$\int _0^\infty \mathrm{d} x; x^\alpha \mathrm{e}^{-\beta x} = \Gamma (1+\alpha )\beta ^{-(1+\alpha )}$
$x^{-1} = \int _0^\infty \mathrm{d} \alpha ; \mathrm{e}^{-\alpha x}$
$\int _0^\infty \mathrm{d} x; x^\alpha (1-x)^\beta = \Gamma (1+\alpha )\Gamma (1+\beta ) / \Gamma (1 + \alpha + \beta )$
$\int _0^\infty \mathrm{d} a_1; \cdots \int _0^\infty \mathrm{d} a_n; f\left(sum a_i \right) = frac{1}{(n-1)!} \int _0^\infty \mathrm{d} s; s^{n-1} f(s)$ | 304 | 749 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-40 | latest | en | 0.3796 |
https://www.doubtnut.com/question-answer/how-many-permutations-of-the-letters-of-the-word-madhubani-do-not-begin-with-m-but-end-with-i--21257 | 1,627,660,006,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153971.20/warc/CC-MAIN-20210730154005-20210730184005-00642.warc.gz | 773,901,168 | 78,849 | Home
>
English
>
Class 11
>
Maths
>
Chapter
>
Permutations
>
How many permutations of the l...
# How many permutations of the letters of the word MADHUBANI do not begin with M but end with I ?
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 23-12-2020
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
176.2 K+
85.1 K+
Text Solution
Solution :
Required arrangements=8!-7!=7!(8-1)=35280.
Image Solution
Find answer in image to clear your doubt instantly:
30620460
1.3 K+
26.2 K+
2:28
43959298
3.1 K+
29.7 K+
4:00
61736641
2.4 K+
48.9 K+
1:49
72790752
1.5 K+
29.8 K+
1:44
25877
7.3 K+
145.6 K+
1:24
21242
4.1 K+
81.7 K+
3:46
1447803
10.9 K+
218.1 K+
7:59
43959332
3.1 K+
62.6 K+
2:41
43959252
2.9 K+
58.5 K+
1:30
43959254
28.1 K+
29.1 K+
1:18
21357
7.2 K+
143.9 K+
1:24
72790753
5.7 K+
9.2 K+
2:09
76129278
1.9 K+
38.0 K+
6:53
72790636
25.2 K+
26.1 K+
2:23
43959296
40.6 K+
45.3 K+
2:39 | 423 | 1,076 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-31 | latest | en | 0.52446 |
http://www.pinterest.com/ebyka/math/ | 1,409,245,172,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500830903.34/warc/CC-MAIN-20140820021350-00154-ip-10-180-136-8.ec2.internal.warc.gz | 545,145,995 | 42,732 | Come on in! Join Pinterest today...it only takes like a second or so.
# Math
## Related Boards
Use Legos to introduce kids to adding with fractions
### Want an Easy Way to Teach Kids Math? Try Using LEGO
babble.com
Dont-Feed-The-Raccoon-a-fast-paced-game-of-number-identification-Stay-At-Home-Educator-791x1000
### Don't Feed The Raccoon!
stayathomeeducator.com
### Mrs. Bohaty's Kindergarten Kingdom: Math tools
mrsbohatyskindergartenkingdom.blogspot.com
Walk the Plank: practicing addition facts to 20- Both players line up cubes, roll dice and add, take the opponent's sum cube. First person to take all of the other player's cubes wins.
### Primarily Speaking: A Fun {and Easy} Math Game
primarily-speaking.blogspot.com
Trace hands, cut out glue down, except for the fingers. Make sums to 10 record underneath.
### Number Sense Craftivity
reliefteachingideas.wordpress.com
Exponent foldables...perfect for when my kids get to this!
### Exponent Foldables
tothesquareinch.wordpress.com
Probability anchor chart. I would change it slightly by using the same colour in the image for the writing (eg., "more likely" would be in blue because the pie chart shows blue is more likely).
Distributive Property Combo Meal Activity I don't want to buy it but I like the idea.
Teaching the distributive property using combo meals.
Distributive Property | Dave May Teaches
Angry Birds used in a math art center to illustrate the distributive property of multiplication over addition.
I've got a foldable for that!: Math Important Book
Great center activity using fractions on a number line. It can be adapted for 3rd or 4th grade.
FRACTION VIDEO~ Easy mini-lesson for teaching fractions on a number line. Well done! (approx. 6:38 min.)
PowerPoint Game Templates
Geometric Attribute War; Attributes of 3-D Figures. - War game with Figure Cards, Battle Cards, and Math Talk cards which are great for generating mathematical conversations and building in accountability to your workstations - Teacher notes - Student instructions for War game - Student reference sheet showing how to count edges, vertices, and faces - Printable practice sheet for identifying geometric attributes - 10 question quiz on identifying attributes \$
Divide fractions by whole numbers using models - 6.NS.1
Free! Creating a witches brew recipe using fractions. Students work on multiplying and dividing fractions to create new recipes! Fun for Halloween.
Flocabulary - Dividing Fractions song: Keep, Change, Flip
Here's a great anchor chart on dividing fractions.
Teaching With a Mountain View: Anchor Charts
Math Helper - 2 Sided Card...make something like this for Deb or OS?
2 + 1 Math Rocks! The Decimal Song | 612 | 2,714 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2014-35 | latest | en | 0.824101 |
https://www.thequint.com/tech-and-auto/tech-news/wordle-651-word-of-the-day-for-today-check-the-hints-and-clues-for-saturday-know-the-final-answer | 1,713,163,681,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816942.33/warc/CC-MAIN-20240415045222-20240415075222-00444.warc.gz | 926,969,252 | 187,115 | # Wordle 651 Puzzle Today: Check Hints, Clues, and Word of the Day for 1 April
## Wordle 651 word of the day for today, 1 April 2023: Read the hints and clues to find the final answer.
Published
Tech News
i
Aa
Aa
Small
Aa
Medium
Aa
Large
Begin your Saturday by solving the Wordle puzzle and learn a new term today. Wordle 651 puzzle for today, Saturday, 1 April 2023, is updated for all interested players. You can visit the official website of the New York Times - nytimes.com to find the puzzle for today. Players are advised to read the hints available on different online platforms before they start using their limited chances in the game. We also help our readers in all the ways possible.
Wordle 651 puzzle for today, Saturday, 1 April, can be solved by anyone who wants to play the game. People from all age groups have access to the puzzles. The words of the day are updated at midnight so players can start their day by solving them. Most people like to solve the words right in the morning.
Sometimes the words of the day are too difficult. Players are forced to look for online help when they get stuck while solving the puzzles because everyone wants to get the score. Different platforms state certain hints that players can use while solving the terms.
Wordle is a simple online game that has extremely easy rules. This game became a hit in 2022, especially among the millennial generation. The word game comes up with a five-letter word and each player gets only six chances to solve it.
## Wordle 651 Hints and Clues: Saturday, 1 April 2023
Let's take a look at the Wordle 651 hints and clues for today, Saturday, 1 April 2023:
• The word of the day starts with letter M.
• The second letter is a vowel.
• The word of the day ends with alphabet H.
• The word of the day has the alphabet C.
These are all the hints we had for the day. We hope these clues will help you find the answer in no time. Now, you can keep reading if you are here for the final term today.
## Wordle 651 Word of the Day Today: 1 April
Readers who follow this space daily are aware that we state the word of the day for all players who are struggling to finish their game.
Wordle 651 word of the day for today, Saturday, 1 April 2023, is stated below:
MARCH
(At The Quint, we are answerable only to our audience. Play an active role in shaping our journalism by becoming a member. Because the truth is worth it.)
×
× | 579 | 2,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-18 | latest | en | 0.957318 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.