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https://www.uvm.edu/~summer/course/202206/61821/summer-2022/mathematics/qr-pre-calculus-mathematics/ | 1,656,538,276,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103645173.39/warc/CC-MAIN-20220629211420-20220630001420-00736.warc.gz | 1,115,312,809 | 10,192 | ### Registration
View Summer 2022 Courses
# Mathematics: QR: Pre-Calculus Mathematics
## MATH 010 OL2 (CRN: 61821)
### 3 Credit Hours—Seats Available!
Skills in working with numerical, algebraic, and trigonometric expressions are developed in preparation for MATH 021. May not be taken for credit concurrently with, or following receipt of, credit for any mathematics course numbered MATH 021 or above. Prerequisite: Two years of secondary school algebra; one year of secondary school geometry.
### Related Program
High School & Pre-College Programs
### Notes
Dates: July 5 - August 12, 2022 Open to Degree and CDE students; Prerequisite: Two years of secondary school algebra; one year of secondary school geometry.
### Section Description
Math 010 OL2 is an asynchronous online class. This means that we do not have specific meeting days and times for the class. All resources will be available online and you can work on the course material whenever it is convenient for you. There will be several deadlines each week that need to be met, but you can complete the coursework flexibly to meet these deadlines. This 3-credit course covers topics in algebra including polynomial, rational, inverse, exponential, and logarithmic functions. Trigonometry topics include triangle trigonometry, the laws of sines and cosines, and an introduction to trigonometry as a collection of periodic functions. Upon successful completion of the course, students will be prepared to take a course in Calculus. After completing this course students will be able to: • Represent functions algebraically, numerically, graphically and verbally, including absolute value, polynomial, rational, piecewise-defined, root/radical, exponential, logarithmic, trigonometric and inverse trigonometric functions. • Analyze the algebraic structure and graph of a function to identify domain, range, intercepts, asymptotes, intervals on which the function is increasing, decreasing or constant, local extrema, the inverse function, etc. • Perform operations on functions and transformations on graphs of functions. • Solve a variety of equations, systems of equations, and inequalities in a variety of contexts and applications, including those involving absolute value, polynomial, rational, piecewise-defined, root/radical, exponential, logarithmic, trigonometric and inverse trigonometric functions.
### Section Expectation
We will be using the ALEKS online learning program in this course. You are required to purchase access to ALEKS, which comes with an electronic textbook: Precalculus, second edition, by John Coburn. The eBook version of the text contains all of the material contained within a hardcopy of the text but also includes links to demonstrations, tutorials, videos, and other resources. There is no need to purchase a personal hardcopy of the text. You can purchase the access code at the UVM Bookstore or directly from the ALEKS website, www.aleks.com. You are expected to watch all lesson videos and to spend 6-8 hours per week working on ALEKS Objectives, Quizzes or Tests.
### Evaluation
Homework/Objectives (on ALEKS) 25% Quizzes (on ALEKS) 25% Pie Progress (on ALEKS) 10% Midterm Exam (on ALEKS) 20% Final Exam (on ALEKS) 20%
### Meetings
#### July 5, 2022 to August 12, 2022
##### Location
Online (View Campus Map)
### Important Dates
Courses may be cancelled due to low enrollment. Show your interest by enrolling.
Last Day to Add July 7, 2022 July 11, 2022 July 12, 2022 July 14, 2022 August 1, 2022
### Other courses you may be interested in…
SectionTitleInstructorsDatesDaysTimesCreditsCRN
MATH 009 OL1
• Nichole Caisse
to N/ASee Notes 361819
MATH 017 OL1
• Karla Karstens
to N/ASee Notes 361822
MATH 019 OL1
• Nichole Caisse
to N/ASee Notes 3, 61823
MATH 020 OL2
• Karla Karstens
to N/ASee Notes 361830
There are no courses that meet this criteria.
QUESTIONS? | 929 | 3,886 | {"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-2022-27 | latest | en | 0.889463 |
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# The Arithmetic Peddle Explained
If you’ve ever wondered why your bicycle spins, you may have wondered how it’s made. The motion of a bicycle is actually explained by a mathematical formula. This article will explore the formula used to calculate the speed of a bicycle and the time it takes to complete a ride.
## Mathematical formula explains motion of a bicycle
The motion of a bicycle is complicated, and scientists have spent years trying to figure out its exact physics. It involves inertia forces, centrifugal forces, and leaning of the rider’s body, along with torque applied to the handlebars Digital Marketing
In the free body diagram, forces acting on the bicycle and rider are represented by arrows whose lengths correspond to their relative magnitudes. Long arrows represent strong forces, while short ones represent weak forces. The forces that cancel each other out in the vertical direction are the weight and the normal.
Bicycles behave like gyroscopes, so scientists have been trying to understand how they balance. As early as the nineteenth century, a group of scientists at Nottingham University led by J.P. Meijaard had been puzzling over the balance of a bicycle.
As the rider pedals, a force acts on the main crank that turns the bike in a clockwise direction. This force is balanced by a lesser force in the opposite direction. As the force increases in the left direction, the bike moves to the left.
## Speed of a bicycle
The arithmetic peddle is an important component of cycling. This pedal allows you to increase the speed of your bicycle. To increase the speed of your bicycle, you need to know how much force you can apply to your pedals. The pedals are made of four-inch-sprockets and two-inch-sprockets. The pedals rotate at a rate of one revolution per second.
The formula for calculating the arithmetic peddle for speed of a bike is relatively simple. You must know the weight in kilos and the speed in meters per second in order to find the correct power output. The amount of power needed will depend on the slope, which means that the steeper it is, the more power you will need. You will also need to know the slope’s percentage.
Cadence is another important factor. In cycling, cadence is a measurement of the speed at which you pedal. It is a ratio of the crank rpm to the speed of the wheel. The ideal cadence is 80 to 100 rpm. Beginner cyclists will usually pedal between 60 and 85 rpm. More experienced cyclists will pedal between 75 and 95 rpm. Professional cyclists, on the other hand, can pedal more than 100 rpm during sprints and attacks.
The concept of calculating the arithmetic peddle for speed of a bike isn’t new, and it’s actually been around for quite some time. The theory was developed by Jason Moore at the University of California, Davis. He was inspired by the research he had done on aircraft pilots. He was able to develop a model of human control by using sensors and a rigid upper-body harness.
Bicycles use no fossil fuels, emit no pollution and use natural resources. The power you use to pedal a bicycle is converted into kinetic energy by using your muscles, not by fossil fuels. In fact, 90 percent of the pedal energy is converted into kinetic energy.
If you want to increase the speed of your bicycle, you need to increase the power you exert. At 20 miles per hour, four-fifths of your energy is used to overcome air resistance. This means that you need to increase your power by 83% if you want to accelerate to 25 miles per hour. If you’re an occasional biker, 3 miles per hour is enough to balance out the equation.
The bicycle wheel is another key component that multiplies the speed of the bicycle. It works much like a car’s gears. Pedals turn the center axle of the bicycle, but the leverage from the wheel turns the outer rim much further. This helps the bicycle accelerate faster than it would without gears.
## Time it takes to ride a bicycle
Cycling is an aerobic exercise that requires a great deal of energy. The distance covered by a cyclist depends on several factors, including speed and resistance. For example, a cyclist can pedal at eight to ten kilometers per hour on a flat surface, but if they are cycling up a slope, they can reach speeds of twenty to forty kilometers per hour. A cyclist can also reach speeds of up to 50 kilometers per hour on steep incline.
Papadopoulos’s first goal was to understand the mechanism of bicycle stability. To accomplish this, he scrutinized over 30 published equations for bicycle motion. He was appalled by the ‘bad science’ found in these equations, which had numerous errors. He decided to rewrite his equations from scratch.
The distance travelled by each cyclist depends on their fitness level. Xavier must cycle for eleven and a half kilometers, while Yves must cycle for eight kilometers. Both riders are able to complete their journey in a single go, but they can also cycle in segments.
In general, a bicycle ride of ten miles takes about 45 minutes. However, the average ride time for a beginner is closer to an hour. The duration of a bicycle ride can also vary greatly depending on the level of fitness, gear, and route. | 1,084 | 5,202 | {"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} | 4.4375 | 4 | CC-MAIN-2023-40 | latest | en | 0.938269 |
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2.1.3 Using Deductive Reasoning to Verify Conjectures − Worksheet
Example 1: Does each conclusion use inductive or deductive reasoning? 1. At Bell High School, students must take Biology before they take Chemistry. Sam is in Chemistry, so Marcia concludes that he has taken Biology.
2. A detective learns that his main suspect was out of town the day of the crime. He concludes that the suspect is innocent.
Example 2: Determine if each conjecture is valid by the Law of Detachment. 3. Given: If you want to go on a field trip, you must have a signed permission slip. Zola has a signed permission slip. Conjecture: Zola wants to go on a field trip.
4. Given: If the side lengths of a rectangle are 3 ft and 4 ft, then its area is 12 ft 2 . A rectangle has side lengths of 3 ft and 4 ft. Conjecture: The area of the rectangle is 12 ft 2 .
Example 3: Determine if each conjecture is valid by the Law of Syllogism. 5. Given: If you fly from Texas to California, you travel from the central to the Pacific time zone. If you travel from the central to the Pacific time zone, then you gain two hours. Conjecture: If you fly from Texas to California, you gain two hours.
87
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# Cost Volume Profit is a powerful tool that helps managers understand the relationships among cost, volume
and profit. CVP focuses on how profits are affected by the following five elements:
1. Prices of products.
2. Volume or level of activity.
3. Per unit variable costs.
4. Total fixed costs.
5. Mix of products sold.
Contribution income statement emphasizes the behavior of costs and is prepared for management use and is not
ordinarily available to those outside the company.
Contribution Margin
## Contribution Margin: the amount remaining
from sales revenue after variable expenses
have been deducted.
- The amount that is available to cover
fixed expenses and then then to
provide profits for the period.
- To break even, the total CM in
dollars must equal the fixed costs.
- If each unit sold provides \$100 in CM then, The number of units needed to break even = Fixed -
Expenses/ CM per unit
- Once the break-even point is reached, operating income will increase by the unit CM for each
- To summarize, if sales are zero, the companys operating loss equals its fixed expenses. Each unit that is
sold reduces the loss by the amount of the unit CM. Once the break-even point has been reached, each
additional unit sold increases the companys operating profit by the amount of the unit CM.
## - The relationships among revenues, costs, and level of
activity presented in graphic form.
- 3 Steps:
1. Plot the line parallel to the volume axis that
represents total fixed expenses. For Acoustic
Concepts, total fixed expenses are \$35,000.
2. Plot the line representing total expenses (fixed
plus variable) at various activity levels. For
example, in Exhibit 41, total expenses at an
activity level of 600 speakers are calculated as
follows: Total expenses at other activity levels
are calculated using this approach.
3. Plot the line representing total sales dollars at
various activity levels. For example, in Exhibit
41, sales at an activity level of 600 speakers
are \$150,000 (600 speakers \$250 per
speaker). Total sales at other activity levels are
calculated using this approach.
4. The anticipated profit or loss at any given level
of sales is measured by the vertical distance
between the total revenue line (sales) and the
total expenses line (variable expenses plus
fixed expenses).
5. The break-even point is where the total
revenue and total expenses lines intersect.
## A simpler form of the CVP graph, called a profit graph. That
graph is based on the following equation: (graph pic not shown here but its just an upward sloping vertical line)
## Profit = Unit CM Q Fixed expenses
Rearrange the above equation to isolate Q and let profit = 0, to determine the break even number of units.
## Contribution Margin Ratio
The CM expressed as a
percentage of total sales is
referred to as the contribution
margin (CM) ratio. This ratio is
computed as follows:
## - The CM ratio shows how the CM is affected by a change in total sales.
- Change in CM = CM ratio Change in sales
- Operating income will also increase by the amount of change in the CM assuming FC remain the same.
- The effect on operating income of any dollar change in total sales can be computed by simply applying
the CM ratio to the dollar change.
## Key Formulas for Contribution Format Income Statements
Variable expense ratio is the ratio of variable expenses to sales.
This leads to a useful equation that relates the CM ratio to the variable expense ratio as follows:
## Change in Fixed Cost and Sales Volume
Incremental Analysis; An analytical approach that focuses only on those items of revenue, cost, and volume
that will change as a result of a decision.
Suppose the sales manager thinks that a \$10,000 increase in the monthly advertising budget (FC) would
increase monthly sales by \$30,000. Should the advertising budget be increased?
## Assuming no other factors need to
be considered, the increase in the
approved since it would lead to an
increase in operating income of
\$2,000.
## Since in this case only the fixed
costs and the sales volume
change, the solution can be
presented in an even shorter
format, as follows
Change in Variable Costs and Sales Volume
Suppose management wants to increase variable costs by \$10 to increase overall quality of products and expects
sales to increase from 400 units to 480 units.
- If variable costs go up, the CM per unit decreases from \$100 to \$90
- Higher quality
parts should be
used.
## Change in Fixed Costs, Selling Price, and Sales Volume
Suppose management wants to decrease selling price by \$20 and increase the advertising budget by \$15,000 to
increase sales by 50% from 400 units to 600 units.
## Changes should not be
incremental analysis to
the left.
\$700 decrease in
operating income can be
verified by creating a
contribution income
statement as the one in
the left.
## Change in Cariable Cost, Fixed Cost, and Sales Volume
Pay sales commission by \$15 per unit sold, rather than pay flat wages of \$6000. We expect sales to increase by
15% to 460 units from the current 400.
Changing the sales staff from
a salaried basis to a
commission basis will affect
both fixed and variable costs.
Fixed costs will decrease by
\$6,000, from \$35,000 to
\$29,000. Variable costs per
unit will increase by \$15,
from \$150 to \$165, and the
unit CM will decrease from
\$100 to \$85:
## Change in Regular Selling Price
Can sell in bulk 150 speaker to a wholesaler if a price is worked out.
If the company wants to increase profits by \$3,000, what should the sale price be per unit.
- No fixed expenses are included in the computation. This is because fixed expenses are not affected by
the bulk sale, so all of the additional revenue that is in excess of variable costs goes to increasing the
profits of the company.
## Break Even Computations
The break-even point can be computed using either the equation method or the formula methodthe two
methods are equivalent.
Equation Method:
## - \$250Q = \$150Q + \$35,000 + \$0
The break-even point in sales dollars can be computed by multiplying the break- even level of unit sales by the
selling price per unit 350 speakers \$250 = \$87,500
The break-even point in total sales dollars, X, can also be directly computed as follows:
## Sales = Variable expenses + Fixed expenses + Profits
X = 0.60X + \$35,000 + \$0
X = \$87,500
## \$87,500 \$250 = 350 speakers
Note: In Break Even Analysis, the numbers should always be rounded up for sales level and units.
The Formula Method: The approach centers on the idea discussed earlier that each unit sold provides a certain
amount of CM that goes toward covering fixed costs.
## To find how many units must be sold to
break even, we simply rearrange the profit
equation to divide total fixed costs by the unit
CM.
A variation of this method uses the CM ratio instead of the unit CM. The result is the break-even in total sales
dollars rather than in total units sold:
## Target Operating Profit Analysis
Suppose management wants to earn a target net income of \$40,000. How many speakers have to be sold.
## The Equation Method:
Sales = Variable expenses + Fixed expenses + Profits
\$250Q = \$150Q + \$35,000 + \$40,000
Q = 750 speakers
Thus, the target operating profit can be achieved by selling 750 speakers per month, which represents \$187,500
in total sales (\$250 750 speakers).
The Formula Method: The second approach involves expanding the formula used to determine break-even
units to include the target operating profit as follows:
The dollar sales needed to attain the target operating profit can be computed as follows:
## After Tax Analysis:
In general, operating profit after taxes can be computed as a fixed percentage of income before taxes. To
calculate income taxes, we multiply the tax rate (t) by the operating profit before taxes (B). Therefore, after-tax
profit is equal to profit before taxes (1 t) and is derived as follows:
## Whenever target operating profit is
expressed on an after-tax basis, the
equation method can be used as
described earlier, except that target
operating profit must be restated to a
pre-tax basis.
Margin of Safety: The excess of budgeted (or actual) sales over the break- even volume of sales.
It states the amount by which sales can drop before losses begin to be incurred.
## Margin of safety = Total budgeted (or actual) sales Break-even sales
Cost Structure: The relative proportion of fixed and variable costs in an organization.
## Cost Structure and Profit Stability:
- Comparing high fixed cost oriented companies vs. high variable cost oriented companies.
- Bogside farm has a high variable cost, and sterling farm has a high fixed costs.
If sales are expected to be greater than \$100,000 then sterling farms (low VC) is better because it has a
higher CM and its operating income will therefore increase more rapidly as sales increase.
## What if sales drop below \$100,000.
Bogside Farm is less vulnerable to downturns than Sterling Farm for two reasons.
- First, due to its lower fixed expenses, Bogside Farm has a lower break-even point and a higher margin of
safety, as shown by the computations above. Therefore, it will not incur losses as quickly as Sterling
Farm in periods of sharply declining sales.
- Second, due to its lower CM ratio, Bogside Farm will not lose CM as rapidly as Sterling Farm when
sales fall off. Thus, Bogside Farms income will be less volatile.
Operating Leverage: a measure of how sensitive operating income is to percentage changes in sales. Operating
leverage acts as a multiplier.
- If operating leverage is high, a small percentage increase in sales can produce a much larger percentage
increase in operating income.
The degree of operating leverage is a measure, at a given level of sales, of how a percentage change in sales
volume will affect profits. It is computed by the following formula:
Bogside = 4
Sterling = 7
In general, this relation between the percentage change in operating income is given by the following formula:
Percentage change in operating income = Degree of operating leverage Percentage change in sales
- If two companies have the same total revenue and same total expense but different cost structures, then
the company with the higher proportion of fixed costs in its cost structure will have higher operating
leverage.
- The degree of operating leverage is greatest at sales levels near the break-even point and decreases as
sales and profits rise.
o Because as you get closer to the break even point, net income which is the denominator of the
degree of operating leverage formula, approaches 0.
Indifference Analysis: At which point will a company be indifferent between using a Labour intensive
program (high VC and low fixed) and a Capital intensive production system (high FC and low VC) assuming
the quality is the same.
We can calculate the point at which Goodwin will be indifferent about using a LIP system versus a CIP system
as follows:
1. Determine the unit CM multiplied by the number of units (Q) minus the total fixed costs of each
alternative.
2. Set up an equation with each alternative on opposite sides of the equal sign.
3. Solve for Q, the indifference point:
\$12Q \$1,800,000 = \$18Q \$3,600,000
\$6Q = \$1,800,000
Q = 300,000 units
Shortcut:
## The Definition of Sales Mix
Sales Mix: the relative proportions in which a companys products are sold.
- Profits will be greater if high-margin rather than low-margin items make up a relatively large proportion
of total sales.
- A shift in the sales mix from high-margin items to low-margin items can cause total profits to decrease
even though total sales may increase.
- Conversely, a shift in the sales mix from low-margin items to high-margin items can cause the opposite
effecttotal profits may increase even though total sales decrease.
Sales Mix and Break-Even Analysis:
The break-even point is \$X in sales, which was computed by dividing the fixed costs by the companys overall
CM ratio.
Calculating the break-even point in units when a company sells more than one product. The approach is based
on calculating a weighted-average CM per unit for the multiple products based on the existing unit sales mix
and the individual CM per unit for each product.
Note that the sales mix percentages are based on the unit sales of each product as a percentage of total sales.
The final step is to divide total fixed costs by the weighted-average CM per unit.
Assumptions of the CVP Analysis
## A number of assumptions typically underlie CVP analysis:
1. Selling price is constant throughout the entire relevant range. The price of a product or service does not
change as volume changes.
2. Costs are linear throughout the entire relevant range, and they can accurately be divided into variable
and fixed elements. The variable element is constant per unit, and the fixed element is constant in total
over the entire relevant range.
3. In multi-product companies, the sales mix is constant.
4. In manufacturing companies, inventories do not change. The number of units produced equals the
number of units sold (this assumption is considered further in Chapter 8).
SUMMARY
1. Costvolumeprofit (CVP) analysis is based on a simple model of how contribution margin (CM) and
operating income respond to changes in selling prices, costs, and vol- ume. The analysis is based on the
contribution income statement approach and requires a detailed understanding of cost behaviour. [LO1]
2. A CVP graph depicts the relationships between sales volume in units and fixed expenses, variable
expenses, total expenses, total sales, and profits. The CVP graph is useful for de- veloping intuition
about how costs and profits respond to changes in sales volume. [LO2]
3. The CM ratio is the ratio of the total CM to total sales. This ratio can be used to estimate the effect of a
change in total sales on operating income. [LO3]
4. The techniques of CVP analysis can be used to estimate the effects on CM and operating profit of
changes to sales volume, fixed costs, variable costs per unit, and selling prices. A useful aspect of the
analysis is that managers can evaluate the profit impact of the trade- offs inherent to many operating
decisions, such as increasing advertising costs to boost sales volumes. [LO4]
5. The break-even point is the level of sales (in units or in dollars) at which the company generates zero
profits. The break-even point can be computed using several different tech- niques that are all based on
the simple profit equation. [LO5]
6. The profit equation can also be used to compute the level of sales required to attain a target profit. [LO6]
7. The margin of safety is the amount by which the companys current sales exceed break- even sales.
[LO7]
8. The degree of operating leverage measures the effect of a percentage change in sales on the companys
operating income. The higher the degree of operating leverage, the more sensitive operating income will
be to a change in sales. The degree of operating leverage is not constantit depends on the companys
current level of sales. [LO8]
9. The profits of a multi-product company are affected by its sales mix. Changes in the sales mix can affect
the break-even point, margin of safety, and other critical measures. [LO9] | 3,364 | 15,324 | {"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} | 3.171875 | 3 | CC-MAIN-2019-26 | latest | en | 0.926702 |
https://studylib.net/doc/9999472/vocabulary---doral-academy-preparatory | 1,610,724,342,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495901.0/warc/CC-MAIN-20210115134101-20210115164101-00205.warc.gz | 665,377,492 | 13,029 | # Vocabulary - Doral Academy Preparatory
NAME _____________________________________________ DATE ____________________________ PERIOD _____________
Chapter 1 Vocabulary
algebraic expression
coefficient
continuous function
coordinate system
dependent variable
discrete function
domain
end behavior
exponent
factors
function
identity
independent variable
intercept
like terms
multiplicative inverses
open sentences
order of operations
power
range
replacement set
reciprocal
solution set
symmetry
variable
Choose a term from the vocabulary list above to complete the sentence.
1. In the algebraic expression 8q, the letter q is called a(n)
________________.
1.________________________
2. An expression like c3 is an example of a(n) __________________
and is read “c cubed.”
2. ________________________
3. A function graphed with a line or smooth curve is
called a(n) __________________.
3. ________________________
4. The process of finding a value for a variable that results
in a true sentence is called solving the ___________________ .
4. ________________________
5. ____________ are terms that contain the same variables,
with corresponding variables having the same power.
5. ________________________
6. The ____________ of a term is the numerical factor.
6. ________________________
7. The set of the first number of the ordered pairs of a function
is the _________________ .
7. ________________________
8. In a(n) ______________ , there is exactly one output for each input.
8. ________________________
9. The set of second numbers of the ordered pairs in a relation
is the _____________ of the relation.
9. ________________________
Define each term in your own words.
10. end behavior
10. _______________________
11. solution set
11. _______________________
Chapter 1
60
Glencoe Algebra 1 | 394 | 1,794 | {"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} | 2.765625 | 3 | CC-MAIN-2021-04 | longest | en | 0.717212 |
https://www.coursehero.com/file/6680909/option-strategies-handout/ | 1,512,960,356,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512054.0/warc/CC-MAIN-20171211014442-20171211034442-00236.warc.gz | 694,530,104 | 25,731 | option_strategies_handout
# option_strategies_handout - BASIC OPTION STRATEGIES Example...
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Example 1 : ABC trades at \$97.50. ABC calls with an exercise price of \$95 trade at \$6.25 and ABC puts with an exercise price of \$93 trade for \$0.90. What are the intrinsic values and time values for these options? Intrinsic Value Time Value ABC 95 Calls P-X = (97.50 – 95)=\$2.50 Vc – IV = (6.25 – 2.50)=\$3.75 ABC 93 Puts X-P = \$0 Vp – IV = \$0.90 Example 2 : An investor buys a call option with a strike price of \$35. The option price is \$6.50. Draw a payoff table and a profit/loss diagram. What is the breakeven price? What is the maximum gain on this position? What is the maximum loss? If at maturity the stock price is \$43, what is the total profit/loss? Price of stock @ expiration 25 30 35 40 45 Cost of Call Gain/loss Net payoff Example 3 : An investor writes a call option with a strike price of \$35. The option price is \$6.50. Draw a payoff table and a profit/loss diagram. What is the breakeven price? What is the maximum gain on this position? What is the maximum loss? If at maturity the stock price is \$43, what is the total profit/loss? Price of stock @ expiration 25 30 35 40 45 Cost of Call Gain/loss Net payoff Example 4 : An investor buys a put option with a strike price of \$20. The option price is \$4.00. Draw a payoff table and a profit/loss diagram. What is the breakeven price? What is the maximum gain on this position? What is the maximum loss? If at maturity the stock price is \$22, what is the total profit/loss? Price of stock
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Ask a homework question - tutors are online | 538 | 2,054 | {"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} | 3.484375 | 3 | CC-MAIN-2017-51 | latest | en | 0.81901 |
http://talk.elementsofprogramminginterviews.com/t/bounds-for-11-5-variant/1178 | 1,547,731,848,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658981.19/warc/CC-MAIN-20190117123059-20190117145059-00447.warc.gz | 216,361,307 | 2,701 | # Bounds for 11.5 variant
#1
I understand the logic that’s supposed to be applied, but I’m confused on how best to represent the bounds. in particular where x > 1 and y < 1. Say x = 8 and y = 0.0000001, we’re looking a huge number to be the answer, let alone find the upper bound. I’m starting to trip myself up on the various combinations now. There’s got to be an easier way? | 104 | 379 | {"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} | 2.953125 | 3 | CC-MAIN-2019-04 | latest | en | 0.88488 |
https://modcalculator.com/molar-mass-of/AlF | 1,696,273,720,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511002.91/warc/CC-MAIN-20231002164819-20231002194819-00654.warc.gz | 443,524,695 | 6,346 | ## Tool Overview: Molar Mass of Aluminium monofluoride (AlF)
### Solving for the atomic mass of Aluminium monofluoride (AlF)
Need to know the atomic mass of a Aluminium monofluoride molecule? Our molar mass calculator uses the periodic table and the chemical formula to solve for the molar mass of a chemical compound based on the compound's empirical formula. The calculator takes the elemental composition of the compound and weighs the elements to get an empirical formula mass. Note that the calculator assumes a pure substance - if you're aware of dilution or impurities, make appropriate adjustments for the molarity of a given substance.
This project started with as a molar mass calculator for chemical reactions. You can use our calculator to solve for the theoretical yield of an experiment. We also have a percent yield calculator which can help you apply this to actual experiments. Use the mole ratio and empirical formula to understand the limits of the reactants.
Other terms: atomic mass of Aluminium monofluoride, molar mass of Aluminium monofluoride, molecular mass,
### How Does The Molar Mass Calculator Work?
We take the formula you provide (NaCl - common table salt - in our default example) and unpack it into the component elements. Then we compare each atom against a table of the standard atomic weights for that element. We present the results in a table at the bottom of the molar mass calculator - it will show the count of atoms, the atomic weight of each element, and the molecular weight for the molecule. It solves for total mass of a molecular formula (average molecular weight).
From there we break the formula for Aluminium monofluoride into parts - a Aluminum atom, a Fluorine atom, etc.
We don't have brackets implemented (yet), so you will need to unpack any bracketed expressions. They don't affect the weight anyhow. Simply take each element and multiple it by the number of times the bracketed structure occurs. For example: (C6H5)3PCCO => C18H15PCCO
### Finding Molar Mass for Other Chemical Compounds
Our molar mass calculator has this for a variety of other compounds: sodium chloride, carbon dioxide, sulfuric acid, glucose...
### Bookmarking, Save, and Share Results
The tool is designed so you can flip between different parts of a problem set. We recommend you bookmark it so you can refer back to it. You can also share results with a study partner or tutor by hitting calculate and copying the URL for this page. When your study partner opens up the URL, they will see your calculations. It's easy share & save results via email. (Be sure to hit calculate first, however)
You also have the option of saving links to the calculations in your research notes files, so you can quickly re-open or check them later. Again - hit calculate first so the URL is updated with your most recent changes. Then copy and save the url.
### What is Molar Mass in Chemistry?
Molar mass is an important concept in adapting chemical formulas to real world conditions. We may be able to balance a chemical equation and determine that one molecule of hydrogen combines with two of oxygen to make water (or the compound of your choice). But how would you set up the materials in the laboratory? Or if you were, for example, buying oxygen for a process, how would you determine how much to use to make a given quantity of water? Molar mass allows us to convert a chemical reaction into specific amounts of reagents required for the process. By converting the atomic interaction into grams, we can measure and use an appropriate amount of the necessary reagents. Formula mass helps us solve for this.
What Is Relative Atomic Mass / Relative Molecular Mass / Average Molecular Weight?
The relative atomic mass of a compound is the ratio of the average mass of the elements in a chemical compound to the atomic mass constant, which is defined as 1/12 the mass of a carbon 12 atom. For a single sample, the relative atomic mass of the sample is the weighted arithmetic mean of the masses of the individual atoms present in the sample (also known as the average atomic mass). This will vary by isotope of the element (carbon-12 vs. carbon-13, for example, since the two isotopes have a different atomic mass due to additional neutrons). In the real world, this can vary based on where the sample was collected - due to variances in the specific isotopes of the elements present (driven by differences in radioactive decay and how the material was aggregated to begin with).
### How to find Molar Mass
Take a standard chemistry formula for a molecule, split it up into the component atoms, and look up the molar weight of each atom. Add the weight of the atoms in the molecule and you have the molar mass for the molecule.
## Mathematics & Science Workbench
Content: theoretical yield calculator | 1,012 | 4,837 | {"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-2023-40 | latest | en | 0.867377 |
https://web2.0calc.com/questions/help_99497 | 1,532,235,976,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593010.88/warc/CC-MAIN-20180722041752-20180722061752-00269.warc.gz | 799,229,461 | 5,205 | +0
# Help
+1
237
2
+322
How to simplify the equation -3sqrt(5)/sqrt(15)
Davis Mar 2, 2017
#1
+7155
+6
$$\frac{-3\sqrt{5}}{\sqrt{15}} \\ = \frac{-3\sqrt{5}(\sqrt{15})}{\sqrt{15}(\sqrt{15})} \\ = \frac{-3\sqrt{75}}{15} \\ = \frac{-3(5\sqrt{3})}{15} \\ = \frac{-15\sqrt{3}}{15} \\ = -\sqrt{3}$$
:)
hectictar Mar 2, 2017
#1
+7155
+6
$$\frac{-3\sqrt{5}}{\sqrt{15}} \\ = \frac{-3\sqrt{5}(\sqrt{15})}{\sqrt{15}(\sqrt{15})} \\ = \frac{-3\sqrt{75}}{15} \\ = \frac{-3(5\sqrt{3})}{15} \\ = \frac{-15\sqrt{3}}{15} \\ = -\sqrt{3}$$
:)
hectictar Mar 2, 2017
#2
+322
+5
Thx
Davis Mar 2, 2017 | 312 | 593 | {"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} | 4.25 | 4 | CC-MAIN-2018-30 | latest | en | 0.371962 |
https://www.acmicpc.net/problem/20649 | 1,709,492,577,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476397.24/warc/CC-MAIN-20240303174631-20240303204631-00690.warc.gz | 615,943,120 | 8,986 | 시간 제한메모리 제한제출정답맞힌 사람정답 비율
1 초 512 MB28513511248.908%
## 문제
Farmer John has recently expanded the size of his farm, so from the perspective of his cows it is effectively now infinite in size! The cows think of the grazing area of the farm as an infinite 2D grid of square "cells", each filled with delicious grass (think of each cell as a square in an infinite chessboard). Each of Farmer John's $N$ cows ($1\le N\le 1000$) starts out in a different cell; some start facing north, and some start facing east.
Every hour, every cow either
• Stops (and then remains stopped from that point on) if the grass in her current cell was already eaten by another cow.
• Eats all the grass in her current cell and moves one cell forward according to the direction she faces.
Over time, each cow therefore leaves a barren "rut" of empty cells behind her.
If two cows move onto the same grassy cell in the same move, they share the cell and continue moving in their respective directions in the next hour.
Farmer John isn't happy when he sees cows that stop grazing, and he wants to know who to blame for his stopped cows. If cow $b$ stops in a cell that cow $a$ originally ate, then we say that cow $a$ stopped cow $b$. Moreover, if cow $a$ stopped cow $b$ and cow $b$ stopped cow $c$, we say that cow $a$ also stopped cow $c$ (that is, the "stopping" relationship is transitive). Each cow is blamed in accordance with the number of cows she stopped. Please compute the amount of blame assigned to each cow -- that is, the number of cows she stopped.
## 입력
The first line of input contains $N$. Each of the next $N$ lines describes the starting location of a cow, in terms of a character that is either N (for north-facing) or E (for east-facing) and two nonnegative integers $x$ and $y$ ($0\le x\le 10^9$, $0\le y\le 10^9$) giving the coordinates of a cell. All $x$-coordinates are distinct from each-other, and similarly for the $y$-coordinates.
To be as clear as possible regarding directions and coordinates, if a cow is in cell $(x,y)$ and moves north, she ends up in cell $(x,y+1)$. If she instead had moved east, she would end up in cell $(x+1, y)$.
## 출력
Print $N$ lines of output. Line $i$ in the output should describe the blame assigned to the $i$th cow in the input.
## 예제 입력 1
6
E 3 5
N 5 3
E 4 6
E 10 4
N 11 1
E 9 2
## 예제 출력 1
0
0
1
2
1
0
In this example, cow 3 stops cow 2, cow 4 stops cow 5, and cow 5 stops cow 6. By transitivity, cow 4 also stops cow 6. | 702 | 2,478 | {"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} | 3.078125 | 3 | CC-MAIN-2024-10 | latest | en | 0.952621 |
https://www.educationworld.com/a_tsl/archives/02-1/lesson023.shtml | 1,725,786,560,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650976.41/warc/CC-MAIN-20240908083737-20240908113737-00107.warc.gz | 731,890,609 | 32,729 | >> A Tsl >> Archives >> 02 1 >>
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Home > Teacher Lesson Plans > Mathematics > Lesson Plan
L E S S O N P L A N
• Mathematics
Applied Math
3-5
Brief Description
Students blow bubbles, measure them, predict their staying power, and graph the results.
Objectives
Students
• predict how long their bubbles will stay "alive."
• graph the results of the bubble competition.
Keywords
bubble, graph, math, predict
Materials Needed
• bubble solution
• graph paper
Lesson Plan
Students begin the lesson by filling in an experiment procedure plan. The plan should include
• a list of materials used
• the purpose of the experiment
• a prediction about how long (in seconds) bubbles will "survive" before they pop
• experiment procedures
• results and conclusions
When the students write their predictions, have them predict the longest amount of time a bubble will last. Teachers should provide students with a graph. The students will graph how long their bubbles last. The best way to handle this would be to have the students pick one bubble and count the seconds until it hits the ground or choose five bubbles and graph the results of those five. Then have students graph the results of their classmates' bubbles or their own five bubbles.
Assessment
After the activity, discuss the results and fill in the conclusion part of the experiment.
Submitted By
Kellie Slaughter, Belmont Hills Elementary School, Smyrna, Georgia
Originally published 5/31/2002
Last updated 03/18/2007
To help us keep our Lesson Plan Database as current as possible, please e-mail us to report any links that are not working. | 355 | 1,630 | {"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-2024-38 | latest | en | 0.909384 |
http://greenemath.com/Algebra1/29/AddingPolynomialsLesson.html | 1,553,282,374,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202688.89/warc/CC-MAIN-20190322180106-20190322202106-00438.warc.gz | 89,622,360 | 11,379 | Lesson Objectives
• Demonstrate an understanding of the definition of a term and like terms
• Learn the definition of a polynomial, monomial, binomial, and trinomial
• Learn how to find the degree of a term and the degree of a polynomial
• Learn how to add polynomials
• Learn how to subtract polynomials
## How to Add and Subtract Polynomials
In this lesson, we will learn how to add and subtract polynomials. Let’s begin by learning about polynomials in general. First and foremost, let’s review the concept of a term. A term is a number, variable, or the product of a number and one or more variables.
3xyz » is a term
-2x2y4 » is a term
9 » is a term
In each example above: 3xyz, -2x2y4, and 9, we have a single term. When we combine terms together using the addition "+" or subtraction "-" operations we obtain an algebraic expression:
3x3 + 7x2 - 5 » algebraic expression
-9x2y - xy + 9 » algebraic expression
When we have a number that multiplies a variable or variables, the number is referred to as a coefficient. When a number is not multiplying a variable, it is known as a constant.
2x + 7 » 2 - coefficient, 7 - constant
-3y + 5 » -3 - coefficient, 5 - constant
When we talk about "like terms", these are terms with the exact same variable parts. This means the variable or variables are the same and each variable is raised to the same power:
3x, 7x » like terms - same variable (x), raised the same power (1)
-5x3, 2x3 » like terms - same variable (x), raised to the same power (3)
9x, 3y » not like terms - different variables (x, y)
11x4, 11x3 » not like terms - same variable (x), raised to different powers (4, 3)
We can combine like terms by performing operations with the coefficients and leaving the variable part unchanged. We can show this with the distributive property:
5x + 3x
Add the coefficients (5 and 3) and leave the variable part (x) unchanged:
5x + 3x = (5 + 3)x = 8x
9xy - 13xy
Subtract the coefficients (9 - 13) and leave the variable part (xy) unchanged:
9xy - 13xy = (9 - 13)xy = -4xy
### What is a Polynomial?
A polynomial in x (or some other variable) is a single term or the sum of a finite amount of terms (axn) where:
a is any real number
n is any whole number
It is very important to understand that n, the exponent cannot be negative. This will come up early in your studies of polynomials.
5x3 - 2x2 + 1 » polynomial
Each term matches the format of axn where a is a real number and n is a whole number. When we think about the constant 1, we can write this as:
1x0 = 1 • 1 = 1
Recall raising a number or variable to the power of 0 results in 1. We will see this trick used often in our study of algebra.
5x-3 - 2x2 + 1 » not a polynomial
Notice the exponent of (-3) on x, this violates the definition of a polynomial.
We have special names for polynomials with one term, two terms, and three terms. A polynomial with one term only is known as a monomial. A polynomial with two terms only is known as a binomial. Lastly, a polynomial with three terms is known as a trinomial.
2x2 - 11 » binomial (two terms)
-7x5 + x2 - 5 » trinomial (three terms)
9 » monomial (one term)
### Degree of a Polynomial
The degree of a term is the sum of the exponents on all variables of the term.
9x3 » degree of 3
-4x2y3 » degree of 5
7x4y8z11 » degree of 23
The degree of a polynomial is the largest degree of any non-zero term of the polynomial.
12x5 - 2x2 + 1 » degree of 5
x9y12 + 2x3y4 - 7 » degree of 21
### Writing Polynomials in Standard Form
When we work with polynomials, it is custom to write them in standard form. This means the term with the largest degree is first or leftmost, followed by the term with the next largest degree and so on. Let's write the following polynomial in standard form:
7x2 + 3 + 2x9
We want the term with the highest degree (2x9) first:
2x9
We follow this by the term with the next largest degree (7x2):
2x9 + 7x2
We continue to follow this format, our next and final term will be 3:
2x9 + 7x2 + 3
When we add two or more polynomials together, we simply combine like terms. Let's look at a few examples.
Example 1: Simplify each
(1 + 5x) + (1 + 4x)
To add polynomials, we will just combine like terms. To make this easy, let's drop the parentheses and rearrange the addition:
5x + 4x + 1 + 1
5x + 4x = (5 + 4)x = 9x
1 + 1 = 2
(1 + 5x) + (1 + 4x) = 9x + 2
Example 2: Simplify each
(2x5 - 4) + (x5 - 1)
To add polynomials, we will just combine like terms. To make this easy, let's drop the parentheses and rearrange the addition:
2x5 + 1x5 + (-4) + (-1)
2x5 + 1x5 = (2 + 1)x5 = 3x5
(-4) + (-1) = (-5)
(2x5 - 4) + (x5 - 1) = 3x5 - 5
Example 3: Simplify each
(11x + 11x2) + (7x2 - 3x) + (x - x2)
To add polynomials, we will just combine like terms. To make this easy, let's drop the parentheses and rearrange the addition:
11x2 + 7x2 + (-1x2) + 11x + (-3x) + x
11x2 + 7x2 + (-1x2) = (11 + 7 + (-1))x2 = 17x2
11x + (-3x) + x = (11 + (-3) + 1)x = 9x
(11x + 11x2) + (7x2 - 3x) + (x - x2) = 17x2 + 9x
### How to Subtract Polynomials
When we subtract one polynomial form another, we change the subtraction operation to addition and change each term inside of the parentheses into its opposite. Once this is done, we can add our polynomials by combining like terms. Many teachers will perform this operation by changing the subtraction operation to addition and then placing a (-1) outside of the parentheses. The (-1) will be distributed to each term inside of the parentheses and change the sign of each term.
Example 4: Simplify each
(14 - 7x) - (2 + 12x)
Let's change our subtraction to addition and change each term inside of the parentheses to its opposite:
(14 - 7x) + (-2 - 12x)
Now we can add, let's drop our parentheses and rearrange terms:
-7x + (-12x) + 14 + (-2)
-7x + (-12x) = (-7 + (-12))x = -19x
14 + (-2) = 12
(14 - 7x) + (-2 - 12x) = -19x + 12
Example 5: Simplify each
(6x2 + 9 - 10x4) - (4x4 + 6) - (-13 - 9x4)
Let's change our subtraction to addition and change each term inside of the parentheses to its opposite:
(6x2 + 9 - 10x4) + (-4x4 - 6) + (13 + 9x4)
Now we can add, let's drop our parentheses and rearrange terms:
-10x4 + (-4x4) + 9x4 + 6x2 + 9 + (-6) + 13
-10x4 + (-4x4) + 9x4 = (-10 + (-4) + 9)x4 = -5x4
6x2 » no like terms to combine with
9 + (-6) + 13 = 16
-5x4 + 6x2 + 16 | 2,007 | 6,272 | {"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} | 5.03125 | 5 | CC-MAIN-2019-13 | latest | en | 0.873704 |
https://www.resavr.com/comment/dice-product-probability-113788 | 1,550,647,181,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247494485.54/warc/CC-MAIN-20190220065052-20190220091052-00056.warc.gz | 952,691,975 | 7,471 | ### Dice product probability.
Let P(n, p) be the probability that the product of n dice is <= p. You want to find P(3, 30).
We can use the Bayes formula to break this out based on the result of the first roll.
``````P(3, 30) =
P(3, 30|first roll = 1)P(first roll = 1) +
P(3, 30|first roll = 2)P(first roll = 2) +
P(3, 30|first roll = 3)P(first roll = 3) +
P(3, 30|first roll = 4)P(first roll = 4) +
P(3, 30|first roll = 5)P(first roll = 5) +
P(3, 30|first roll = 6)P(first roll = 6)
``````
P(first roll = 1) = P(first roll = 2) = ... = 1/6.
So we just need to figure out P(3, 30|first roll = 1), P(3, 30|first roll = 2), ... etc.
The key is to realize that if the first roll is (say) 4, then the product of all three dice is <= 30 if the product of the remaining two dice <= (30/4).
So P(3, 30|first roll = 4) = P(2, 7).
Reasoning similarly:
``````P(3, 30) =
P(2, 30)/6 +
P(2, 15)/6 +
P(2, 10)/6 +
P(2, 7)/6 +
P(2, 6)/6 +
P(2, 5)/6
``````
We can calculate P(2, ...) by making a table of the 36 possibilities and simply counting... or we can apply Bayes again!
``````P(3, 30) =
(P(1, 30)/6 + P(1, 15)/6 + P(1, 10)/6 + P(1, 7)/6 + P(1, 6)/6 + P(1, 5)/6)/6 +
(P(1, 15)/6 + P(1, 7)/6 + P(1, 5)/6 + P(1, 4)/6 + P(1, 3)/6 + P(1, 2)/6)/6 +
(P(1, 10)/6 + P(1, 5)/6 + P(1, 3)/6 + P(1, 2)/6 + P(1, 2)/6 + P(1, 1)/6)/6 +
(P(1, 7)/6 + P(1, 3)/6 + P(1, 2)/6 + P(1, 1)/6 + P(1, 1)/6 + P(1, 1)/6)/6 +
(P(1, 6)/6 + P(1, 3)/6 + P(1, 2)/6 + P(1, 1)/6 + P(1, 1)/6 + P(1, 1)/6)/6 +
(P(1, 5)/6 + P(1, 2)/6 + P(1, 1)/6 + P(1, 1)/6 + P(1, 1)/6 + P(1, 0)/6)/6
``````
Pulling out a common factor of 1/216, we get:
``````P(3, 30) = (1/216)(
(6 + 6 + 6 + 6 + 6 + 5) +
(6 + 6 + 5 + 4 + 3 + 2) +
(6 + 5 + 3 + 2 + 2 + 1) +
(6 + 3 + 2 + 1 + 1 + 1) +
(6 + 3 + 2 + 1 + 1 + 1) +
(5 + 2 + 1 + 1 + 1 + 0)
)
``````
Or finally:
``````P(3, 30) = (1/216)(29 + 19 + 14 + 14 + 10) = 86/216 = 43/108 = about 39.8%
`````` | 974 | 1,892 | {"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} | 4.53125 | 5 | CC-MAIN-2019-09 | latest | en | 0.596056 |
https://sellfy.com/p/4ybh/ | 1,521,305,376,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645248.22/warc/CC-MAIN-20180317155348-20180317175348-00062.warc.gz | 695,771,291 | 9,284 | .RAR
# Project 3
If ε belongs in the set, represent it as #
If ε belongs in the set, it should be listed before any other elements
All other elements of the set should be sorted in dictionary order (specifically, the C function strcmp from string.h defines the sorting order)
OutputThe output for the example grammar given previously when run with the command line argument of 1 like ./a.out 1:FIRST(decl) = { ID } FIRST(idList1) = { #, COMMA } FIRST(iDList) = { ID } Task 2: FOLLOW SetsFor each of the non-terminals of the input grammar, in the order that they appear in the non-terminals section of the input, compute the FOLLOW set for that non-terminal and output one line in the following format:FOLLOW(<symbol) = { <set_items } Where <symbol should be replaced by the non-terminal and <set_items should be replaced by the comma-separated list of elements of the FOLLOW set ordered in the following manner:
If EOF belongs in the set, represent it as \$
If EOF belongs in the set, it should be listed before any other elements
All other elements of the set should be sorted in dictionary order (specifically, the C function strcmp from string.h defines the sorting order)
OutputThe output for the example grammar given previously when run with the command line argument of 2 like ./a.out 2:FOLLOW(decl) = { \$ } FOLLOW(idList1) = { colon } FOLLOW(idList) = { colon } EvaluationYour submission will be graded on passing the automated test cases.The test cases (there will be multiple test cases in each category, each with equal weight) will be broken down in the following way (out of 105 points):
Task 0 for all grammars: 20 points
Calculating FIRST sets for grammars without ε: 30 points
Calculating FIRST sets for grammars with ε: 20 points
Calculating FOLLOW sets for grammars without ε: 25 points
Calculating FOLLOW sets for grammars with ε: 10 points
Note that the output produced by your program should exactly match the expected output, and no partial credit will be given.For every test case, there are 3 expected output files respectively for each task. For example, for the input file test01.txt, there is test01.txt.expected0 for the expected output of task 0, test01.txt.expected1 for the expected output of task 1 and test01.txt.expected2 for the expected output of task 2. There is also a modified version of the test script test1.sh that you should use to test your code with the provided test cases for this project. You can find this test script along with the project material.SubmissionSubmit your code on the course submission website:https://cse340.fulton.asu.edu/cse340/ | 595 | 2,599 | {"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-2018-13 | latest | en | 0.808506 |
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03-24-2010, 04:12 PM #1
nathan
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partigyle, getting started
I'm doing a big ass Hoppy Valley Double IPA again soon. Last time I had a lot of sugar left in my mash tun.
I'm thinking this time I would like to use some of that.
Eventually I'd like to work it out so I can run the recipe as a grain-only partigyle, but I need to inch into it, as I'll need a better handle on how to calculate my gravities.
My plan is to take my original recipe for the Hoppy Valley DIPA and add 10-20% more base malt.
I'll do the math to determine how many points will end up in the kettle at boil volume (based on pre-boil gravity), then try and put together a SS (if I have time, if not I'll just use a calculator) to input current volume and current gravity while running off. It will output the total points in kettle. When I hit that number from my target, I will stop runoff and top off with water and get that boil going.
Then I will run off the remainder, tracking the sugars down to 2 degrees plato. I stop at 2 degrees, stir and take a measure of the gravity and volume. I can then calculate an extra-light DME addition to achieve a desired Pale Ale or ESB pre-boil gravity, and water to get to volume.
Having learned how much sugar I can get from the second runnings (I fly sparge) I can adjust upward with my DIPA mash in future batches to introduce more sugars from the mash and work towards an all-grain end result.
Sound reasonable?
My DIPA has a specific recipe and I'm okay with the base malt adjustments, but I need to hit specific volumes and numbers for it. I'm a bit more open with the 2nd batch, as I'm just trying to figure out ways of doing the partigyle for fun (and beer).
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03-24-2010, 04:17 PM #2
nealf
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Sounds like a plan to me... some thoughts would be you could just drain off 7 gallons or whatever your preboil amount... add some more 2row and maybe a touch more crystal (mash it for a little while) to increase the mouthfeel & gravity. I have to assume you are going to be mashing your IIPA a few degrees lower than you would normally mash an APA and doing this might make up for it a bit.
Here is a really simple calculator that I used when doing my partigyles:
http://www.astrocaver.com/java/Parti-Gyle.html
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03-24-2010, 04:18 PM #3
david_42
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Can you quantify how much sugar you left behind before? I batch sparge and my numbers run 50%, 33% and 16%, so it is very easy for me to set up a partigyle. I just use half of the first sparge or all of the second.
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03-24-2010, 04:21 PM #4
nathan
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my pre-boil volume for DIPA is 18.5 gallons. Post boil gravity is 1.090 (this time).
I try to keep everything in 15.5 gallon batches now (everyone gets a carboy).
I think my runoff was at about 8 plato when I stopped the runoff on the last DIPA.
I guess there's no reason why I could not batch sparge it... I'll have to consider the options.
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03-24-2010, 04:27 PM #5
nathan
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that calculator suggests that with a 50/50 (two 15 gallon batches), for a 1.090/1.045 I would start with a 1.067. That is planning for a 30 gallon batch, right?
Calculating my recipe to that has 78 pounds of grain. I've got a blichmann 20G, so I'm not going to fit that much in it to mash. If I track my runoff down to 2P, I can take a volume measurement on this time through and see what it is. Is it close to a 10g batch? If so I can pull out my keggle, dust it off, and plan on doing the 2nd batch as a 10g in the future. I'll need to plan the first one based on carboys people can take home, but knowing it in the future will mean I can say 3 carboys of DIPA and 2 carboys of ESB, or whatever.
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Similar Threads Thread Thread Starter Forum Replies Last Post maskednegator Recipes/Ingredients 19 03-11-2010 04:49 PM nealf Recipes/Ingredients 14 03-11-2010 01:58 PM bbrim All Grain & Partial Mash Brewing 1 03-04-2009 12:46 PM MNBugeater All Grain & Partial Mash Brewing 10 08-27-2008 08:36 PM Orfy Cooking & Pairing 12 02-24-2008 08:23 PM | 1,410 | 5,219 | {"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-2015-14 | latest | en | 0.935194 |
https://testbook.com/question-answer/in-the-given-circuithaving-a-current-depende--611b6be4c293ca13d00eaf38 | 1,638,670,240,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363134.25/warc/CC-MAIN-20211205005314-20211205035314-00092.warc.gz | 635,517,256 | 31,054 | # In the given circuit having a current-dependent voltage source, what is the voltage of the dependent source?
This question was previously asked in
PGCIL DT Electrical 13 Aug 2021 Official Paper (NR I)
View all PGCIL Diploma Trainee Papers >
1. 8 V
2. 6 V
3. 5 V
4. 7.2 V
Option 2 : 6 V
## Detailed Solution
CONCEPT:
There are two types of Kirchoff’s Laws:
Kirchoff’s first law:
• This law is also known as junction rule or current law (KCL). According to it the algebraic sum of currents meeting at a junction is zero i.e. Σ i = 0.
• In a circuit, at any junction, the sum of the currents entering the junction must be equal the sum of the currents leaving the junction i.e., i1 + i3 = i2 + i4
• This law is simply a statement of “conservation of charge” as if current reaching a junction is not equal to the current leaving the junction, charge will not be conserved.
Kirchoff’s second law:
• This law is also known as loop rule or voltage law (KVL) and according to it “the algebraic sum of the changes in potential in the complete traversal of a mesh (closed-loop) is zero”, i.e. Σ V = 0.
• This law represents “conservation of energy” as if the sum of potential changes around a closed loop is not zero, unlimited energy could be gained by repeatedly carrying a charge around a loop.
• If there are n meshes in a circuit, the number of independent equations in accordance with loop rule will be (n - 1).
Application:
Moving in the clockwise direction ie from DABC.
- 6 V + v1 - 5i V + v2 = 0
- 6 V + 4i - 5i + 6i = 0
i = 6 / 5 = 1.2 A
Voltage across dependent source is 5 i = 5 × 1.2 = 6 V | 485 | 1,611 | {"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-2021-49 | latest | en | 0.811488 |
http://mathhelpforum.com/calculus/165143-vector-fields.html | 1,480,988,696,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541864.44/warc/CC-MAIN-20161202170901-00211-ip-10-31-129-80.ec2.internal.warc.gz | 160,707,487 | 9,919 | 1. ## Vector Fields
Verify that the vector field 6 F(x,y) = (x-y)i + (x+y)j has the flow lines given by x(t) = Ae^(t)cos(t+α)i + Ae^(t)sin(t+α)j where A and α are arbitrary real constants.
Show that the flow lines described follow a generally spiral path, in one case degenerating into a point.
2. Originally Posted by Playthious
Verify that the vector field 6 F(x,y) = (x-y)i + (x+y)j has the flow lines given by x(t) = Ae^(t)cos(t+α)i + Ae^(t)sin(t+α)j where A and α are arbitrary real constants.
We only have to verify $\vec{x}\;'(t)=F[\vec{x(t)}]$ for all $t\in\mathbb{R}$ .
Show that the flow lines described follow a generally spiral path, in one case degenerating into a point.
If $A=0$ , then $\vec{x}(t)=(0,0)$ for all $t\in\mathbb{R}$ .
Regards.
Fernando Revilla | 251 | 777 | {"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": 0, "img_math": 0, "codecogs_latex": 5, "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-2016-50 | longest | en | 0.85529 |
https://www.physicsforums.com/threads/magnetism-in-infinite-conducting-slab.38818/ | 1,542,444,124,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743351.61/warc/CC-MAIN-20181117082141-20181117104141-00143.warc.gz | 963,214,177 | 12,647 | # Homework Help: Magnetism in infinite conducting slab
1. Aug 9, 2004
### daveed
"a conducting slab has infinite extent in the x and y directions and thickness L in the z direction. The slab is centered at z=0 and carries a uniform current density J=Ji where i, j, and k are unit vectors in the x, y, and z directiosn."
-Find the magnetic field B at all points.
-A square loop of side a is placed distance b above the slab. The loop has unit normal vector n=sin(q)i+cos(q)j and applied current J. what is the net force and net torque to the loop as a function of q?
-the applied current I is now removed from the loop and the current density in the slab J=Ji is reduced to zero over time T. The wire used to construct the loop has resistance/unit S. How much charge flows through each cross section of the loop wire due to the reduction in current density.
2. Aug 10, 2004
### Locrian
What have you tried so far?
It seems to me ampere's law would work well in determining the magnetic field; use a square loop such that you know exactly how much current passing through it and the only field is at the top and bottom. Integrating a line of wires across the surface and then again through the depth is another option, but sounds horrible to me.
Once you have the magnetic field, the second part should come together easily enough.
3. Dec 9, 2004
### daveed
i really dont know how to go about doing this when its not a single wire... sorry =(
4. Dec 9, 2004
### Tide
Have you considered the integral form of Ampere's Law?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 390 | 1,618 | {"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 | 3 | CC-MAIN-2018-47 | latest | en | 0.938271 |
https://getacho.com/what-is-1-54-as-a-percentage/ | 1,679,869,105,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946535.82/warc/CC-MAIN-20230326204136-20230326234136-00377.warc.gz | 320,349,220 | 12,607 | # What is 1.54 as a Percentage?
The number 1.54 is a decimal representation of a fractional number. It is an expression of a fraction where the numerator (the number at the top of the fraction) is equal to 1 and the denominator (the number at the bottom of the fraction) is equal to 100. As such, the fraction can be written in the form of 1/100 or 0.01. In order for the fraction to be expressed as a percentage, it must be multiplied by 100. This will result in the fraction 1/100 becoming equal to 1.54%.
## Calculating 1.54 as a Percentage
In order to calculate 1.54 as a percentage, the fraction must be converted into a decimal number. To do this, the numerator (1) is divided by the denominator (100). This will give a result of 0.01. This number is then multiplied by 100 to obtain the percentage of 1.54%. This calculation can also be done by dividing 1.54 by 100 to get 0.0154 and then multiplying this result by 100 to get 1.54%.
## Uses for 1.54 as a Percentage
The number 1.54 as a percentage can be used in many different contexts. For example, it can be used to calculate the rate of interest on a loan or investment. It can also be used to calculate taxes, discounts, and sale prices. In addition, it can be used to calculate the cost of goods and services as a percentage of their original price.
## 1.54 as a Percentage Compared to Other Percentages
1.54 as a percentage is a relatively small amount when compared to other common percentages. For example, 25% is much higher than 1.54% and 10% is much lower. This means that 1.54% is closer to 0% than it is to any other percentage. As such, it is usually used for smaller calculations or for calculations that involve a small rate of change.
##
In conclusion, 1.54 as a percentage is a fraction expressed as a decimal number which is then multiplied by 100 to obtain the percentage. It can be used for various calculations such as calculating interest rates, taxes, discounts, sale prices, and the cost of goods and services. It is a relatively small percentage when compared to other common percentages and is usually used for smaller calculations or calculations that involve a small rate of change. | 521 | 2,181 | {"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} | 4.59375 | 5 | CC-MAIN-2023-14 | latest | en | 0.947503 |
https://www.jobilize.com/physics-ap/terms/electric-current-current-by-openstax?qcr=www.quizover.com | 1,610,765,542,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703499999.6/warc/CC-MAIN-20210116014637-20210116044637-00366.warc.gz | 829,364,086 | 19,396 | # 20.1 Current
Page 1 / 8
• Define electric current, ampere, and drift velocity
• Describe the direction of charge flow in conventional current.
• Use drift velocity to calculate current and vice versa.
## Electric current
Electric current is defined to be the rate at which charge flows. A large current, such as that used to start a truck engine, moves a large amount of charge in a small time, whereas a small current, such as that used to operate a hand-held calculator, moves a small amount of charge over a long period of time. In equation form, electric current $I$ is defined to be
$I=\frac{\Delta Q}{\Delta t}\text{,}$
where $\Delta Q$ is the amount of charge passing through a given area in time $\Delta t$ . (As in previous chapters, initial time is often taken to be zero, in which case $\Delta t=t$ .) (See [link] .) The SI unit for current is the ampere (A), named for the French physicist André-Marie Ampère (1775–1836). Since $I=\Delta Q/\Delta t$ , we see that an ampere is one coulomb per second:
$\text{1 A}=\text{1 C/s}$
Not only are fuses and circuit breakers rated in amperes (or amps), so are many electrical appliances.
## Calculating currents: current in a truck battery and a handheld calculator
(a) What is the current involved when a truck battery sets in motion 720 C of charge in 4.00 s while starting an engine? (b) How long does it take 1.00 C of charge to flow through a handheld calculator if a 0.300-mA current is flowing?
Strategy
We can use the definition of current in the equation $I=\Delta Q/\Delta t$ to find the current in part (a), since charge and time are given. In part (b), we rearrange the definition of current and use the given values of charge and current to find the time required.
Solution for (a)
Entering the given values for charge and time into the definition of current gives
$\begin{array}{lll}I& =& \frac{\Delta Q}{\Delta t}=\frac{\text{720 C}}{\text{4.00 s}}=\text{180 C/s}\\ & =& \text{180 A.}\end{array}$
Discussion for (a)
This large value for current illustrates the fact that a large charge is moved in a small amount of time. The currents in these “starter motors” are fairly large because large frictional forces need to be overcome when setting something in motion.
Solution for (b)
Solving the relationship $I=\Delta Q/\Delta t$ for time $\Delta t$ , and entering the known values for charge and current gives
$\begin{array}{lll}\Delta t& =& \frac{\Delta Q}{I}=\frac{\text{1.00 C}}{0.300×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{C/s}}\\ & =& \text{3.33}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{s.}\end{array}$
Discussion for (b)
This time is slightly less than an hour. The small current used by the hand-held calculator takes a much longer time to move a smaller charge than the large current of the truck starter. So why can we operate our calculators only seconds after turning them on? It’s because calculators require very little energy. Such small current and energy demands allow handheld calculators to operate from solar cells or to get many hours of use out of small batteries. Remember, calculators do not have moving parts in the same way that a truck engine has with cylinders and pistons, so the technology requires smaller currents.
[link] shows a simple circuit and the standard schematic representation of a battery, conducting path, and load (a resistor). Schematics are very useful in visualizing the main features of a circuit. A single schematic can represent a wide variety of situations. The schematic in [link] (b), for example, can represent anything from a truck battery connected to a headlight lighting the street in front of the truck to a small battery connected to a penlight lighting a keyhole in a door. Such schematics are useful because the analysis is the same for a wide variety of situations. We need to understand a few schematics to apply the concepts and analysis to many more situations.
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Car speedometer measures the rate of change of distance per unit time.
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is higher resolution of microscope using red or blue light?.explain
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Unlike transverse waves such as electromagnetic waves, longitudinal waves such as sound waves cannot be polarized. ... Since sound waves vibrate along their direction of propagation, they cannot be polarized
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An American traveler in New Zealand carries a transformer to convert New Zealand’s standard 240 V to 120 V so that she can use some small appliances on her trip.
What is the ratio of turns in the primary and secondary coils of her transformer?
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what is energy
Yusuf
How electric lines and equipotential surface are mutually perpendicular?
The potential difference between any two points on the surface is zero that implies È.Ŕ=0, Where R is the distance between two different points &E= Electric field intensity. From which we have cos þ =0, where þ is the angle between the directions of field and distance line, as E andR are zero. Thus
sorry..E and R are non zero...
By how much leeway (both percentage and mass) would you have in the selection of the mass of the object in the previous problem if you did not wish the new period to be greater than 2.01 s or less than 1.99 s?
hello
Chichi
Hi
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hello
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Hi I'm Matthew, and the answer is Lee weighs in mass 0.008kg OR 0.009kg
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14 year old answers college physics and the crowd goes wild!
Matthew
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# theorem
THE'OREM, n. [Gr. to see.]
1. In mathematics, a proposition which terminates in theory,and which considers the properties of things already made or done; or it is a speculative proposition deduced from several definitions compared together.
A theorem is a proposition to be proved by a chain of reasoning. A theorem is something to be proved; a problem is something to be done.
2. In algebra or analysis, it is sometimes used to denote a rule, particularly when that rule is expressed by symbols.
A universal theorem, extends to any quantity without restriction.
A particular theorem, extends only to a particular quantity.
A negative theorem, expresses the impossibility of any assertion.
A local theorem, is that which relates to a surface.
A solid theorem, is that which considers a space terminated by a solid, that is, by any of the three conic sections.
## Evolution (or devolution) of this word [theorem]
1828 Webster1844 Webster1913 Webster
THE'OREM, n. [Gr. to see.]
1. In mathematics, a proposition which terminates in theory,and which considers the properties of things already made or done; or it is a speculative proposition deduced from several definitions compared together.
A theorem is a proposition to be proved by a chain of reasoning. A theorem is something to be proved; a problem is something to be done.
2. In algebra or analysis, it is sometimes used to denote a rule, particularly when that rule is expressed by symbols.
A universal theorem, extends to any quantity without restriction.
A particular theorem, extends only to a particular quantity.
A negative theorem, expresses the impossibility of any assertion.
A local theorem, is that which relates to a surface.
A solid theorem, is that which considers a space terminated by a solid, that is, by any of the three conic sections.
THE'O-REM, n. [Fr. theoreme; Sp. and It. teorema; Gr. θεωρημα, from θεωρεω, to see.]
1. In mathematics, a proposition which terminates in theory, and which considers the properties of things already made or done; or it is a speculative proposition deduced from several definitions compared together. A theorem is a proposition to be proved by a chain of reasoning. A theorem is something to be proved; a problem is something to be done. Day.
2. In algebra or analysis, it is sometimes used to denote a rule, particularly when that rule is expressed by symbols. Cyc. A universal theorem, extends to any quantity without restriction. A particular theorem, extends only to a particular quantity. A negative theorem, expresses the impossibility of any assertion. A local theorem, is that which relates to a surface. A solid theorem, is that which considers a space terminated by a solid, that is, by any of the three conic sections.
The"o*rem
1. That which is considered and established as a principle; hence, sometimes, a rule.
Not theories, but theorems ((?)), the intelligible products of contemplation, intellectual objects in the mind, and of and for the mind exclusively. Coleridge.
By the theorems,
Which your polite and terser gallants practice,
I re-refine the court, and civilize
Their barbarous natures.
Massinger.
2. To formulate into a theorem.
3. A statement of a principle to be demonstrated.
* A theorem is something to be proved, and is thus distinguished from a problem, which is something to be solved. In analysis, the term is sometimes applied to a rule, especially a rule or statement of relations expressed in a formula or by symbols; as, the binomial theorem; Taylor's theorem. See the Note under Proposition, n., 5.
Binomial theorem. (Math.) See under Binomial. -- Negative theorem, a theorem which expresses the impossibility of any assertion. -- Particular theorem (Math.), a theorem which extends only to a particular quantity. -- Theorem of Pappus. (Math.) See Centrobaric method, under Centrobaric. -- Universal theorem (Math.), a theorem which extends to any quantity without restriction.
1828 Webster1844 Webster1913 Webster
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Theorem
THE'OREM, noun [Gr. to see.]
1. In mathematics, a proposition which terminates in theory, and which considers the properties of things already made or done; or it is a speculative proposition deduced from several definitions compared together.
A theorem is a proposition to be proved by a chain of reasoning. A theorem is something to be proved; a problem is something to be done.
2. In algebra or analysis, it is sometimes used to denote a rule, particularly when that rule is expressed by symbols.
A universal theorem extends to any quantity without restriction.
A particular theorem extends only to a particular quantity.
A negative theorem expresses the impossibility of any assertion.
A local theorem is that which relates to a surface.
A solid theorem is that which considers a space terminated by a solid, that is, by any of the three conic sections.
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### Word of the Day
IMPORT'ANCE, n.
1. Weight; consequence; a bearing on some interest; that quality of any thing by which it may affect a measure, interest or result. The education of youth is of great importance to a free government. A religious education is of infinite importance to every human being.
2. Weight or consequence in the scale of being.
Thy own importance know.
Nor bound thy narrow views to things below.
3. Weight or consequence in self-estimation.
He believes himself a man of importance.
4. Thing implied; matter; subject; importunity. [In these senses, obsolete.]
### Random Word
FLANK, n. [Eng. flag. Gr. probably connected with lank, and so called from its laxity, or from breadth.]
1. The fleshy or muscular part of the side of an animal, between the ribs and the hip. Hence,
2. The side of an army, or of any division of an army, as of a brigade, regiment or battalion. To attack an enemy in flank, is to attack them on the side.
3. In fortification, that part of a bastion which reaches from the curtain to the face, and defends the opposite face, the flank and the curtain; or it is a line drawn from the extremity of the face towards the inside of the work.
FLANK, v.t.
1. To attack the side or flank of an army or body of troops; or to place troops so as to command or attack the flank.
2. To post so as to overlook or command on the side; as, to flank a passage.
3. To secure or guard on the side; as flanked with rocks.
FLANK, v.i.
1. To border; to touch.
2. To be posted on the side.
### Noah's 1828 Dictionary
First dictionary of the American Language!
Noah Webster, the Father of American Christian education, wrote the first American dictionary and established a system of rules to govern spelling, grammar, and reading. This master linguist understood the power of words, their definitions, and the need for precise word usage in communication to maintain independence. Webster used the Bible as the foundation for his definitions.
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http://msgroups.net/greatplains/rounding-in-dexterity/587203 | 1,600,596,982,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400197946.27/warc/CC-MAIN-20200920094130-20200920124130-00489.warc.gz | 86,475,469 | 11,491 | #### Rounding in Dexterity
```Hi,
I am trying to use the round function in Dexterity which has the syntax
round(field,side,expression{,mode}).
I see that GP does the rounding such that
12.119 >> 12.12
12.111 >> 12.11
What option should I use for the "mode" in round function to get these
results?
Regards,
Sup
```
0
sup (57)
10/3/2006 10:25:01 PM
greatplains 29623 articles. 6 followers.
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[PageSpeed] 2
```You'll note that "mode" is optional but for the different methods that it'll
round the help file does give complete examples. For your example regular
"rounding" is what you need. Just set the "expression" to 2 so it rounds at
2 decimals. Done.
patrick
mbs dev support
--
This posting is provided "AS IS" with no warranties, and confers no rights.
"sup" <sup@discussions.microsoft.com> wrote in message
news:94B3E640-1DA1-4EF2-B694-1919B280081F@microsoft.com...
> Hi,
> I am trying to use the round function in Dexterity which has the syntax
> round(field,side,expression{,mode}).
> I see that GP does the rounding such that
> 12.119 >> 12.12
> 12.111 >> 12.11
> What option should I use for the "mode" in round function to get these
> results?
>
> Regards,
> Sup
```
0
prot1 (1345)
10/5/2006 12:17:07 AM
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I'm making price list. 150 articles and I want that every price have last number 9. Ex. 1651,00 = 1659,00 1255,00 = 1259,00 etc... Help! =3DINT(A1/10)*10+9 HIH On 26 Mar, 09:49, Vladek <Vla...@discussions.microsoft.com> wrote: > I'm making price list. 150 articles and I want that every price have last > number 9. > Ex. 1651,00 =3D 1659,00 > =A0 =A0 =A01255,00 =3D 1259,00 > etc... > > Help! Hi, Try this =CEILING(A1,10)-1 -- Mike When competing hypotheses are otherwise equal, adopt the hypothesis that introduces th...
Dexterity error after upgrade when selecting vendor
We recently upgraded to Version 8.0.0.69. Our main site is working fine, but a branch office employee gets an error message when selecting a vendor. I have a screen shots of the error and version info. All other functions of the program are fine. Since this is my first time supporting this product, I'm having a hard time running this down. Thanks for your help! John Hi What is the error message? Do you have a Var to log a support call with? Regards James "John" <John@discussions.microsoft.com> wrote in message news:5558E29D-051A-4937-9961-6281C53F56EA@micro...
How can I turn off rounding for numbers???
Excel 2003: I've got a column of cells. I highlight the entire column, do a format / cells / number then I choose 0 decimal places I have a list of serial numbers that I want to input, and then "fill down"... because they're all in sequence. However, when I input. They are 14 digits long. Every time I input a 14-digit number, Excel forces it to round up to the 5th number from the right. How can I get it to stop doing this? Hi Excel should not round in this case. Maybe it shows the number in the scientific notation. Try the folowing custom format: 00000000000000 >...
round
Here is what I am trying to do I have 2 columns that have been formated and formula added to calculate the difference between 2 times. e.g 22:00-01:00 equals 3 hours. The problem I am having is when the times are 22:00 to midnight or 00:00. It will not show anything. So I have to input either 23:59 or 00:01, the problem with this is that the resulting time shown is either 2.98 or something similar How do I get it to round to the nearest quarter hour Bobby "bobby todd" <anonymous@discussions.microsoft.com> wrote in message news:A47EF5F0-ABC7-405B-B3E8-86F24CD5DBBA@microsoft....
Rounding Errors Help
Is there any way to display a rounded number in a cell, but use th original unrounded number when other cells reference that cell in calculation? It would be a pain to use the components of the rounde cell in the calculation in which i would need the unrounded number. I need a function that displays a rounded number, but uses th unrounded number in calculations -- mattflo ----------------------------------------------------------------------- mattflow's Profile: http://www.excelforum.com/member.php?action=getinfo&userid=2512 View this thread: http://www.excelforum.com/showthread.ph...
Dexterity runtime error
I am receiving a Dexterity runtime error - you do not have permission to open this file when attempting to launch enterprise. The whole company is having this problem. I do not believe any changes have been made to the system. What would cause this error? The last time I encountered this there was a problem with the reports.dic. It's either a permissions/access issue (the file is not available) or the file got corrupted (this sometimes happens when reports are imported or edited while other users are in Great Plains). You can try pointing to a local reports.dic as a test to troubl...
Issues with Rounding (or lack thereof)
Hello, I've got a question with an Access 2000 Database. I've got a query with a calculated field: Der_Percent_Good. Der_Percent_Good: Nz((100*(1-[Num_Out_Tol]/[Shipment_Sample_Size])),"N/ A") This field is supposed to be a percentage, and I'd like this field to be able to round to 2 decimal places, for right now it does not. Right now it has a large amount of decimal places if the situation arises, like 1/3 (goes out of view on my textbox on the form), and it will have 0 decimal places if it turns out to be a whole number (0/100). I've tried selecting Properties-... | 3,964 | 15,818 | {"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-2020-40 | latest | en | 0.854016 |
https://turningtooneanother.net/2019/11/26/what-is-the-application-of-knot-theory/ | 1,685,973,563,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652116.60/warc/CC-MAIN-20230605121635-20230605151635-00438.warc.gz | 635,593,845 | 9,512 | # What is the application of knot theory?
## What is the application of knot theory?
Knot theory provides insight into how hard it is to unknot and reknot various types of DNA, shedding light on how much time it takes the enzymes to do their jobs.
What is a knot in physics?
In quantum physics, a knot may be regarded as the orbit in spacetime of a charged particle. One way of calculating the Jones polynomial in quantum theory involves using the Chern-Simons function for gauge fields. The term “knot” as it is used by mathematicians is abstracted from this experience just a little bit.
What is a knot in science?
Knot theory is the mathematical branch of topology that studies mathematical knots, which are defined as embeddings of a circle in 3-dimensional Euclidean space, R3. This is basically equivalent to a conventional knotted string with the ends joined together to prevent it from becoming undone.
### Why are mathematical knots important?
It is an ultimate purpose of knot theory to clarify a topological difference of knot phenomena in mathematics and in science. In this study, a building power and a computational ability in mathematics are needed in addition to the intuition power having to do with a figure.
How does the knot theory work?
Knot theory, in mathematics, the study of closed curves in three dimensions, and their possible deformations without one part cutting through another. Knots may be regarded as formed by interlacing and looping a piece of string in any fashion and then joining the ends.
What are DNA knots?
Just like any long polymer chain, DNA tends to form knots. Using technology that allows them to stretch DNA molecules and image the behavior of these knots, MIT researchers have discovered, for the first time, the factors that determine whether a knot moves along the strand or “jams” in place.
## How do muscles knot up?
Muscle knots usually happen because a muscle has been irritated by a repetitive motion. Athletes will notice muscle knots after training one group of muscles for a long period of time. A muscle might also knot up when it’s in an awkward position for too long.
How do knots work?
First, the more times the strands cross, the stronger the knot. And the twisting of strands as they cross one another also plays a role: If the strands are twisted in opposite directions, the twist balances out, locking the knot into place.
What are knots in statistics?
Knots are cutpoints that defines different regions (or partitions) for a variable. In each regions, a fitting must occurs. The definition of different regions is a way to stay local in the fitting process.
### What does the knot symbol mean?
Knots have been a popular symbol of love and marriage for centuries, with variations of knots used around the globe to symbolize eternal love. The phrase ‘tying the knot’ literally means to get married. The true love knot, in the same way, symbolizes the following: An unbreakable bond. Eternal connection.
What is a knot knot theory?
What is difference between knot and cycle?
A cycle is a necessary condition for deadlock. If the graph is expedient, then a knot is a sufficient condition for deadlock. | 652 | 3,190 | {"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-2023-23 | latest | en | 0.954908 |
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Post subject: Variants of permute the central elements on the big cubesPosted: Sun Dec 08, 2013 9:24 am
Joined: Mon Sep 30, 2013 2:16 am
Hello everybody!
In this video I showed different variants of permute the central elements on the big cubes.
I used modification of algorithm r U' l' U r' U' l U (see step 6 in http://www.speedsolving.com/wiki/index.php/Cage_Method)
Can to somebody this will interesting.
And I will be glad if who shows other ways for permute of the centers.
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Post subject: Re: Variants of permute the central elements on the big cubePosted: Sun Dec 08, 2013 11:40 am
Joined: Sat Jan 15, 2011 10:22 am
Good job if you came up with that on your own, but basically this is just normal commutators for center pieces.
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Post subject: Re: Variants of permute the central elements on the big cubePosted: Sun Dec 08, 2013 11:22 pm
Joined: Mon Sep 30, 2013 2:16 am
EMI94100 wrote:
Good job if you came up with that on your own, but basically this is just normal commutators for center pieces.
Thanks!
Yes, some variants I received as a result of my practice, for example, the permute of centers from 3 sides.
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Post subject: Re: Variants of permute the central elements on the big cubePosted: Sun Dec 08, 2013 11:37 pm
Joined: Wed Jun 19, 2013 2:50 pm
Location: Deep in the Heart of Texas
There is an algorithm that I came up with based on the corner flip algorithm: [(r' u r u' or r u' r' u), (U, U', F, F', R, or R')]. This acts as a three cycle between two colors, so it is almost the same as your algorithm, but it can be done 12 different ways instead of 2, so it will be slightly more useful, but less efficient. So overall, the two algorithms are about equal.
_________________
For all of you that bought a KO 8x8x8: You should have bought a V8!
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Post subject: Re: Variants of permute the central elements on the big cubePosted: Mon Dec 09, 2013 12:52 am
Joined: Mon Sep 30, 2013 2:16 am
Yes, if to look narrowly, all this is based on the commutator [x, y, x', y'] with intermediate moves:
l' U r U' l U r' U'
l' U' r U l U' r' U
r2 f2 r2 f2
l2 U r2 U' l2 U r2 U'
U r2 U' l' U r2 U' l
b D' l' D b' D' l D
(l' r u l r') U2 (l' r u' l r') U2
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Jump to: Select a forum ------------------ Announcements General Puzzle Topics New Puzzles Puzzle Building and Modding Puzzle Collecting Solving Puzzles Marketplace Non-Twisty Puzzles Site Comments, Suggestions & Questions Content Moderators Off Topic | 932 | 3,220 | {"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-2014-23 | latest | en | 0.896681 |
https://www.moodlemonkey.com/solution/hi6007-statistics-and-research-methods-for-business-decision-making-assignment/ | 1,652,940,687,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662525507.54/warc/CC-MAIN-20220519042059-20220519072059-00028.warc.gz | 1,048,767,322 | 31,739 | # HI6007 Statistics And Research Methods For Business Decision Making Assignment
(a). Arrange the data of 20 student’s result
Student number Results 1 42 2 53 3 54 4 61 5 61 6 61 7 62 8 63 9 64 10 66 11 67 12 67 13 68 14 69 15 71 16 71 17 76 18 78 19 81 20 83 Total 1318
Data has arranged in ascending order for getting correct results.
• Compute Mean, Median and Mode
To calculate mean , following formula will implement:
Σ xi /9Total of marks) =1318; n (number of students)=20
131820= 65.9 is the mean value of student’s results.
Median calculation
Median = N2+1=202+1=11th Item of the above table will be median
Median is 67
• Compute 1st and 3rd Quartile
1st Quartile = 14 N+1 = 20+1/4 = 5th Item = 61
3rd Quartile =34N+1=3420+1=634=15th item=71
• Compute and Intercept 90th Percentile
Formula = 90% of total number of students(observation) = 0.90*20 = 18th Item = 78
(b) Inferential Statistics:
Inferential statistics include select sample from the available observation, in order to identify or solve the issue through applying appropriate tests. These tests are also helpful in hypothesis testing and prove the outcome with valid numbers.
(i) Prepare Joint Probability Table
Applied for More than 1 University Age Group Yes No 23 and Under 207 207/808*100=25.62% 201 201/1210*100=16.61% 24-26 299 299/808*100=37.0% 379 379/1210*100=31.32% 27-30 185 185/808*100 = 22.90% 268 268/1210*100=22.15% 31-35 66 66/808*100=8.17% 193 193/1210*100= 15.95% 36 and over 51 51/808*100=6.31% 169 169/1210*100= 13.97% Total / Joint Probability(%) 808 100% 1210 100%
(ii) Given that a student applied to more than 1 university, what is the probability that the student is 24-26 years old.
Probability of student is 24-26 years old = 299/808 =37.00%
• Is the number of universities applied to independent of student age? Explain
Student age is an independent variable against number of observation collected for application made for enrolment in more than one university at a time. Any student at any age can take enrolment of in more than one university or they can adopt only one university at a time.
(b)
Information provided-
x f(x) 10 0.05 20 0.1 30 0.1 40 0.2 50 0.35 60 0.2 Total
X represent number of new clients for counselling cases in the year 2021.
Formula of calculating Expected value = 𝐸(𝑥) = 𝜇 = ∑𝑥 𝑓(𝑥)
x f(x) (𝑥 − 𝜇) 2 (𝑥 − 𝜇) 2*f(x) 10 0.05 -33 1089 54.45 20 0.1 -23 529 52.9 30 0.1 -13 169 16.9 40 0.2 -3 9 1.8 50 0.35 7 49 17.15 60 0.2 17 289 57.8 Total 201
Expected Value= (10*0.05+20*0.1+30*0.1+40*0.2+50*0.35+60*0.2) =43
Formula of Variance of a discrete random variable 𝑉𝑎𝑟(𝑥) = ∑(𝑥𝜇) 2 𝑓(𝑥
Variance = 201 (calculation shows in table)
1. Formulate Hypothesis :
Problem statement: Population annual expenditure on prescription drugs per person is lower in the Midwest than the Northeast.
Hypothesis Statement:
Ho: µ ≤ \$838 or
Ho: µ = \$838
Ha: µ > \$838
Problem statement can test on one tail test from left tail as it requires testing of lower limit.
• Suitable test Statistics
One (Left) tail test
Formula:
• Calculate value of relevant test statistics and P- value
Sample Mean (x) = \$745
Null Hypothesis Mean = \$838
SD = 300
Sample size = 60
Applying Formula (745-838)/300/sqrt(n)
Z= -93/38.75 = -2.40
From the table given of Z score , at significance level of 0.05 ,
P value = 0.0071
• Based on the p value in part (III), at 99% confidence level, decide the decision criteria.
If the confidence level is 99% then there is 1% of significance level for this problem and at this level the critical value is 2.326 , for this Z-score is -2.4 which is less than critical value (2.326> -2.4). Null hypothesis shall be rejected.
• Make the conclusion Based on the analysis.
As per rejection of null hypothesis, it is concluded that the prescribed drugs expenditure is not lower in Midwest as comparison to Northwest.
• State the null and alternative hypothesis for single factor ANOVA to test for any significant difference in the mean price of gasoline for the three brands.
Hypothesis
H0 = µ1= µ2= µ3
H1 = µ1≠µ2≠ µ3
(ii) State the decision rule at 5% significance level.
Reject the H0 id t stat > Z critical value, Otherwise accept the null hypothesis
(iii) Calculate the test statistics
A B C 3.77 3.83 3.78 3.72 3.83 3.87 3.87 3.85 3.89 3.76 3.77 3.79 3.83 3.84 3.87 3.85 3.84 3.87 3.93 4.04 3.99 3.79 3.78 3.79 3.78 3.84 3.79 3.81 3.84 3.86 Sample Mean 3.811 3.846 3.85 Varience 0.003349 0.004844 0.00382
ANOVA one- way test Formula
Formula F= MSTR / MSE
MSTR = 𝑆𝑆𝑇𝑅 / 𝑘 – 1
MSE = SSE /𝑛r – k
𝑥Ӗ= (3.81 + 3.84 + 3.85)/3 = 3.83
SSTR= 10(3.81- 3.83)+ 10(3.84-3.83)2 + 10(3.85-3.83)= 0.009
MSTR = 0.009/ (3-1) = 0.0045
P-value and critical value approaches
Value of test statistic
SSE = 9(0.003) +9(0.005) + 9(0.004) =0.108
MSE = 0.108/(30-3) = 0.004
F= 0.0045/0.004 =1.125
ANOVA Table
Source of variation Sum of Squares Degrees of Freedom Mean Square F P- value Treatment 0.009 2 0.0045 1.125 0.044 Error 0.108 27 0.004 Total 0.117 29 0.0085
P- value calculation
Here Numerator df = 2; Denominator Df = 27 then the value of F at 0.01 = 5.49
Decision on the basis of test
The p-value < .05, So null hypothesis shall be rejected
Decision as per critical value approach
Based on an F distribution with 2 numerator d.f. and 27 denominator d.f., F.05 = 3.35.
Reject H0 if F < 3.35
Here F = 1.125 <3.35 which is evidence for rejection of null hypothesis.
(d) Based on the calculated test statistics decide whether any significant difference in the mean price of gasoline for three bands.
The value of F is 1.125 which is lower than F critical value this means hypothesis has been rejected that means there is significant difference in the mean price of gasoline in all the three brands.
• Complete the missing entries from A to H in this output
A= R Square = SSR/SST = 35250755.68/ 42699148.82 = 0.82
B= Observation = 50 (provided in Answer)
C= residual = Total- Regression = 49-2 = 47
D= 42699148.82-7448393.14 = 35250755.68
E= SSRegression / dfreg. = 35250755.68/2 = 17625377.8
F= SSR /(50-3) = 7448393.148/47 = 158476.45
G= 17625377.8/158476.45 = 111.217647
H= Coefficient of income /Standard error of income = 8.36
• Estimate the annual credit card charges for a three-person household with an annual income of \$40,000
To estimate charges of credit card , intercept, household value and size has been considered from the ANOVA table.
The annual credit card charges for three person family is \$3700 where annual income is \$40000.
• Did the estimated regression equation provide a good fit to the data? Explain
No, The reason behind the same is high variability between two variables X and Y which fails in establishing good fit to the data.
1. Using linear trend equation forecast the sales of face masks for October 2020
Month Sales (\$) 1 17000 2 18000 3 19500 4 22000 5 21000 6 23000
Linear Trend Equation = Y= Mx+B
M= Y2-Y1X1-X2 = 23000-21000/6-5 = 2000/1 = 2000
Y = Mx+B
Y= ?
X=1
B= 23000
Y = 2000*1+23000
Y = 2000+23000= 25000
\$25000 will be the forecasted sale in the month of October.
• Sales forecast will be
Sales Weight Weighted Sale July 22,000 0.2 4400 August 21,000 0.3 6300 September 23,000 0.5 11500 Total 22200
So, the expected sale for the next month will be \$22200 | 2,543 | 7,354 | {"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} | 4.53125 | 5 | CC-MAIN-2022-21 | latest | en | 0.827071 |
https://quizlet.com/explanations/questions/find-the-indicated-sides-of-the-right-triangle-b-5600-c-6500-a-3a8c682e-f59493de-8795-4d00-9fe0-a8988db0ce3d | 1,702,113,527,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100873.6/warc/CC-MAIN-20231209071722-20231209101722-00001.warc.gz | 520,951,986 | 77,911 | Try the fastest way to create flashcards
Question
# Find the indicated sides of the right triangle. b = 5600, c = 6500, a = ?
Solutions
Verified
Step 1
1 of 3
In this problem, use the Pythagorean Theorem:
$c^2=a^2+b^2\tag{1}$
where $a$, $b$, and $c$ are the sides of the triangle.
Since the unknown value is $a$, rewrite the equation $\text{(1)}$ in terms of $a$ by subtracting $b^2$ on both sides of the equation.
\begin{aligned} a^2&=c^2-b^2 \end{aligned}
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View Full Version : Position of an orbiting sphere
drew_jewel
07-07-2015, 01:44 PM
I'm trying to get the position of a sphere that is rotating around an idle object in my opengl application. This is how I perform the orbiting:
glTranslatef(positions[i].getPosX(), //center of rotation (yellow ball)
positions[i].getPosY(),
positions[i].getPosZ());
glRotatef(rotation_angle,0.0,1.0,0.0); //angle of rotation
glTranslatef(distance[i].getPosX(), //distance from the center of rotation
distance[i].getPosY(),
distance[i].getPosZ());
Variable rotation_angle loops from 0 to 360 endlessly. In the distance vector I'm only changing the z-distance of the object, for example let's say the idle object is in (0,0,0), the distance vector could be (0,0,200). I need the position of the object that is orbiting to perform collision detection.
GClements
07-07-2015, 04:06 PM
I need the position of the object that is orbiting to perform collision detection.
In which case, forget about using any OpenGL matrix functions other than glLoadMatrix(). Just construct the matrices in the application (either use GLM or write your own code).
For a single object and a simple program, you can get the matrices out of OpenGL using glGetDoublev() and use gluUnProject(). But if you have many objects or performance matters, that's highly inefficient.
Carmine
07-09-2015, 12:16 PM
I'm trying to get the position of a sphere that is rotating around an idle object in my opengl application.
The first translate is o.k. I'd take out the glRotate and replace the second glTranslate with something like this -
float r = 200.0,
x = r * cos (rotation_angle), // This assumes rotation_angle is in units of radians
y = 0.0,
z = r * sin (rotation_angle);
glTranslatef (x, y, z); | 448 | 1,757 | {"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-2018-30 | latest | en | 0.77543 |
https://python.quantecon.org/chang_ramsey.html | 1,571,336,225,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986675598.53/warc/CC-MAIN-20191017172920-20191017200420-00059.warc.gz | 683,936,622 | 194,278 | # Competitive Equilibria of Chang Model¶
Co-author: Sebastian Graves
In addition to what’s in Anaconda, this lecture will need the following libraries:
In [1]:
!pip install polytope
## Overview¶
This lecture describes how Chang [Cha98] analyzed competitive equilibria and a best competitive equilibrium called a Ramsey plan.
He did this by
• characterizing a competitive equilibrium recursively in a way also employed in the dynamic Stackelberg problems and Calvo model lectures to pose Stackelberg problems in linear economies, and then
• appropriately adapting an argument of Abreu, Pearce, and Stachetti [APS90] to describe key features of the set of competitive equilibria
Roberto Chang [Cha98] chose a model of Calvo [Cal78] as a simple structure that conveys ideas that apply more broadly.
A textbook version of Chang’s model appears in chapter 25 of [LS18].
This lecture and Credible Government Policies in Chang Model can be viewed as more sophisticated and complete treatments of the topics discussed in Ramsey plans, time inconsistency, sustainable plans.
Both this lecture and Credible Government Policies in Chang Model make extensive use of an idea to which we apply the nickname dynamic programming squared.
In dynamic programming squared problems there are typically two interrelated Bellman equations
• A Bellman equation for a set of agents or followers with value or value function $v_a$.
• A Bellman equation for a principal or Ramsey planner or Stackelberg leader with value or value function $v_p$ in which $v_a$ appears as an argument.
We encountered problems with this structure in dynamic Stackelberg problems, optimal taxation with state-contingent debt, and other lectures.
In [2]:
import numpy as np
import polytope
import quantecon as qe
import matplotlib.pyplot as plt
%matplotlib inline
polytope failed to import cvxopt.glpk.
will use scipy.optimize.linprog
### The Setting¶
First, we introduce some notation.
For a sequence of scalars $\vec z \equiv \{z_t\}_{t=0}^\infty$, let $\vec z^t = (z_0, \ldots , z_t)$, $\vec z_t = (z_t, z_{t+1}, \ldots )$.
An infinitely lived representative agent and an infinitely lived government exist at dates $t = 0, 1, \ldots$.
The objects in play are
• an initial quantity $M_{-1}$ of nominal money holdings
• a sequence of inverse money growth rates $\vec h$ and an associated sequence of nominal money holdings $\vec M$
• a sequence of values of money $\vec q$
• a sequence of real money holdings $\vec m$
• a sequence of total tax collections $\vec x$
• a sequence of per capita rates of consumption $\vec c$
• a sequence of per capita incomes $\vec y$
A benevolent government chooses sequences $(\vec M, \vec h, \vec x)$ subject to a sequence of budget constraints and other constraints imposed by competitive equilibrium.
Given tax collection and price of money sequences, a representative household chooses sequences $(\vec c, \vec m)$ of consumption and real balances.
In competitive equilibrium, the price of money sequence $\vec q$ clears markets, thereby reconciling decisions of the government and the representative household.
Chang adopts a version of a model that [Cal78] designed to exhibit time-inconsistency of a Ramsey policy in a simple and transparent setting.
By influencing the representative household’s expectations, government actions at time $t$ affect components of household utilities for periods $s$ before $t$.
When setting a path for monetary expansion rates, the government takes into account how the household’s anticipations of the government’s future actions affect the household’s current decisions.
The ultimate source of time inconsistency is that a time $0$ Ramsey planner takes these effects into account in designing a plan of government actions for $t \geq 0$.
## Setting¶
### The Household’s Problem¶
A representative household faces a nonnegative value of money sequence $\vec q$ and sequences $\vec y, \vec x$ of income and total tax collections, respectively.
The household chooses nonnegative sequences $\vec c, \vec M$ of consumption and nominal balances, respectively, to maximize
$$\sum_{t=0}^\infty \beta^t \left[ u(c_t) + v(q_t M_t ) \right] \tag{1}$$
subject to
$$q_t M_t \leq y_t + q_t M_{t-1} - c_t - x_t \tag{2}$$
and
$$q_t M_t \leq \bar m \tag{3}$$
Here $q_t$ is the reciprocal of the price level at $t$, which we can also call the value of money.
Chang [Cha98] assumes that
• $u: \mathbb{R}_+ \rightarrow \mathbb{R}$ is twice continuously differentiable, strictly concave, and strictly increasing;
• $v: \mathbb{R}_+ \rightarrow \mathbb{R}$ is twice continuously differentiable and strictly concave;
• $u'(c)_{c \rightarrow 0} = \lim_{m \rightarrow 0} v'(m) = +\infty$;
• there is a finite level $m= m^f$ such that $v'(m^f) =0$
The household carries real balances out of a period equal to $m_t = q_t M_t$.
Inequality (2) is the household’s time $t$ budget constraint.
It tells how real balances $q_t M_t$ carried out of period $t$ depend on income, consumption, taxes, and real balances $q_t M_{t-1}$ carried into the period.
Equation (3) imposes an exogenous upper bound $\bar m$ on the household’s choice of real balances, where $\bar m \geq m^f$.
### Government¶
The government chooses a sequence of inverse money growth rates with time $t$ component $h_t \equiv {M_{t-1}\over M_t} \in \Pi \equiv [ \underline \pi, \overline \pi]$, where $0 < \underline \pi < 1 < { 1 \over \beta } \leq \overline \pi$.
The government faces a sequence of budget constraints with time $t$ component
$$-x_t = q_t (M_t - M_{t-1})$$
which by using the definitions of $m_t$ and $h_t$ can also be expressed as
$$-x_t = m_t (1-h_t) \tag{4}$$
The restrictions $m_t \in [0, \bar m]$ and $h_t \in \Pi$ evidently imply that $x_t \in X \equiv [(\underline \pi -1)\bar m, (\overline \pi -1) \bar m]$.
We define the set $E \equiv [0,\bar m] \times \Pi \times X$, so that we require that $(m, h, x) \in E$.
To represent the idea that taxes are distorting, Chang makes the following assumption about outcomes for per capita output:
$$y_t = f(x_t), \tag{5}$$
where $f: \mathbb{R}\rightarrow \mathbb{R}$ satisfies $f(x) > 0$, is twice continuously differentiable, $f''(x) < 0$, and $f(x) = f(-x)$ for all $x \in \mathbb{R}$, so that subsidies and taxes are equally distorting.
Calvo’s and Chang’s purpose is not to model the causes of tax distortions in any detail but simply to summarize the outcome of those distortions via the function $f(x)$.
A key part of the specification is that tax distortions are increasing in the absolute value of tax revenues.
Ramsey plan: A Ramsey plan is a competitive equilibrium that maximizes (1).
Within-period timing of decisions is as follows:
• first, the government chooses $h_t$ and $x_t$;
• then given $\vec q$ and its expectations about future values of $x$ and $y$’s, the household chooses $M_t$ and therefore $m_t$ because $m_t = q_t M_t$;
• then output $y_t = f(x_t)$ is realized;
• finally $c_t = y_t$
This within-period timing confronts the government with choices framed by how the private sector wants to respond when the government takes time $t$ actions that differ from what the private sector had expected.
This consideration will be important in lecture credible government policies when we study credible government policies.
The model is designed to focus on the intertemporal trade-offs between the welfare benefits of deflation and the welfare costs associated with the high tax collections required to retire money at a rate that delivers deflation.
A benevolent time $0$ government can promote utility generating increases in real balances only by imposing sufficiently large distorting tax collections.
To promote the welfare increasing effects of high real balances, the government wants to induce gradual deflation.
### Household’s Problem¶
Given $M_{-1}$ and $\{q_t\}_{t=0}^\infty$, the household’s problem is
\begin{aligned} \mathcal{L} & = \max_{\vec c, \vec M} \min_{\vec \lambda, \vec \mu} \sum_{t=0}^\infty \beta^t \bigl\{ u(c_t) + v(M_t q_t) + \lambda_t [ y_t - c_t - x_t + q_t M_{t-1} - q_t M_t ] \\ & \quad \quad \quad + \mu_t [\bar m - q_t M_t] \bigr\} \end{aligned}
First-order conditions with respect to $c_t$ and $M_t$, respectively, are
\begin{aligned} u'(c_t) & = \lambda_t \\ q_t [ u'(c_t) - v'(M_t q_t) ] & \leq \beta u'(c_{t+1}) q_{t+1} , \quad = \ {\rm if} \ M_t q_t < \bar m \end{aligned}
The last equation expresses Karush-Kuhn-Tucker complementary slackness conditions (see here).
These insist that the inequality is an equality at an interior solution for $M_t$.
Using $h_t = {M_{t-1}\over M_t}$ and $q_t = {m_t \over M_t}$ in these first-order conditions and rearranging implies
$$m_t [u'(c_t) - v'(m_t) ] \leq \beta u'(f(x_{t+1})) m_{t+1} h_{t+1}, \quad = \text{ if } m_t < \bar m \tag{6}$$
Define the following key variable
$$\theta_{t+1} \equiv u'(f(x_{t+1})) m_{t+1} h_{t+1} \tag{7}$$
This is real money balances at time $t+1$ measured in units of marginal utility, which Chang refers to as ‘the marginal utility of real balances’.
From the standpoint of the household at time $t$, equation (7) shows that $\theta_{t+1}$ intermediates the influences of $(\vec x_{t+1}, \vec m_{t+1})$ on the household’s choice of real balances $m_t$.
By “intermediates” we mean that the future paths $(\vec x_{t+1}, \vec m_{t+1})$ influence $m_t$ entirely through their effects on the scalar $\theta_{t+1}$.
The observation that the one dimensional promised marginal utility of real balances $\theta_{t+1}$ functions in this way is an important step in constructing a class of competitive equilibria that have a recursive representation.
A closely related observation pervaded the analysis of Stackelberg plans in lecture dynamic Stackelberg problems.
## Competitive Equilibrium¶
Definition:
• A government policy is a pair of sequences $(\vec h,\vec x)$ where $h_t \in \Pi \ \forall t \geq 0$.
• A price system is a nonnegative value of money sequence $\vec q$.
• An allocation is a triple of nonnegative sequences $(\vec c, \vec m, \vec y)$.
It is required that time $t$ components $(m_t, x_t, h_t) \in E$.
Definition:
Given $M_{-1}$, a government policy $(\vec h, \vec x)$, price system $\vec q$, and allocation $(\vec c, \vec m, \vec y)$ are said to be a competitive equilibrium if
• $m_t = q_t M_t$ and $y_t = f(x_t)$.
• The government budget constraint is satisfied.
• Given $\vec q, \vec x, \vec y$, $(\vec c, \vec m)$ solves the household’s problem.
## Inventory of Objects in Play¶
Chang constructs the following objects
1. A set $\Omega$ of initial marginal utilities of money $\theta_0$
• Let $\Omega$ denote the set of initial promised marginal utilities of money $\theta_0$ associated with competitive equilibria.
• Chang exploits the fact that a competitive equilibrium consists of a first period outcome $(h_0, m_0, x_0)$ and a continuation competitive equilibrium with marginal utility of money $\theta _1 \in \Omega$.
2. Competitive equilibria that have a recursive representation
• A competitive equilibrium with a recursive representation consists of an initial $\theta_0$ and a four-tuple of functions $(h, m, x, \Psi)$ mapping $\theta$ into this period’s $(h, m, x)$ and next period’s $\theta$, respectively.
• A competitive equilibrium can be represented recursively by iterating on
$$\begin{split} h_t & = h(\theta_t) \\ m_t & = m(\theta_t) \\ x_t & = x(\theta_t) \\ \theta_{t+1} & = \Psi(\theta_t) \end{split} \tag{8}$$ starting from $\theta_0$
The range and domain of $\Psi(\cdot)$ are both $\Omega$
3. A recursive representation of a Ramsey plan
• A recursive representation of a Ramsey plan is a recursive competitive equilibrium $\theta_0, (h, m, x, \Psi)$ that, among all recursive competitive equilibria, maximizes $\sum_{t=0}^\infty \beta^t \left[ u(c_t) + v(q_t M_t ) \right]$.
• The Ramsey planner chooses $\theta_0, (h, m, x, \Psi)$ from among the set of recursive competitive equilibria at time $0$.
• Iterations on the function $\Psi$ determine subsequent $\theta_t$’s that summarize the aspects of the continuation competitive equilibria that influence the household’s decisions.
• At time $0$, the Ramsey planner commits to this implied sequence $\{\theta_t\}_{t=0}^\infty$ and therefore to an associated sequence of continuation competitive equilibria.
4. A characterization of time-inconsistency of a Ramsey plan
• Imagine that after a ‘revolution’ at time $t \geq 1$, a new Ramsey planner is given the opportunity to ignore history and solve a brand new Ramsey plan.
• This new planner would want to reset the $\theta_t$ associated with the original Ramsey plan to $\theta_0$.
• The incentive to reinitialize $\theta_t$ associated with this revolution experiment indicates the time-inconsistency of the Ramsey plan.
• By resetting $\theta$ to $\theta_0$, the new planner avoids the costs at time $t$ that the original Ramsey planner must pay to reap the beneficial effects that the original Ramsey plan for $s \geq t$ had achieved via its influence on the household’s decisions for $s = 0, \ldots, t-1$.
## Analysis¶
A competitive equilibrium is a triple of sequences $(\vec m, \vec x, \vec h) \in E^\infty$ that satisfies (2), (3), and (6).
Chang works with a set of competitive equilibria defined as follows.
Definition: $CE = \bigl\{ (\vec m, \vec x, \vec h) \in E^\infty$ such that (2), (3), and (6) are satisfied $\bigr\}$.
$CE$ is not empty because there exists a competitive equilibrium with $h_t =1$ for all $t \geq 1$, namely, an equilibrium with a constant money supply and constant price level.
Chang establishes that $CE$ is also compact.
Chang makes the following key observation that combines ideas of Abreu, Pearce, and Stacchetti [APS90] with insights of Kydland and Prescott [KP80].
Proposition: The continuation of a competitive equilibrium is a competitive equilibrium.
That is, $(\vec m, \vec x, \vec h) \in CE$ implies that $(\vec m_t, \vec x_t, \vec h_t) \in CE \ \forall \ t \geq 1$.
(Lecture dynamic Stackelberg problems also used a version of this insight)
We can now state that a Ramsey problem is to
$$\max_{(\vec m, \vec x, \vec h) \in E^\infty} \sum_{t=0}^\infty \beta^t \left[ u(c_t) + v(m_t) \right]$$
subject to restrictions (2), (3), and (6).
Evidently, associated with any competitive equilibrium $(m_0, x_0)$ is an implied value of $\theta_0 = u'(f(x_0))(m_0 + x_0)$.
To bring out a recursive structure inherent in the Ramsey problem, Chang defines the set
$$\Omega = \left\{ \theta \in \mathbb{R} \ \text{ such that } \ \theta = u'(f(x_0)) (m_0 + x_0) \ \text{ for some } \ (\vec m, \vec x, \vec h) \in CE \right\}$$
Equation (6) inherits from the household’s Euler equation for money holdings the property that the value of $m_0$ consistent with the representative household’s choices depends on $(\vec h_1, \vec m_1)$.
This dependence is captured in the definition above by making $\Omega$ be the set of first period values of $\theta_0$ satisfying $\theta_0 = u'(f(x_0)) (m_0 + x_0)$ for first period component $(m_0,h_0)$ of competitive equilibrium sequences $(\vec m, \vec x, \vec h)$.
Chang establishes that $\Omega$ is a nonempty and compact subset of $\mathbb{R}_+$.
Definition: $\Gamma(\theta) = \{ (\vec m, \vec x, \vec h) \in CE | \theta = u'(f(x_0))(m_0 + x_0) \}$.
Thus, $\Gamma(\theta)$ is the set of competitive equilibrium sequences $(\vec m, \vec x, \vec h)$ whose first period components $(m_0, h_0)$ deliver the prescribed value $\theta$ for first period marginal utility.
If we knew the sets $\Omega, \Gamma(\theta)$, we could use the following two-step procedure to find at least the value of the Ramsey outcome to the representative household
1. Find the indirect value function $w(\theta)$ defined as
$$w(\theta) = \max_{(\vec m, \vec x, \vec h) \in \Gamma(\theta)} \sum_{t=0}^\infty \beta^t \left[ u(f(x_t)) + v(m_t) \right]$$
1. Compute the value of the Ramsey outcome by solving $\max_{\theta \in \Omega} w(\theta)$.
Thus, Chang states the following
Proposition:
$w(\theta)$ satisfies the Bellman equation
$$w(\theta) = \max_{x,m,h,\theta'} \bigl\{ u(f(x)) + v(m) + \beta w(\theta') \bigr\} \tag{9}$$
where maximization is subject to
$$(m,x,h) \in E \ {\rm and} \ \theta' \in \Omega \tag{10}$$
and
$$\theta = u'(f(x)) (m+x) \tag{11}$$
and
$$-x = m(1-h) \tag{12}$$
and
$$m \cdot [ u'(f(x)) - v'(m) ] \leq \beta \theta' , \quad = \ {\rm if} \ m < \bar m \tag{13}$$
Before we use this proposition to recover a recursive representation of the Ramsey plan, note that the proposition relies on knowing the set $\Omega$.
To find $\Omega$, Chang uses the insights of Kydland and Prescott [KP80] together with a method based on the Abreu, Pearce, and Stacchetti [APS90] iteration to convergence on an operator $B$ that maps continuation values into values.
We want an operator that maps a continuation $\theta$ into a current $\theta$.
Chang lets $Q$ be a nonempty, bounded subset of $\mathbb{R}$.
Elements of the set $Q$ are taken to be candidate values for continuation marginal utilities.
Chang defines an operator
$$B(Q) = \theta \in \mathbb{R} \ \text{ such that there is } \ (m,x,h, \theta') \in E \times Q$$
such that (11), (12), and (13) hold.
Thus, $B(Q)$ is the set of first period $\theta$’s attainable with $(m,x,h) \in E$ and some $\theta' \in Q$.
Proposition:
1. $Q \subset B(Q)$ implies $B(Q) \subset \Omega$ (‘self-generation’).
2. $\Omega = B(\Omega)$ (‘factorization’).
The proposition characterizes $\Omega$ as the largest fixed point of $B$.
It is easy to establish that $B(Q)$ is a monotone operator.
This property allows Chang to compute $\Omega$ as the limit of iterations on $B$ provided that iterations begin from a sufficiently large initial set.
### Some Useful Notation¶
Let $\vec h^t = (h_0, h_1, \ldots, h_t)$ denote a history of inverse money creation rates with time $t$ component $h_t \in \Pi$.
A government strategy $\sigma=\{\sigma_t\}_{t=0}^\infty$ is a $\sigma_0 \in \Pi$ and for $t \geq 1$ a sequence of functions $\sigma_t: \Pi^{t-1} \rightarrow \Pi$.
Chang restricts the government’s choice of strategies to the following space:
$$CE_\pi = \{ {\vec h} \in \Pi^\infty: \text{ there is some } \ (\vec m, \vec x) \ \text{ such that } \ (\vec m, \vec x, \vec h) \in CE \}$$
In words, $CE_\pi$ is the set of money growth sequences consistent with the existence of competitive equilibria.
Chang observes that $CE_\pi$ is nonempty and compact.
Definition: $\sigma$ is said to be admissible if for all $t \geq 1$ and after any history $\vec h^{t-1}$, the continuation $\vec h_t$ implied by $\sigma$ belongs to $CE_\pi$.
Admissibility of $\sigma$ means that anticipated policy choices associated with $\sigma$ are consistent with the existence of competitive equilibria after each possible subsequent history.
After any history $\vec h^{t-1}$, admissibility restricts the government’s choice in period $t$ to the set
$$CE_\pi^0 = \{ h \in \Pi: {\rm there \ is } \ \vec h \in CE_\pi \ {\rm with } \ h=h_0 \}$$
In words, $CE_\pi^0$ is the set of all first period money growth rates $h=h_0$, each of which is consistent with the existence of a sequence of money growth rates $\vec h$ starting from $h_0$ in the initial period and for which a competitive equilibrium exists.
Remark: $CE_\pi^0 = \{h \in \Pi: \text{ there is } \ (m,\theta') \in [0, \bar m] \times \Omega \ \text{ such that } \ m u'[ f((h-1)m) - v'(m)] \leq \beta \theta' \ \text{ with equality if } \ m < \bar m \}$.
Definition: An allocation rule is a sequence of functions $\vec \alpha = \{\alpha_t\}_{t=0}^\infty$ such that $\alpha_t: \Pi^t \rightarrow [0, \bar m] \times X$.
Thus, the time $t$ component of $\alpha_t(h^t)$ is a pair of functions $(m_t(h^t), x_t(h^t))$.
Definition: Given an admissible government strategy $\sigma$, an allocation rule $\alpha$ is called competitive if given any history $\vec h^{t-1}$ and $h_t \in CE_\pi^0$, the continuations of $\sigma$ and $\alpha$ after $(\vec h^{t-1},h_t)$ induce a competitive equilibrium sequence.
### Another Operator¶
At this point it is convenient to introduce another operator that can be used to compute a Ramsey plan.
For computing a Ramsey plan, this operator is wasteful because it works with a state vector that is bigger than necessary.
We introduce this operator because it helps to prepare the way for Chang’s operator called $\tilde D(Z)$ that we shall describe in lecture credible government policies.
It is also useful because a fixed point of the operator to be defined here provides a good guess for an initial set from which to initiate iterations on Chang’s set-to-set operator $\tilde D(Z)$ to be described in lecture credible government policies.
Let $S$ be the set of all pairs $(w, \theta)$ of competitive equilibrium values and associated initial marginal utilities.
Let $W$ be a bounded set of values in $\mathbb{R}$.
Let $Z$ be a nonempty subset of $W \times \Omega$.
Think of using pairs $(w', \theta')$ drawn from $Z$ as candidate continuation value, $\theta$ pairs.
Define the operator
$$D(Z) = \Bigl\{ (w,\theta): {\rm there \ is } \ h \in CE_\pi^0$$
$$\text{ and a four-tuple } \ (m(h), x(h), w'(h), \theta'(h)) \in [0,\bar m]\times X \times Z$$
such that
$$w = u(f(x( h))) + v(m( h)) + \beta w'( h) \tag{14}$$
$$\theta = u'(f(x( h))) ( m( h) + x( h)) \tag{15}$$
$$x(h) = m(h) (h-1) \tag{16}$$
$$m(h) (u'(f(x(h))) - v'(m(h))) \leq \beta \theta'(h) \tag{17}$$
$$\quad \quad \ \text{ with equality if } m(h) < \bar m \Bigr\}$$
It is possible to establish.
Proposition:
1. If $Z \subset D(Z)$, then $D(Z) \subset S$ (‘self-generation’).
2. $S = D(S)$ (‘factorization’).
Proposition:
1. Monotonicity of $D$: $Z \subset Z'$ implies $D(Z) \subset D(Z')$.
2. $Z$ compact implies that $D(Z)$ is compact.
It can be shown that $S$ is compact and that therefore there exists a $(w, \theta)$ pair within this set that attains the highest possible value $w$.
This $(w, \theta)$ pair i associated with a Ramsey plan.
Further, we can compute $S$ by iterating to convergence on $D$ provided that one begins with a sufficiently large initial set $S_0$.
As a very useful by-product, the algorithm that finds the largest fixed point $S = D(S)$ also produces the Ramsey plan, its value $w$, and the associated competitive equilibrium.
## Calculating all Promise-Value Pairs in CE¶
Above we have defined the $D(Z)$ operator as:
$$D(Z) = \{ (w,\theta): \exists h \in CE^0_\pi \text{ and } (m(h),x(h),w'(h),\theta'(h)) \in [0,\bar m] \times X \times Z$$
such that
$$w = u(f(x(h))) + v(m(h)) + \beta w'(h)$$$$\theta = u'(f(x(h)))(m(h) + x(h))$$$$x(h) = m(h)(h-1)$$$$m(h)(u'(f(x(h))) - v'(m(h))) \leq \beta \theta'(h) \text{ (with equality if } m(h) < \bar m) \}$$
We noted that the set $S$ can be found by iterating to convergence on $D$, provided that we start with a sufficiently large initial set $S_0$.
Our implementation builds on ideas in this notebook.
To find $S$ we use a numerical algorithm called the outer hyperplane approximation algorithm.
It was invented by Judd, Yeltekin, Conklin [JYC03].
This algorithm constructs the smallest convex set that contains the fixed point of the $D(S)$ operator.
Given that we are finding the smallest convex set that contains $S$, we can represent it on a computer as the intersection of a finite number of half-spaces.
Let $H$ be a set of subgradients, and $C$ be a set of hyperplane levels.
We approximate $S$ by:
$$\tilde S = \{(w,\theta)| H \cdot (w,\theta) \leq C \}$$
A key feature of this algorithm is that we discretize the action space, i.e., we create a grid of possible values for $m$ and $h$ (note that $x$ is implied by $m$ and $h$). This discretization simplifies computation of $\tilde S$ by allowing us to find it by solving a sequence of linear programs.
The outer hyperplane approximation algorithm proceeds as follows:
1. Initialize subgradients, $H$, and hyperplane levels, $C_0$.
2. Given a set of subgradients, $H$, and hyperplane levels, $C_t$, for each subgradient $h_i \in H$:
• Solve a linear program (described below) for each action in the action space.
• Find the maximum and update the corresponding hyperplane level, $C_{i,t+1}$.
3. If $|C_{t+1}-C_t| > \epsilon$, return to 2.
Step 1 simply creates a large initial set $S_0$.
Given some set $S_t$, Step 2 then constructs the set $S_{t+1} = D(S_t)$. The linear program in Step 2 is designed to construct a set $S_{t+1}$ that is as large as possible while satisfying the constraints of the $D(S)$ operator.
To do this, for each subgradient $h_i$, and for each point in the action space $(m_j,h_j)$, we solve the following problem:
$$\max_{[w',\theta']} h_i \cdot (w,\theta)$$
subject to
$$H \cdot (w',\theta') \leq C_t$$$$w = u(f(x_j)) + v(m_j) + \beta w'$$$$\theta = u'(f(x_j))(m_j + x_j)$$$$x_j = m_j(h_j-1)$$$$m_j(u'(f(x_j)) - v'(m_j)) \leq \beta \theta'\hspace{2mm} (= \text{if } m_j < \bar m)$$
This problem maximizes the hyperplane level for a given set of actions.
The second part of Step 2 then finds the maximum possible hyperplane level across the action space.
The algorithm constructs a sequence of progressively smaller sets $S_{t+1} \subset S_t \subset S_{t-1} \cdots \subset S_0$.
Step 3 ends the algorithm when the difference between these sets is small enough.
We have created a Python class that solves the model assuming the following functional forms:
$$u(c) = log(c)$$$$v(m) = \frac{1}{500}(m \bar m - 0.5m^2)^{0.5}$$$$f(x) = 180 - (0.4x)^2$$
The remaining parameters $\{\beta, \bar m, \underline h, \bar h\}$ are then variables to be specified for an instance of the Chang class.
Below we use the class to solve the model and plot the resulting equilibrium set, once with $\beta = 0.3$ and once with $\beta = 0.8$.
(Here we have set the number of subgradients to 10 in order to speed up the code for now - we can increase accuracy by increasing the number of subgradients)
In [3]:
"""
Author: Sebastian Graves
Provides a class called ChangModel to solve different
parameterizations of the Chang (1998) model.
"""
import numpy as np
import quantecon as qe
import time
from scipy.spatial import ConvexHull
from scipy.optimize import linprog, minimize, minimize_scalar
from scipy.interpolate import UnivariateSpline
import numpy.polynomial.chebyshev as cheb
class ChangModel:
"""
Class to solve for the competitive and sustainable sets in the Chang (1998)
model, for different parameterizations.
"""
def __init__(self, β, mbar, h_min, h_max, n_h, n_m, N_g):
# Record parameters
self.β, self.mbar, self.h_min, self.h_max = β, mbar, h_min, h_max
self.n_h, self.n_m, self.N_g = n_h, n_m, N_g
# Create other parameters
self.m_min = 1e-9
self.m_max = self.mbar
self.N_a = self.n_h*self.n_m
# Utility and production functions
uc = lambda c: np.log(c)
uc_p = lambda c: 1/c
v = lambda m: 1/500 * (mbar * m - 0.5 * m**2)**0.5
v_p = lambda m: 0.5/500 * (mbar * m - 0.5 * m**2)**(-0.5) * (mbar - m)
u = lambda h, m: uc(f(h, m)) + v(m)
def f(h, m):
x = m * (h - 1)
f = 180 - (0.4 * x)**2
return f
def θ(h, m):
x = m * (h - 1)
θ = uc_p(f(h, m)) * (m + x)
return θ
# Create set of possible action combinations, A
A1 = np.linspace(h_min, h_max, n_h).reshape(n_h, 1)
A2 = np.linspace(self.m_min, self.m_max, n_m).reshape(n_m, 1)
self.A = np.concatenate((np.kron(np.ones((n_m, 1)), A1),
np.kron(A2, np.ones((n_h, 1)))), axis=1)
# Pre-compute utility and output vectors
self.euler_vec = -np.multiply(self.A[:, 1], \
uc_p(f(self.A[:, 0], self.A[:, 1])) - v_p(self.A[:, 1]))
self.u_vec = u(self.A[:, 0], self.A[:, 1])
self.Θ_vec = θ(self.A[:, 0], self.A[:, 1])
self.f_vec = f(self.A[:, 0], self.A[:, 1])
self.bell_vec = np.multiply(uc_p(f(self.A[:, 0],
self.A[:, 1])),
np.multiply(self.A[:, 1],
(self.A[:, 0] - 1))) \
+ np.multiply(self.A[:, 1],
v_p(self.A[:, 1]))
# Find extrema of (w, θ) space for initial guess of equilibrium sets
p_vec = np.zeros(self.N_a)
w_vec = np.zeros(self.N_a)
for i in range(self.N_a):
p_vec[i] = self.Θ_vec[i]
w_vec[i] = self.u_vec[i]/(1 - β)
w_space = np.array([min(w_vec[~np.isinf(w_vec)]),
max(w_vec[~np.isinf(w_vec)])])
p_space = np.array([0, max(p_vec[~np.isinf(w_vec)])])
self.p_space = p_space
# Set up hyperplane levels and gradients for iterations
def SG_H_V(N, w_space, p_space):
"""
This function initializes the subgradients, hyperplane levels,
and extreme points of the value set by choosing an appropriate
origin and radius. It is based on a similar function in QuantEcon's
Games.jl
"""
# First, create a unit circle. Want points placed on [0, 2π]
inc = 2 * np.pi / N
degrees = np.arange(0, 2 * np.pi, inc)
# Points on circle
H = np.zeros((N, 2))
for i in range(N):
x = degrees[i]
H[i, 0] = np.cos(x)
H[i, 1] = np.sin(x)
# Then calculate origin and radius
o = np.array([np.mean(w_space), np.mean(p_space)])
r1 = max((max(w_space) - o[0])**2, (o[0] - min(w_space))**2)
r2 = max((max(p_space) - o[1])**2, (o[1] - min(p_space))**2)
r = np.sqrt(r1 + r2)
# Now calculate vertices
Z = np.zeros((2, N))
for i in range(N):
Z[0, i] = o[0] + r*H.T[0, i]
Z[1, i] = o[1] + r*H.T[1, i]
# Corresponding hyperplane levels
C = np.zeros(N)
for i in range(N):
C[i] = np.dot(Z[:, i], H[i, :])
return C, H, Z
C, self.H, Z = SG_H_V(N_g, w_space, p_space)
C = C.reshape(N_g, 1)
self.c0_c, self.c0_s, self.c1_c, self.c1_s = np.copy(C), np.copy(C), \
np.copy(C), np.copy(C)
self.z0_s, self.z0_c, self.z1_s, self.z1_c = np.copy(Z), np.copy(Z), \
np.copy(Z), np.copy(Z)
self.w_bnds_s, self.w_bnds_c = (w_space[0], w_space[1]), \
(w_space[0], w_space[1])
self.p_bnds_s, self.p_bnds_c = (p_space[0], p_space[1]), \
(p_space[0], p_space[1])
# Create dictionaries to save equilibrium set for each iteration
self.c_dic_s, self.c_dic_c = {}, {}
self.c_dic_s[0], self.c_dic_c[0] = self.c0_s, self.c0_c
def solve_worst_spe(self):
"""
Method to solve for BR(Z). See p.449 of Chang (1998)
"""
p_vec = np.full(self.N_a, np.nan)
c = [1, 0]
# Pre-compute constraints
aineq_mbar = np.vstack((self.H, np.array([0, -self.β])))
bineq_mbar = np.vstack((self.c0_s, 0))
aineq = self.H
bineq = self.c0_s
aeq = [[0, -self.β]]
for j in range(self.N_a):
# Only try if consumption is possible
if self.f_vec[j] > 0:
# If m = mbar, use inequality constraint
if self.A[j, 1] == self.mbar:
bineq_mbar[-1] = self.euler_vec[j]
res = linprog(c, A_ub=aineq_mbar, b_ub=bineq_mbar,
bounds=(self.w_bnds_s, self.p_bnds_s))
else:
beq = self.euler_vec[j]
res = linprog(c, A_ub=aineq, b_ub=bineq, A_eq=aeq, b_eq=beq,
bounds=(self.w_bnds_s, self.p_bnds_s))
if res.status == 0:
p_vec[j] = self.u_vec[j] + self.β * res.x[0]
# Max over h and min over other variables (see Chang (1998) p.449)
self.br_z = np.nanmax(np.nanmin(p_vec.reshape(self.n_m, self.n_h), 0))
"""
Method to solve for E(Z). See p.449 of Chang (1998)
"""
# Pre-compute constraints
aineq_C_mbar = np.vstack((self.H, np.array([0, -self.β])))
bineq_C_mbar = np.vstack((self.c0_c, 0))
aineq_C = self.H
bineq_C = self.c0_c
aeq_C = [[0, -self.β]]
aineq_S_mbar = np.vstack((np.vstack((self.H, np.array([0, -self.β]))),
np.array([-self.β, 0])))
bineq_S_mbar = np.vstack((self.c0_s, np.zeros((2, 1))))
aineq_S = np.vstack((self.H, np.array([-self.β, 0])))
bineq_S = np.vstack((self.c0_s, 0))
aeq_S = [[0, -self.β]]
# Update maximal hyperplane level
for i in range(self.N_g):
c_a1a2_c, t_a1a2_c = np.full(self.N_a, -np.inf), \
np.zeros((self.N_a, 2))
c_a1a2_s, t_a1a2_s = np.full(self.N_a, -np.inf), \
np.zeros((self.N_a, 2))
c = [-self.H[i, 0], -self.H[i, 1]]
for j in range(self.N_a):
# Only try if consumption is possible
if self.f_vec[j] > 0:
# COMPETITIVE EQUILIBRIA
# If m = mbar, use inequality constraint
if self.A[j, 1] == self.mbar:
bineq_C_mbar[-1] = self.euler_vec[j]
res = linprog(c, A_ub=aineq_C_mbar, b_ub=bineq_C_mbar,
bounds=(self.w_bnds_c, self.p_bnds_c))
# If m < mbar, use equality constraint
else:
beq_C = self.euler_vec[j]
res = linprog(c, A_ub=aineq_C, b_ub=bineq_C, A_eq = aeq_C,
b_eq = beq_C, bounds=(self.w_bnds_c, \
self.p_bnds_c))
if res.status == 0:
c_a1a2_c[j] = self.H[i, 0] * (self.u_vec[j] \
+ self.β * res.x[0]) + self.H[i, 1] * self.Θ_vec[j]
t_a1a2_c[j] = res.x
# SUSTAINABLE EQUILIBRIA
# If m = mbar, use inequality constraint
if self.A[j, 1] == self.mbar:
bineq_S_mbar[-2] = self.euler_vec[j]
bineq_S_mbar[-1] = self.u_vec[j] - self.br_z
res = linprog(c, A_ub=aineq_S_mbar, b_ub=bineq_S_mbar,
bounds=(self.w_bnds_s, self.p_bnds_s))
# If m < mbar, use equality constraint
else:
bineq_S[-1] = self.u_vec[j] - self.br_z
beq_S = self.euler_vec[j]
res = linprog(c, A_ub=aineq_S, b_ub=bineq_S, A_eq = aeq_S,
b_eq = beq_S, bounds=(self.w_bnds_s, \
self.p_bnds_s))
if res.status == 0:
c_a1a2_s[j] = self.H[i, 0] * (self.u_vec[j] \
+ self.β*res.x[0]) + self.H[i, 1] * self.Θ_vec[j]
t_a1a2_s[j] = res.x
idx_c = np.where(c_a1a2_c == max(c_a1a2_c))[0][0]
self.z1_c[:, i] = np.array([self.u_vec[idx_c]
+ self.β * t_a1a2_c[idx_c, 0],
self.Θ_vec[idx_c]])
idx_s = np.where(c_a1a2_s == max(c_a1a2_s))[0][0]
self.z1_s[:, i] = np.array([self.u_vec[idx_s]
+ self.β * t_a1a2_s[idx_s, 0],
self.Θ_vec[idx_s]])
for i in range(self.N_g):
self.c1_c[i] = np.dot(self.z1_c[:, i], self.H[i, :])
self.c1_s[i] = np.dot(self.z1_s[:, i], self.H[i, :])
def solve_sustainable(self, tol=1e-5, max_iter=250):
"""
Method to solve for the competitive and sustainable equilibrium sets.
"""
t = time.time()
diff = tol + 1
iters = 0
print('### --------------- ###')
print('Solving Chang Model Using Outer Hyperplane Approximation')
print('### --------------- ### \n')
print('Maximum difference when updating hyperplane levels:')
while diff > tol and iters < max_iter:
iters = iters + 1
self.solve_worst_spe()
diff = max(np.maximum(abs(self.c0_c - self.c1_c),
abs(self.c0_s - self.c1_s)))
print(diff)
# Update hyperplane levels
self.c0_c, self.c0_s = np.copy(self.c1_c), np.copy(self.c1_s)
# Update bounds for w and θ
wmin_c, wmax_c = np.min(self.z1_c, axis=1)[0], \
np.max(self.z1_c, axis=1)[0]
pmin_c, pmax_c = np.min(self.z1_c, axis=1)[1], \
np.max(self.z1_c, axis=1)[1]
wmin_s, wmax_s = np.min(self.z1_s, axis=1)[0], \
np.max(self.z1_s, axis=1)[0]
pmin_S, pmax_S = np.min(self.z1_s, axis=1)[1], \
np.max(self.z1_s, axis=1)[1]
self.w_bnds_s, self.w_bnds_c = (wmin_s, wmax_s), (wmin_c, wmax_c)
self.p_bnds_s, self.p_bnds_c = (pmin_S, pmax_S), (pmin_c, pmax_c)
# Save iteration
self.c_dic_c[iters], self.c_dic_s[iters] = np.copy(self.c1_c), \
np.copy(self.c1_s)
self.iters = iters
elapsed = time.time() - t
print('Convergence achieved after {} iterations and {} \
seconds'.format(iters, round(elapsed, 2)))
def solve_bellman(self, θ_min, θ_max, order, disp=False, tol=1e-7, maxiters=100):
"""
Continuous Method to solve the Bellman equation in section 25.3
"""
mbar = self.mbar
# Utility and production functions
uc = lambda c: np.log(c)
uc_p = lambda c: 1 / c
v = lambda m: 1 / 500 * (mbar * m - 0.5 * m**2)**0.5
v_p = lambda m: 0.5/500 * (mbar*m - 0.5 * m**2)**(-0.5) * (mbar - m)
u = lambda h, m: uc(f(h, m)) + v(m)
def f(h, m):
x = m * (h - 1)
f = 180 - (0.4 * x)**2
return f
def θ(h, m):
x = m * (h - 1)
θ = uc_p(f(h, m)) * (m + x)
return θ
# Bounds for Maximization
lb1 = np.array([self.h_min, 0, θ_min])
ub1 = np.array([self.h_max, self.mbar - 1e-5, θ_max])
lb2 = np.array([self.h_min, θ_min])
ub2 = np.array([self.h_max, θ_max])
# Initialize Value Function coefficients
# Calculate roots of Chebyshev polynomial
k = np.linspace(order, 1, order)
roots = np.cos((2 * k - 1) * np.pi / (2 * order))
# Scale to approximation space
s = θ_min + (roots - -1) / 2 * (θ_max - θ_min)
# Create a basis matrix
Φ = cheb.chebvander(roots, order - 1)
c = np.zeros(Φ.shape[0])
# Function to minimize and constraints
def p_fun(x):
scale = -1 + 2 * (x[2] - θ_min)/(θ_max - θ_min)
p_fun = - (u(x[0], x[1]) \
+ self.β * np.dot(cheb.chebvander(scale, order - 1), c))
return p_fun
def p_fun2(x):
scale = -1 + 2*(x[1] - θ_min)/(θ_max - θ_min)
p_fun = - (u(x[0],mbar) \
+ self.β * np.dot(cheb.chebvander(scale, order - 1), c))
return p_fun
cons1 = ({'type': 'eq', 'fun': lambda x: uc_p(f(x[0], x[1])) * x[1]
* (x[0] - 1) + v_p(x[1]) * x[1] + self.β * x[2] - θ},
{'type': 'eq', 'fun': lambda x: uc_p(f(x[0], x[1]))
* x[0] * x[1] - θ})
cons2 = ({'type': 'ineq', 'fun': lambda x: uc_p(f(x[0], mbar)) * mbar
* (x[0] - 1) + v_p(mbar) * mbar + self.β * x[1] - θ},
{'type': 'eq', 'fun': lambda x: uc_p(f(x[0], mbar))
* x[0] * mbar - θ})
bnds1 = np.concatenate([lb1.reshape(3, 1), ub1.reshape(3, 1)], axis=1)
bnds2 = np.concatenate([lb2.reshape(2, 1), ub2.reshape(2, 1)], axis=1)
# Bellman Iterations
diff = 1
iters = 1
while diff > tol:
# 1. Maximization, given value function guess
p_iter1 = np.zeros(order)
for i in range(order):
θ = s[i]
res = minimize(p_fun,
lb1 + (ub1-lb1) / 2,
method='SLSQP',
bounds=bnds1,
constraints=cons1,
tol=1e-10)
if res.success == True:
p_iter1[i] = -p_fun(res.x)
res = minimize(p_fun2,
lb2 + (ub2-lb2) / 2,
method='SLSQP',
bounds=bnds2,
constraints=cons2,
tol=1e-10)
if -p_fun2(res.x) > p_iter1[i] and res.success == True:
p_iter1[i] = -p_fun2(res.x)
# 2. Bellman updating of Value Function coefficients
c1 = np.linalg.solve(Φ, p_iter1)
# 3. Compute distance and update
diff = np.linalg.norm(c - c1)
if bool(disp == True):
print(diff)
c = np.copy(c1)
iters = iters + 1
if iters > maxiters:
print('Convergence failed after {} iterations'.format(maxiters))
break
self.θ_grid = s
self.p_iter = p_iter1
self.Φ = Φ
self.c = c
print('Convergence achieved after {} iterations'.format(iters))
# Check residuals
θ_grid_fine = np.linspace(θ_min, θ_max, 100)
resid_grid = np.zeros(100)
p_grid = np.zeros(100)
θ_prime_grid = np.zeros(100)
m_grid = np.zeros(100)
h_grid = np.zeros(100)
for i in range(100):
θ = θ_grid_fine[i]
res = minimize(p_fun,
lb1 + (ub1-lb1) / 2,
method='SLSQP',
bounds=bnds1,
constraints=cons1,
tol=1e-10)
if res.success == True:
p = -p_fun(res.x)
p_grid[i] = p
θ_prime_grid[i] = res.x[2]
h_grid[i] = res.x[0]
m_grid[i] = res.x[1]
res = minimize(p_fun2,
lb2 + (ub2-lb2)/2,
method='SLSQP',
bounds=bnds2,
constraints=cons2,
tol=1e-10)
if -p_fun2(res.x) > p and res.success == True:
p = -p_fun2(res.x)
p_grid[i] = p
θ_prime_grid[i] = res.x[1]
h_grid[i] = res.x[0]
m_grid[i] = self.mbar
scale = -1 + 2 * (θ - θ_min)/(θ_max - θ_min)
resid_grid[i] = np.dot(cheb.chebvander(scale, order-1), c) - p
self.resid_grid = resid_grid
self.θ_grid_fine = θ_grid_fine
self.θ_prime_grid = θ_prime_grid
self.m_grid = m_grid
self.h_grid = h_grid
self.p_grid = p_grid
self.x_grid = m_grid * (h_grid - 1)
# Simulate
θ_series = np.zeros(31)
m_series = np.zeros(30)
h_series = np.zeros(30)
# Find initial θ
def ValFun(x):
scale = -1 + 2*(x - θ_min)/(θ_max - θ_min)
p_fun = np.dot(cheb.chebvander(scale, order - 1), c)
return -p_fun
res = minimize(ValFun,
(θ_min + θ_max)/2,
bounds=[(θ_min, θ_max)])
θ_series[0] = res.x
# Simulate
for i in range(30):
θ = θ_series[i]
res = minimize(p_fun,
lb1 + (ub1-lb1)/2,
method='SLSQP',
bounds=bnds1,
constraints=cons1,
tol=1e-10)
if res.success == True:
p = -p_fun(res.x)
h_series[i] = res.x[0]
m_series[i] = res.x[1]
θ_series[i+1] = res.x[2]
res2 = minimize(p_fun2,
lb2 + (ub2-lb2)/2,
method='SLSQP',
bounds=bnds2,
constraints=cons2,
tol=1e-10)
if -p_fun2(res2.x) > p and res2.success == True:
h_series[i] = res2.x[0]
m_series[i] = self.mbar
θ_series[i+1] = res2.x[1]
self.θ_series = θ_series
self.m_series = m_series
self.h_series = h_series
self.x_series = m_series * (h_series - 1)
In [4]:
ch1 = ChangModel(β=0.3, mbar=30, h_min=0.9, h_max=2, n_h=8, n_m=35, N_g=10)
ch1.solve_sustainable()
### --------------- ###
Solving Chang Model Using Outer Hyperplane Approximation
### --------------- ###
Maximum difference when updating hyperplane levels:
[1.9168]
[0.66782]
[0.49235]
[0.32412]
[0.19022]
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-4-d19a06b35f4c> in <module>
1 ch1 = ChangModel(β=0.3, mbar=30, h_min=0.9, h_max=2, n_h=8, n_m=35, N_g=10)
----> 2 ch1.solve_sustainable()
<ipython-input-3-ae015887b922> in solve_sustainable(self, tol, max_iter)
271 iters = iters + 1
272 self.solve_worst_spe()
274 diff = max(np.maximum(abs(self.c0_c - self.c1_c),
275 abs(self.c0_s - self.c1_s)))
233 res = linprog(c, A_ub=aineq_S, b_ub=bineq_S, A_eq = aeq_S,
234 b_eq = beq_S, bounds=(self.w_bnds_s, \
--> 235 self.p_bnds_s))
236 if res.status == 0:
237 c_a1a2_s[j] = self.H[i, 0] * (self.u_vec[j] \
~/anaconda3/lib/python3.7/site-packages/scipy/optimize/_linprog.py in linprog(c, A_ub, b_ub, A_eq, b_eq, bounds, method, callback, options, x0)
539 x, fun, slack, con, status, message = _postprocess(
540 x, c_o, A_ub_o, b_ub_o, A_eq_o, b_eq_o, bounds,
--> 541 complete, undo, status, message, tol, iteration, disp)
542
543 sol = {
~/anaconda3/lib/python3.7/site-packages/scipy/optimize/_linprog_util.py in _postprocess(x, c, A_ub, b_ub, A_eq, b_eq, bounds, complete, undo, status, message, tol, iteration, disp)
1412 status, message = _check_result(
1413 x, fun, status, slack, con,
-> 1414 lb, ub, tol, message
1415 )
1416
~/anaconda3/lib/python3.7/site-packages/scipy/optimize/_linprog_util.py in _check_result(x, fun, status, slack, con, lb, ub, tol, message)
1325 # nearly basic feasible solution. Postsolving can make the solution
1326 # basic, however, this solution is NOT optimal
-> 1327 raise ValueError(message)
1328
1329 return status, message
ValueError: The algorithm terminated successfully and determined that the problem is infeasible.
In [5]:
def plot_competitive(ChangModel):
"""
Method that only plots competitive equilibrium set
"""
poly_C = polytope.Polytope(ChangModel.H, ChangModel.c1_c)
ext_C = polytope.extreme(poly_C)
fig, ax = plt.subplots(figsize=(7, 5))
ax.set_xlabel('w', fontsize=16)
ax.set_ylabel(r"$\theta$", fontsize=18)
ax.fill(ext_C[:,0], ext_C[:,1], 'r', zorder=0)
ChangModel.min_theta = min(ext_C[:, 1])
ChangModel.max_theta = max(ext_C[:, 1])
# Add point showing Ramsey Plan
idx_Ramsey = np.where(ext_C[:, 0] == max(ext_C[:, 0]))[0][0]
R = ext_C[idx_Ramsey, :]
ax.scatter(R[0], R[1], 150, 'black', 'o', zorder=1)
w_min = min(ext_C[:, 0])
# Label Ramsey Plan slightly to the right of the point
ax.annotate("R", xy=(R[0], R[1]), xytext=(R[0] + 0.03 * (R[0] - w_min),
R[1]), fontsize=18)
plt.tight_layout()
plt.show()
plot_competitive(ch1)
In [6]:
ch2 = ChangModel(β=0.8, mbar=30, h_min=0.9, h_max=1/0.8,
n_h=8, n_m=35, N_g=10)
ch2.solve_sustainable()
### --------------- ###
Solving Chang Model Using Outer Hyperplane Approximation
### --------------- ###
Maximum difference when updating hyperplane levels:
[0.06369]
[0.02476]
[0.02153]
[0.01915]
[0.01795]
[0.01642]
[0.01507]
[0.01284]
[0.01106]
[0.00694]
[0.0085]
[0.00781]
[0.00433]
[0.00492]
[0.00303]
[0.00182]
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-6-1970f7c91f36> in <module>
1 ch2 = ChangModel(β=0.8, mbar=30, h_min=0.9, h_max=1/0.8,
2 n_h=8, n_m=35, N_g=10)
----> 3 ch2.solve_sustainable()
<ipython-input-3-ae015887b922> in solve_sustainable(self, tol, max_iter)
271 iters = iters + 1
272 self.solve_worst_spe()
274 diff = max(np.maximum(abs(self.c0_c - self.c1_c),
275 abs(self.c0_s - self.c1_s)))
233 res = linprog(c, A_ub=aineq_S, b_ub=bineq_S, A_eq = aeq_S,
234 b_eq = beq_S, bounds=(self.w_bnds_s, \
--> 235 self.p_bnds_s))
236 if res.status == 0:
237 c_a1a2_s[j] = self.H[i, 0] * (self.u_vec[j] \
~/anaconda3/lib/python3.7/site-packages/scipy/optimize/_linprog.py in linprog(c, A_ub, b_ub, A_eq, b_eq, bounds, method, callback, options, x0)
539 x, fun, slack, con, status, message = _postprocess(
540 x, c_o, A_ub_o, b_ub_o, A_eq_o, b_eq_o, bounds,
--> 541 complete, undo, status, message, tol, iteration, disp)
542
543 sol = {
~/anaconda3/lib/python3.7/site-packages/scipy/optimize/_linprog_util.py in _postprocess(x, c, A_ub, b_ub, A_eq, b_eq, bounds, complete, undo, status, message, tol, iteration, disp)
1412 status, message = _check_result(
1413 x, fun, status, slack, con,
-> 1414 lb, ub, tol, message
1415 )
1416
~/anaconda3/lib/python3.7/site-packages/scipy/optimize/_linprog_util.py in _check_result(x, fun, status, slack, con, lb, ub, tol, message)
1325 # nearly basic feasible solution. Postsolving can make the solution
1326 # basic, however, this solution is NOT optimal
-> 1327 raise ValueError(message)
1328
1329 return status, message
ValueError: The algorithm terminated successfully and determined that the problem is infeasible.
In [7]:
plot_competitive(ch2)
## Solving a Continuation Ramsey Planner’s Bellman Equation¶
In this section we solve the Bellman equation confronting a continuation Ramsey planner.
The construction of a Ramsey plan is decomposed into a two subproblems in Ramsey plans, time inconsistency, sustainable plans and dynamic Stackelberg problems.
• Subproblem 1 is faced by a sequence of continuation Ramsey planners at $t \geq 1$.
• Subproblem 2 is faced by a Ramsey planner at $t = 0$.
The problem is:
$$J(\theta) = \max_{m,x,h,\theta'} u(f(x)) + v(m) + \beta J(\theta')$$
subject to:
$$\theta \leq u'(f(x))x + v'(m)m + \beta \theta'$$$$\theta = u'(f(x))(m + x )$$$$x = m(h-1)$$$$(m,x,h) \in E$$$$\theta' \in \Omega$$
To solve this Bellman equation, we must know the set $\Omega$.
We have solved the Bellman equation for the two sets of parameter values for which we computed the equilibrium value sets above.
Hence for these parameter configurations, we know the bounds of $\Omega$.
The two sets of parameters differ only in the level of $\beta$.
From the figures earlier in this lecture, we know that when $\beta = 0.3$, $\Omega = [0.0088,0.0499]$, and when $\beta = 0.8$, $\Omega = [0.0395,0.2193]$
In [8]:
ch1 = ChangModel(β=0.3, mbar=30, h_min=0.99, h_max=1/0.3,
n_h=8, n_m=35, N_g=50)
ch2 = ChangModel(β=0.8, mbar=30, h_min=0.1, h_max=1/0.8,
n_h=20, n_m=50, N_g=50)
/home/ubuntu/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:35: RuntimeWarning: invalid value encountered in log
In [9]:
ch1.solve_bellman(θ_min=0.01, θ_max=0.0499, order=30, tol=1e-6)
ch2.solve_bellman(θ_min=0.045, θ_max=0.15, order=30, tol=1e-6)
/home/ubuntu/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:311: RuntimeWarning: invalid value encountered in log
Convergence achieved after 15 iterations
Convergence achieved after 72 iterations
First, a quick check that our approximations of the value functions are good.
We do this by calculating the residuals between iterates on the value function on a fine grid:
In [10]:
max(abs(ch1.resid_grid)), max(abs(ch2.resid_grid))
Out[10]:
(6.463130481471069e-06, 6.875604405820468e-07)
The value functions plotted below trace out the right edges of the sets of equilibrium values plotted above
In [11]:
fig, axes = plt.subplots(1, 2, figsize=(12, 4))
for ax, model in zip(axes, (ch1, ch2)):
ax.plot(model.θ_grid, model.p_iter)
ax.set(xlabel=r"$\theta$",
ylabel=r"$J(\theta)$",
title=rf"$\beta = {model.β}$")
plt.show()
The next figure plots the optimal policy functions; values of $\theta',m,x,h$ for each value of the state $\theta$:
In [12]:
for model in (ch1, ch2):
fig, axes = plt.subplots(2, 2, figsize=(12, 6), sharex=True)
fig.suptitle(rf"$\beta = {model.β}$", fontsize=16)
plots = [model.θ_prime_grid, model.m_grid,
model.h_grid, model.x_grid]
labels = [r"$\theta'$", "$m$", "$h$", "$x$"]
for ax, plot, label in zip(axes.flatten(), plots, labels):
ax.plot(model.θ_grid_fine, plot)
ax.set_xlabel(r"$\theta$", fontsize=14)
ax.set_ylabel(label, fontsize=14)
plt.show()
With the first set of parameter values, the value of $\theta'$ chosen by the Ramsey planner quickly hits the upper limit of $\Omega$.
But with the second set of parameters it converges to a value in the interior of the set.
Consequently, the choice of $\bar \theta$ is clearly important with the first set of parameter values.
One way of seeing this is plotting $\theta'(\theta)$ for each set of parameters.
With the first set of parameter values, this function does not intersect the 45-degree line until $\bar \theta$, whereas in the second set of parameter values, it intersects in the interior.
In [13]:
fig, axes = plt.subplots(1, 2, figsize=(12, 4))
for ax, model in zip(axes, (ch1, ch2)):
ax.plot(model.θ_grid_fine, model.θ_prime_grid, label=r"$\theta'(\theta)$")
ax.plot(model.θ_grid_fine, model.θ_grid_fine, label=r"$\theta$")
ax.set(xlabel=r"$\theta$", title=rf"$\beta = {model.β}$")
axes[0].legend()
plt.show()
Subproblem 2 is equivalent to the planner choosing the initial value of $\theta$ (i.e. the value which maximizes the value function).
From this starting point, we can then trace out the paths for $\{\theta_t,m_t,h_t,x_t\}_{t=0}^\infty$ that support this equilibrium.
These are shown below for both sets of parameters
In [14]:
for model in (ch1, ch2):
fig, axes = plt.subplots(2, 2, figsize=(12, 6))
fig.suptitle(rf"$\beta = {model.β}$")
plots = [model.θ_series, model.m_series, model.h_series, model.x_series]
labels = [r"$\theta$", "$m$", "$h$", "$x$"]
for ax, plot, label in zip(axes.flatten(), plots, labels):
ax.plot(plot)
ax.set(xlabel='t', ylabel=label)
plt.show()
### Next Steps¶
In Credible Government Policies in Chang Model we shall find a subset of competitive equilibria that are sustainable in the sense that a sequence of government administrations that chooses sequentially, rather than once and for all at time $0$ will choose to implement them.
In the process of constructing them, we shall construct another, smaller set of competitive equilibria.
• Share page | 15,613 | 50,182 | {"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": 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} | 2.796875 | 3 | CC-MAIN-2019-43 | longest | en | 0.849887 |
https://www.numbersaplenty.com/30734 | 1,670,542,589,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711368.1/warc/CC-MAIN-20221208215156-20221209005156-00231.warc.gz | 976,063,478 | 3,248 | Search a number
30734 = 2112127
BaseRepresentation
bin111100000001110
31120011022
413200032
51440414
6354142
7155414
oct74016
946138
1030734
1121100
1215952
1310cb2
14b2b4
15918e
hex780e
30734 has 12 divisors (see below), whose sum is σ = 51072. Its totient is φ = 13860.
The previous prime is 30727. The next prime is 30757. The reversal of 30734 is 43703.
It is a Curzon number.
It is a nialpdrome in base 11.
It is a congruent number.
It is an unprimeable number.
30734 is an untouchable number, because it is not equal to the sum of proper divisors of any number.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written in 5 ways as a sum of consecutive naturals, for example, 179 + ... + 305.
It is an arithmetic number, because the mean of its divisors is an integer number (4256).
230734 is an apocalyptic number.
30734 is a deficient number, since it is larger than the sum of its proper divisors (20338).
30734 is a wasteful number, since it uses less digits than its factorization.
30734 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 151 (or 140 counting only the distinct ones).
The product of its (nonzero) digits is 252, while the sum is 17.
The square root of 30734 is about 175.3111519556. The cubic root of 30734 is about 31.3236980584.
The spelling of 30734 in words is "thirty thousand, seven hundred thirty-four".
Divisors: 1 2 11 22 121 127 242 254 1397 2794 15367 30734 | 465 | 1,550 | {"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} | 2.96875 | 3 | CC-MAIN-2022-49 | latest | en | 0.881153 |
https://www.coursehero.com/file/6393179/Math151HW6/ | 1,516,140,384,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886739.5/warc/CC-MAIN-20180116204303-20180116224303-00167.warc.gz | 890,619,936 | 43,059 | Math151HW6
# Math151HW6 - 13.2 3-9 odd 11-17 odd(do only ab not c 21 23...
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MATH 151 HW #6: CHAPTER 13 ADDITIONAL “MISC. HOMEWORK” MAY BE ASSIGNED LATER! REMEMBER, HW #5 IS THE CH.12 HW ON CONICS! SPRING 2004 Write your name (encoded, if you like) and clearly separate sections! Consider photocopying your HW if you don't expect to collect it from me later. Show work where appropriate, and use “good form and procedure,” as in class! This (and HW #5) are due on the last day of class (when you take the Final). Graded out of 10 points (including possible additional HW). “*” denotes “See Hint below.” 13.1: 3-17 odd, 25-31 odd Look at the Examples. Example 5 demonstrates the usefulness of parametric equations (as opposed to more traditional ways of describing a curve).
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Unformatted text preview: 13.2: 3-9 odd, 11-17 odd (do only ab, not c), 21, 23, 25 Read Example 5 on pp.656-7 regarding the surface area of a sphere. 13.3: 1-11 odd, 13*, 15, 17, 19, 23, 27, 29, 31, 37-47* odd, 51, 53, 55 Hint on 13: Manipulate the equation first; the solution is short! Hint on 47: Multiply both sides of the equation by something first. Look at 57: You would use implicit differentiation to crack this one. Look at 66. 13.4: 1-9* odd, 17, 19*, 23, 27, 29 Hint on 9: Use 13.3 #19. Hint on 19: Use Figure 13.35 on p.670. 13.5: Skip, but skim this sometime. Polar Equations of Conics (mixes Chs.12, 13). Misc. Homework may be assigned later....
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Let W be a subspace of R^n, and P the orthogonal projection onto W. Then Ker P is W^perp.
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If T : R 3 → R 3 is projection onto a line through the origin, describe geometrically the eigenvalues and eigenvectors of T. | 1,297 | 3,739 | {"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} | 3.640625 | 4 | CC-MAIN-2023-50 | longest | en | 0.847891 |
https://www.jiskha.com/display.cgi?id=1264705388 | 1,503,062,064,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104636.62/warc/CC-MAIN-20170818121545-20170818141545-00013.warc.gz | 937,559,398 | 4,221 | # College Algebra
posted by .
Translate the problem into a pair of linear equations in two variables. Solve the equations using either elimination or substitution. State your answer for both variables.
The perimeter of a rectangle is 72m. If the width were doubled and the length were increasd by 9m, then the perimeter would be 114m. What are the dimentions?
• College Algebra -
2L+2W=72
2(2W) + 2(L+9)=114
check that.
• College Algebra -
Does this mean that the dimentions are 72 and 114 or do you take it father to find what w & l are which I think might be w=12 and l=24. does that sound right?
• College Algebra -
Of course you have to find width w, and length L.
I would go back and check W=12, L=24 in the original equations. If they check, you are home bound.
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Homework 8 29
# Homework 8 29 - 29.1 The first step in this problem is to...
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Unformatted text preview: 29.1) The first step in this problem is to create a system of first order equations. This can be found on the attached sheet as hand calculations. The numerical solution problem is then broken up into parts: a) interval 1 to 2, b) interval 1 to 2.1 a) >> f = inline('[y(2);y(3);‐y(3).^2+3.*y(2).^3‐cos(y(1)).^2+exp(‐t).*sin(3.*t)]','t','y') f = Inline function: f(t,y) = [y(2);y(3);‐y(3).^2+3.*y(2).^3‐cos(y(1)).^2+exp(‐t).*sin(3.*t)] >> [T Y] = ode45(f,[1 2],[1;2;0]); *Command took approximately 12 seconds. >> theta = Y(:,1); >> plot(T,theta) b) The second part portion of the problem, over the interval from 1 to 2.1, was not able to be solved. When the interval was set to the limits of 1.0 and 2.1 the computation took an extremely long period of time. As stated above the computation time for the interval of 1 to 2 from part “a” took only around 12 seconds to complete. As the limits approached 1.0 to 2.1 the time required for the computation grew at an exponential rate. For the interval 1.00 to 2.01 below the computation time took approximately 1 minute. >> [T Y] = ode45(f,[1 2.01],[1;2;0]); *Computation took approximately 1 minute >> theta = Y(:,1); >> plot(T,theta) When the limits were expanded to 1.00 to 2.02 the computation was still not concluded after 40 minutes. >> [T Y] = ode45(f,[1 2.02],[1;2;0]); ??? Operation terminated by user during ==> inlineeval at 11 In ==> inline.feval at 36 INLINE_OUT_ = inlineeval(INLINE_INPUTS_, INLINE_OBJ_.inputExpr, INLINE_OBJ_.expr); In ==> ode45 at 324 f(:,2) = feval(odeFcn,t+hA(1),y+f*hB(:,1),odeArgs{:}); The reason MATLab cannot calculate the graph as the time approaches 2.1 is because of the exponential nature of the equation as time approaches the asymptote. The value of theta will go to infinity before time 2.1 is reached. ...
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Fund Quantum Mechanics Lect & HW Solutions 108
# Fund Quantum Mechanics Lect & HW Solutions 108 - 1...
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90 CHAPTER 7. TIME EVOLUTION Classical physics understands the wave nature of light well, and not its particle nature. This is the opposite of the situation for an electron, where classical physics understands the particle nature, and not the wave nature. 7.1.2.2 Solution schrodsol-b Question: For the one-dimensional harmonic oscillator, the energy eigenvalues are E n = 2 n + 1 2 ω Write out the coefficients c n (0) e i E n t/ ¯ h for those energies. Now classically, the harmonic oscillator has a natural frequency ω . That means that whenever ωt is a whole multiple of 2 π , the harmonic oscillator is again in the same state as it started out with. Show that the coefficients of the energy eigenfunctions have a natural frequency of
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Unformatted text preview: 1 2 ω ; 1 2 ωt must be a whole multiple of 2 π for the coe±cients to return to their original values. Answer: The coe±cients are c n (0) e − i (2 n +1) 2 ωt Now if ωt is 2 π , the argument of the exponential equals i times an odd multiple of π . That makes the exponential equal to minus one. It takes until ωt = 4 π until the exponential returns to its original value one. 7.1.2.3 Solution schrodsol-c Question: Write the full wave function for a one-dimensional harmonic oscillator. Formulae are in chapter 3.6.2. Answer: Using the give formulae Ψ( x, t ) = ∞ s n =0 c n (0) e − i (2 n +1) 2 ωt h n ( x ) 7.1.3 Energy conservation...
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http://purrr.tidyverse.org/reference/rerun.html | 1,513,120,959,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948520042.35/warc/CC-MAIN-20171212231544-20171213011544-00389.warc.gz | 231,890,061 | 3,558 | This is a convenient way of generating sample data. It works similarly to replicate(..., simplify = FALSE).
rerun(.n, ...)
Arguments
.n Number of times to run expressions Expressions to re-run.
Value
A list of length .n. Each element of ... will be re-run once for each .n. It
There is one special case: if there's a single unnamed input, the second level list will be dropped. In this case, rerun(n, x) behaves like replicate(n, x, simplify = FALSE).
Examples
10 %>% rerun(rnorm(5))#> [[1]]
#> [1] 0.5802748 -0.5941795 -0.3718553 -1.4230352 -1.1404838
#>
#> [[2]]
#> [1] 1.2685047 -0.2801044 0.9110540 2.0043945 0.2471404
#>
#> [[3]]
#> [1] -0.4290761 -2.1837551 -0.1236349 0.2873839 -0.3715993
#>
#> [[4]]
#> [1] 1.6265623 0.3099083 -1.5640407 -1.2306235 -1.3701975
#>
#> [[5]]
#> [1] 0.1065558 -0.3988255 -0.3534061 0.2345011 1.7697899
#>
#> [[6]]
#> [1] 0.45787638 1.14375811 -1.30772262 -0.94254960 -0.09496125
#>
#> [[7]]
#> [1] -1.130047 -1.456208 -1.698718 -1.240827 -1.058299
#>
#> [[8]]
#> [1] -0.75437857 -0.58510694 0.55814208 -0.01900658 -0.16935602
#>
#> [[9]]
#> [1] 1.295331 -1.426060 1.754104 -2.115575 1.009702
#>
#> [[10]]
#> [1] 0.2545351 -0.3775541 0.3724693 -1.4626811 0.9037093
#> 10 %>%
rerun(x = rnorm(5), y = rnorm(5)) %>%
map_dbl(~ cor(.x$x, .x$y))#> [1] 0.3884734 -0.2797106 -0.1499915 -0.1213636 0.1875155 -0.1596498
#> [7] 0.1293135 0.5927749 0.7689696 -0.4564721 | 678 | 1,428 | {"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} | 2.5625 | 3 | CC-MAIN-2017-51 | longest | en | 0.450822 |
https://www.physicsforums.com/threads/tachyons-the-double-slit-experiment-and-a-bit-of-fiction.845976/ | 1,519,177,265,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813187.88/warc/CC-MAIN-20180221004620-20180221024620-00473.warc.gz | 925,387,290 | 17,128 | Tachyons, the Double Slit Experiment, and a bit of Fiction
1. Dec 1, 2015
StephenCutter
Hello, All!
I'm writing a bit of fiction, and as a newcomer to the world of Quantum Physics, I thought I'd clarify a few concepts here, to avoid lapsing into technobabble. I'm trying to make something along the lines of the film Primer, which tries to incorporate as much realism as possible within the premise.
For the purpose of this discussion, please understand that I'm not asking if what I'm suggesting is possible in reality. I'm ONLY asking whether, if each of the suppositions I'm making (the existence of tachyons, etc.) WERE true, what I'm suggesting would be plausible.
Let's start at the beginning moving forward-
To my understanding, the double slit experiment works as follows: When an electron or other particle is fired at an intervening object with two slits in it, and impacts against a backstop capable of detecting it, it's impact appears to be determined by probability. That is to say, rather than simply landing in one of 2 areas corresponding to the slots like a solid particle, it instead lands as though it were a wave- in an interference pattern determined by probability.
If the experiment is modified by placing a detector over one or both of the slits, it seems to affect how it behaves. It is as if the act of measuring which slit has altered the probability of where the particle lands, reducing it so that rather than appearing to fall as an interference pattern, it lands more directly on the far side of either slot.
According to theories supporting wave function collapse of superimposed states, this is because the act of observing it defines its path, causing it to fall where the observer would expect. With the variations on detector placement, some even question whether it causes retroactive changes in the behavior of the electron in question, causing it to behave as though it had been following that clearly defined trajectory all along. Regardless, they say that once it collapses, that is reality, and the other states simply cease to exist.
Proponents of 'many worlds' style theories say that the wave function does not collapse to a single reality, and that it simply decoheres, and we're only capable of seeing one possible outcome.
Would it be fair to say that those outcomes are still determined by probability?
And now, for the sci-fi part...
Suppose that tachyonic particles existed, possibly electrons somehow accelerated to superluminal speed. If we could somehow perform the double slit experiment with them, would we see them generated in an interference pattern, then travel back through the slits to the source?
If we had a monitor, as with the variant on the double slit experiment, would they appear to emanate from the two areas directly beyond the slit rather than in an interference pattern?
Thanks for your time! Like I said, this is fiction, so please no responses informing me that tachyons can't exist, etc. Just looking to make sure I've described the theories correctly and that my hypotheticals make some kind of sense!
2. Dec 1, 2015
Hornbein
By my rather primitive knowledge, you seem to have an understanding that is reasonably close to the way many people think.
The part that bothers me is this
If we could somehow perform the double slit experiment with them, would we see them generated in an interference pattern, then travel back through the slits to the source?
We wouldn't be able to see them traveling at all. In the double slit experiment we can't ever see the particles or whatever traveling. If we could detect them while traveling, then no interference pattern could be generated.
3. Dec 1, 2015
.Scott
So far, so good.
Completely covering a lit with a detector will do that, but the "magic" happens when you allow the particle to pass through the slit - but detect which slit it went through. Then you get two blurry blobs, one from each slit, and no interference pattern.
If you determine which path (which slit) the particle passed through, there will be no interference pattern. If you do not determine its path, it will behave as though it went both ways.
There is an experimental setup called "delayed choice erasure" where you are free to interpret the result as an interference pattern and a non-interference pattern being formed before a "decision" is made on whether the which way information will be collected. However, that's only an interpretation someone would make after studying the setup and results closely, recognizing the paradox being demonstrated, and then doing the best they could to describe the result in classical (non-QM) terms.
Ok. But one particle does not an interference pattern make. You collect scores of particles and the accumulated result of these many collapses show the interference pattern.
Yes.
Of course, there is no known road to superluminal speed - but acceleration isn't it. Let's just say you have a detectable tachyon - manufactured by whatever story you prefer.
A detectable tachyon could appear to be travelling backward through time from some reference frames. So that would be a possible viewpoint.
Yes.
4. Dec 1, 2015
StephenCutter
Thanks to those who have replied so far!
To give a clearer picture of what I'm trying to deal with, the story revolves around a group of physicists trapped in a room with a specialized 'camera' and several monitors. The monitors are each displaying an alternate version of reality occurring approximately 10 seconds in the future.
The idea is that someone has designed a way to fire tachyons, then use the 'camera' to read them and decode them as an image.
Since the tachyons have a single source, but have each struck a different 'probability' point in the future, hence the reason why the signal is decoded and split into several images showing different outcomes limited by probability based on what is occurring in the present.
5. Dec 1, 2015
Hornbein
Sure, that's good enough for science fiction. It's cool. | 1,266 | 6,000 | {"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-2018-09 | longest | en | 0.967416 |
https://mathhothouse.me/2020/10/10/limits-and-continuity-part-5-iitjee-math-tutorial-problems-for-practice/ | 1,642,697,750,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302355.97/warc/CC-MAIN-20220120160411-20220120190411-00503.warc.gz | 432,546,982 | 21,870 | ## Limits and Continuity: part 5: IITJEE Math: Tutorial problems for practice
Problem 1:
Find the value of the following limit:
$\lim_{x \rightarrow 0} \frac{\sin{a} - \tan{a}}{\sin^{a}}$
Problem 2:
Find the values of the constant a and b such that the following limit is zero:
$\lim_{x \rightarrow \infty} [\frac{x^{2}+1}{x+1} -ax-b]$
Problem 3:
Find the value of the following limit:
$\lim_{\alpha \rightarrow \beta} \frac{\sin^{2}{\alpha}-\sin^{2}{\beta}}{\alpha^{2}-\beta^{2}}$
Problem 4:
If a, b, c, d are positive, then find the value of the following limit:
$\lim_{x \rightarrow \infty}(1+\frac{1}{a+bx})^{c+dx}$
Problem 5:
Find the value of the following limit:
$\lim_{x \rightarrow 0} \frac{(1-\cos{(2x)})\sin{(5x)}}{x^{2}\sin{(3x)}}$
Problem 6:
Find the value of the following limit:
$\lim_{x \rightarrow \infty} \frac{\sqrt{x^{2}-1}}{2x+1}$
Problem 7:
Find the value of the following limit:
$\frac{\log{(1+x+x^{2})}+\log{(1-x+x^{2})}}{\sec{x}-\cos{x}}$
Problem 8:
Find the value of the following limit:
$\lim_{x \rightarrow \infty} (\frac{2+x}{1+x})^{2x+1}$
Problem 9:
Find the value of f(0) such that the following function is continuous at zero:
$f(x) = (x+1)^{\cot{x}}$
Problem 10:
Let $f^{''}(x)$ be continuous at zero and $f^{''}(0)=4$. Then, find the numerical value of the following limit:
$\lim_{x \rightarrow 0}\frac{2f(x)-3f(2x)+f(4x)}{x^{2}}$
Problem 11:
Find the value of the following limit:
$\lim_{n \rightarrow \infty} (\frac{n^{3}}{3n^{2}-4} - \frac{n^{2}}{3n+2})$
Problem 12:
Find the values of x where the following function is discontinuous:
$f(x) = \frac{\sin{x} \log{(x-2)}}{(x^{2}-4x+3)}$
Problem 13:
The value of p for which the following function may be continuous at zero is what:
$f(x) = \frac{(4x-1)^{3}}{(\sin{\frac{x}{p}})(\log{(1+\frac{x^{2}}{3})})}$, when $x \neq 0$, and
$f(x) = 12(\log{4})^{3}$, when $x = 0$.
Problem 14:
Find the value of the following limit:
$\lim_{x \rightarrow 0} \frac{1-\cos{(mx)}}{1-\cos{(nx)}}$
Problem 15:
If $f(x) = \frac{4-7x}{3x+4}$ and $\lim_{x \rightarrow 2}f(x) = k$, and $\lim_{x \rightarrow 0}f(x) = m$, then the equation whose roots are $\frac{1}{k}, \frac{1}{m}$ is (a) $x^{2}+x=0$ (b) $x^{2}-1=0$ (c) $x^{2}+1=0$ (d) $x^{2}+2x=0$
Problem 16:
Find the value of the following limit:
$\lim_{x \rightarrow 1} \frac{x+x^{2}+x^{3}+\ldots + x^{n}-n}{x-1}$
Problem 17:
Find the value of the following limit:
$\lim_{x \rightarrow 1} \frac{\sqrt[n]{x^{m}}-1}{\sqrt[m]{x^{n}}-1}$
Problem 18:
Find the value of the following limit:
$\lim_{x \rightarrow a} \frac{\tan{x} - \tan{a}}{\sin{a} - \sin{x}}$
Regards,
Nalin Pithwa
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,005 | 2,729 | {"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": 0, "img_math": 30, "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} | 4.5 | 4 | CC-MAIN-2022-05 | latest | en | 0.510747 |
http://betterlesson.com/lesson/resource/3055226/tttt-student-work-samples | 1,488,057,549,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171834.68/warc/CC-MAIN-20170219104611-00269-ip-10-171-10-108.ec2.internal.warc.gz | 26,289,630 | 20,821 | ## TTTT Student Work Samples - Section 3: Homework Review
TTTT Student Work Samples
# Problem Solving with Right Triangles and Trig
Unit 12: Triangle Similarity and Trigonometric Ratios
Lesson 7 of 11
## Big Idea: By playing Trig Tool Tag, students increase participation and confidence in their trigonometry problem solving skills.
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### Jessica Uy
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Environment: Urban | 278 | 1,154 | {"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-2017-09 | latest | en | 0.846634 |
https://www.convert-measurement-units.com/convert+kcal+h+to+erg+per+second.php | 1,660,855,914,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573399.40/warc/CC-MAIN-20220818185216-20220818215216-00409.warc.gz | 609,954,927 | 12,855 | Convert kcal/h to erg/s (kcal/h to erg per second)
## kcal/h into erg per second
numbers in scientific notation
Direct link to this calculator:
https://www.convert-measurement-units.com/convert+kcal+h+to+erg+per+second.php
## How many erg per second make 1 kcal/h?
1 kcal/h = 11 630 000 erg per second [erg/s] - Measurement calculator that can be used to convert kcal/h to erg per second, among others.
# Convert kcal/h to erg per second (kcal/h to erg/s):
1. Choose the right category from the selection list, in this case 'Power'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), 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 'kcal/h'.
4. Finally choose the unit you want the value to be converted to, in this case 'erg per second [erg/s]'.
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, '707 kcal/h'. In so doing, either the full name of the unit or its abbreviation can be used. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Power'. 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: '62 kcal/h to erg/s' or '54 kcal/h into erg/s' or '82 kcal/h -> erg per second' or '99 kcal/h = erg/s' or '13 kcal/h to erg per second' or '82 kcal/h into erg per second'. 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, '(14 * 51) kcal/h'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '707 kcal/h + 2121 erg per second' or '27mm x 62cm x 65dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 4.742 716 006 224 ×1023. For this form of presentation, the number will be segmented into an exponent, here 23, and the actual number, here 4.742 716 006 224. 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 4.742 716 006 224 E+23. 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: 474 271 600 622 400 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 871 | 3,763 | {"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} | 3.53125 | 4 | CC-MAIN-2022-33 | latest | en | 0.841297 |
http://www.mablee.com/caoxsh19851/19881dFHpTY/ | 1,544,838,147,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826686.8/warc/CC-MAIN-20181215014028-20181215040028-00140.warc.gz | 419,926,214 | 12,755 | Learning Songs Kids Dvd Vcd Cover Ben Page Children Free Super Simple Child Study Video Baby Educational Videos Childrens And Famous Websites Easy Preschool Esl Funny Song English
Counting Method / June 6, 2018 / Penelope Chow
: #1985119881
: Image/JPEG
: 1000x1224 px
: June 6, 2018
Unitizing: Our number system groups objects into 10 once 9 is reached. We use a base 10 system whereby a 1 will represent ten, one hundred, one thousand etc. Of the counting principles, this one tends to cause the greatest amount of difficulty for children.
Like the natural numbers, Z is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers, and, importantly, 0, Z (unlike the natural numbers) is also closed under subtraction. The integers form a unital ring which is the most basic one, in the following sense: for any unital ring, there is a unique ring homomorphism from the integers into this ring. This universal property, namely to be an initial object in the category of rings, characterizes the ring Z. | 263 | 1,131 | {"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-2018-51 | latest | en | 0.900238 |
http://www.readbag.com/nps-history-history-online-books-npsg-explosives-chapter8 | 1,596,969,935,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738552.17/warc/CC-MAIN-20200809102845-20200809132845-00024.warc.gz | 172,614,595 | 10,777 | `Chapter 8 BLAST DESIGNHOLE PATTERNS Hole array is the arrangement of blastholes (both in plan and section). The basic blasthole arrays are single-row, square, or rectangular and staggered arrays. Irregular arrays are also used to take in irregular areas at the edge of a regular array. The term SPACING denotes the lateral distance on centers between holes in a row. The BURDEN is the distance from a single row to the face of the excavation, or between rows in the usual case where rows are fired in sequence.(Figure 8-1) Square or rectangular pattern(Figure 8-2) Staggered pattern(Figure 8-3) Sing113 DELAY PATTERNS Delay patterns, and varying the hole array to fit natural excavation topography, allow for more efficient use of the explosive energy in the blast. Benches may be designed and carried forth with more than one face so that simple blasting patterns can be used to remove the rock. In the illustration that follows (Figure 8-4) shows a typical bench cut with two free faces and fired with one delay per row.(Figure 8-4) Typical bench cut with two free faces and fired with one delay per row.Figure 8-5 indicates that the direction of throw of the blasted rock can be controlled by varying the delay pattern. The rock will move forward normally to the rows of holes. If the holes are fired in oblique rows as in Figure 8-5, the rock mass would be thrown to the right during blasting.(Figure 8-5) Direction of throw of blasted rock.114 POWDER FACTOR Calculating Powder Factors The POWDER FACTOR is a relationship between how much rock is broken and how much explosive is used to break it. It can serve a variety of purposes, such as an indicator of how hard the rock is, or the cost of the explosives needed, or even as a guide to planning a shot. Powder factor can be expressed as a quantity of rock broken by a unit weight of explosives. Or, alternatively, it can be the amount of explosives required to break a unit measure of rock. Since rock is usually measured in pounds, there are several possible combinations that can express the powder factor. Powder Factor = Tons of rock (or cubic yards) per pounds of explosive. Normal range = 4 to 7 Shallow holes = 1 to 2 External loads = .3 Tons of Rock = Powder Factor lbs of Explosives The higher the powder factor, the lighter the load. Lower powder factor means more explosives. Example: 1.5 tons = PF of 6 .25lbs BURDEN-SPACING CALCULATION From Powder Factor of 1 lb./c.y. 1. 2. 3. 4. 5. 6. Determine borehole size. Determine stemming: 24 x borehole diameter; Divide by 12 to get the number of feet. Determine subdrilling: 1/3 x stemming. Determine amount of hole to be loaded. Use bench height plus subdrilling minus stemming. From Table 4 of Blaster's Guide, determine pounds/foot of explosive. Determine total load. Multiply amount of hole to be loaded (Step #4) by the pounds per foot of explosives (Step #5). 7. Divide the total load (Step #6) by the bench height. This will equal the number of cubic yards that can be broken at 1 lb/cy. 8. Determine approximate square pattern from Table 1 of Blaster's Guide, or multiply the number ob 9. 10. Adjust to a rectangular pattern of the same total cubic yards. Adjust stemming and subdrilling amounts.NOTE: For powder factor other than 1 lb/cy: divide the resultant number of cubic yards obtained in Step #7 by the powder factor desired. 115 Formula for determination of resultant height of water when using cartridges to "dry up" the borehole. H Dh W Dc H = = = = Resultant height of water in feet Borehole diameter in inches Water in hole in feet Cartridge diameter in inches 2 = Dh x W 2 2 Dh DcBOREHOLE COUPLING(Figure 8-6) Borehole coupling.AIR: THE ENEMY OF AN EXPLOSIVE Borehole coupling is critical to good fragmentation of rock. The borehole should never exceed the diameter of the explosive by more than one-half inch. The air gap around an explosive charge absorbs the shock energy and results in poor fragmentation. The explosive column illustrated in Figure 8-6 on the right will produce the best fragmentation. EXPLOSIVES ECONOMICS The economic analysis of the use of explosives is an important part of blasting operations in mining and construction. Explosives are energy, and the efficient use of this energy is a major factor in keeping rock blasting costs under control. High-energy explosives enhance fragmentation, which ultimately produces a positive effect on production costs. The degree of fragmentation or movement obtained is directly related to the amount of operation. This relationship is illustrated in Figure 8-7. 116 explosive energy applied to the surrounding rock. Analysis of the cost of explosives requires that the effects of explosive energy be placed into proper perspective within the entire drilling, blasting, handling and processing operation. This relationship is illustrated in Figure 8-7.(Figure 8-7) Analysis of efficient blast design..Efficient blast designs combined with the proper choice of explosive can produce better fragmentation with associated lower operating costs compared to blast designs and explosives used under adverse conditions. As a result, the efficient use of explosives, along with the proper borehole diameter selection, are the keys to a successful blasting program. COST OF ENERGY The only way to evaluate accurately the cost of explosives, is to examine the effects of blasting and to determine the optimum degree of fragmentation. In most cases, the productivity rate is influenced by the degree of fragmentation. To obtain well-fragmented rock by blasting, explosive energy must be well distributed throughout the rock. To be effective in rock blasting, this energy must be applied at the proper millisecond delay interval to allow for optimum rock movement. The type and cost of explosives will vary from one operation to another, dependent upon many conditions. The geologic formation, such as hard seams, cap rock, hard bottom, or large toes, dictate the use of high-energy explosives. Water-filled boreholes require the use of water-resistant products at a premium cost. The cost of a product upgrade to cope with wet conditions is an obvious input. Other variables, such as the size of mucking equipment and drilling equipment, fragmentation tolerance, and production demands, will also influence the choice of explosives. Although a significant recurrent expense, the cost of explosives is usually only a small percentage of the total costs encountered in breaking, moving, and processing rock and ore. The small difference in the cost of a higher energy explosive is insignificant compared to a decrease in production caused by insufficient fragmentation. ENERGY FACTORS The energy factor describes the energy distribution within a given unit of rock. Energy distribution within a shot is measured by the energy factor, which compares the explosive energy to a quantity of rock broken. The explosive energy distribution within the entire blast is then evaluated along with its resulting fragmentation and 117 its effect on operating costs. Blasting analysis next becomes a function of the energy factor, explosives cost, fragmentation results, and subsequent production. Proper energy distribution is important in obtaining the desired fragmentation and movement of the bottom or toe portion of the shot. Energy distribution becomes an important factor when wet holes are encountered, as cartridged explosive products must be smaller than the borehole diameter to allow for easier loading. The resulting decrease in the diameter of the explosives column, reduces the amount of explosive energy within the borehole. The blaster must use higher energy explosives to balance the lost energy. Necessary explosive energy adjustments at the borehole can be made to compensate for excessive toe, hard bottom, or cap rock. In addition, higher energy explosives can be substituted for lower energy explosives to increase the energy distribution within the rock, thereby increasing fragmentation. However, if fragmentation was satisfactory before the introduction of additional explosive energy, the improved energy distribution within the shot will allow for an expansion of the drilling pattern, with resultant decrease in overall drilling costs. Improved production rates and consequent cost reduction in digging, hauling, crushing, or moving rock are the major benefits obtained from the efficient application of explosive energy. There are other benefits from better fragmentation, such as reduced secondary blasting, reduced power consumption at the crusher, and less wear and maintenance on equipment with less down time. Explosive efficiency is the ratio of the amount of energy released to the calculated thermochemical energy. Emulsions are highly efficient explosives, due primarily to their microscopic particle size. In contrast, explosives with varying particle size, such as ANFO or water gels, will not have a uniform burning rate, and therefore, will not be as efficient. Studies comparing the calculated thermochemical energy to the measured energy by the underwater bubble energy technique, have shown that the emulsions released 93 percent of the calculated thermochemical energy. Water gels with varying particle sizes achieved only 55 to 70 percent of their calculated thermochemical energy. The explosive efficiencies of ANFO, and particularly of high-density ANFO, range from 50 to 80 percent of their calculated energies. As a result, emulsion explosives are not only thermochemically efficient, but are cost-efficient as well. DRILLING AND LOADING CONSIDERATIONS While the relative rock hardness has an effect both on drilling and explosives performance, environmental factors exert their influence as well. Among the factors to consider in studying drilling costs are: bit costs, labor, fuel consumption, penetration rates, maintenance, machine life, and machine cost. For example, severe water conditions in the borehole will require more expensive explosives resulting in a higher energy cost than would be experienced with a maximum use of ANFO. In semiarid regions of the southwest United States, drilling costs may represent as much as 80 percent of the total drilling and blasting costs, mainly because of utilization of the lower cost explosive. By contrast, in the northeastern United States, hard rock formations exist in a relatively wet environment, where the explosives costs can be as much as 70 percent of the total drilling and blasting expense. Blasting vibrations and air blast concerns may have a direct influence on which blasting program the operator may select. Because of such constraints, the blaster may need to impose limits on quantity of explosives per delay, which relates directly to hole sizes and depths. Borehole diameter and the depth of the hole must always be evaluated in any cost analysis including explosives. Relatively expensive explosives required for efficient performance in a three-inch diameter hole may be unrealistic from a cost standpoint for a six-inch diameter hole. Though drilling cost in a six-inch hole in similar material will be at least twice that for the smaller hole, less expensive explosives usually can be utilized in the larger hole with no sacrifice in explosive performance. When contemplating a change in drill hole size, compare the area of influence associated with the respective 118 holes. Whereas a three-inch diameter hole has an area of 7.07 square inches, the six-inch diameter hole is four times greater, with an area of 28.27 square inches. The same relationship exists with respect to energy of similar explosives. Different blasting conditions may indicate both a change in bench designs as well as explosive selection. An increase in hole size, along with the same explosive, may still fall short of providing the desired results. Such situations may also require the adoption of a higher velocity, more energetic explosive. Where drilling and blasting conditions are both severe, the operator should use premium explosives, at least in the bottom one third of the drill hole. Clearing the toe, and therefore contributing to a smoother floor, is a major asset to improve overall economy because of the favorable impact on excavation and haulage costs. BLAST DESIGN RULES OF THUMB FOR BLAST DESIGN Nearly every occupation or discipline has its own particular "rules of thumb," and blasting is no exception. These are rules, not law. The following is a set of guidelines based on practical experience and technical information. Blasting projects vary so much that there can be no set of rules to cover every possible contingency. Be sure that the advantage gained from breaking the rule is greater than the penalty to be paid for that violation. RULE 1: The detonation velocity of the explosive should match, as closely as possible, the sonic velocity of the rock to be blasted. The rock's sonic velocity (VSO) is a reliable indicator of its structural integrity and resistance to fragmentation. As the detonation velocity of the explosive (VOD) increases toward the rock's VSO, fragment size decreases and uniformity of fragmentation increases. If large fragments are desired from a blast, such as for ballast rock, then the VOD should be considerably lower than the rock VSO. Obviously, if the VOD is too low relative to the rock VSO, the result will be huge irregular blocks or chunks. If detonation proceeds at the same, or close to the same, speed as the transmission of sound waves, the resulting breakage will be optimum, more uniform, and equal along the entire explosive column. Maximum borehole spacings and burdens can then be used. There is no value in using an explosive that has a VOD greatly in excess of the VSO of the rock, since there is little or no improvement in fragmentation above the VSO. When selecting an explosive to match up the VSO of a rock mass, there is no need to match the VOD of the explosive to within a few feet per second of the VSO of the rock. Variance in the velocities below 10 percent either way should be more than sufficient. RULE 2: Generally select the most dense explosive possible, consistent with water, loading conditions, and desired results. The more dense the explosive, the more of it that can be placed in a borehole of a given size. Explosive density is expressed as its weight in g/cc, and since water has a density of 1.0 g/cc, it is obvious that an explosive with a density less than 1.0 will float, whereas, if the density is more than 1.0, it will sink. Less obvious is that the higher the explosive density, the greater the weight of explosive material (potential energy) that can be placed within the borehole. For example: A single borehole 3.5 inches in diameter and 33 feet deep will hold, if stemmed 8 feet with 127.5 pounds of an explosive with a density of 1.25 g/cc. If an explosive with the same VOD, but with a density of 1.4 g/cc, was used in that borehole, the amount of explosive which can be loaded in the hole increases to 142.5 lbs. If a powder factor of one pound per cubic yard is required, the spacing and burden for the explosive with 119 the density of 1.4 will be greater than that for the explosive with a density of 1.25. RULE 3: Select explosives according to the characteristics of the rock formation to be blasted. Although Rule 1 states that explosives should be selected on the basis of matching VOD to VSO, and Rule 2 stresses high density, there are many instances where the structural characteristics of the rock formation allow, or even require, use of lower density, lower velocity explosives (i.e., ANFO). In those instances where the planes of separation in the rock are smaller than the degree of fragmentation required, the rock can often be blasted by merely "bumping" the rock with explosives. A close study of the breakage planes in the rock mass will indicate whether or not a lower velocity and density explosive should be used. This subjective determination hinges upon knowledge of the rock and experience with varying rock formations. RULE 4 - When using slurry or water gel explosives, always determine the critical temperature below which the explosive will fail to reliably detonate. Almost all slurry explosives have a critical temperature below which they may not detonate, or may not sustain detonation in elongated columns. This critical temperature is usually noted in the technical data sheets supplied by the manufacturer or distributor. Under no circumstances should the explosive be used when the temperature of the explosive at time of loading is below that critical temperature. In addition to the temperature of the explosive itself, consider the temperature of water that may be in the borehole, since slurry products are often used in wet boreholes. In many parts of the United States, particularly in the northern states during winter months, it is not unusual for water temperature in the boreholes to reach below the critical temperature of the explosive. Guidelines FOR BLASTING GEOMETRY RULE 1: The distance between holes (spacing) should not be greater than one-half the depth of the borehole. When the effect of a blast is simulated on graph paper using an assumed or idealized angle of breakage of 90 degrees, the diagrams indicate that in each instance where the distance between holes in a row is greater than one-half the depth of the hole, the angles of breakage intersect so far above the bottom of the holes that the primary relief for each hole is to the surface. This causes both a great deal of vertical throw and a very uneven bottom. The greater the disparity between depth and spacing, the more pronounced the effect will be, to the point where the angles of breakage intersect above the surface of the shot. RULE 2: In any blast where there is hole-for-hole delay, the spacing to burden relationship should be seven to five. Several investigations about the science of rock mechanics have suggested that the optimum spacing to burden relationship should be 2:1, with the burden equal to one-half the spacing. Field experience shows that this relationship has two drawbacks. First, the blaster may assume that this relationship will apply when there is no delay system at work, when in fact, the optimum spacing to burden relationship in all instantaneous blasts should be 1:1 to ensure equal distribution of explosives in the blast. Second, if an instantaneous blast is fired with a spacing to burden relationship of 2:1, the back wall of the blasted area will, in most cases, be "sawtoothed." When delays are used, particularly when there is hole-for-hole and row-for-row delay, with no two holes firing on the same period, the angle of breakage approaches the idealized ratio of 2:1. The slight addition of burden avoids the possibility of "blowout," or violent throw from relieved burdens during the shifting of burden from one hole to another. 120 RULE 3: Stemming should be equal to the burden. The purpose of stemming, it has long been assumed, is to return the borehole to its original condition as much as possible in order to reduce noise, and possibly rifling at the top portion of the hole. Stemming also serves to confine and maximize efficient use of the explosive's energy. If the explosive detonation process takes place up the borehole, the surface of the rock above the stemming is as much a free face (assuming there is a free face) as the free face that is parallel to the boreholes. If the stemming is greater than the burden, the rock at the top of the borehole will have less cracking from reflection and refraction of compressive and tensile waves. Then stemming should equal burden, and be of such material as to return the rock close to its original condition. Drill fines, tamped into the hole are ideal. RULE 4: Subdrill (if necessary) should be between .3 and .5 of spacing. Some investigators state that subdrill should be equal to .3 of burden. This is true in instances where spacing and burden are equal, such as with instantaneous blasts. It will also work when there is row-for-row delay. In blasts where the delay system is both row-for-row and hole-for-hole, however, the subdrill should be determined by the largest dimension, which is the spacing. An average subdrill of .4 of spacing is best to use for planning purposes.121 122 `
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TECHNICAL LEAFLET EXPLOGELA4 | 4,268 | 20,725 | {"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} | 3.171875 | 3 | CC-MAIN-2020-34 | latest | en | 0.929151 |
https://www.aqua-calc.com/one-to-one/density/gram-per-us-teaspoon/grain-per-cubic-foot/1 | 1,571,706,329,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987795403.76/warc/CC-MAIN-20191022004128-20191022031628-00124.warc.gz | 799,679,285 | 9,070 | # 1 gram per US teaspoon [g/tsp] in grains per cubic foot
## g/tsp to gr/ft³ unit converter of density
1 gram per US teaspoon [g/tsp] = 88 659.5 grains per cubic foot [gr/ft³]
### grams per US teaspoon to grains per cubic foot density conversion cards
• 1
through
25
grams per US teaspoon
• 1 g/tsp to gr/ft³ = 88 659.5 gr/ft³
• 2 g/tsp to gr/ft³ = 177 319 gr/ft³
• 3 g/tsp to gr/ft³ = 265 978.5 gr/ft³
• 4 g/tsp to gr/ft³ = 354 638 gr/ft³
• 5 g/tsp to gr/ft³ = 443 297.5 gr/ft³
• 6 g/tsp to gr/ft³ = 531 957 gr/ft³
• 7 g/tsp to gr/ft³ = 620 616.5 gr/ft³
• 8 g/tsp to gr/ft³ = 709 276 gr/ft³
• 9 g/tsp to gr/ft³ = 797 935.5 gr/ft³
• 10 g/tsp to gr/ft³ = 886 595 gr/ft³
• 11 g/tsp to gr/ft³ = 975 254.5 gr/ft³
• 12 g/tsp to gr/ft³ = 1 063 914 gr/ft³
• 13 g/tsp to gr/ft³ = 1 152 573.5 gr/ft³
• 14 g/tsp to gr/ft³ = 1 241 233 gr/ft³
• 15 g/tsp to gr/ft³ = 1 329 892.5 gr/ft³
• 16 g/tsp to gr/ft³ = 1 418 552 gr/ft³
• 17 g/tsp to gr/ft³ = 1 507 211.5 gr/ft³
• 18 g/tsp to gr/ft³ = 1 595 871 gr/ft³
• 19 g/tsp to gr/ft³ = 1 684 530.5 gr/ft³
• 20 g/tsp to gr/ft³ = 1 773 190 gr/ft³
• 21 g/tsp to gr/ft³ = 1 861 849.5 gr/ft³
• 22 g/tsp to gr/ft³ = 1 950 509 gr/ft³
• 23 g/tsp to gr/ft³ = 2 039 168.5 gr/ft³
• 24 g/tsp to gr/ft³ = 2 127 828 gr/ft³
• 25 g/tsp to gr/ft³ = 2 216 487.5 gr/ft³
• 26
through
50
grams per US teaspoon
• 26 g/tsp to gr/ft³ = 2 305 147 gr/ft³
• 27 g/tsp to gr/ft³ = 2 393 806.5 gr/ft³
• 28 g/tsp to gr/ft³ = 2 482 466 gr/ft³
• 29 g/tsp to gr/ft³ = 2 571 125.5 gr/ft³
• 30 g/tsp to gr/ft³ = 2 659 785 gr/ft³
• 31 g/tsp to gr/ft³ = 2 748 444.5 gr/ft³
• 32 g/tsp to gr/ft³ = 2 837 104 gr/ft³
• 33 g/tsp to gr/ft³ = 2 925 763.5 gr/ft³
• 34 g/tsp to gr/ft³ = 3 014 423 gr/ft³
• 35 g/tsp to gr/ft³ = 3 103 082.5 gr/ft³
• 36 g/tsp to gr/ft³ = 3 191 742 gr/ft³
• 37 g/tsp to gr/ft³ = 3 280 401.5 gr/ft³
• 38 g/tsp to gr/ft³ = 3 369 061 gr/ft³
• 39 g/tsp to gr/ft³ = 3 457 720.5 gr/ft³
• 40 g/tsp to gr/ft³ = 3 546 380 gr/ft³
• 41 g/tsp to gr/ft³ = 3 635 039.5 gr/ft³
• 42 g/tsp to gr/ft³ = 3 723 699 gr/ft³
• 43 g/tsp to gr/ft³ = 3 812 358.5 gr/ft³
• 44 g/tsp to gr/ft³ = 3 901 018 gr/ft³
• 45 g/tsp to gr/ft³ = 3 989 677.5 gr/ft³
• 46 g/tsp to gr/ft³ = 4 078 337 gr/ft³
• 47 g/tsp to gr/ft³ = 4 166 996.5 gr/ft³
• 48 g/tsp to gr/ft³ = 4 255 656 gr/ft³
• 49 g/tsp to gr/ft³ = 4 344 315.5 gr/ft³
• 50 g/tsp to gr/ft³ = 4 432 975 gr/ft³
• 51
through
75
grams per US teaspoon
• 51 g/tsp to gr/ft³ = 4 521 634.5 gr/ft³
• 52 g/tsp to gr/ft³ = 4 610 294 gr/ft³
• 53 g/tsp to gr/ft³ = 4 698 953.5 gr/ft³
• 54 g/tsp to gr/ft³ = 4 787 613 gr/ft³
• 55 g/tsp to gr/ft³ = 4 876 272.5 gr/ft³
• 56 g/tsp to gr/ft³ = 4 964 932 gr/ft³
• 57 g/tsp to gr/ft³ = 5 053 591.5 gr/ft³
• 58 g/tsp to gr/ft³ = 5 142 251 gr/ft³
• 59 g/tsp to gr/ft³ = 5 230 910.5 gr/ft³
• 60 g/tsp to gr/ft³ = 5 319 570 gr/ft³
• 61 g/tsp to gr/ft³ = 5 408 229.5 gr/ft³
• 62 g/tsp to gr/ft³ = 5 496 889 gr/ft³
• 63 g/tsp to gr/ft³ = 5 585 548.5 gr/ft³
• 64 g/tsp to gr/ft³ = 5 674 208 gr/ft³
• 65 g/tsp to gr/ft³ = 5 762 867.5 gr/ft³
• 66 g/tsp to gr/ft³ = 5 851 527 gr/ft³
• 67 g/tsp to gr/ft³ = 5 940 186.5 gr/ft³
• 68 g/tsp to gr/ft³ = 6 028 846 gr/ft³
• 69 g/tsp to gr/ft³ = 6 117 505.5 gr/ft³
• 70 g/tsp to gr/ft³ = 6 206 165 gr/ft³
• 71 g/tsp to gr/ft³ = 6 294 824.5 gr/ft³
• 72 g/tsp to gr/ft³ = 6 383 484 gr/ft³
• 73 g/tsp to gr/ft³ = 6 472 143.5 gr/ft³
• 74 g/tsp to gr/ft³ = 6 560 803 gr/ft³
• 75 g/tsp to gr/ft³ = 6 649 462.5 gr/ft³
• 76
through
100
grams per US teaspoon
• 76 g/tsp to gr/ft³ = 6 738 122 gr/ft³
• 77 g/tsp to gr/ft³ = 6 826 781.5 gr/ft³
• 78 g/tsp to gr/ft³ = 6 915 441 gr/ft³
• 79 g/tsp to gr/ft³ = 7 004 100.5 gr/ft³
• 80 g/tsp to gr/ft³ = 7 092 760 gr/ft³
• 81 g/tsp to gr/ft³ = 7 181 419.5 gr/ft³
• 82 g/tsp to gr/ft³ = 7 270 079 gr/ft³
• 83 g/tsp to gr/ft³ = 7 358 738.5 gr/ft³
• 84 g/tsp to gr/ft³ = 7 447 398 gr/ft³
• 85 g/tsp to gr/ft³ = 7 536 057.5 gr/ft³
• 86 g/tsp to gr/ft³ = 7 624 717 gr/ft³
• 87 g/tsp to gr/ft³ = 7 713 376.5 gr/ft³
• 88 g/tsp to gr/ft³ = 7 802 036 gr/ft³
• 89 g/tsp to gr/ft³ = 7 890 695.5 gr/ft³
• 90 g/tsp to gr/ft³ = 7 979 355 gr/ft³
• 91 g/tsp to gr/ft³ = 8 068 014.5 gr/ft³
• 92 g/tsp to gr/ft³ = 8 156 674 gr/ft³
• 93 g/tsp to gr/ft³ = 8 245 333.5 gr/ft³
• 94 g/tsp to gr/ft³ = 8 333 993 gr/ft³
• 95 g/tsp to gr/ft³ = 8 422 652.5 gr/ft³
• 96 g/tsp to gr/ft³ = 8 511 312 gr/ft³
• 97 g/tsp to gr/ft³ = 8 599 971.5 gr/ft³
• 98 g/tsp to gr/ft³ = 8 688 631 gr/ft³
• 99 g/tsp to gr/ft³ = 8 777 290.5 gr/ft³
• 100 g/tsp to gr/ft³ = 8 865 950 gr/ft³
#### Foods, Nutrients and Calories
TABATCHNICK, ROASTED RED PEPPER AND TOMATO SOUP, UPC: 071262041841 contain(s) 44 calories per 100 grams or ≈3.527 ounces [ price ]
THIN SLICED LEAN CHICKEN, UPC: 075450086676 contain(s) 161 calories per 100 grams or ≈3.527 ounces [ price ]
Foods high in Vitamin D2 (ergocalciferol), foods low in Vitamin D2 (ergocalciferol), and Recommended Dietary Allowances (RDAs) for Vitamin D
#### Gravels, Substances and Oils
CaribSea, Freshwater, Eco-Complete Planted, Red weighs 865 kg/m³ (54.00019 lb/ft³) with specific gravity of 0.865 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Xylene [C24H30] weighs 864 kg/m³ (53.93776 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-422A, liquid (R422A) with temperature in the range of -40°C (-40°F) to 60°C (140°F)
#### Weights and Measurements
tonne per square micrometer (t/μm²) is a metric measurement unit of surface or areal density
A force that acts upon an object can cause the acceleration of the object.
g/US gal to g/metric c conversion table, g/US gal to g/metric c unit converter or convert between all units of density measurement.
#### Calculators
Calculate volume of a hollow cylinder and its surface area | 2,908 | 5,984 | {"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-2019-43 | latest | en | 0.171797 |
https://www.photo.net/discuss/threads/dof-and-circle-of-confusion.31713/ | 1,653,533,003,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662595559.80/warc/CC-MAIN-20220526004200-20220526034200-00135.warc.gz | 1,075,091,278 | 16,528 | DOF and circle of confusion
Discussion in 'Large Format' started by roy_feldman, Aug 28, 1998.
1. roy_feldman
A while back I read a chart wich would give the correct F- stop setting based on near to far standards movement. In other words focusing on the nearest point then measuring the distance on the monorail to the farthest point entering that distance to a scale which would gine me the proper F stop. Since purchasing the Rodenstock Studio calculator I have learned that monorail angle and repro ratio are also factors. My problem is that the Rodenstock calculator is"based on a Circle of C.. of .03mm for 35mm.IYO is this sufficent for large enlargement? Are there other scales out there? I use 4X5
The circle of confusion for 4x5 is usually set at .15mm or 0.09 mm.
You will find some online calculators here:
http://home.sol.no/~gjon/depth.htm
3. roy_feldman
Thanks for the response, I had no idea there were so many DOF scales out there but none really answer my question. Based on measurment of movement of lens standards ONLY is there a DOF scale out there? Mr. Grepstad's answer was overwhelming, could you narrow it down?
4. ron_shaw
An article in Photo Techniques (Mar/Apr 96) on view camera focus used this method. Basically, you tape or glue a mm scale on your bed, and then focus on the nearest object of interest, read the position on the mm scale, then focus on the farthest object of interest, and note its position on the mm scale, and then, using this focus spread (in mm), refer to a chart for optimum f stop to use, based on line pairs per mm resolution. I photocopied the chart and taped it to the back of my camera for reference. I hope this helps.
5. alan_gibson|6
The scale mentioned by Ron sounds as if it is based on the formula t=2.C.N', where t=depth of focus (not field), C is diameter of Circle of confusion, and N' is effective aperture, probably assumed here to be equal to the marked aperture. Thus the scale is independant of focal length (neat, huh?). The repro factor is relevant for closeups.
<p>
Of course, one advantage of LF is movements, especially puting the lens at an angle to the film. See http://fox.nstn.ca/~hmmerk/ for the gory details.
6. roy_feldman
Mr.Shaw has gone "above and beyond" by offering to snail mail the article mentioned in his response. Mr. Gibson your formula sounds great but having attended art school I'm afraid I don't have the where with all to figure it out.I know that in 35mm a COF of .03 is low end and .01 equals what film can be resolved at,I have no idea how or why this translates to 4x5.Is it possible,Mr. G. to give a example a art school grad can follow? ( I know I should have listened to Dad and taken some "real" classes, but that was 20 years ago).
7. alan_gibson|6
If you look at the DoF marking on a 35mm lens, it will look something like this:
+------+------+------+------+------+------+
| | | | | | |
22 16 8 V 8 16 22
I've only shown marks for f/8 and f/16.
On a 35mm lens, the scale is on a fixed part of the lens, and distance marks on the focusing ring can be compared to the f/numbers to see the DoF.
On a 5x4 camera, you may not have a distance scale. Suppose you focus by racking the front back and forwards. With a ruler and pen, you can create a paper scale that looks much like the one above, and hold it against the camera as you rack back and forth. If the total racking distance is the same as the distance between the two f/8 marks, then you have to stop down to f/8 or smaller.
Now for a bit of theory. Ignore this paragraph if you like. Depth of focus refers to the total distance we could move the film backwards and forwards, while the image still looks sharp. Call this distance "t". For distant objects, the value of t is approximately twice the diameter of the circle of confusion, multiplied by the aperture number. "t = 2.C.N". The value of C is often taken to be 0.03mm for 35mm cameras, or 0.1mm for 5x4. These numbers are not set in stone: Nikon used different values for their cheap (consumer) "E" series lenses than their more expensive regular lenses.
Using C = 0.1mm, here are the value of t for various values of N.
N t (mm)
=== ======
5.6 1.1
8 1.6
11 2.2
16 3.2
22 4.4
t is the total distance, so the two f/8 marks should be 1.6mm apart, so each of them is 0.8mm from the central mark. So your scale will look like this:
+------+------+------+------+------+------+
| | | | | | |
22 16 8 V 8 16 22
Note that these distances are small; you will need a small pen.
Of course, if you are photographing a landscape, or other plane object, you would probably tilt the lens or film, rather than relying on depth-of-field to keep things in focus. | 1,214 | 4,680 | {"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-2022-21 | latest | en | 0.936551 |
http://dspace.luguniv.edu.ua/xmlui/handle/123456789/4380/browse?rpp=20&order=ASC&sort_by=1&etal=-1&type=title&starts_with=K | 1,618,507,354,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038087714.38/warc/CC-MAIN-20210415160727-20210415190727-00343.warc.gz | 30,482,171 | 6,667 | Digital Repository of Luhansk Taras Shevchenko National University
# Browsing Algebra and Discrete Mathematics. - № 1 (27). - 2019 by Title
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This paper is about sparse numerical semigroups and applications in the Weierstrass semigroups theory. We describe and find the genus of certain families of sparse numerical semigroups with Frobenius number even and we ...
• (ДЗ "ЛНУ імені Тараса Шевченка", 2019)
A vector balleans is a vector space over R en- dowed with a coarse structure in such a way that the vector opera- tions are coarse mappings. We prove that, for every ballean (X, E), there exists the unique free vector ...
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A 2-monomial matrix over a commutative ring R is by definition any matrix of the form M(t, k, n) = Φ Ik 0 0 tIn−k , 0 < k < n, where t is a non-invertible element of R, Φ the companion matrix to λ n − 1 and Ik ...
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We study formulas for eigenvectors of strongly connected simply laced quivers in terms of eigenvalues. The relation of these formulas to the isomorphism of quivers is investigated.
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In this paper, we characterize all finitely gene- rated multiplication R-modules whose the first nonzero Fitting ideal of them is contained in only finitely many maximal ideals. Also, we prove that a finitely generated ...
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We denote the number of distinct topologies which can be defined on a set X with n elements by T(n). Similarly, T0(n) denotes the number of distinct T0 topologies on the set X. In the present paper, we prove that for ...
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The rings we consider in this article are com- mutative with identity 1 6= 0 and are not fields. Let R be a ring. We denote the collection of all proper ideals of R by I(R) and the collection I(R) \ {(0)} by I(R) ∗ . ...
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In 1995 D. V. Belkin described the lattice of quasivarieties of modules over principal ideal domains [1]. The following paper provides a description of the lattice of subquasivarieties of the variety of modules over a given ... | 633 | 2,335 | {"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-2021-17 | latest | en | 0.803505 |
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# Jane Ullahjanoulle
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Reversing a String
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/** * Author: Jane Ullah * Purpose: Reversing a string provided by the user. * Date: 12/18/2011 */ import java.util.Scanner; public class ReversingString { public static void main(String[] args) {
Created Dec 19, 2011
Problem 2 - Project Euler in Java by Jane Ullah
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//Author: Jane Ullah //Date: 12/19/2011 //Site: janetalkscode.com long a1 = 0L, a2 = 1L; long nextTerm = 0L; long sumEvenTerms = 0L; boolean fourMillionthReached = false; while ( !fourMillionthReached ) { if (nextTerm >= 4000000) {
Created Dec 19, 2011
Problem 1 - Project Euler in Python by Jane Ullah
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x = 0 for i in range(0,1000): if (i%3 == 0) or (i%5 == 0): x += i; print('total sum is: ' + str(x));
Created Dec 20, 2011
Problem 3 - Project Euler in Java by Jane Ullah
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Created Dec 20, 2011
Problem 4 - Project Euler in Java by Jane Ullah
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/** * * @author Jane Ullah * @date 12/19/2011 * @purpose Solving Problem 4 from ProjectEuler.net * */ public static void main(String[] args) { /*if (isPalindrome(1222321)) { System.out.println("Yes.");
Created Dec 27, 2011
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/** * @author Jane Ullah * @purpose Problem 5 - Project Euler * @date 12/26/2011 * @site http://janetalkscode.com */ public class Problem5 { public static void main(String[] args) { int sum = 1, primeNum = 0;
Created Dec 27, 2011
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Created Dec 27, 2011
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/** * @author Jane Ullah * @purpose Problem 7 - Project Euler * @date 12/27/2011 * @site http://janetalkscode.com */ public class Problem7 { public static void main(String[] args) { int primeNum = 0, index = 0;
Created Dec 28, 2011
Problem 8 - Project Euler in Java by Jane Ullah
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/** * @author Jane Ullah * @purpose Problem 8 - Project Euler * @date 12/28/2011 * @site http://janetalkscode.com */ public class Problem8 { public static void main(String[] args) { String digits = "73167176531330624919225119674426574742355349194934"
Created Dec 29, 2011
Multiplying Digits in a String
View parseString.java
public static int parseString(String digits) { int product = 1; //Strips out non-word characters digits = digits.replaceAll("\\W",""); for (int i = 0; i < digits.length(); i++){ product *= Integer.parseInt(digits.substring(i,(i+1))); } return (digits.length() > 0) ? product:0; /*
You can’t perform that action at this time. | 919 | 2,973 | {"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-2021-04 | latest | en | 0.572707 |
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Use this task to see if your students can work with ratios and proportions in a real world application. It works great as partner work, a quiz, homework, or practice. Answer key is included.
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CCSS.Math.Content.6.RP.A.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities.
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A112180 McKay-Thompson series of class 40a for the Monster group. 1
%I
%S 1,0,3,4,4,4,7,12,13,16,22,28,38,44,55,72,83,104,129,156,187,220,273,
%T 328,384,452,539,652,757,880,1041,1220,1428,1652,1924,2244,2585,2992,
%U 3458,3992,4581,5244,6053,6936,7910,9024,10303,11784,13380,15176
%N McKay-Thompson series of class 40a for the Monster group.
%H G. C. Greubel, <a href="/A112180/b112180.txt">Table of n, a(n) for n = 0..1000</a>
%H D. Ford, J. McKay and S. P. Norton, <a href="http://dx.doi.org/10.1080/00927879408825127">More on replicable functions</a>, Comm. Algebra 22, No. 13, 5175-5193 (1994).
%H <a href="/index/Mat#McKay_Thompson">Index entries for McKay-Thompson series for Monster simple group</a>
%F Expansion of A - q/A, where A = q^(1/2)*(eta(q^4)*eta(q^5)/( eta(q)* eta(q^20))), in powers of q. - _G. C. Greubel_, Jun 26 2018
%F a(n) ~ exp(Pi*sqrt(2*n/5)) / (2^(5/4) * 5^(1/4) * n^(3/4)). - _Vaclav Kotesovec_, Jun 27 2018
%e T40a = 1/q +3*q^3 +4*q^5 +4*q^7 +4*q^9 +7*q^11 +12*q^13 +...
%t eta[q_] := q^(1/24)*QPochhammer[q]; A:= q^(1/2)*(eta[q^4]*eta[q^5]/( eta[q]*eta[q^20])); a := CoefficientList[Series[A - q/A, {q, 0, 60}], q]; Table[a[[n]], {n, 1, 50}] (* _G. C. Greubel_, Jun 26 2018 *)
%o (PARI) q='q+O('q^50); A = eta(q^4)*eta(q^5)/(eta(q)*eta(q^20)); Vec(A - q/A) \\ _G. C. Greubel_, Jun 26 2018
%K nonn
%O 0,3
%A _Michael Somos_, Aug 28 2005
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Last modified March 18 19:58 EDT 2019. Contains 321293 sequences. (Running on oeis4.) | 756 | 1,846 | {"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 | 3 | CC-MAIN-2019-13 | latest | en | 0.524157 |
http://www.britgo.org/bgj/06625.html | 1,469,771,863,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829972.19/warc/CC-MAIN-20160723071029-00087-ip-10-185-27-174.ec2.internal.warc.gz | 342,428,397 | 5,999 | ## Spot The Move - Answers
British Go Journal No. 66. November 1985. Page 25.
Richard Granville
The problems for these solutions are on page 5 of BGJ 65.
This month's problems are on page 5.
The answer to problem 1 may be found on page 24.
The original article used coordinates (such as K10) for most of this article. It has been altered to use marked up diagrams for the EBGJ.
The answers in BGJ 66 are numbered differently to the problems in BGJ 65. This second answer here relates to question 3!
You might wish to open a second window beside the first one to view Dia X4 whilst reading the text in the first window.
### Problem 3 - Marks
Marks for various black moves.
### Problem 3 - Commentary
Despite the five different moves preferred by the panel, there is general agreement about what to do.
#### Panelists replies:
• H - Jeff Ansell 2k
• G - Brian Chandler 2d
• H - Matthew Macfadyen 6d
• L - Toby Manning 2d
• B - John Rickard 3d
• C - Francis Roads 3d
• L - Piers Shepperson 3d
Shepperson: "Although there are several big points on the board, there seems to me to be only one urgent area. If White gets the chance to play M (threatening Q), Black will be almost forced to make the N-J exchange, giving White a very favourable result. Hence Black must prevent this; L is the obvious way to do this (considering , H would be misplaced)."
Manning agrees, and so do I, but other panelists preferred H:
Ansell: "H prevents white M, which would be overwhelming, and puts pressure on the running white group, a possible continuation would be white E, black F, white D, black A."
Macfadyen: "H is the vital point of white's shape - if black plays M, white can force him down immediately with H, P and J, building eyeshape in the process, and then attack at A."
The other possibility is to play in the centre immediately:
Rickard: "If White were to play around C, black would be forced to move into the uninteresting area near white's thickness. I am not good at judging shape in such situations, but for some reason B feels better than C."
Roads prefers C, and Chandler considers D in the centre, but ends up preferring G.
In the game, my opponent played H, and I immediately extended to K - Black seems to have no devastating attack in the centre.
The answer to problem 2 can be found on page 26. | 577 | 2,322 | {"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} | 3.5625 | 4 | CC-MAIN-2016-30 | longest | en | 0.926767 |
https://www.techiedelight.com/find-correct-order-alphabets-dictionary-ancient-origin/ | 1,548,114,493,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583822341.72/warc/CC-MAIN-20190121233709-20190122015709-00608.warc.gz | 958,184,937 | 26,447 | Find the correct order of alphabets in a given dictionary of ancient origin
Given a dictionary of ancient origin where the words are arranged alphabetically, find the correct order of alphabets in the ancient language.
For example,
Input: Ancient dictionary { ¥€±, €±€, €±‰ð, ðß, ±±ð, ±ßß }
Output: The correct order of alphabets in the ancient language are: {¥ € ‰ ð ± ß}
Since the input is small, more than one ordering is possible. Another such ordering is {¥ € ð ± ß ‰}
Input: Ancient dictionary { ÿ€±š, €€€ß, €€‰ð, ðß, ±ß¥š }
Output: The correct order of alphabets in the ancient language are: {ÿ € ‰ ð ±}
The alphabets {š, ß, ¥} are not included in the order as they are not properly defined.
If we carefully analyze the problem, we can see that it is variation of topological sorting of a DAG. A Topological Sorting of a directed acyclic graph is a linear ordering of its vertices such that for every directed edge (u -> v) from vertex u to vertex v, u comes before v in the ordering.
For example, consider the dictionary { ¥€±, €±€, €±‰ð, ðß, ±±ð, ±ßß } of ancient words. For each of the edge (x -> y) shown below, x should appear before y in the final ordering.
¥ --> €
€ --> ð, ‰
± --> ß
ð --> ±
If we go a topological sorting on above graph, we get the correct order of alphabets in the ancient language: {¥ € ‰ ð ± ß} or {¥ € ð ± ß ‰}.
The idea is to iterate through the complete dictionary and compare adjacent words for a character mismatch. If a mismatch between adjacent words is seen, we insert such pair into a graph. The resultant graph is a DAG since the all words the dictionary are arranged alphabetically. Now since the graph has no directed cycles, we can perform topological sorting on it which gives correct order of alphabets in the ancient language.
The algorithm can be implemented as follows in C++:
Output:
The correct order of alphabets in the ancient language are: ¥ € ‰ ð ± ß
The time complexity of above solution is O(N*M) where N is the dictionary size and M is the maximum length of a word in the dictionary.
(1 votes, average: 5.00 out of 5) | 572 | 2,095 | {"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-2019-04 | longest | en | 0.88623 |
https://kr.mathworks.com/matlabcentral/answers/708593-time-series-analysis-problems-with-date | 1,723,746,889,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641311225.98/warc/CC-MAIN-20240815173031-20240815203031-00108.warc.gz | 261,335,764 | 29,367 | # Time Series analysis problems with date
조회 수: 2 (최근 30일)
mike 2021년 1월 4일
답변: Biral Pradhan 2022년 1월 25일
Hi all-and a happy new year (lets hope its better than the last one......)
My query concerns time series.
I have a dataset which consists of 7 seperate excel files, first column is date and second column is a scalar number.
The date is a period of 8 months, jan-aug for 10 years (repeated in each excel file).
I have plotted the first file and the first 8 months plot fine for the ten year period, however, the sep-dec period is also plotted which shows no data, therefore there is a straight line until the next year starts on the graph.
I have used the code
= datevec(cam{:,1}); and also tried
= datenum (cam{:,1});
both do not work!!
So my question is how do I plot the data for just the dates shown which Jan 1st-Aug 31st 2010-2020?
Im fairly new to all this, I have tried using the mathworks help section and tried several other pieces of code, but I get the dreaded red writing, I apologise if this is a billy basic question!
Mike
##### 댓글 수: 4이전 댓글 2개 표시이전 댓글 2개 숨기기
Is the goal to remove the straight line from the data while maintaining the sept-dec dates along the x axis? Or is the goal to remove the gap completly and jump from Aug to Jan along the x axis?
mike 2021년 1월 17일
Ideally to remove the lines and the gaps from the x axis so that I can them smooth and detrend.
Cheers
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### 답변 (1개)
I understand, you want to remove the straight line as well as x-axis labels corresponding to those 4 months (Sep-Dec) for each year. I assume, you plan on interpolating the data for these 4 months later. While plotting monthly data using datetime values from different tables (as in your case), the axis spacing is determined uniformly, so you get labels separated by 1 calendar month.
I do not think it is possible to abruptly change the spacing from 1 calender month to 4 abruptly. A workaround would be to use a timetable and resample your data.
댓글을 달려면 로그인하십시오.
### 카테고리
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Translated by | 583 | 2,199 | {"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} | 2.8125 | 3 | CC-MAIN-2024-33 | latest | en | 0.882314 |
https://kristenskindergarten.com/2011/10/ | 1,500,799,205,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424296.90/warc/CC-MAIN-20170723082652-20170723102652-00702.warc.gz | 683,665,762 | 59,516 | # Monthly Archives: October 2011
## Math work stations
I am in the process of introducing some new activities into our math work station tubs. The children were becoming bored with three or four of the activities, so it was time to change some of them out.
This is an activity from a site called K-5 math resources. I printed out the dinosaur pictures and then cut them apart. The children need to put the numbers back in order to complete the puzzles correctly.
This game is another from the same site. It’s called race to trace. The children each have one of these sheets and they also have a die and a dry erase marker. The first child rolls the die and traces one of the numbers that they roll. They keep taking turns until one of them has traced all their numbers.
This last activity is one I created. It’s called roll a pumpkin. Again the children share one large die but they each have their own pumpkin mat (these can be found on my blog under pattern blocks). They take turns rolling the die and consulting the answer sheet. The answer sheet tells them what block each number corresponds with. They then put that block on their pumpkin sheet and then the other player takes a turn.
Kristen 🙂
## Centers/Discovery time
Here is a look at the some of the centers we have going at the end of our day that the children can visit.
## Meaningful morning messages
I’ve decided that this is the year I want to work on making my morning messages more meaningful each day. I’ve been working different skills into my morning messages so that it helps me and the children focus in on what our target skill will be for that day/week. This week, we have been focusing on rhyming words.
The children love coming up with words that rhyme with our sight words and try to guess what the word will be that I pick to rhyme with on that day.
Kristen 🙂
## Anchor charts
One thing I have promised myself that I would do more of this year is to create anchor charts. I’ve been creating them for our reading, writing, and literacy work station times/areas. Here are some examples
I’ve found that not only do these charts help the students remember what to do, they cut down on lots of behavior problems because the children know exactly what they are supposed to be doing because they suggested the ideas on the chart.
We create these charts together after lots of modeling from the children and myself. Writers workshop has been great so far because the children can tell you exactly what they should be doing and how they should be doing it.
How do you store your anchor charts?
Kristen 🙂
## Another idea from Debbie Diller
Another idea that Debbie Diller shared was this one about punctuation and how to help children understand what the various marks mean. I really liked this stop light version because my children can visualize the punctuation mark and then remember the color associated with it.
I tried this idea out today so that I could link the idea of finding commas in our big books with using commas in writing. I also like that it is a ready made anchor chart that the children can refer back to many times in the year. Now I just need more wall space to hang them all up! 🙂
Kristen 🙂
## So much to share!
I had the pleasure of seeing Debbie Diller again when she came to talk in my district. She had some great ideas (as always) that I was able to go try in my classroom right away.
This chart is a schema chart. We listed all the things we knew about apples and such before we went to the apple orchard. I told the children that their job at the orchard was to listen for some new information they could add to their schema. Boy, did they do their jobs!
Today we took the copies of the pictures I took on our trip and glued them around the edges to make a border. Thanks to Debbie for that awesome suggestion!!!
I can see using this for so many different topics! The children were/are so proud of their work, both all the things they already knew and all of the things they learned! | 864 | 4,023 | {"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-2017-30 | longest | en | 0.968491 |
http://simonstechblog.blogspot.com/2012/10/ | 1,540,252,594,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515555.58/warc/CC-MAIN-20181022222133-20181023003633-00416.warc.gz | 335,288,970 | 14,881 | ### Angle based SSAO
Introduction
SSAO (Screen space ambient occlusion) is a common post processing effect that approximate how much light is occluded in a given surface by the surrounding objects. In this year SIGGRAPH, there are a few slides in "The Technology behind the Unreal Engine 4 Elemental Demo" about how they implement SSAO. Their technique can either use only the depth buffer or with the addition of per-pixel normal. And I tried to implement both version with a slight modification:
Using only the depth buffer
The definition of ambient occlusion is to calculate the visibility integral over the hemisphere of a given surface:
To approximate this in screen space, we design our sampling pattern as paired samples:
paired sample pattern
So for each pair of samples, we can approximate how much the shading point is occluded in 2D instead of integrating over the hemisphere:
The AO term for each given pair of samples will be min( (θleft + θright)/π, 1). Then by averaging the AO terms of all the sample pairs (in my case, there are 6 pairs), we achieve the following result:
Dealing with large depth differences
As seen from the above screen shot, there is dark halos around the knight. But the knight should not contribute AO to the castle as he is too far away. So to deal with the large depth differences. I adopt the approach used in Toy Story 3. If one of the paired sample is too far away from the shading point, say the red point in the following figure, it will be replace by the pink point, which is on the same plane as the other valid paired sample:
So we can interpolate between the red point and the pink point for dealing with the large depth difference. Now the dark halo has gone:
The above treatment only handle if one of the paired sample is far away from shading point. What if both of the samples have large depth differences?
dark halo artifact is shown around the sword
AO strength of this pic is increased to high light the artifact
In this case, it will result in the dark halo around the sword in the above screen shot. Remember we are averaging the all the paired samples to compute the final AO value. So to deal with this artifact, we just assign a weight to each paired samples and then re-normalize the final result. Say, for each paired sample, if both of the samples are within a small depth differences, that sample pair will have a weight of 1. If only 1 sample is far away, that pair will have a weight of 0.5. And finally if both of the samples is far away, the weight will be 0. This can eliminate most(but not all) of the artifacts:
Approximating arc-cos function
In this approach, the AO is calculated by using the angle between the paired samples, which need to evaluate the arc-cos function which is a bit expensive. We can approximate acos(x) with a linear function: π(1-x)/2.
And the resulting AO looks much darker with this approximation:
computed with the arc-cos function
computed with the linear approximation
Note that the maximum error between the two function is around 18.946 degree.
This may affect the AO for the area of a curved surface with low tessellation. You may either need to increase the bias angle threshold or switch to a more accurate function. So my second attempt is to approximate it with a quadratic function: π(1- sign(x) * x * x)/2.
And this approximation shows a much similar result to the one using the arc-cos function.
computed with the arc-cos function
And the maximum error of this function is around 9.473 degree.
Using per-pixel normal
We can enhance the details of AO by making use of the per-pixel normal. The per-pixel normal is used for further restricting the angle to compute the AO where the angle θleft, θright are clamped to the tangent plane :
And here is the final result:
Conclusion
The result of this AO is pleasant by taking total 12 samples per pixel and with 16 rotation in 4x4 pixel block at half resolution. I did not apply bilateral blur to the AO result, but applying the blur may gives a softer AO look. Also approximating the arc-cos function with a linear function although is not accurate, but it gives a good enough result for me. Finally more time are need to spend on generating the sampling pattern in the future where the pattern I currently used is nearly uniform distributed (with some jittering).
References
[1] The Technology behind the Unreal Engine 4 Elemental Demo http://advances.realtimerendering.com/s2012/Epic/The%20Technology%20Behind%20the%20Elemental%20Demo%2016x9.pptx
[2] Rendering techniques in Toy Story 3 | 1,022 | 4,573 | {"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-2018-43 | longest | en | 0.921402 |
https://skyciv.com/el/quick-calculators/flooring-joist-span-calculator/ | 1,718,920,251,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862006.96/warc/CC-MAIN-20240620204310-20240620234310-00392.warc.gz | 471,675,164 | 26,471 | # Υπολογιστής ανοίγματος δοκών δαπέδου
The SkyCiv floor joist span calculator allows you to calculate the maximum span of timber joints. The calculations in this tool are based on the American Wood Council's standards. The floor joist calculator considers factors including the joist size, πάχος, ύψος, wood type, and wood grade as well as loading and deflection factors. Various common types of wood and grades are available within the below floor joist spacing calculator.
Starting Flooring Joist Span Calculator...
# Need a Custom Floor Joist Span Calculator?
Εξερευνήστε τη δυνατότητα της δικής σας προσαρμοσμένης αριθμομηχανής SkyCiv
για ταχύτερη ανάκαμψη του σχεδιασμού και σημαντική εξοικονόμηση χρόνου και κόστους.
## How to use the floor joist span calculator?
To begin using the floor joist calculator enter the details of the joist on the left-hand side. These details will include:
• Joist Size
• Joist Thickness
• Joist Height
• On-Center Spacing
Then complete the advanced details including:
• Ξύλο & Grad
• Deflection Limit
• Load per Area Unit
Run the floor joist spacing calculator to determine the maximum span
## Flooring joist calculator assumptions and limitations:
• This tool supports only natural solid lumber joists
• This tool uses the American Wood Council's standards for reference. "
• These calculations are only approximations and should be used as the starting point to get an idea about the Joist Span.
## Συχνές ερωτήσεις
### What is a floor joist?
A floor joist is a member that supports the weight of a floor and loads placed on a floor. These are horizontal structural members that are found parallel to the ground and help distribute the weight to the underlying structural foundations. Flooring joists are vital to the stability and strength of a floor structure. The above flooring joist calculator can be used to determine the appropriate size of joists depending on loading and other factors.
### What is a floor joist span table?
Floor joist span tables are references used by engineers, builders, and construction workers to estimate the appropriate size of joists based on factors such as the joist parameters and floor loading. The above floor joist calculator can be used in place of a floor joist span table.
### How does on-center spacing affect floor joists' maximum span?
The on-center spacing is the spacing between each joist. Γενικά, the larger the spacing (distance between each joist) then the lower the maximum allowable span is.
### Do floor joist span tables and calculators work for wood like LVL (Ξυλεία από πλαστικοποιημένο καπλαμά)?
Floor joist span calculators and tables can be used for LVL timber as long as considerations around the material properties and behavior of the lumber are taken into account.
### Do floor joist span calculators and tables work for deck joists as well?
Ναί, floor joist span calculators and tables can be used for deck design as long as all considerations are taken into account.
### Can floor joist tables be used to calculate the maximum allowable span for timber floor joists?
Ναί, floor joist tables are generally available for timber floor joists. The correct table needs to be used based on the type of wood, lumber, or timber joist. Εναλλακτικά, the above floor joist spacing calculator contains options for many types of wood including:
• Alaska Yellow Cedar
• Aspen
• Baldcypress
• Beech-Birch-Hickory
• Coast Sitka Spruce
• Cottonwood
• Douglas Fir-Larch
• Douglas Fir-Larch (North)
• Douglas Fir-South
• Eastern Hemlock-Balsam Fir
• Eastern Hemlock-Tamarack
• Eastern Softwoods
• Eastern White Pine
• Hem-For
# Get a custom calculator that's as unique as your business
Εξερευνήστε τη δυνατότητα της δικής σας προσαρμοσμένης αριθμομηχανής SkyCiv
για ταχύτερη ανάκαμψη του σχεδιασμού και σημαντική εξοικονόμηση χρόνου και κόστους. | 905 | 3,862 | {"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-2024-26 | latest | en | 0.782063 |
http://www.creativeakademy.org/2015/06/how-can-we-specify-position-of-object-physics-class9.html | 1,513,594,603,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948615810.90/warc/CC-MAIN-20171218102808-20171218124808-00383.warc.gz | 332,342,184 | 32,072 | ### How can We Specify the Position of an object | Physics Class 9
To specify the position of an object, we need a point with respect to which we identify the position of that object. This point is known as the origin or reference point.
Consider a man starts walking from point O. First he goes to point P and then to point Q. as we can see in the given image.
We can say that the position of the object is OP form the origin O in direction of x-axis and PQ from origin O in direction of y-axis. In term of co-ordinate system we say it as (OP, PQ) or (5,4) in term of actual value of axis. | 143 | 593 | {"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} | 3.078125 | 3 | CC-MAIN-2017-51 | latest | en | 0.919816 |
https://casoilresource.lawr.ucdavis.edu/software/r-advanced-statistical-package/using-lm-and-predict-apply-standard-curve-analytical-data/ | 1,656,748,058,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103989282.58/warc/CC-MAIN-20220702071223-20220702101223-00722.warc.gz | 213,915,338 | 12,088 | # Using lm() and predict() to apply a standard curve to Analytical Data
```#first the sample data
#note that field sep might be different based on pre-formatting
#then the standards:
# comput simple linear models from standards
# "mg_nitrogen as modeled by area under curve"
lm.N <- lm(mg_N ~ area_N, data=cn_std)
lm.C <- lm(mg_C ~ area_C, data=cn_std)
# check std curve stats:
summary(lm.N)
Multiple R-Squared: 0.9999, Adjusted R-squared: 0.9999
summary(lm.C)
Multiple R-Squared: 1, Adjusted R-squared: 1```
Apply the standard curve to the raw measurements
```# note that the predict method is looking for column names that where originally
# used in the creation of the lm object
# i.e. area_N for lm.N and area_C for lm.C
# therefore it is possible to pass the original data matrix with both
# values to predict(), while specifiying the lm object
cn\$mg_N <- predict(lm.N, cn)
cn\$mg_C <- predict(lm.C, cn)```
Merge sample mass to calculate percent C/N by mass
```#read in the initial mass data, note that by default string data will be read in as a factor
# i.e. factors are like treatments, and this data type will not work in some functions
#take a look at how the mass data was read in by read.table()
str(cn.mass)
'data.frame': 75 obs. of 5 variables:
\$ id : <b>Factor</b> w/ 26 levels</b> "004K","007K",..: 15 16 17 18 19 20 21 22 23 24 ...
\$ pedon_id : <b>Factor</b> w/ 18 levels "004K","007K",..: 15 15 15 18 18 18 17 17 17 16 ...
\$ horizon_num: int 2 5 7 2 4 6 2 4 5 2 ...
\$ sample_id : <b>Factor</b> w/ 75 levels "A1","A10","A11",..: 23 24 14 15 16 25 29 30 31 32 ...
\$ sample_mg : num 24.6 27.5 33.3 25.9 25.8 ...
# use the merge() function to join the two dataframes based on the cell_id column
#merge() does not work with columns of type "level"
# convert them to characters in upper case, and append them to the original dataframe:
# note that merge is case sensitive !!!
cn\$cell_id <- toupper(as.character(cn\$sample_id))
cn.mass\$cell_id <- toupper(as.character(cn.mass\$sample_id))
#only keep our pedon data, leave behind the checks
cn.complete <- merge(x=cn, y=cn.mass, by.x="cell_id", by.y="cell_id", sort=FALSE, all.y=TRUE)
##calculate percent N and C, appending to the cn.complete dataframe
cn.complete\$pct_N <- (cn.complete\$mg_N / cn.complete\$sample_mg) * 100
cn.complete\$pct_C <- (cn.complete\$mg_C / cn.complete\$sample_mg) * 100
#look at the results:
str(cn.complete)
'data.frame': 75 obs. of 13 variables:
\$ cell_id : chr "B8" "B9" "B10" "B11" ...
\$ sample_id.x: Factor w/ 81 levels "A1","A10","A11",..: 24 25 15 16 17 26 30 31 32 33 ...
\$ area_N : num 2225431 208028 341264 1377688 168328 ...
\$ area_C : num 85307240 8296664 14624760 50879560 6690868 ...
\$ mg_N : num 0.09261 0.01096 0.01635 0.05830 0.00935 ...
\$ mg_C : num 1.2609 0.1204 0.2141 0.7510 0.0967 ...
\$ id : Factor w/ 26 levels "004K","007K",..: 15 16 17 18 19 20 21 22 23 24 ...
\$ pedon_id : Factor w/ 18 levels "004K","007K",..: 15 15 15 18 18 18 17 17 17 16 ...
\$ horizon_num: int 2 5 7 2 4 6 2 4 5 2 ...
\$ sample_id.y: Factor w/ 75 levels "A1","A10","A11",..: 23 24 14 15 16 25 29 30 31 32 ...
\$ sample_mg : num 24.6 27.5 33.3 25.9 25.8 ...
\$ pct_N : num 0.3759 0.0398 0.0491 0.2254 0.0363 ...
\$ pct_C : num 5.117 0.438 0.643 2.903 0.375 ...
#save the data for further processing:
write.table(cn.complete, file="cn.complete.table", col.names=TRUE, row.names=FALSE)```
Measure the accuracy of the sensor in the machine with simple correlation
```### get a measure of how accurate the sensor was, based on our checks:
#just the first 5 columns, in case there is extra
cn.checks <- cn[c(13,26,39,52,65,78),][1:5]
#make a list of the mg of ACE in each check
checks.mg_ACE <- c(0.798, 1.588, 1.288, 1.574, 1.338, 1.191)
#make a column of the REAL mg_N based on the percent N in ACE
cn.checks\$real_mg_N <- checks.mg_ACE * 0.104
#make a column of the REAL mg_C based on the percent C in ACE
cn.checks\$real_mg_C <- checks.mg_ACE * 0.711
# check with cor()```
Create a mutli-figure diagnostic plot
```layout(mat=matrix(c(1, 4, 2, 3), nc = 2, nr = 2), width=c(1,1), height=c(1,2))
#first the std curves
par(mar = c(4,4,2,2))
#Nitrogen
plot(mg_N ~ area_N, data=cn_std, xlab="Area Counts", ylab="mg", main="Std Curve for N", cex=0.7, pch=16, cex.axis=0.6)
rug(cn\$area_N, ticksize=0.02, col="gray")
rug(cn\$mg_N, ticksize=0.02, col="gray", side=2)
abline(lm.N, col="gray", lty=2)
points(cn\$area_N, cn\$mg_N, col="blue", cex=0.2, pch=16)
#Carbon
plot(mg_C ~ area_C, data=cn_std, xlab="Area Counts", ylab="mg", main="Std Curve for C", cex=0.7, pch=16, cex.axis=0.6)
rug(cn\$area_C, ticksize=0.02, col="gray")
rug(cn\$mg_C, ticksize=0.02, col="gray", side=2)
abline(lm.C, col="gray", lty=2)
points(cn\$area_C, cn\$mg_C, col="blue", cex=0.2, pch=16)
#possible problems
points(cn\$area_C[which(cn\$area_C > 1.0e+08)], cn\$mg_C[which(cn\$area_C > 1.0e+08)] , col="red")
#sample plot of carbon distributions within each pedon:
#note that las=2 makes axis labels perpendicular to axis
par(mar = c(7,4,4,2))
#boxplot illustrating the within-pedon variation of Carbon
boxplot(cn.complete\$pct_C ~ cn.complete\$pedon_id , cex.axis=0.6, boxwex=0.2, las=2, main="Percent Total Carbon", ylab="% C", xlab="Pedon ID", cex=0.4)
#boxplot illustrating the within-pedon variation of Nitrogen
boxplot(cn.complete\$pct_N ~ cn.complete\$pedon_id , cex.axis=0.6, boxwex=0.2, las=2, main="Percent Total Nitrogen", ylab="% N", xlab="Pedon ID", cex=0.4)```
R: Multi-figure plot of Carlo-Erba Data | 2,027 | 5,606 | {"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-2022-27 | latest | en | 0.757495 |
http://www.docstoc.com/docs/73002706/Chaper-Uncertainty-and-Reasoning | 1,369,486,114,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705953421/warc/CC-MAIN-20130516120553-00060-ip-10-60-113-184.ec2.internal.warc.gz | 434,598,430 | 18,035 | # Chaper Uncertainty and Reasoning
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Document Sample
``` Part II
Methods of AI
Chapter 5
Uncertainty and Reasoning
Part II: Methods of AI
Chapter 5 – Uncertainty and Reasoning
5.1 Uncertainty
5.2 Probabilistic Reasoning
5.3 Probabilistic Reasoning over Time
5.4 Making Decisions
Chapter 5 – Uncertainty and Reasoning
5.1 Uncertainty
Chapter 5.1 – Uncertainty
Uncertainty – Introduction (1)
Let action At = leave for airport t minutes before flight
Will At get me there in time?
Problems:
1) partial observability (road state, other drivers’ plans, etc.)
2) noisy sensors (KCBS traffic reports)
3) uncertainty in action outcomes (flat tire, etc.)
4) immense complexity of modeling and predicting traffic
Chapter 5.1 – Uncertainty
Uncertainty – Introduction (2)
Hence a purely logical approach either
1) risks falsehood: “A25 will get me there on time” or
2) leads to conclusions that are too weak for decision making:
“A25 will get me there on time if there’s no accident on the
bridge and it doesn’t rain and my tires remain intact etc.
etc.”
(A1440 might reasonably be said to get me there on time
but I’d have to stay overnight in the airport…)
Chapter 5.1 – Uncertainty
Methods for handling uncertainty
Default or nonmonotonic logic:
Assume my car does not have a flat tire
Assume A25 works unless contradicted by evidence
Issues: What assumptions are reasonable?
Rules with fudge factors:
A25 0.3 get there on time
Sprinkler 0.99 WetGrass
WetGrass 0.7 Rain
Chapter 5.1 – Uncertainty
Methods for handling uncertainty (2)
Probability
Given the available evidence,
A25 will get me there on time with probability 0.04
Mahaviracarya (9th C.), Cardamo (1565) theory of gambling
(Fuzzy logic handles degree of truth NOT uncertainty e.g.,
WetGrass is true to degree 0.2)
Chapter 5.1 – Uncertainty
Probability (1)
Probability assertions summarize effects of
laziness: failure to enumerate exceptions, qualifications, etc.
ignorance: lack of relevant facts, initial conditions, etc.
Subjective or Bayesian probability:
Probabilities relate propositions to one’s own state of knowledge
e.g., P(A25no reported accident) = 0.06
Chapter 5.1 – Uncertainty
Probability (2)
These are not claims of some probabilistic tendency in the
current situation (bit might be learned from past experience of
similar situations)
Probabilities of propositions change with new evidence:
e.g., P(A25no reported accident, 5 a.m.) = 0.15
(Analogous to logical entailment status KB= α, not truth.)
Chapter 5.1 – Uncertainty
Making decisions under uncertainty
Suppose I believe the following:
P(A25 gets me there in time…) = 0.04
P(A90 gets me there in time…) = 0.70
P(A120 gets me there in time…) = 0.95
P(A1440 gets me there in time…) = 0.9999
Which action to choose?
Depends on my preferences for missing flight vs. airport cuisine, etc.
Utility theory is used to represent and infer preferences
Decision theory = utility theory + probability theory
Chapter 5.1 – Uncertainty
Probability basics
Begin with a set Ω – the sample space
e.g., 6 possible rolls of a die.
ω Ω is a sample point/possible world/atomic event
A probability space or probability model is a sample space
with an assignment P(ω) for every ω Ω s.t.
0 P(ω) 1
ω P(ω) = 1
e.g., P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6.
An event A is any subset of Ω
P(A) = { ω A} P(ω)
E.g., P(die roll < 4) = 1/6 + 1/6 + 1/6 = ½
Chapter 5.1 – Uncertainty
Random variables
A random variable is a function from sample points to some
range, e.g., the reals or Booleans e.g., Odd(1) = true
P induces a probability distribution for any r.v. X:
P(X x i ) {: X ( ) i }( )
e.g., P(Odd = true) = 1/6 + 1/6 + 1/6 = 1/2
Chapter 5.1 – Uncertainty
Propositions (1)
Think of a proposition as the event (set of sample points)
where the proposition is true
Given Boolean random variables A and B:
event a = set of sample points where A(ω) = true
event a = set of sample points where A(ω) = false
event a b = points where A(ω) = true and B(ω) = true
Chapter 5.1 – Uncertainty
Propositions (2)
Often in AI applications, the sample points are defined by the
values of a set of random variables, i.e., the sample space is
the Cartesian product of the ranges of the variables
With Boolean variables, sample point = propositional logic model
e.g., A = true, B = false, or a b
Proposition = disjunction of atomic events in which it is true
e.g., (a b) (a b) (a b) (a b)
P(a b) = P(a b) + P(a b) + P(a b)
Chapter 5.1 – Uncertainty
Why use probability?
The definitions imply that certain logically related events must have
related probabilities
E.g., P(a b) = P(a) + P(b) - P(a b)
de Finetti (1931): an agent who bets according to probabilities that
violate these axioms can be forced to bet so as to lose money
regardless of outcome.
Chapter 5.1 – Uncertainty
De Finetti‘s Argument (Example)
Agent 1 Agent 2 Outcome for Agent 1
Proposition Belief Bet Stakes A B A B A B A B
A 0.4 A 4:6 -6 -6 4 4
B 0.3 B 3:7 -7 3 -7 3
AB 0.8 (A B) 2 : 8 2 2 2 -8
-11 -1 -1 -1
Agent 1 always looses
Chapter 5.1 – Uncertainty
Syntax for propositions
Propositional or Boolean random variables
e.g., Cavity (do I have a cavity?)
Discrete random variables (finite or infinite)
e.g., Weather is one of sunny, rain, cloudy, snow
Weather = rain is a proposition
Values must be exhaustive and mutually exclusive
Continuous random variables (bounded or unbounded)
e.g., Temp = 21.6; also allow, e.g. Temp < 22.0
Arbitrary Boolean combinations of basic propositions
Chapter 5.1 – Uncertainty
Prior probability (1)
Prior or unconditional probabilities of propositions
e.g., P(Cavity = true) = 0.1 and P(Weather = sunny) = 0.72
correspond to belief to arrival of any (new) evidence
Probability distribution gives values for all possible assignments:
P(Weather) = 0.72, 0.1, 0.08, 0.1 (normalized, i.e., sums to 1)
Chapter 5.1 – Uncertainty
Prior probability (2)
Joint probability distribution for a set of r.v. s. gives the
probability of every atomic event on those r.v.s. (i.e., every
sample point)
P(Weather, Cavity) = a 4 x 2 matrix of values:
Weather = Sunny rain cloudy snow
Cavity = true 0.144 0.02 0.016 0.02
Cavity = false 0.576 0.08 0.064 0.08
distribution because every event is a sum of sample points
Chapter 5.1 – Uncertainty
Probability for continuous variables
Express distribution as a parameterized function of value:
P(X = x) = U[18.26](x) = uniform density between 18 and 26
Here P is a density, integrates to 1.
P(X = 20.5) = 0.125 really means
lim P(20.5 X 20.5 dx) / dx 0.125
dx0
Chapter 5.1 – Uncertainty
Gaussian density
1 ( x ) 2 / 2 2
P( x) e
2
Chapter 5.1 – Uncertainty
Conditional probability (1)
Conditional or posterior probabilities
e.g., P(cavitytoothache) = 0.8
i.e., given that toothache is all I know
NOT “if toothache then 80% chance of cavity”
(Notation for conditional distributions:
P(CavityToothache) = 2-element vector of 2-element
vectors)
Chapter 5.1 – Uncertainty
Conditional probability (2)
If we know more, e.g., cavity is also given, then we have
P(cavitytoothache, cavity) = 1
Note: the less specific belief remains valid after more evidence
arrives, but is not always useful.
New evidence may be irrelevant, allowing simplification, e.g.
P(cavitytoothache, 49ersWin) = P(cavitytoothache) = 0.8
This kind of inference, sanctioned by domain knowledge, is
crucial.
Chapter 5.1 – Uncertainty
Conditional probability (3)
Definition of conditional probability:
P ( a b)
P ( a | b) if P(b) 0
P(b)
Product rule gives an alternative formulation:
P(a b) P(a | b) P(b) p(a | b) P(b)
A general version holds for whole distributions, e.g.,
P(Weather, Cavity) = P(Weather Cavity) P(Cavity)
(View as a 4 x 2 set of equations, not matrix mult.)
Chapter 5.1 – Uncertainty
Conditional Probability vs. Implication
Take care: P(B\A) P(A B)
Example: P(A, B) = 0.25
P(A, B) = 0.25
P(A, B) = 0.25
P(A, B) = 0.25
A, B A, B P(A B) = P(A, B) + P(A, B)+ P(A, B)
A, B A, B = 0.75
0.25
A, B (A, B) P(B|A) = P(A, B) =
P(A) 0.5
A, B (A, B) = 0.5
Chapter 5.1 – Uncertainty
Conditional probability (4)
Chain rule is derived by successive application of product
rule:
P( X1 ,..., X n ) P( X1 ,..., X n1 ) P( X nX1 ,..., X n1 )
P( X 1 ,..., X n 2 ) P( X n1X 1 ,..., X n 2 ) P( X nX 1 ,..., X n 1 )
...
i 1 P( X iX 1 ,..., X i 1 )
n
Chapter 5.1 – Uncertainty
Inference by enumeration (1)
toothache toothache
catch catch catch catch
cavity .108 .012 .072 .008
cavity .016 .064 .144 .576
For any proposition Θ, sum the atomic events where it is true:
P() {: | }( )
P(toothache) = 0.108+0.012+0.016+0.064 = 0.2
Chapter 5.1 – Uncertainty
Inference by enumeration (2)
toothache toothache
catch catch catch catch
cavity .108 .012 .072 .008
cavity .016 .064 .144 .576
For any proposition Θ, sum the atomic events where it is true:
P() {: | }( )
P(toothache) = 0.108+0.012+0.072+0.008+0.016+0.064 = 0.28
Chapter 5.1 – Uncertainty
Inference by enumeration (3)
toothache toothache
catch catch catch catch
cavity .108 .012 .072 .008
cavity .016 .064 .144 .576
Can also compute conditonal probabilities:
P( cavity toothache)
P( cavity|toothache) =
P(toothache)
0.016+0.064
= = 0.4
0.108+0.012+0.016+0.064
Chapter 5.1 – Uncertainty
Normalization
toothache toothache
catch catch catch catch
cavity .108 .012 .072 .008
cavity .016 .064 .144 .576
Dominator can be viewed as a normalization constant
P(Cavity|toothache) = α P(Cavity, toothache)
= α [P(Cavity, toothache,catch) + P(Cavity, toothache, catch)]
= α [<0.108,0.016> + <0.012,0.064>] = α <0.12,0.08> = <0.6,0.4>
General idea: compute distribution on query variable by fixing evidence
variables and summing over hidden variables
Chapter 5.1 – Uncertainty
Inference by enumeration, contd.
Typically, we are interested in
the posterior joint distribution of the query variables Y
given specific values e for the evidence variables E
Let the hidden variables be H = X – Y – E
Then the required summation of joint entries is done by summing out the
hidden variables:
P(Y | E e) P(Y | E e) h P(Y, E e, H h)
The terms in the summation are joint entries because Y, E, and H together
exhaust the set of random variables.
Obvious problems:
1) Worst-case time complexity O(dn) where d is the largest arity
2) Space complexity O(dn) to store the joint distribution
3) How to find the numbers for O(dn) entries???
Chapter 5.1 – Uncertainty
Independence
A and B are independent iff
P(A|B) = P(A) or P(B|A)=P(B) or P(A,B)=P(A)P(B)
P(Toothache, Catch, Cavity, Weather)
= P(Toothache, Catch, Cavity) P(Weather)
32 entries reduced to 12; for n independent biased coins, 2n → n
Absolute independence powerful but rare
Dentistry is a large field with hundreds of variables, non of which are independent.
What to do?
Chapter 5.1 – Uncertainty
Conditional independence (1)
P(Toothache, Cavity, Catch) has 23 –1 = 7 independent entries
If I have a cavity, the probability that the probe catches in it doesn’t depend
on whether I have o toothache:
(1) P(catchtoothache, cavity) = P(catchcavity)
The same independence holds if I haven’t got a cavity:
(2) P(catchtoothache, cavity) = P(catchcavity)
Catch is conditionally independent of Toothache given Cavity:
P(CatchToothache, Cavity) = P(CatchCavity)
Chapter 5.1 – Uncertainty
Conditional independence (2)
Equivalent statements:
P(ToothacheCatch, Cavity) = P(Toothache Cavity)
P(Toothache, CatchCavity) = P(Toothache Cavity) P(CatchCavity)
Write out full joint distribution using chain rule:
P(Toothache Catch, Cavity)
= P(ToothacheCatch, Cavity) P(Catch, Cavity)
= P(ToothacheCatch, Cavity) P(CatchCavity) P(Cavity)
= P(Toothache Cavity) P(CatchCavity) P(Cavity)
l.e., 2 + 2 + 1 = 5 independent numbers (equation 1 and 2 remove2)
In most cases, the use of conditional independence reduces the size
of the representation of the joint distribution from exponential in n to
linear in n.
Conditional independence is our most basic and robust form of
Chapter 5.1 – Uncertainty
Bayes’ Rule
Product rule P(a b) P(b | a) P(a)
Bayes’ rule
P(b a ) P(a )
P ( a b)
P(b)
or in distribution form
P( X | Y ) P(Y )
P(Y | X ) P( X | Y ) P(Y )
P( X )
Useful for assessing diagnostic probability from causal probability:
P( Effect | Cause) P(Cause)
P(Cause | Effect) P( X | Y ) P(Y )
P( Effect)
E.g., let M be meningitis, S be stiff neck:
P( s | m) P(m) 0.8 0.0001
P(m | s ) 0.0008
P( s ) 0.1
Note: posterior probability of meningitis still very small.
Chapter 5.1 – Uncertainty
Bayes’ Rule and conditional independence
P(cavitytoothache catch)
= P(toothache catch Cavity) P(Cavity)
= P(toothache Cavity) P(catchCavity) P(Cavity)
This is an example of a naïve Bayes model:
P(Cause, Effect1 ,..., Effectn ) P(Cause)i P( EffectiCause)
Total number of parameter is linear in n.
Chapter 5.1 – Uncertainty
Summary
Probability is a rigorous formalism for uncertain knowledge.
Joint probability distribution specifies probability of every atomic
event.
Queries can be answered by summing over atomic events.
For nontrivial domains, we must find a way to reduce the joint
size.
Independence and conditional independence provide the tools.
Chapter 5.1 – Uncertainty
Chapter 5.1 – Uncertainty
Normalisation
Relative comparison of probabilities is often sufficient:
M := Meningitis, N := Nackensteife, S := Schleudertrauma
(whiplash)
P(M|N) = P(N|M) P(M) P(S|N) = P(N|S) P(S)
P(N) P(N)
P(M|N) P(N|M) P(M)
=
P(S|N) P(N|S) P(S)
Comparison of both diagnoses possible without knowledge
on P(N)
Often decisions can be based on relative comparison of
probabilities
Chapter 5.1 – Uncertainty
Normalisation
Sometimes relative probabilities are weak for careful diagnoses;
Nevertheless, knowledge about basic probabilities like P (N) can
often be avoided.
P(N|M) P(M) P(M|N) = P(N|M) P(M)
P(M|N) =
P(N) P(N)
P(N|M) P(M) P(N|M) P(M)
P(M|N) + P(M|N) = +
P(N) P(N)
= 1/ P(N) (P(N|M) P(M) + P(N|M) P(M)) = 1
P(N) = P(N|M) P(M) + P(N|M) P(M)
P(N|M) P(M)
P(M|N) =
P(N|M) P(M) + P(N|M) P(M))
In general:
P(M|N) = P(N|M) P(M) where is normalisation constant, such that
CPT entries for PCMIN sum up to 1.
Chapter 5.1 – Uncertainty
```
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# answer_22 - 1.3.1(c p T T T T F F F F q T T F F T T F F r T...
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p q r A:"B and C are both lying" B:"Only one of the other two is lying" C:"At least one of us is lying" T T T F F F T T F F T T T F T F F T T F F T T T F T T F T T F T F F F T F F T F T T F F F T F T 1.3.1 (c) 1.3.4 (a) ¬p∧(q∨r) (b) (¬p∧¬q)∨(p∧r) (c) (p∧q)∨(p∧r)∨(q∧r)∨(p∧q∧r) or (p∧q)∨(p∧r)∨(q∧r) (d) [(p∧q)∨(p∧r)∨(q∧r)]∧¬(p∧q∧r) or [(p∧q∧Ør)∨(p∧r∧Øq)∨(q∧r∧Øp)] 1.3.5 (b) f∧m∧a (d) ¬f∨a 1.3.6 (b) This person is not female and age is not over 30. or This person is male who is not over 30. (d) Either this person is a female over age 30 or this person is a male with math major. 1.3.11 (c) p q p∨q ¬p∨q (p∨q)∧(¬p∨q) T T T T T T F T F F F T T T T F F F T F (e) p q r p∧q (p∧q)∨r T T T T T T T F T T T F T F T T F F F F F T T F T F T F F F F F T F T F F F F F
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1.3.12 (b) f: The staff is friendly t: they very well paid. f t ¬f ¬f∧t f∨(¬f∧t) T T F F T T F F F T F T T T T F F T F F (d) f: You play sports t: You play mahjong h: Nobody knows your name f t h f∨t
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{[ snackBarMessage ]} | 604 | 1,347 | {"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-2018-17 | latest | en | 0.642533 |
https://mega-commercials.com/qa/what-does-5-000-words-look-like.html | 1,611,010,632,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703517159.7/warc/CC-MAIN-20210118220236-20210119010236-00025.warc.gz | 477,518,054 | 7,842 | # What Does 5 000 Words Look Like?
## How many pages is a 100000 words?
200.0 pagesThe answer is 100,000 words is 200.0 pages single-spaced or 400.0 pages double-spaced..
## How long does it take to write a 50000 word book?
5-8 monthsHere’s how long it takes to write a book based on our daily writing guidelines: 50,000-word book — 5-8 months.
## What is the longest book in the world?
A la recherche du temps perduA la recherche du temps perdu by Marcel Proust contains an estimated 9,609,000 characters (each letter counts as one character. Spaces are also counted, as one character each). The title translates to “Remembrance of Things Past”.
## Is 20 000 words enough for a book?
Ask yourself the following: Do you want a small, intimate book? … This trim size is good for only 200 to 250 words per printed book page. This translates into just 20,000 to 25,000 words for a 100 page book, 30,000 to 42,500 words for a 150-page book, perhaps just 40,000 words for a 200-page book.
## How many pages is 750 words?
1.5 pagesA word count of 750 words will equal about 1.5 pages single spaced or 3 pages double spaced. Of course it will depend on the word processor settings, what font and font size you’re using and page margins.
## Is 5000 words enough for a book?
The average short story should run anywhere from 5,000 to 10,000 words, but they can be anything above 1,000 words. Flash fiction is a short story that is 500 words or less.
## What does 4000 words look like?
Answer: 4000 words is 8 pages single spaced or 16 pages double spaced.
## What does 1000 words look like?
1000 words is around 2 & 1/3rds of a page visually, single-spaced, and 4 pages double-spaced.
## Is a 4000 word essay long?
How long does it take to write a 4,000 word essay? It takes about 13 hours and 20 minutes to write a 4,000 word essay.
## What does 3000 words look like?
Answer: 3000 words is 6 pages single spaced or 12 pages double spaced.
## How many pages is 50000 words?
If you’re thinking bigger and wondering, for example, how many pages is 50,000 words, simply divide your target word count (50,000) by 500 (since that’s the average words per page). Your answer here is 100 pages. Don’t let those commas instill fear.
## Can you write 5000 words in a day?
Originally Answered: Is it possible to write a 5,000-word essay in a day? Yes, it is possible to write a 5,000-word essay in a day. One needs to spend some time for research and go to a silent place where he can brainstorm easily. He may then write the entire essay at a stretch most probably in the late hours of night.
## How long does it take to write 3000 words?
Writing 3,000 words will take about 1.3 hours for the average writer typing on a keyboard and 2.5 hours for handwriting. However, if the content needs to include in-depth research, links, citations, or graphics such as for a blog article or high school essay, the length can grow to 10 hours.
## How many pages is 40000 words?
80.0 pagesThe answer is 40,000 words is 80.0 pages single-spaced or 160.0 pages double-spaced. Typical documents that are 40,000 words include include novels, novellas, and other published books.
## Can you do a 3000 word essay in one day?
We wouldn’t recommend writing an essay in such a short period of time, but the good news is that 3,000 words in a day is totally doable. Get your head down and you could meet the deadline, and even produce an essay you are proud of.
## How many pages is 90000 words?
180.0 pagesA 90,000 word count will create about 180.0 pages single-spaced or 360.0 pages double-spaced when using normal margins (1″) and 12 pt.
## How many words is Harry Potter?
1,084,170 wordsHarry Potter and the Order of the Phoenix – 257,045 words. Harry Potter and the Half-Blood Prince – 168,923 words. Harry Potter and the Deathly Hallows – 198,227 words. The entire Harry Potter series – 1,084,170 words.
## How long will it take to write 4000 words?
Writing 4,000 words will take about 1.7 hours for the average writer typing on a keyboard and 3.3 hours for handwriting. However, if the content needs to include in-depth research, links, citations, or graphics such as for a blog article or high school essay, the length can grow to 13.3 hours. | 1,113 | 4,244 | {"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} | 2.875 | 3 | CC-MAIN-2021-04 | latest | en | 0.886762 |
https://darkhavenbookreviews.com/slide/ch-2-scientific-method-x7mqva | 1,611,178,762,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522133.33/warc/CC-MAIN-20210120213234-20210121003234-00593.warc.gz | 287,618,642 | 49,812 | # Ch. 2 - Scientific Method
Ch. 2 Scientific Method A systematic approach to solve problems through investigations by 1. Observing, 2. Generalizing, 3. Theorizing, 4. Testing, 5. Conclusion. Whats done in an observation Collect data, communicate Generalizing Organize data, hypothesis Theorizing Planning and predicting
Test Design experiment, test, graph and interpret results Conclusion May change hypothesis and retest Explain what happened in your results Why Measurements do we need measurements? If I told you to go get me a pop would you
know what size? The International System of Units (SI) is used throughout the world so we dont confuse figures and measurements. SI is metric Written the following way: 21 400 or 0.157 82 7 SI fundamental units Mass, time, temperature, length, amount of substance. Pg. 34 Unit SI Unit Prefixes of measurements for length
KHDMDCM Kilo, hecto, deka, meter, deci, centi, milli Ex. Kilogram (1000 grams) or Kilometer (1000 meters) 1 gram is about 1 pencil mass Your book is about 2 Kg Derived Unit are combinations of fundamental units. Area Derived Units length X width (m2) Volume length X width X height (m3)
Volume is the amount of space occupied by an object. 1ml = 1cm3 1 liter = 1000-ml 1 ml = how many liters? Density is mass divided by volume (D=m/v) How heavy something is for its size. Ex. Pg. 38 (Density Lab) Temperature is the measure of the average kinetic energy of the particles in a sample of matter. What is the SI unit for temp.?
Kelvin (K) Heat is the sum of the kinetic energy of the particles in a sample of matter. How does a thermometer work? We use Celsius more than Kelvin 0 C = 273 K freezing point 100 C = 373 K- boiling point Joule is force times length (f X l) 1 J = 1 n-m Joule is the SI unit for energy A calorie is the amount of heat needed to raise the temperature of 1 g of water 1 degrees C.
1 calorie = 4.184 J Kcal or C = KJ Food Calories are kilocalories. Conversion Factors Sample Problem 2-2 Specific heat is the amount of heat energy required to raise the temp. of 1 g of a substance 1 degree C. Label is J/(g x C) Water heats up and cools down slowly, thus it has a high specific heat. Accuracy vs. Precision Accuracy is determined by how close your results are to the correct answer.
Precision is determined by how close your results are to each other. Ex. Dart Board pg. 44 Sample Problem 2-3 (Percent Error) SIGNIFICANT FIGURES Significant figures are all the digits known with certainty plus one final digit, which is uncertain or estimated. Read pg. 46 bottom 2 paragraphs
Read pg. 47 Table 2-5 Do Sample Problems on pg. 47 & 48 Rounding off Addition & Subtraction Problems Rounding off Multiplication & Division Problems This Scientific Notation is very helpful to determine the correct number of significant figures. Ex. 50 890 5.089 X 104 Ex. 0.005031 5.031 X 10-3 Ex. 74 621 000 7.462 1 X 107 Using Adding
& Subtracting Scientific Notation Using Multiplying & Dividing Scientific Notation
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Unformatted text preview: t f (x) = (continued on the other side) 1 1 [12] 5. Find the derivatives of the following functions (you don’ need to simplify the …nal answer, but you must show how t you calculate it): (a) f (x) = 3x2=3 1 + x3 (b) f (x) = ex (sin x cos x) q p (c) f (x) = x + x2 + 1 (d) f (x) = ln(1 + arctan x) 1 . x (a) Use the de…nition of derivative [6] 6. Let f (x) = f (x+h) f (x) h h!0 f 0 (x) = lim to …nd f 0 (2). (b) Find the equation of the tangent line T (x) to the graph of f (x) at the point (2; f (2)). [3] Bonus Question. Use the de…nition jxj = p x2 and the Chain Rule to show that if f (x) = j sin xj sin x cos x whenever sin x 6= 0. j sin xj What can you say about f 0 (x) when sin x = 0 ? then f 0 (x) = 2...
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Ask a homework question - tutors are online | 380 | 1,044 | {"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-2017-51 | latest | en | 0.80371 |
https://www.scienceforums.net/topic/115398-general-equation-of-the-circle/#comment-1061605 | 1,696,114,805,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510730.6/warc/CC-MAIN-20230930213821-20231001003821-00238.warc.gz | 1,069,855,725 | 19,960 | # General equation of the circle
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How we can derive the general equation of the circle?
X square + Y square + 2gx + 2fy + c = 0
Thanks & Regards,
Prashant S Akerkar
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Derive from what?
Another way to view this equation is
(x - f)2 +(y - g)2 = r2
This give a circle, centre (f, g) radius r.
Multiply it out and compare.
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You can probably get there from trigonometry. Given a centre position X and Y and radius r, you could integrate tan over 360 degrees...
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You don't need trigonometry to obtain this standard result, just good old Pythagoras.
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Thanks.
1 Is
X square + Y square + 2gx + 2fy + c = 0
and
(x-a) square + (y-b) square = r square
same?
2 What will be the general equation of the semicircle?.
Is it
(x-a) square + (y-b) square = r square / 2 ?
Thanks & Regards,
Prashant S Akerkar
##### Share on other sites
3 hours ago, prashantakerkar said:
1 Is
X square + Y square + 2gx + 2fy + c = 0
and
(x-a) square + (y-b) square = r square
same?
Yes
3 hours ago, prashantakerkar said:
2 What will be the general equation of the semicircle?.
Is it
(x-a) square + (y-b) square = r square / 2 ?
No that is a circle with a smaller radius (70.7% of the original radius)
For a semicircle, you need to define a line through the centre of the circle and add the condition that the included points have to be on one side of the line.
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Thanks.
What will be then the general equation of the Semicircle?
Thanks & Regards,
Prashant S Akerkar
##### Share on other sites
27 minutes ago, prashantakerkar said:
Thanks.
What will be then the general equation of the Semicircle?
Thanks & Regards,
Prashant S Akerkar
5 hours ago, prashantakerkar said:
Thanks.
1 Is
X square + Y square + 2gx + 2fy + c = 0
and
(x-a) square + (y-b) square = r square
same?
2 What will be the general equation of the semicircle?.
Is it
(x-a) square + (y-b) square = r square / 2 ?
Thanks & Regards,
Prashant S Akerkar
You need to show you understand more basic algebra before tackling the next stage.
Did you do the algebraic comparison I suggested or just guess?
A semicirircle does not have a single equation.
It is described by the intersection of a circle and a straight line along with some conditions.
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Thank you.
Will there be more than one general equations of the Semicircle ?
How to derive the general equation of a Semicircle by drawing a Semicircle figure?
Thanks & Regards,
Prashant S Akerkar
##### Share on other sites
You need to do more than nod wisely when someone tells you something.
Your responses tell me that you are not understanding what we are telling you.
I am not saying this to be nasty, but because you need to think before you answer.
In particular you need to answer the questions we are asking, they are all meant to help.
You are not doing this at the moment.
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• 3 weeks later...
On 7/25/2018 at 5:10 AM, prashantakerkar said:
Thank you.
Will there be more than one general equations of the Semicircle ?
How to derive the general equation of a Semicircle by drawing a Semicircle figure?
Thanks & Regards,
Prashant S Akerkar
What semi-circle? You have already been told that the equation of a circle with center at (a, b) and radius r is (x- a)^2+ (y- b)^2= r^2. The equation of the semi-circle consisting of the right half of that circle is (x- a)^2+ (y- b)^2= r^2 with the restriction that x> a. Similarly, the left half is the same but with x< a, the top is with y> b and the bottom y< b.
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Register a new account | 1,041 | 3,878 | {"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} | 3.859375 | 4 | CC-MAIN-2023-40 | latest | en | 0.823181 |
http://www.convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=British+pint&To=volume | 1,526,955,960,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864622.33/warc/CC-MAIN-20180522014949-20180522034949-00476.warc.gz | 365,788,292 | 3,877 | Partner with ConvertIt.com
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Conversion Result: ```British pint = 5.6826128524935E-04 volume (volume) ``` Related Measurements: Try converting from "British pint" to amphora (Greek amphora), bath (Israeli bath), beer gallon (English beer gallon), bushel (dry bushel), cc (cubic centimeters), dry quart, ephah (Israeli ephah), fifth, gallon, hekat (Israeli hekat), magnum, minim, nebuchadnezzar, noggin, omer (Israeli omer), register ton, Roman amphora, tou (Chinese tou), tun (English tun), UK quart (British quart), or any combination of units which equate to "length cubed" and represent capacity, section modulus, static moment of area, or volume. Sample Conversions: British pint = .01612589 bushel (dry bushel), .12500001 Canadian gallon, .00057337 displacement ton, .1290071 dry gallon, .75059375 fifth, .15011875 gallon, .18764844 jeroboam, .00020157 last, .00039119 load, .03002375 nebuchadnezzar, .14334122 omer (Israeli omer), .06450355 peck (dry peck), .00119142 pipe, .0053753 sack, .00056826 stere, 3.2 tea cup, 115.29 teaspoon, .06249988 UK peck (British peck), .04619038 vedro (Russian vedro), .75000031 wine bottle.
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Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 474 | 1,715 | {"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-2018-22 | longest | en | 0.641627 |
http://math.stackexchange.com/questions/214619/convolution-converges-in-infinity-norm | 1,469,385,717,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824133.26/warc/CC-MAIN-20160723071024-00133-ip-10-185-27-174.ec2.internal.warc.gz | 156,173,917 | 17,446 | # Convolution converges in infinity norm?
Assume $\phi$ to be a nonnegative continuous function on the real line with compact support. Also assume that integral of $\phi$ over $\mathbb{R}$ is normalized to $1$. Let $\phi_e(x) = \frac{1}{e}\phi\left(\frac{x}{e}\right)$.
I succeed to prove that $\phi_e$ is an approximation to the identity & the convoluton of $\phi_e$ with $L^p$ function $g$ converges to $g$ in $L^p$ norm.
What I really need help with is the following. Prove or disprove when $p=\infty$. I thought my proof didn't work for $p = \infty$, so I tried all night to find counterexamples. But I failed. Can anyone give an answer or idea for my question?
-
Hint: Uniform limits of continuous functions are continuous. – Jose27 Oct 16 '12 at 5:35
Take $g(x)=\chi_{(0,+\infty)}(x)$. Then $h_e(x):= g\star\phi_e(x)=\int_{-\infty}^{x/e}\phi(t)dt$. We have, if $x\in (0,\delta/k)$, that $$|h(x)-h_{k^{-1}}(x)|=\int_{xk}^{+\infty}\phi(t)dt\geq \int_{\delta}^{+\infty}\phi(t)dt.$$ We choose $\delta$ such that $\int_{\delta}^{+\infty}\phi(t)dt\geq 1/2$ (if it's not possible, consider $\psi(t)=\phi(-t)$.
As the measure of $(0,\delta/k)$ is not $0$, $\lVert h-h_{k^{-1}}\rVert\geq 1/2$.
Davide, how do you define $h(x)$？ – Topoguy Feb 13 '15 at 7:22 | 441 | 1,258 | {"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} | 3.53125 | 4 | CC-MAIN-2016-30 | latest | en | 0.720272 |
http://quant.stackexchange.com/questions/tagged/fixed-income?page=1&sort=active&pagesize=30 | 1,469,771,679,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829972.19/warc/CC-MAIN-20160723071029-00018-ip-10-185-27-174.ec2.internal.warc.gz | 199,313,823 | 26,761 | # Tagged Questions
Securities which obligate the borrower/issuer to make payments on a fixed schedule. Fixed income securities include sovereign, corporate and municipal bonds, corporate loans, and securitized lending (e.g., ABS). "Fixed" refers only to the schedule of obligatory payments, not the amount, and may ...
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Suppose quantified duration (like Macaulay duration with changing intervals) $Dur = \frac{\sum t_{i} PV_{i}}{\sum PV_{i}}$ and two funds having durations $D_{a}$ and $D_{b}$. You own them in the ...
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### Deriving the yield curve from the HJM dynamics
If I know that my model follows a no-arbitrage HJM model: $$df(\tau) = \left(\sigma(\tau)\int_0^{\tau}\sigma(u)du\right)dt +\sigma(\tau)dW_{\tau}$$ (where $\tau:=T-t$, ...
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### Derive an expression for the value of the asset as a function of time, V(t), t>=0
An investor deposits USD 300 in a bank account at time 0, reinvests all interest payments and continuously invests USD 300 per annum, until the total value of the deposits reaches USD 3312. At that ... | 2,946 | 13,048 | {"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": 1, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2016-30 | latest | en | 0.910788 |
https://www.physicsforums.com/threads/volume-of-air-flow-through-an-orifice-at-speed.967421/ | 1,555,927,148,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578551739.43/warc/CC-MAIN-20190422095521-20190422121521-00134.warc.gz | 769,298,996 | 17,600 | # B Volume of air flow through an orifice at speed
#### motolectric
Trying to figure out how much air can (or would) flow through an air intake on a vehicle.
I have a rectangular intake opening that is 13.52 square inches, the vehicle is traveling at 60 MPH for 1 hour.
Obviously more air is being forced into the intake due to the speed but I can't sort out how to calculate that.
M./
Related General Physics News on Phys.org
#### russ_watters
Mentor
Trying to figure out how much air can (or would) flow through an air intake on a vehicle.
I have a rectangular intake opening that is 13.52 square inches, the vehicle is traveling at 60 MPH for 1 hour.
Obviously more air is being forced into the intake due to the speed but I can't sort out how to calculate that.
Air intake for what? The backpressure from whatever the intake is to will determine the [change in] flow rate.
#### motolectric
Hi,
Essentially there is no (or little) back pressure as the housing has vents for any excess pressure.
I have attached 3 graphics.
The first shows the calculations for the amount of air/fuel mixture the engine would injest.
The 2nd shows the orifices I am dealing with.
The 2 red boxes represent the 2 intakes at the front of the hollow cast aluminum frame of a motorcycle.
The 54 mm red circle is the intake to the fuel injection system.
The housing has both the 3 vents seen to the lower left and a tubular vent to the rear that is not seen in this image.
The circles and rectangle at the top are from the conduit from the frame to the airbox housing and it is plain to see that they are large enough to flow enough air for the 54 mm fuel injection system.
The issue I am trying to resolve is whether there is excess pressure (and how much pressure) in the housing (it is sealed other than the inlet at the top and the vents I described.
I say there is and others say there isn't. This is as far as I have gotten.
On the bike, at 3,000 RPM in 4th gear you are doing 60 MPH so that is the common variable.
Some people encourage buyers of the vehicle to change to a fully open airbox stating that the OEM is "restrictive".
I claim that not only is it not restrictive but that in order to make use of the excess airflow the engineers put in the vents to use the excess airflow to cool the cylinders.
The air would not be expected to flow outward unless there was some pressurization inside the airbox enclosure.
Again, thanks for any tips or info on solving my problem.
WD./
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"Volume of air flow through an orifice at speed"
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Release 5.1
Contents:
Contents:
Computes the rolling unique kth largest value forward or backward of the current row.
For purposes of this calculation, two instances of the same value are treated at one value for k. So, if your dataset contains four rows with column values `10``9``9`, and `8`, then `KTHLARGESTUNIQUE` returns `9` for `k=2` and `8` for `k=3`.
`ROLLINGKTHLARGESTUNIQUE` computes the `KTHLARGESTUNIQUE` value across a defined window of values within a column.
• If an input value is missing or null, it is not factored in the computation. For example, for the first row in the dataset, the rolling calculation of previous values is undefined.
• The row from which to extract a value is determined by the order in which the rows are organized based on the `order` parameter.
If you are working on a randomly generated sample of your dataset, the values that you see for this function might not correspond to the values that are generated on the full dataset during job execution.
• Inputs:
• Required column name
• Required kth value, which is a positive integer
• Two optional integer parameters that determine the window backward and forward of the current row. The default integer parameter values are `-1` and `0`, which computes the rolling function from the current row back to the first row of the dataset.
• This function works with the following transforms:
For more information on a non-rolling version of this function, see KTHLARGESTUNIQUE Function.
## Basic Usage
Column example:
`derive type:single value:ROLLINGKTHLARGESTUNIQUE(myCol, 2)`
Output: Generates a new column containing the rolling second largest unique value in the `myCol` column from the first row of the dataset to the current one.
Rows before example:
`window value:ROLLINGKTHLARGESTUNIQUE(myNumber, 2, 3)`
Output: Generates the new column, which contains the rolling second largest unique value of the current row and the two previous row values in the `myNumber` column.
Rows before and after example:
`window value:value:ROLLINGKTHLARGESTUNIQUE(myNumber, 4, 3, 2)`
Output: Generates the new column, which contains the rolling fourth largest unique value of the two previous row values, the current row value, and the two rows after the current one in the `myNumber` column.
## Syntax and Arguments
`window value:value:ROLLINGKTHLARGESTUNIQUE(col_ref, rowsBefore_integer, rowsAfter_integer) order: order_col [group: group_col]`
ArgumentRequired?Data TypeDescription
col_refYstringName of column whose values are applied to the function
k_integerYinteger (positive)The ranking of the unique value to extract from the source column
rowsBefore_integerNintegerNumber of rows before the current one to include in the computation
rowsAfter_integerNintegerNumber of rows after the current one to include in the computation
For more information on the `order` and `group` parameters, see Window Transform.
### col_ref
Name of the column whose values are used to compute the function.
• Multiple columns and wildcards are not supported.
Usage Notes:
Required?Data TypeExample Value
YesString (column reference to Integer or Decimal values)`myColumn`
### k_integer
Integer representing the ranking of the unique value to extract from the source column. Duplicate values are treated as a single value for purposes of this function's calculation.
NOTE: The value for `k` must be an integer between 1 and 1,000 inclusive.
• `k=1` represents the maximum value in the column.
• If k is greater than or equal to the number of values in the column, the minimum value is returned.
• Missing and null values are not factored into the ranking of `k`.
Usage Notes:
Required?
Data Type
Example Value
YesInteger (positive)```4 ```
### rowsBefore_integer, rowsAfter_integer
Integers representing the number of rows before or after the current one from which to compute the rolling function, including the current row. For example, if the first value is `5`, the current row and the four rows after it are used in the computation. Negative values for `k` compute the rolling average from rows preceding the current one.
• `rowBefore=1` generates the current row value only.
• `rowBefore=-1` uses all rows preceding the current one.
• If `rowsAfter` is not specified, then the value `0` is applied.
• If a `group` parameter is applied, then these parameter values should be no more than the maximum number of rows in the groups.
Usage Notes:
Required?Data TypeExample Value
NoInteger`4`
## Examples
### Example - ROLLINGKTHLARGEST functions
This example describes how to use the following rolling computational functions:
• `ROLLINGKTHLARGEST` - computes the kth largest value from a rolling window of rows before and after the current row. Duplicate values are treated as having the same k values. See ROLLINGKTHLARGEST Function.
• `ROLLINGKTHLARGESTUNIQUE` - computes the unique kth largest value from a rolling window of rows before and after the current row. Duplicate values are treated as having different k values. See ROLLINGKTHLARGESTUNIQUE Function.
The following dataset contains daily counts of server restarts across three servers over the preceding week. High server restart counts can indicate poor server health. In this example, you are interested in knowing for each server the rolling highest and second highest count of restarts per server over the previous week.
Source:
DateServerRestarts
2/21/18s014
2/21/18s020
2/21/18s030
2/22/18s014
2/22/18s021
2/22/18s032
2/23/18s012
2/23/18s023
2/23/18s034
2/24/18s011
2/24/18s020
2/24/18s032
2/25/18s015
2/25/18s020
2/25/18s034
2/26/18s011
2/26/18s022
2/26/18s031
2/27/18s011
2/27/18s022
2/27/18s032
Transform:
First, you want to maintain the row information as a separate column. Since data is ordered already by the `Date` column, you can use the following:
`derive type:single value:ROWNUMBER() as:'entryId'`
Use the following function to compute the rolling kth largest value of server restarts per server over the previous week. In this case, you can use the `ROLLINGKTHLARGEST` function, setting k=1. Uniqueness doesn't matter for calculating the highest value:
`derive type: multiple value: rollingkthlargest(Restarts, 1, 6, 0) group: Server order: Server as: 'rollingkthlargest_1'`
Use the following function to compute the rolling second highest value. In this case, you can use `ROLLINGKTHLARGESTUNIQUE`
`derive type: multiple value: rollingkthlargestunique(Restarts, 2, 6, 0) group: Server order: Server as: 'rollingKthLargestUnique_2'`
Results:
entryIdDateServerRestartsrollingKthLargestUnique_2rollingkthlargest_Restarts
32/21/18s02000
62/22/18s02101
92/23/18s02313
122/24/18s02013
152/25/18s02013
182/26/18s02223
212/27/18s02223
42/21/18s03000
72/22/18s03202
102/23/18s03424
132/24/18s03224
162/25/18s03424
192/26/18s03124
222/27/18s03224
22/21/18s01444
52/22/18s01444
82/23/18s01224
112/24/18s01124
142/25/18s01545
172/26/18s01145
202/27/18s01145
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# Would 8ozs be too much for a 2 month old?
I feed her 6 now, but she still seems as though she is hungry. I burp her well and give her a binki to see if it will curve it but it doesnt. So I give her two more ozs and she is fine. Is that two much? She is 2 1/2 months.
Asked by SabrenaLeigh at 8:40 AM on Nov. 9, 2009 in Babies (0-12 months)
Level 22 (14,998 Credits)
• it's only too much if she pukes it up. My son was on 8oz by that age too
Answer by Zakysmommy at 8:45 AM on Nov. 9, 2009
• To tell how much formula to use a day take your baby's weight in pounds and multiply by 2.5. If your 2 month old baby weighs 10 pounds, 10 X 2.5 = 33 ounces for the day. If you give 6 bottles a day then you sould be giving a little less than 6 ounces a bottle.
http://www.babycenter.com/0_how-to-tell-how-much-formula-your-baby-needs_9136.bc
Babies have a strong suckling need. They spend hours at the breast. That's why you can't let babies eat as much as they want at a bottle, they can overfeed and become overweight. At least overweight for what they would have been if they were breastfeed. This can lead to a lifetime of weight problems.
You probably heard breastfed babies are healthier and this is one of the lifelong health benefits breastfed babies have. I think moms should understand as much as possible so they can minimize the disadvantages of formula feeding.
Answer by Gailll at 9:00 AM on Nov. 9, 2009
• Oh Good GOD, let your baby eat as much as she wants to eat. My son was formula fed from the time he was 1 month old. He had 8 ozs of formula from the time he was 2 months old. When he was 3 months he had enough cereal in it to spoon feed. He was healthy, Yes he was a very fat and adorable baby. He weighed 22lbs when he was 6 months old. He did NOT use the bottle as a need to suck, he had a pacifier. The bottle was FOOD only, and he didn't eat at night. Hate to break the mold, but weight problems are due to genetics and eating habbits that usually come from the parents, AND inactivity. My son, who was sooo fat as a baby, is 6'1" and is only 135lbs. We dont' have weight issues in my family. If your daughter wants 8oz of formula, then give it to her. She will tell you if it's too much
Answer by Zakysmommy at 9:07 AM on Nov. 9, 2009
• Personally I think it's way too much at a time for a baby of any age.... 6 oz at a time should be plenty. I would change her to a slower flow nipple so that her tummy has time to know that it's full.
I think that when people say she can have as much as she wants as long as she doesn't throw up... that's just stupid. I mean, really think of how OVER full you would have to be to throw up... she can be full and even too full and keep everything down. (I watched my nephew take 14 oz at a time because of this theory)
Your baby should take around 6 6 oz bottles/day at this age... but if she is acting like she is hungry then feeding her smaller amounts more often would probably be better for her..... try 8 4 oz bottles in a day (more similar to a breastfed baby) and she might be happier.
Answer by AmiJanell at 9:34 AM on Nov. 9, 2009
• its on the slowest flow nipple thats made so that one is out.
Answer by SabrenaLeigh at 9:38 AM on Nov. 9, 2009
• Hun, if your daughter wants 8oz then give it to her. Or maybe just try 7oz. It won't hurt her. Some babies are more hungary than others. my daughter NEVER took more than 6oz, where my son was a pig lol
Answer by Zakysmommy at 9:41 AM on Nov. 9, 2009
• If she wants it then give it. The CRAP that Gailll quotes is bullshit. That is a GUIDELINE for feeding formula, not the gospel. Like she would know, anyway.
Mine took 8oz every 2 hours starting at 6 weeks. They didn't puke it back up, they weren't hungry for more (until they started taking 10oz at 5 months) and they are certainly not having any health issues due to their taking of formula.
You can't overfeed a baby. If they take too much, they will puke it back up. You certainly can't make them swallow it if they don't want it. Yes they have a strong sucking need....that's why they either use pacifiers or their thumbs. You can always let a baby have as much as it wants with a bottle! They're hungry!!! Why should a formula fed baby go hungry just because Gailll's sources say it "shouldn't" eat over a certain amount.
Feed on demand. Breast or bottle. And ignore gailll.
Answer by Anonymous at 9:50 AM on Nov. 9, 2009
• I BF my daughter until she was 6 months old. I would let her nurse as often as she wanted. Who am I to say when she's full ya know? She's gonna be 8 months in a few days and now she prefers actual food over formula. She eats at least 3 meals a day and only takes about 2 bottles a day on the average. She's a happy and healthy girl.
Answer by piercedbeauty21 at 10:36 AM on Nov. 9, 2009
• i asked my doc the same thing they said let them eat as much as they want....the doc said shell letcha know when shes done.
Answer by photogrypher at 11:23 AM on Nov. 9, 2009
• Ignore Gaill. All of what she said is just nonsense. Your baby will eat as much as she feels she needs. Sometimes putting infant cereal in the baby formula will fill her up with the six ounces. If your baby does infact require eight ounces of formula, then that's perfectly fine. Every baby requires a different amount of formula. Talk with her Pedi if using cereal would help more then more ounces in the bottle.
Answer by JazzlikeMraz at 12:12 PM on Nov. 9, 2009
Join CafeMom now to contribute your answer and become part of our community. It's free and takes just a minute. | 1,528 | 5,670 | {"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-2018-13 | latest | en | 0.988239 |
https://www.digglicious.com/tips-for-writing/what-are-the-7-types-of-lines/ | 1,680,200,636,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00300.warc.gz | 804,630,272 | 12,717 | ## What are the 7 types of lines?
There are many types of lines: thick, thin, horizontal, vertical, zigzag, diagonal, curly, curved, spiral, etc. and are often very expressive.
## How do you describe a function?
A function is a relation in which each possible input value leads to exactly one output value. We say “the output is a function of the input.” The input values make up the domain, and the output values make up the range.
## What is origin of graph?
On a two-dimensional graph it is where the X axis and Y axis cross, such as on the graph here: Sometimes written as the letter O. In three dimensions it is the point (0, 0, 0) where the x, y and z axis cross: Cartesian Coordinates.
## What is a real life example of a line?
Real-world examples of line segments are a pencil, a baseball bat, the cord to your cell phone charger, the edge of a table, etc. Think of a real-life quadrilateral, like a chessboard; it is made of four line segments. Unlike line segments, examples of line segments in real life are endless.
## Where is the origin of a shape?
Usually, you will be asked to rotate a shape around the origin, which is the point (0, 0) on a coordinate plane. You can rotate shapes 90, 180, or 270 degrees around the origin using three basic formulas.
## How do you graph 0 0 on a number line?
On a number line graph, we would start with an open circle on 0, because 0 IS NOT included in the solution. Draw a line extending to the left, indicating that x can be any value to the left of 0. On ax-y- grid, we would have a vertical line to represent x = 0, but as 0 IS NOT included, the line would be dotted.
## What are the different types of graph?
Popular graph types include line graphs, bar graphs, pie charts, scatter plots and histograms. Graphs are a great way to visualize data and display statistics. For example, a bar graph or chart is used to display numerical data that is independent of one another.
## What are characteristics of a graph?
Interpret key features of a graph—the intercepts, maximums, minimums, and the intervals when the function is increasing or decreasing—in terms of a situation. Understand and be able to use the terms “horizontal intercept,” “vertical intercept,” “maximum,” and “minimum” when talking about graphs of functions.
## What are 3 things a graph must have?
Essential Elements of Good Graphs:
• A title which describes the experiment.
• The graph should fill the space allotted for the graph.
• Each axis should be labeled with the quantity being measured and the units of measurement.
• Each data point should be plotted in the proper position.
• A line of best fit.
## How do you tell if a graph represents a function?
Use the vertical line test to determine whether or not a graph represents a function. If a vertical line is moved across the graph and, at any time, touches the graph at only one point, then the graph is a function. If the vertical line touches the graph at more than one point, then the graph is not a function.
## How do you introduce a graph?
To catch your audience’s attention from the very beginning, you can use the following phrases for introduction:
1. Let me show you this bar graph…
2. Let’s turn to this diagram…
3. I’d like you to look at this map…
4. If you look at this graph, you will notice…
5. Let’s have a look at this pie chart…
## What are the 4 sections of a graph called?
The intersecting x- and y-axes divide the coordinate plane into four sections. These four sections are called quadrants. Quadrants are named using the Roman numerals I, II, III, and IV beginning with the top right quadrant and moving counter clockwise. Locations on the coordinate plane are described as ordered pairs.
## What 5 things do all graphs need?
There are five things about graph that need our attention when designing graphs:
• visual structures,
• axes and background,
• scales and tick marks,
• grid lines,
• text.
## Are lines always straight?
A line can be straight or curved. In geometry, the word line means a straight line. A straight line is the shortest distance between two points. A straight line is the line traced by a point moving in a direction that does not change.
## How do you explain a graph?
In math, a graph can be defined as a pictorial representation or a diagram that represents data or values in an organized manner. The points on the graph often represent the relationship between two or more things.
## Whats is a function?
A technical definition of a function is: a relation from a set of inputs to a set of possible outputs where each input is related to exactly one output. We can write the statement that f is a function from X to Y using the function notation f:X→Y. …
## How do you describe a function from a graph?
Defining the Graph of a Function. The graph of a function f is the set of all points in the plane of the form (x, f(x)). We could also define the graph of f to be the graph of the equation y = f(x). So, the graph of a function if a special case of the graph of an equation.
## What are the 4 parts of a graph called?
The following pages describe the different parts of a bar graph.
• The Title. The title offers a short explanation of what is in your graph.
• The Source. The source explains where you found the information that is in your graph.
• X-Axis. Bar graphs have an x-axis and a y-axis.
• Y-Axis.
• The Data.
• The Legend.
## What is an example of a non function?
The equations y=±√x and x2+y2=9 are examples of non-functions because there is at least one x-value with two or more y-values.
## What is the origin of a straight line?
The equation of a straight line is y = mx + c. If c, which is the intercept, is = 0, this means that y = mx. This is the origin of that line. If the coordinates of a line are given as (x1, y1) = (4, 12) and (x2, y2) = (12, 34), this means that the line does not pass through the origin at any point all.
## What is the meaning of line?
a continuous extent of length, straight or curved, without breadth or thickness; the trace of a moving point. something arranged along a line, especially a straight line; a row or series: a line of trees. a number of persons standing one behind the other and waiting their turns at or for something; queue.
## What are the key components of a graph?
CARMALT – Basic parts of graphs
5 components of a good graph are: TITLE, AXES, INCREMENTS, LABELS, SCALE
tells what graph is about TITLE
changing variable is known as _____ INDEPENDENT
Dependent variable is on which axis that is vertical? Y
## Is the origin of a graph always 0 0?
In a Cartesian coordinate system, the origin is the point where the axes of the system intersect. The coordinates of the origin are always all zero, for example (0,0) in two dimensions and (0,0,0) in three.
## What are 5 ways to describe a graph?
Describing language of a graph
• UP: increase / rise / grow / went up / soar / double / multiply / climb / exceed /
• DOWN: decrease / drop / fall / decline / plummet / halve / depreciate / plunge.
• UP & DOWN: fluctuate / undulated / dip /
• SAME: stable (stabilised) / levelled off / remained constant or steady / consistent.
## What is 0 0 on a graph?
The center of the coordinate system (where the lines intersect) is called the origin. The axes intersect when both x and y are zero. The coordinates of the origin are (0, 0).
## What are the 4 main features of a graph?
Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity.
## What is a function in your own words?
A function is a relation that maps a set of inputs, or the domain, to the set of outputs, or the range. Note that for a function, one input cannot map to more than one output, but one output may be mapped to more than one input.
## What happens when an equation is 0 0?
If the elimination produces the equation 0=0, then the two equations are for the same line. This means that there are an infinite number of solutions. | 1,884 | 8,078 | {"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} | 4.34375 | 4 | CC-MAIN-2023-14 | latest | en | 0.915905 |
https://essayzilla.org/a-researcher-records-the-following-data-for-the-number-of-times-an-interviewer-is-interrupted-during-a-series-of-interviews-0-0-3-2-5-3-0-8-0-2-1-and-1-is-the-mode-equal-to-the-median-in/ | 1,653,140,769,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662539101.40/warc/CC-MAIN-20220521112022-20220521142022-00274.warc.gz | 277,069,617 | 9,653 | Select Page
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## A researcher records the following data for the number of times an interviewer is interrupted during a series of interviews: 0, 0, 3, 2, 5, 3, 0, 8, 0, 2, 1, and 1. Is the mode equal to the median in this example?.
A researcher records the following data for the number of times an interviewer is interrupted during a series of interviews: 0, 0, 3, 2, 5, 3, 0, 8, 0, 2, 1, and 1. Is the mode equal to the median in this example?…
A researcher records the following data for the number of different food items chosen by seven participants in a buffet-style setting: 1, 6, 2, 5, 4, 3, and 7. Is the mean equal to the median in this example?…
A researcher measures the mean time (in seconds) it takes two groups of children to complete an activity task. She finds that Group A (M = 22 seconds) completed the task more quickly than Group B (M = 36 seconds).She then computes a weighted mean for both groups combined and calculates = 26. Based on the information provided, which group had a larger sample size?…
The ________ is the value that occurs most often or at the highest frequency in a distribution…
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6. 24/7 Customer Support: At Eminence Papers, we have put in place a team of experts who answer all customer inquiries promptly. The best part is the ever-availability of the team. Customers can make inquiries anytime. | 716 | 3,195 | {"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} | 2.546875 | 3 | CC-MAIN-2022-21 | latest | en | 0.933154 |
http://mathhelpforum.com/differential-geometry/167730-how-does-cos-sin-affect-convergence-series.html | 1,524,215,796,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937193.1/warc/CC-MAIN-20180420081400-20180420101400-00213.warc.gz | 205,675,528 | 11,678 | # Thread: How does cos/sin affect convergence of series
1. ## How does cos/sin affect convergence of series
{sum 1, inf} cos(r+1)/r^2+1
or
{sum 1, inf} sin(r^2 +1)/r^2+1
Do i try get rid of the cos/sine? if so how
Or is there a general rule im missing?
Thanks
2. Originally Posted by sj247
{sum 1, inf} cos(r+1)/r^2+1
or
{sum 1, inf} sin(r^2 +1)/r^2+1
Do i try get rid of the cos/sine? if so how
Or is there a general rule im missing?
Thanks
Use Dirichlet's test to prove that both these sums converge. Note that since $\displaystyle \displaystyle \sum \cos(r)$ is bounded and $\displaystyle \displaystyle \frac{1}{r^2+1}\to 0$ for example.
3. Originally Posted by sj247
{sum 1, inf} cos(r+1)/r^2+1
or
{sum 1, inf} sin(r^2 +1)/r^2+1
Do i try get rid of the cos/sine? if so how
Or is there a general rule im missing?
Thanks
Because is...
$\displaystyle \displaystyle \frac{|\cos (r+1)|}{r^{2}+1} < \frac{1}{r^{2}}$ (1)
... and ...
$\displaystyle \displaystyle \frac{|\sin (r^{2}+1)|}{r^{2}+1} < \frac{1}{r^{2}}$ (2)
... and the series $\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{1}{r^{2}}$ converges, also the two series You proposed converge...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
4. Originally Posted by chisigma
Because is...
$\displaystyle \displaystyle \frac{|\cos (r+1)|}{r^{2}+1} < \frac{1}{r^{2}}$ (1)
... and ...
$\displaystyle \displaystyle \frac{|\sin (r^{2}+1)|}{r^{2}+1} < \frac{1}{r^{2}}$ (2)
... and the series $\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{1}{r^{2}}$ converges, also the two series You proposed converge...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Maybe I'm being pedantic, but I don't like to use that here. The fact that the absolute convergence of a series implies the convergence of the series is only true since the real (or complex numbers) is a complete metric space and since this seems like a calculus question I'm not sure if that's ok. I don't know if Dirichlet's test is much easier though :|
5. So in my first example {sum 1, inf} cos(r+1)/r^2+1
(An)=1/(r^2 +1)
(Bn)= cos(r+1)
i have to show (An) is decreasing & {sum}(Bn) is bounded ?
Not sure how to go about showing {sum}(Bn) is bounded. Although obviously it must as (Bn) must be between 1, -1 for any n.
6. Originally Posted by Drexel28
Maybe I'm being pedantic, but I don't like to use that here. The fact that the absolute convergence of a series implies the convergence of the series is only true since the real (or complex numbers) is a complete metric space and since this seems like a calculus question I'm not sure if that's ok. I don't know if Dirichlet's test is much easier though :|
I'm not sure of the above: Dirichlet's test's proof does use completeness of the complex (real) field in a very
clear and decisive way (at least the proofs I know of), whereas absolute convergence ==> convergence is true
without using (at least directly, in one of the proofs I know of) completeness...
Tonio
7. Originally Posted by tonio
I'm not sure of the above: Dirichlet's test's proof does use completeness of the complex (real) field in a very
clear and decisive way (at least the proofs I know of), whereas absolute convergence ==> convergence is true
without using (at least directly, in one of the proofs I know of) completeness...
Tonio
I was just making a remark. There are proofs of Dirichlet's theorem which don't appeal directly to completeness. That said, I do agree that the most common one (the one using summation by parts) does. I guess it's all kind of irrelevant since everything (and I mean this in a very, very loose sense) depends upon completeness if one goes far enough back.
8. Originally Posted by sj247
...i have to show (An) is decreasing & {sum}(Bn) is bounded?... not sure how to go about showing {sum}(Bn) is bounded... although obviously it must as (Bn) must be between 1, -1 for any n...
We now consider the case ot the sums $\displaystyle \displaystyle \sum_{k=1}^{n} \sin k$ and $\displaystyle \displaystyle \sum_{k=1}^{n} \cos k$ and that is valid for all the cases...
$\displaystyle \displaystyle \sum_{k=0}^{n-1} a\ r^{k} = a\ \frac{1-r^{n}}{1-r}$ (1)
... and setting $\displaystyle r=a= e^{i}$ we obtain in some steps...
$\displaystyle \displaystyle S= \sum_{k=1}^{n} e^{i k} = e^{i}\ \frac{1-e^{i n}}{1-e^{i}}$
$\displaystyle \displaystyle = \frac{e^{i}}{2} \ \frac{1- \cos 1 - \cos n + \cos (n-1) + i\ \{\sin 1 - \sin n + \sin (n-1) \}} {1 - \cos 1}$ (2)
Without proceeding further it is clear from (2) that both the quantities...
$\displaystyle \displaystyle \sum_{k=1}^{n} \cos k = \mathcal{R} \{S\}$
$\displaystyle \displaystyle \sum_{k=1}^{n} \sin k = \mathcal{I} \{S\}$
... are bounded...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$ | 1,492 | 4,799 | {"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} | 3.765625 | 4 | CC-MAIN-2018-17 | latest | en | 0.788064 |
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The hidden Golden Ratio
An insight into the golden ratio and where it is hidden
by
Max Jeanneret
on 14 September 2012
Report abuse
Transcript of The hidden Golden Ratio
The Hidden Golden Ratio The golden ratio or Phi is a rectangular angle that is said to be more visually attractive. It is when the length is divided by the width and it equals 1.618. It can continually create more triangles within itself to create more rectangles which include the golden ratio. Another characteristic or the golden ratio is that it can fit the spiral in itself. (refer to image) What is the golden Ratio? Half of the base, the slant height, and the height from the vertex to the center create a right triangle. When that half of the base equal to one, the slant height would equal to the value of Phi and the height would equal to the square root of Phi.
An easier way to find the Golden ratio in the Great Pyramid though is to simply half the base then divide it by the pyramids slanting height. The Great Pyramid of Giza
and The Golden Ratio The Golden Ratio By Max 7W An insight into The: 1.6=chocolate block The Chocolate block Many companies and businesses apply the golden ratio into their products to make them more attractive, one of these being Cadbury chocolate blocks. After taking measurements of a block of chocolate it was easily seen that it closely followed the Golden ratio Why blame the chocolate? The measurements showed the block of chocolates dimensions were roughly 3.8 by 6.2. dividing them equaled the rounded of number of 1.6. The formula would be:
6.2/3.8
=1.6315 The answer is no. The reason is because There are many more hidden Phi around the globe hidden any where from buildings, to the food you eat, to nature, to faces that you see and maybe even the music you hear. All these factors can be hiding the Golden ratio Is that all there is to the golden ratio? But you are going to have to find them yourself The golden ratio equation:
P=SH/ (B/2)
P=356/ (440/2)
P=356/220
P=1.61818181818... Actual Phi =1.618 033988... Bibliography:
Anon., 2012. The beauty pf the golden ratio. [Online] Available at: http://library.thinkquest.org/trio/TTQ05063/phibeauty4.htm [Accessed 03 September 2012].
Anon., 2012. Google images. [Online] Available at: http://www.google.com.au/imghp?hl=en&tab=wi [Accessed 06 September 2012].
Anon., n.d. The golden number. [Online] Available at: http://www.goldennumber.net/ [Accessed 01 September 2012].
Anon., n.d. Measurements of the Great Pyramid. [Online] Available at: http://www.repertorium.net/rostau/measures.html [Accessed 05 September 2012].
Anon., n.d. Vodolstamuseum. [Online] Available at: http://www.valdostamuseum.org/hamsmith/Gpyr.html [Accessed 05 September 2012]. Did the Egyptians ever Realize? taking into account how old the Egyptian race is it seems highly unlikely that they knew that they were building it to the scale of 1.618181818 but it is seen that either it was a tremendous coincidence or the Egyptians knew what they were doing to an extent. It is most likely that the Egyptians simply drew a diagram of sorts and found that it looked perfect before using their mathematical abilities to work out how to achieve this masterpiece It seems almost logical to say that the Egyptians actually knew what they were doing when the build the mighty great pyramid of geeza So that's the Prezi So that's the Prezi
Full transcript | 926 | 3,858 | {"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-2019-04 | latest | en | 0.93516 |
http://thibaultlanxade.com/general/use-the-drop-down-menus-to-complete-the-proof-by-the-unique-line-postulate-you-can-draw-only-one-segment-ac-bc-cd-using-the-definition-of-perpendicular-bisector-reflection-symmetry-reflect-bc-over-l-by-the-definition-of-reflection-c-is-the-ima | 1,712,993,890,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816586.79/warc/CC-MAIN-20240413051941-20240413081941-00022.warc.gz | 30,238,976 | 7,565 | Q:
# Use the drop-down menus to complete the proof. By the unique line postulate, you can draw only one segment, AC BC CD. Using the definition of perpendicular bisector reflection symmetry, reflect BC over l. By the definition of reflection, C is the image of itself and ABC is the image of B. Since reflections preserve angles orientation length, AC = BC.
Accepted Solution
A:
The first one is; BC
The second one is; Reflection
The third one is; A
The fourth one is; Length | 111 | 478 | {"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-2024-18 | latest | en | 0.838629 |
https://spectre-code.org/tutorial_domain_creation.html | 1,620,591,969,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989012.26/warc/CC-MAIN-20210509183309-20210509213309-00613.warc.gz | 554,065,684 | 6,539 | Domain Creation
## Domain Decomposition
The spatial domain on which the solution is computed is divided into non-overlapping cells. For 2D and 3D domains, we use curvilinear quadrilateral and curvilinear hexahedral elements, respectively. The specification of a particular spatial domain is given by the DomainCreator. DomainCreators specify the properties of the initial elements (before AMR takes place), which are then used by Domain to create Blocks. Each Block holds a CoordinateMap (described below) as well as information about its neighbors. For geometrical domains, (e.g. rectilinear, spherical) there are a few shortcuts that can be used such that a user does not need to specify by hand the map and neighbor information for each Block; these methods are explained below.
# CoordinateMaps
Each Block in the Domain must hold a CoordinateMap which describes how to map the logical cube (a domain that goes [-1. 1] in each dimension) to the curvilinear hexahedral element the Block describes. The CoordinateMap also provides the jacobian of the mapping.
## Shortcuts for CoordinateMaps
For spherical domains, Wedge<3> implements the cubed-sphere map, and there exists the method sph_wedge_coordinate_maps in Domain/DomainHelpers.hpp to quickly construct six of these maps at once. For rectilinear multicube domains, Affine and Equiangular map the logical cubes to cartesian cubes. The method maps_for_rectilinear_domains in DomainHelpers allows the user to obtain the CoordinateMaps for all of the Blocks in such a domain at once. These can both be used to provide the map arguments to the Domain constructor.
# Boundary Information
Each Block must know which of the other Blocks in the Domain are its neighbors and which of its logical directions points to an external boundary.
## Shortcuts for Boundary Information
A quick way to encode the neighbor and boundary information for the blocks in a domain is through our convention of numbering and ordering the corners of the blocks in a well-defined way. The corner numbering scheme is described in the OrientationMap tutorial. For spherical domains, DomainHelpers has the methods corners_for_radially_layered_domains and corners_for_biradially_layered_domains, which provides the proper corner numberings for the maps obtained from the method sph_wedge_coordinate_maps. These methods are used in the Shell and Sphere DomainCreators. For rectilinear multicube domains, DomainHelpers has the methods corners_for_rectilinear_domains, which provides the proper corner numberings for the maps obtained from the method maps_for_rectilinear_domains.
# Creating Rectilinear Domains with Shortcuts:
The construction of a rectilinear domain begins with the specification of the total extents of the domain in each cartesian direction, in terms of the number of blocks. These extents are held in an Index object. For an illustrative example, we will explain how to construct a cubical domain which has an extent of two blocks in each dimension:
A 2x2x2 domain.
The first step is to generate the corner numbering for this Domain. For the purposes of this example, we will construct all blocks with their logical directions aligned with one another. As each block must also have an associated block id, we must be aware of the order in which the corners for each block are constructed.
The algorithm corners_for_rectilinear_domains always begins with the block located in the lowest cartesian corner of the domain. The second block is its immediate neighbor in the $$+x$$ direction, and so on until the block in this row with the highest $$x$$ coordinate is reached. The next block is the the immediate neighbor of the lowest corner block in the $$+y$$ direction, and then continues through the neighboring blocks in the $$+x$$ as before. This is the same order in which the global corners numbers are assigned to the vertices of the blocks.
With global corners.
The block corners generated by corners_for_rectilinear_domains are then:
{0,1,3,4,9,10,12,13},
{1,2,4,5,10,11,13,14},
{3,4,6,7,12,13,15,16},
{4,5,7,8,13,14,16,17},
{9,10,12,13,18,19,21,22},
{10,11,13,14,19,20,22,23},
{12,13,15,16,21,22,24,25},
{13,14,16,17,22,23,25,26}
What remains is to specify the CoordinateMaps that each Block will hold. This is handled by maps_for_rectilinear_domains and currently supports both Affine and Equiangular mappings. The coordinate extents of each map is set by the argument to block_demarcations. For a 2x2x1 domain, the call to maps_for_rectilinear_domains could be:
const std::vector<
affine_maps_3d = maps_for_rectilinear_domains<Frame::Inertial>(
Index<3>{2, 2, 1},
{{0.0, 0.5, 2.0}, {0.0, 1.0, 2.0}, {-0.4, 0.3}}},
{Index<3>{}}, {}, false);
For this choice of arguments we obtain the maps for a domain that extends from 0.0 to 2.0 in the $$x$$-direction, from 0.0 to 2.0 in the $$y$$ -direction, and from -0.4 to 0.3 in the $$z$$ direction.
With the corners and maps in hand, we can pass these as arguments to the Domain constructor.
## Non-trivial Domain Example
The aforementioned functions can also take an additional argument to exclude blocks from a domain. For this one needs to know the Index<Dim> for each block they wish to exclude from the domain. For the 2x2x2 domain example, the block indices are:
{0,0,0},
{1,0,0},
{0,1,0},
{1,1,0},
{0,0,1},
{1,0,1},
{0,1,1},
{1,1,1}
In this example, we construct a domain with topology $$S^3$$, which begins with the net for a tesseract. We begin by specifying that the domain extents are three blocks along the $$x$$ and $$y$$ directions, and four blocks along the $$z$$ direction. In addition, we exclude the following block indices:
{0,0,0}, //first block
{1,0,0}, //second block
{2,0,0}, //third block
{0,1,0}, //fourth block
{2,1,0}, //sixth block
{0,2,0}, //seventh block
{1,2,0}, //eight block
{2,2,0}, //ninth block
{0,0,1}, //10th block
{1,0,1}, //11th block
{2,0,1}, //12th block
{0,1,1}, //13th block
{2,1,1}, //15th block
{0,2,1}, //16th block
{1,2,1}, //17th block
{2,2,1}, //18th block
{0,0,2}, //19th block
{2,0,2}, //21st block
{0,2,2}, //25th block
{2,2,2}, //27th block
{0,0,2}, //28rd block
{1,0,2}, //29th block
{2,0,2}, //30th block
{0,1,2}, //31th block
{2,1,2}, //33th block
{0,2,2}, //34th block
{1,2,2}, //35th block
{2,2,2}, //36th block
Alternatively, we can exclude none of the blocks in this way and instead selectively copy the block corners from the returned vector into a new vector that only contains the corners for the desired blocks, if one prefers to work with single-number array indices as opposed to tuples.
Either way, we end up with a vector of block corners corresponding to the following diagram:
A numbered tesseract net.
The maps are obtained similarly. In this case we suggest using the Equiangular maps for this Domain since they adapt the logical coordinates to the angular coordinates on a sphere. We also need to identify corresponding faces with each other to obtain a domain of topology $$S^3$$, so we also need to supply the constructor of Domain with PairsOfFaces. For the $$S^3$$ domain, these faces are:
//Folding highest cube downward:
{{53,54,69,70},{53,54,49,50}},
{{53,57,69,73},{53,57,52,56}},
{{57,58,61,62},{57,58,73,74}},
{{54,58,70,74},{54,58,55,59}},
//Folding cross cubes together:
{{54,55,38,39},{54,50,38,34}},
{{58,59,42,43},{58,62,42,46}},
{{53,52,37,36},{53,49,37,33}},
{{57,56,41,40},{57,61,41,45}},
//Folding second lowest cube upward:
{{38,42,39,43},{38,42,22,26}},
{{33,34,37,38},{21,22,37,38}},
{{36,37,40,41},{21,37,25,41}},
{{41,42,45,46},{41,42,25,26}},
//Folding bottom cube around domain:
{{33,34,49,50},{21,22,5,6}},
{{39,43,55,59},{22,26,6,10}},
{{45,46,61,62},{25,26,9,10}},
{{36,40,52,56},{21,25,5,9}},
{{5,6,9,10},{69,70,73,74}}
std::vector
Index
Definition: Index.hpp:31
std::array
std::unique_ptr | 2,200 | 7,817 | {"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": 1, "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} | 2.8125 | 3 | CC-MAIN-2021-21 | longest | en | 0.874558 |
https://web2.0calc.com/questions/how-many-ways-can-you-arrange-8-pieces-of-fruit | 1,670,121,193,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710953.78/warc/CC-MAIN-20221204004054-20221204034054-00116.warc.gz | 628,520,526 | 6,105 | +0
# How many ways can you arrange 8 pieces of fruit?
0
1052
4
How many ways can you arrange 8 pieces of fruit?
Jul 8, 2014
#2
+576
+8
8!=8*7*6*5*4*3*2*1=40,320.
Think of there being 8 different places that you can put a single piece of fruit. when you first start pick a spot, there are 8 different options you have for putting a piece there. Once you place it you move to the second spot, you now have seven pieces of fruit left to pick from because one was already used. Go to the third spot, now you only have 6 pieces to pick from. This continues until you run out of fruit and places to put it.
Jul 8, 2014
#1
+124676
+8
Assuming that they are all different types.....8! = 40320 ways....!!!!
Jul 8, 2014
#2
+576
+8
8!=8*7*6*5*4*3*2*1=40,320.
Think of there being 8 different places that you can put a single piece of fruit. when you first start pick a spot, there are 8 different options you have for putting a piece there. Once you place it you move to the second spot, you now have seven pieces of fruit left to pick from because one was already used. Go to the third spot, now you only have 6 pieces to pick from. This continues until you run out of fruit and places to put it.
jboy314 Jul 8, 2014
#3
+124676
+5
Let me add one thing to this answer......this is assuming a "linear" arrangement. It the fruit were arranged in a circle, the answer would be just 7! = 5040.
To see this, note that I could "anchor" one piece of fruit in any position on the circle. Then, I have seven ways to choose the next piece of fruit, 6 ways to choose the next, etc.
Jul 8, 2014
#4
+118139
+5
That is assuming that the fruit is all differerent and arranged in a straight line. 8!
If it was arranged around a circular platter then it would be 7!
If they were just randomly placed in the universe then the number of possibilities would be infinite!
I'm sorry - I just realised that Chris said some of this already.
Jul 10, 2014 | 563 | 1,950 | {"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} | 4.28125 | 4 | CC-MAIN-2022-49 | latest | en | 0.950421 |
https://www.softmath.com/tutorials-3/relations/intermediate-algebra-22.html | 1,716,464,367,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058625.16/warc/CC-MAIN-20240523111540-20240523141540-00727.warc.gz | 866,343,410 | 8,608 | # Intermediate Algebra
Course Description:
Through this course students develop various concepts of Algebra.
Students will solve linear , quadratic, rational, and radical equations ; graph
linear equations and inequalities in one variable ; graph linear equations in
two variables; solve and graph systems of linear equations and inequalities
in two variables; simplify rational expressions; simplify expressions
containing rational exponents; simplify complex numbers; solve related
applications.
Prerequisites:
MAT 0024 or MAT 0020 with a grade of S or appropriate placement test score.
Credits: 3
Course Competencies:
Competency 1:
The student will demonstrate knowledge of the slope of
a line
by:
a. Determining the slope of a line given two points that lie on
the line.
b. Determining the slope and intercept(s) of a line given its
equation.
c. Determining the slope of a line from a graph.
d. Finding the slope of a line that is parallel to a given line.
e. Finding the slope of a line that is perpendicular to a given
line.
Competency 2:
The student will demonstrate knowledge of linear
equations and inequalities in two variables by :
a. Solving literal equations.
b. Finding an equation of a line given two points.
c. Finding an equation of a line given a point on the line and
information about the slope of the line.
d. Writing an equation of a line in standard form.
e. Writing an equation of a line in slope-intercept form.
f. Graphing linear equations in two variables using the slope
and y-intercept of the line.
g. Graphing linear inequalities in two variables.
Competency 3:
The student will demonstrate knowledge of equations in two
variables by:
a. Solving direct variation problems.
b. Solving inverse variation problems.
Competency 4:
The student will demonstrate knowledge of systems of
linear equations by:
a. Solving a system of linear equations in two variables
b. Solving a system of linear equations in two variables
using the substitution method .
c. Solving a system of linear equations and inequalities in
two variables by graphing.
d. Solving applications involving systems of linear equations.
Competency 5:
The student will demonstrate knowledge of rational expressions and
equations
by:
a. Performing operations of addition , subtraction,
multiplication and division on rational expressions.
b. Simplifying complex fractions.
c. Solving equations involving rational expressions including
literal equations .
d. Dividing polynomials .
Competency 6:
The student will demonstrate knowledge of radicals and
rational exponents by:
a. Adding , subtracting, multiplying, and dividing expressions
b. Simplifying expressions containing rational exponents.
c. Applying the properties of exponents to expressions with
rational exponents
Competency 7:
The student will demonstrate knowledge of complex
numbers by:
a. Knowing the meaning of i.
b. Writing the square root of a negative number in terms of i.
Competency 8:
The student will demonstrate knowledge of quadratic
equations by:
a. Solving quadratic equations by factoring .
b. Solving quadratic equations by the square root method. | 658 | 3,127 | {"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-2024-22 | latest | en | 0.888075 |
http://encyclopedia.kids.net.au/page/ax/Axiomatic_system | 1,398,327,026,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223205375.6/warc/CC-MAIN-20140423032005-00007-ip-10-147-4-33.ec2.internal.warc.gz | 130,439,442 | 4,897 | ## Encyclopedia > Axiomatic system
Article Content
# Axiomatic system
In mathematics, an axiomatic system is any set of axioms from which some or all axioms can be used in conjunction to logically derive theorems. A mathematical theory consists of an axiomatic system and all its derived theorems.
### Properties of axiomatic systems
An axiomatic system is said to be consistent if it lacks contradiction, i.e the the ability to prove a statement and its negative of being both true based on the same set of axioms.
In an axiomatic system, an axiom is called independent if it is not a theorem that can be derived from other axioms in the system. A system will be called independent if all of its underlying axioms are independent.
Independence is not a necessary requirement for a system, yet consistency is necessary. An axiomatic system will be called complete if no additional axiom can be added to the system without making the new system either dependent or inconsistent.
### Models for axiomatic systems
A mathematical model for an axiomatic system is a well-defined set, which assigns meaning for the undefined terms presented in the system, in a manner that is correct with the relations defined in the system. The existence of a concrete model* proves the consistency of a system.
Models can also be used to show the independence of an axiom in the system. By constructing a valid model for a subsystem without a specific axiom, we show that the omitted axiom is independent if its correctness does not necessarily follow from the subsystem.
Two models are said to be isomorphic if one-to-one correspondence can be found between their elements, in a manner that preserves their relationship. An axiomatic system for which every model is isomorphic to another is called categorial, and the property of categoriallity ensures the completeness of a system.
* A model is called concrete if the meanings assigned are objects and relations from the real world, as opposed to an abstract model which is based on other axiomatic systems.
The first axiomatic system was Euclidean geometry.
All Wikipedia text is available under the terms of the GNU Free Documentation License
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September 11, 2001 Terrorist Attack Plane casualties ... DeLuca, 52, Roxbury, N.J., programer, member of Northern New Jersey Region of SCCA Patrick Joseph Driscoll, 70, Point Pleasant, N.J., retired engineer Edward Porter ... | 518 | 2,529 | {"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} | 3.9375 | 4 | CC-MAIN-2014-15 | longest | en | 0.950598 |
http://mathhelpforum.com/advanced-statistics/225284-dice-throwing.html | 1,498,170,469,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319912.4/warc/CC-MAIN-20170622220117-20170623000117-00261.warc.gz | 265,284,642 | 11,818 | 1. ## dice throwing
Hi,
I would like to calculate the probability of getting the same outcome twice when throwing two dice in the following way. First die i throw 4 times and second die 3 times. So the total sample space is
$S = 6^4 \times 6^3$
the question is how many casaes there are that have exactly 2 matching pairs of outcomes in S.
know this i probably a simple problem but I just cannot wrap my head arround it.
thank you
2. ## Re: dice throwing
Originally Posted by baxy77bax
I would like to calculate the probability of getting the same outcome twice when throwing two dice in the following way. First die i throw 4 times and second die 3 times. So the total sample space is
$S = 6^4 \times 6^3$
the question is how many casaes there are that have exactly 2 matching pairs of outcomes in S.
I for one do not understand the setup.
Tossing a die three times generates a triple. Tossing a die four times generates a 4-tuple.
Now, triples cannot match 4-tuples. So what does "exactly 2 matching pairs" mean"?
Does it mean for example $(2,3,4,3)~\&~(2,3,3)$. There are exactly 2 matching pairs of threes??
Please try to explain this more clearly.
3. ## Re: dice throwing
Please try to explain this more clearly.
Does it mean for example $(2,3,4,3)~\&~(2,3,3)$. There are exactly 2 matching pairs of threes??
no. in this particular example i have 3 pairs (2,2),(3,3) and (3,3)
also in the case [TEX] (3,3,4,3) \& (2,3,3) [\TEX] i have only two matches of 3's since the smallest set dictates the max number of pairs because there will always be possible to maximally create the total of two pairs between set 1 and 2, given [TEX]S_{1} = {3,3,4,3} \mbox{ and } S_{2}={2,3,3} [\TEX] [TEX] (S_{1}[0]=S_{2}[1],S_{1}[1]=S_{2}[2]) [\TEX] is equal to [TEX] (S_{1}[1]=S_{2}[1],S_{1}[3]=S_{2}[2]) [\TEX]
on the other hand given two sets (2,3,4,3) and (1,3,5), here i only have 1 pair (3,3) and this is something i do not care about.
is it more clear now
4. ## Re: dice throwing
Originally Posted by baxy77bax
I would like to calculate the probability of getting the same outcome twice when throwing two dice in the following way. First die i throw 4 times and second die 3 times. So the total sample space is
$S = 6^4 \times 6^3$
the question is how many cases there are that have exactly 2 matching pairs of outcomes in S.
Thank you for the clarification. It is a counting nightmare.
Start with the triple. Say $(1,1,1)$, now to have exactly 2 matching pairs of outcomes the 4-tuple must contain exactly two 1's and two non-1's.
There are $\binom{4}{2}$ places for the two 1's and $5^2$ ways to get two non-1's.
But there are six ways to have a triple with three of the same entry.
Thus in that case there are $6^2\cdot 5^2$ ways to have exactly two matching pairs.
I did say it is a counting nightmare. Say the triple is like $(1,3,3)$.
Now what can the 4-tuples look like?
5. ## Re: dice throwing
well yes, in that case 4-tuple needs to have two threes and two non 3's including non 1, or 1 and a 3 plus two outcomes that do not contain 3 nor 1. yes i know it is a counting nightmare that is why i decided to ask for help since there is noone in my proximity that can help me with this. Therefore , Thank you !! | 957 | 3,231 | {"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": 0, "img_math": 0, "codecogs_latex": 10, "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} | 4.21875 | 4 | CC-MAIN-2017-26 | longest | en | 0.908546 |
http://math.stackexchange.com/questions/208254/first-order-nonlinear-ordinary-differential-equation | 1,469,682,423,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827791.21/warc/CC-MAIN-20160723071027-00003-ip-10-185-27-174.ec2.internal.warc.gz | 153,510,716 | 18,447 | # First-order nonlinear ordinary differential equation
How to solve this differential equation:
$$x\frac{dy}{dx} = y + x\frac{e^x}{e^y}?$$
I tried to rearrange the equation to the form $f\left(\frac{y}{x}\right)$ but I couldn't thus I couldn't use $v = \frac{y}{x}$ to solve it.
-
This differential equation is not homogeneous (and so you can't rearrange it in the form $f(y/x)$.) – Lukas Geyer Oct 6 '12 at 16:15
what is it then? how can you solve it? – David Hoffman Oct 6 '12 at 19:21
If you assume $y>0$, something like $u=\log y$ may help. I haven't checked, so it may also be a waste of time. – Daryl Oct 6 '12 at 20:18
The clue is that $\dfrac{e^x}{e^y}$ should rewrite to $e^{x-y}$ and thus we get the idea that let $u=x-y$ or let $u=y-x$ will convert the ODE whose the terms are not contain any composite functions. – doraemonpaul Oct 7 '12 at 1:31
The substitution $u = e^y$ leads to $$\frac{du}{dx} = \frac{dy}{dx}e^y = \frac yx e^y + e^x = \frac ux \log u + e^x.$$ May be this could lead somewhere. – Sam Oct 7 '12 at 2:06
Rewrite this equation in the form:
$$M(x,y)dx + N(x,y)dy = (xe^x+ye^y)dx-xe^ydy = 0$$
Both $\frac{\partial M}{\partial y} = e^y(1+y)$ and $\frac{\partial N}{\partial x}=-e^y$ are depend on $y$ only. In this case some multiplier $\mu(x,y)$ can be simply found so that
$$\dfrac{\partial (\mu M)}{\partial y} = 0,\quad \dfrac{\partial (\mu N)}{\partial x}=0$$
and you get exact differential equation in form $du(x,y)=0$.
For $\mu$ depending only on $y$ we have $d\ln\mu=\dfrac{dy}{M}\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)$. In our case:
$$\mu(y) = \exp\left(-\int\dfrac{e^y(y+2)}{xe^x+ye^y}dy\right)$$ Solution of our DE is: $$\int \mu M dx + \int \mu N dy = C$$
I do not substitute $\mu$ in the last equation because $\mu$ as I see cannot be expressed in elementary functions and complete solution will be cumbersome.
-
$x\dfrac{dy}{dx}=y+x\dfrac{e^x}{e^y}$
$x\dfrac{dy}{dx}=y+xe^{x-y}$
Let $u=y-x$ ,
Then $y=u+x$
$\dfrac{dy}{dx}=\dfrac{du}{dx}+1$
$\therefore x\left(\dfrac{du}{dx}+1\right)=u+x+xe^{-u}$
$x\dfrac{du}{dx}+x=u+x+xe^{-u}$
$x\dfrac{du}{dx}=xe^{-u}+u$
$(xe^{-u}+u)\dfrac{dx}{du}=x$
Let $v=x+ue^u$ ,
Then $x=v-ue^u$
$\dfrac{dx}{du}=\dfrac{dv}{du}-(u+1)e^u$
$\therefore e^{-u}v\left(\dfrac{dv}{du}-(u+1)e^u\right)=v-ue^u$
$e^{-u}v\dfrac{dv}{du}-(u+1)v=v-ue^u$
$e^{-u}v\dfrac{dv}{du}=(u+2)v-ue^u$
$v\dfrac{dv}{du}=(u+2)e^uv-ue^{2u}$
This belongs to an Abel equation of the second kind.
Let $t=(u+1)e^u$ ,
Then $u=W(et)-1$
$\dfrac{dv}{du}=\dfrac{dv}{dt}\dfrac{dt}{du}=(u+2)e^u\dfrac{dv}{dt}$
$\therefore(u+2)e^uv\dfrac{dv}{dt}=(u+2)e^uv-ue^{2u}$
$v\dfrac{dv}{dt}=v-\dfrac{ue^u}{u+2}$
$v\dfrac{dv}{dt}-v=-\dfrac{t(W(et)-1)}{W(et)(W(et)+1)}$
This belongs to an Abel equation of the second kind in the canonical form. | 1,116 | 2,828 | {"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": 4, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2016-30 | latest | en | 0.815404 |
https://aboutwings.com/2023/05/24/ | 1,709,635,797,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948234904.99/warc/CC-MAIN-20240305092259-20240305122259-00318.warc.gz | 75,571,413 | 13,395 | ## The Math of Blackjack
Blackjack is a casino card game where the player attempts to beat the dealer. Players are dealt two cards and can decide to hit (request more cards) or stand (stop drawing cards). The aim is to get a hand total closer to 21 than the dealer’s. If the player busts, they lose their bet; if the dealer has a blackjack, they win the bet. The game is played in a special area of the casino called the pit, usually overseen by a stern-looking dealer or a pit boss.
#### The math of blackjack
It is important to understand the basic mathematics of blackjack before you play it. As with many casino games, the house has a statistical advantage that will play out over time. However, with proper strategy and a little luck, a knowledgeable player can reduce the house edge to a very small percentage.
A deck of 52 cards is used, with the face cards having a value of 10 and the aces having a value of one or 11 (depending on the type of card). The cards are dealt in a special box called a shoe, and the casino dealer deals himself two cards facing up and then takes a third face up card from the shoe.
The objective of the game is to beat the dealer by getting a higher hand total than the dealer, or by achieving a blackjack. A blackjack pays 3 to 2 or 50% more than any other hand, and is paid when the player receives an Ace with a 10, Jack, Queen, or King.
To achieve a better hand, players have the option to split pairs of cards up to three times, making up to four separate hands. Generally, a pair of 10s should not be split because it will not improve the hand. Players can also re-split any other pair apart from the two original ones, except for the Aces.
A player may buy insurance against the dealer’s potential blackjack, paying an amount equal to their initial wager. This bet is made before the dealer reveals his or her face up card, and players may choose to buy insurance even when the dealer is showing a 10 (a very good sign). Players can surrender their hands in certain situations; they will only receive half of their wager back, but they will avoid any further decisions, and potentially save money.
Other than buying insurance or surrendering, players have the opportunity to make side bets on the basis of information they gather as the game progresses. This can significantly increase their winnings, and it is a key part of the game for card counters.
Aside from basic strategy, the most important element in a blackjack game is counting the cards. Counting allows the player to determine which decisions are best in a given situation, and it helps them to increase their bet size in favorable situations. In addition, the player can use information about the remaining cards to improve their basic strategy rules.
## What is a Horse Race?
A horse race is a sport where horses compete against each other in order to win. The sport has a long history and is practiced in many countries around the world. The horse races are governed by a set of rules and regulations that govern how the race is run. The sport is also popular amongst many people as it offers great betting opportunities.
There are different types of horses that can be used in the horse races. The most common breeds for racing include Thoroughbred, Arabian, and Quarter horses. Different organizations have different regulations about what type of horses can be used in a race. The horses in the races are driven by jockeys who use whips to encourage the horse to go faster. However, the use of the whip can cause pain to the horse so there are rules about how often the jockeys can use it.
In order to win a race, a horse must cross the finish line before the other competitors. If two or more horses cross the finish line at the same time, then the winner is decided based on a photo finish. The stewards will carefully examine a photograph of the finish to determine which horse broke the plane first. If it cannot be determined, the race will be settled by dead heat rules.
The Kentucky Derby is a famous horse race that is held each year in Louisville, Kentucky. It is one of the most famous horse races in the world and is known for its huge prize money. The race has a very rich history, and is usually regarded as the top of the list when it comes to the best horse races in the world.
Other popular horse races include the Prix de l’Arc de Triomphe, which is held each October in Paris, France. It is a large race and attracts competitors from all over the world. The race is named to commemorate the soldiers that died in World War I and is a very popular event for sports betting.
Generally, horse races are conducted on flat tracks and do not include steeplechases. The pedigree of a horse is required in order to participate in a race, which means that it must have a sire and dam that are purebred members of the same breed. The racing season in North America is split into the spring and fall, with the latter having more important races for horses. A horse’s performance will decline as it gets older, but it is not uncommon to see a nine-year old horse compete in a major race. This is due to the escalating price of breeding and sales fees, which has made it profitable to race these older horses. This trend has led to some races being discontinued altogether, but other sponsors have picked up the purses for the more expensive events. | 1,134 | 5,384 | {"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} | 3.515625 | 4 | CC-MAIN-2024-10 | latest | en | 0.969629 |
https://tutorial.techaltum.com/c-program-to-calculate-area-and-circumference-of-circle.html | 1,680,140,563,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949093.14/warc/CC-MAIN-20230330004340-20230330034340-00370.warc.gz | 653,909,326 | 3,587 | C program to calculate Area and Circumference of Circle. In this program we will take radius of circle from user as an input calculate area and circumference of circle using following formula in c language.
• Circumference = 2 * 3.14 * radius
## Area and Circumference of Circle
In the following program we calculate Area and Circumference of Circle. In the following program we will take input radius and store it in radius variable and then put the formul to calculate area and circumference.
``````
#include<stdio.h>
void main()
{
//value of pi is 3.14
printf("The area of Circle is %f",area);
printf("\nThe Circumference of Circle is %f",cf);
}
``````
``` ```
Figure 1
```
```
``` ```
Figure 2
```
``` | 182 | 718 | {"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} | 2.546875 | 3 | CC-MAIN-2023-14 | longest | en | 0.851813 |
https://www.teachoo.com/2153/580/Misc-10---Let-A---1--2--3--4---f---(1--5)--(2--9)--(3--1)/category/Miscellaneous/ | 1,723,676,408,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641137585.92/warc/CC-MAIN-20240814221537-20240815011537-00750.warc.gz | 778,825,670 | 21,565 | Miscellaneous
Chapter 2 Class 11 Relations and Functions
Serial order wise
### Transcript
Misc 10 Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Are the following true? f is a relation from A to B f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)} First elements = 1, 2, 3, 4 All the first elements are in set A So, first element is from set A Second elements = 5, 9, 1, 11 All the second elements are in set B So, second element is from set B Since first element is from set A & second element is from set B Hence, f is a relation from A to B Misc 10 Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Are the following true? (ii) f is a function from A to B f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)} Here, 2 is coming twice Hence, it does not have a unique image So, it is not a function In function, first element is not repeating | 371 | 935 | {"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} | 3.875 | 4 | CC-MAIN-2024-33 | latest | en | 0.919303 |
http://tutorial.math.lamar.edu/Problems/CalcI/TheLimit.aspx | 1,527,224,244,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867041.69/warc/CC-MAIN-20180525043910-20180525063910-00237.warc.gz | 318,297,500 | 23,242 | Paul's Online Math Notes
[Notes]
Calculus I - Practice Problems
Review Previous Chapter Next Chapter Derivatives Tangent Lines and Rates of Change Previous Section Next Section One-Sided Limits
## The Limit
1. For the function answer each of the following questions.
(a) Evaluate the function the following values of x compute (accurate to at least 8 decimal places).
(i) 2.5 (ii) 2.1 (iii) 2.01 (iv) 2.001 (v) 2.0001
(vi) 1.5 (vii) 1.9 (viii) 1.99 (ix) 1.999 (x) 1.9999
(b) Use the information from (a) to estimate the value of .
[Solution]
2. For the function answer each of the following questions.
(a) Evaluate the function the following values of t compute (accurate to at least 8 decimal places).
(i) -0.5 (ii) -0.9 (iii) -0.99 (iv) -0.999 (v) -0.9999
(vi) -1.5 (vii) -1.1 (viii) -1.01 (ix) -1.001 (x) -1.0001
(b) Use the information from (a) to estimate the value of .
[Solution]
3. For the function answer each of the following questions.
(a) Evaluate the function the following values of compute (accurate to at least 8 decimal places). Make sure your calculator is set to radians for the computations.
(i) 0.5 (ii) 0.1 (iii) 0.01 (iv) 0.001 (v) 0.0001
(vi) -0.5 (vii) -0.1 (viii) -0.01 (ix) -0.001 (x) -0.0001
(b) Use the information from (a) to estimate the value of .
[Solution]
4. Below is the graph of . For each of the given points determine the value of and . If any of the quantities do not exist clearly explain why.
(a) (b) (c) (d)
[Solution]
5. Below is the graph of . For each of the given points determine the value of and . If any of the quantities do not exist clearly explain why.
(a) (b) (c) (d)
[Solution]
6. Below is the graph of . For each of the given points determine the value of and . If any of the quantities do not exist clearly explain why.
(a) (b) (c) (d)
[Solution]
Problem Pane
Tangent Lines and Rates of Change Previous Section Next Section One-Sided Limits Review Previous Chapter Next Chapter Derivatives
[Notes]
© 2003 - 2018 Paul Dawkins | 699 | 2,484 | {"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} | 4.34375 | 4 | CC-MAIN-2018-22 | longest | en | 0.583888 |
https://www.physicsforums.com/threads/decidability-of-polynomials-with-integer-coefficients-and-at-least-1-real-root.219884/ | 1,718,391,791,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861568.20/warc/CC-MAIN-20240614173313-20240614203313-00177.warc.gz | 832,493,356 | 14,760 | # Decidability of Polynomials with Integer Coefficients and at Least 1 Real Root?
• Dragonfall
In summary, the question is asking for a proof that the set of polynomials with integer coefficients and at least one real root is decidable. The suggested approach is to use a finite algorithm to determine if an integer coefficient polynomial has a real root, potentially by using the zeros of its derivative and reducing it to a recursive problem. However, it is unclear if this is the correct approach.
Dragonfall
## Homework Statement
Show that the set of polynomials with integer coefficients with at least 1 real root is decidable.
## The Attempt at a Solution
The question did not ask for specific language, just an intuitive finite algorithm will do.
In other words, how do you determine whether an integer coefficient polynomial in one variable has at least one real root?
Anyone?
I was thinking maybe by finding the zeros of the derivative, etc, and thus reducing the problem to a recursive one, but I don't know how to do this precisely, or know whether this is the right approach at all.
## 1. What is a polynomial?
A polynomial is a mathematical expression that consists of variables, coefficients, and exponents. It can have one or more terms and can be added, subtracted, multiplied, and divided.
## 2. What are the different types of polynomials?
The different types of polynomials include monomials, binomials, trinomials, and higher degree polynomials. Monomials have only one term, binomials have two terms, and trinomials have three terms.
## 3. How do you simplify a polynomial?
To simplify a polynomial, you need to combine like terms by adding or subtracting their coefficients. You can also use the distributive property to remove parenthesis and combine terms.
## 4. What is the degree of a polynomial?
The degree of a polynomial is the highest exponent of the variable. For example, in the polynomial 3x^5 + 2x^3 + 4, the degree is 5 because it is the highest exponent.
## 5. How do you solve a polynomial equation?
To solve a polynomial equation, you need to set the polynomial equal to zero and use algebraic methods such as factoring, the quadratic formula, or synthetic division to find the roots or solutions of the equation.
• Calculus and Beyond Homework Help
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2K | 671 | 2,822 | {"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-2024-26 | latest | en | 0.938766 |
https://courses.lumenlearning.com/waymakercollegealgebra/chapter/make-sense-of-the-solution-to-a-system-of-linear-equation/ | 1,660,055,589,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570977.50/warc/CC-MAIN-20220809124724-20220809154724-00261.warc.gz | 214,266,340 | 17,652 | ## Classify Solutions to Systems
### Learning Outcomes
• Determine whether a solution reveals that a system has one, many, or no solutions
• Interpret the solution to a system of equations that represents profits for a business
Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system consists of parallel lines that have the same slope but different $y$ -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as $12=0$.
### Example: Solving an Inconsistent System of Equations
Solve the following system of equations.
$\begin{gathered}&x=9 - 2y \\ &x+2y=13 \end{gathered}$
### Try It
Solve the following system of equations in two variables.
$\begin{gathered}2y - 2x=2\\ 2y - 2x=6\end{gathered}$
## Expressing the Solution of a System of Dependent Equations Containing Two Variables
Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as $0=0$.
### Example: Finding a Solution to a Dependent System of Linear Equations
Find a solution to the system of equations using the addition method.
$\begin{gathered}x+3y=2\\ 3x+9y=6\end{gathered}$
### Writing the general solution
In the previous example, we presented an analysis of the solution to the following system of equations:
$\begin{gathered}x+3y=2\\ 3x+9y=6\end{gathered}$
After a little algebra, we found that these two equations were exactly the same. We then wrote the general solution as $\left(x, -\frac{1}{3}x+\frac{2}{3}\right)$. Why would we write the solution this way? In some ways, this representation tells us a lot. It tells us that x can be anything, x is x. It also tells us that y is going to depend on x, just like when we write a function rule. In this case, depending on what you put in for x, y will be defined in terms of x as $-\frac{1}{3}x+\frac{2}{3}$.
In other words, there are infinitely many (x,y) pairs that will satisfy this system of equations, and they all fall on the line $f(x)-\frac{1}{3}x+\frac{2}{3}$.
### Try It
Solve the following system of equations in two variables.
$\begin{gathered}y - 2x=5 \\ -3y+6x=-15 \end{gathered}$
### try it
Use an online graphing tool to write three different systems:
• A system of equations with one solution
• A system of equations with no solutions
• A system of equations with infinitely many solutions
## Using Systems of Equations to Investigate Profits
Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation $R=xp$, where $x=$ quantity and $p=$ price. The revenue function is shown in orange in the graph below.
The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in the graph below. The $x$ -axis represents quantity in hundreds of units. The y-axis represents either cost or revenue in hundreds of dollars.
The point at which the two lines intersect is called the break-even point. We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also$3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.
The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as $P\left(x\right)=R\left(x\right)-C\left(x\right)$. Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.
### Example: Finding the Break-Even Point and the Profit Function Using Substitution
Given the cost function $C\left(x\right)=0.85x+35{,}000$ and the revenue function $R\left(x\right)=1.55x$, find the break-even point and the profit function.
## Writing a System of Linear Equations Given a Situation
It is rare to be given equations that neatly model behaviors that you encounter in business, rather, you will probably be faced with a situation for which you know key information as in the example above. Below, we summarize three key factors that will help guide you in translating a situation into a system.
### How To: Given a situation that represents a system of linear equations, write the system of equations and identify the solution.
1. Identify the input and output of each linear model.
2. Identify the slope and y-intercept of each linear model.
3. Find the solution by setting the two linear functions equal to another and solving for x, or find the point of intersection on a graph.
Now let’s practice putting these key factors to work. In the next example, we determine how many different types of tickets are sold given information about the total revenue and amount of tickets sold to an event.
The cost of a ticket to the circus is $25.00 for children and$50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets? ### Try It Meal tickets at the circus cost$4.00 for children and $12.00 for adults. If 1,650 meal tickets were bought for a total of$14,200, how many children and how many adults bought meal tickets?
Sometimes, a system of equations can inform a decision. In our next example, we help answer the question, “Which truck rental company will give the best value?”
### Example: Building a System of Linear Models to Choose a Truck Rental Company
Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of$16, then 63 cents a mile.[1] When will Keep on Trucking, Inc. be the better choice for Jamal?
The applications for systems seems almost endless, but we will just show one more. In the next example, we determine the amount 80% methane solution to add to a 50% solution to give a final solution of 60%.
### Example: Solve a Chemical Mixture Problem
A chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane?
## Contribute!
Did you have an idea for improving this content? We’d love your input.
1. Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/ | 1,670 | 7,035 | {"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": 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": 11, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2022-33 | latest | en | 0.903184 |
https://homework.cpm.org/category/CC/textbook/cc3/chapter/6/lesson/6.2.3/problem/6-74 | 1,571,327,906,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986675409.61/warc/CC-MAIN-20191017145741-20191017173241-00336.warc.gz | 559,813,563 | 15,710 | Home > CC3 > Chapter 6 > Lesson 6.2.3 > Problem6-74
6-74.
Simplify each expression below. Homework Help ✎
1. $- 2 \frac { 3 } { 10 } - 1 \frac { 2 } { 5 }$
Rewrite each term as a fraction greater than one.
$- \frac{23}{10}-\frac{7}{5}$
Find the common denominator.
$-\frac{23}{10}-\frac{7}{5}\cdot\frac{2}{2}=-\frac{23}{10}-\frac{14}{10}$
Simplify.
$-\frac{37}{10}$
Remember to show your work.
$-3\frac{7}{10}$
1. $3 \div - \frac { 5 } { 4 }$
Rewrite $3$ as a fraction and then multiply by the reciprocal of the second fraction.
$\frac{3}{1}\cdot-\frac{4}{5}$
$-\frac{12}{5}$
1. $\frac { 3 } { 4 } + 5 \frac { 7 } { 8 }$
Rewrite the second fraction as a fraction greater than one, find the common denominator, and then add.
1. $5 \frac { 1 } { 6 } \cdot ( - \frac { 7 } { 9 } )$
Rewrite the first fraction as a fraction greater than one, then multiply numerators together and denominators together. | 330 | 917 | {"found_math": true, "script_math_tex": 11, "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} | 4.375 | 4 | CC-MAIN-2019-43 | longest | en | 0.652446 |
https://mtp.tools/converters/weight/metric-tonne-to-kilogram-calculator | 1,611,281,896,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529080.43/warc/CC-MAIN-20210122020254-20210122050254-00065.warc.gz | 457,998,155 | 5,543 | # Metric tonne to Kilogram calculator (t to kg)
Convert metric tonnes to kilograms (t to kg) by typing the amount of metric tonnes in the input field below and then clicking in the "Convert" button. If you want to convert from kilograms to metric tonnes, you can use our kilogram to metric tonne converter.
## Formula
Formula used to convert t to kg:
F(x) = x * 1000
For example, if you want to convert 1 t to kg, just replace x by 1 [t]:
1 t = 1 * 1000 = 1000 kg
## Steps
1. Multiply the amount of metric tonnes by 1000.
2. The result will be expressed in kilograms.
## Metric tonne to Kilogram Conversion Table
The following table will show the most common conversions for Metric Tonnes (t) to Kilograms (kg):
Metric Tonnes (t) Kilograms (kg)
0.0001 t 0.1 kg
0.001 t 1 kg
0.01 t 10 kg
0.02 t 20 kg
0.03 t 30 kg
0.04 t 40 kg
0.05 t 50 kg
0.06 t 60 kg
0.07 t 70 kg
0.08 t 80 kg
0.09 t 90 kg
0.1 t 100 kg
0.2 t 200 kg
0.3 t 300 kg
0.4 t 400 kg
0.5 t 500 kg
0.6 t 600 kg
0.7 t 700 kg
0.8 t 800 kg
0.9 t 900 kg
1 t 1000 kg
1.1 t 1100 kg
1.2 t 1200 kg
1.25 t 1250 kg
1.5 t 1500 kg
1.75 t 1750 kg
2 t 2000 kg
3 t 3000 kg
4 t 4000 kg
5 t 5000 kg
6 t 6000 kg
7 t 7000 kg
8 t 8000 kg
9 t 9000 kg
10 t 10000 kg
20 t 20000 kg
30 t 30000 kg
40 t 40000 kg
50 t 50000 kg
60 t 60000 kg
70 t 70000 kg
80 t 80000 kg
90 t 90000 kg
100 t 100000 kg
200 t 200000 kg
300 t 300000 kg
400 t 400000 kg
500 t 500000 kg
600 t 600000 kg
700 t 700000 kg
800 t 800000 kg
900 t 900000 kg
1000 t 1000000 kg
## About the converter
Note that this is a high-precision t to kg calculator, but rounding errors may occur (in a very small percentage of the cases).
Convert metric tonnes to kilograms (t to kg) by pasting or typing the amount of metric tonnes in the metric tonnes input. Check the formula section to manually convert t to kg.
In case you want to convert kilograms to metric tonnes, please use the kg to t converter.
### How many kilograms are in a metric tonne?
1 metric tonne [t] is equal to 1000 kilograms [kg].
## About Metric Tonnes (t)
The tonne (also referred to as the metric ton in the United States and Canada), is widely used non-SI metric unit of mass equal to 1,000 kilograms or one megagram (symbol: Mg). The symbol used for tonne is t. It is equivalent to approximately 2,204.6 pounds, 1.102 short tons (US) or 0.984 long tons (UK).
The tonne is used when you want to express the weight of heavy objects, such as cars, elephants, cars, among others.
## About Kilograms (kg)
A kilogram is a widely used unit of weight, defined on the International System of Units (SI). One kilogram is equal to 1000 grams. The symbol used to represent kilograms is kg.
The kilogram is used to measure almost everything, from food, materials such as wood, your weight, among others.
## FAQs for Metric tonne to Kilogram converter calculator
### What is Metric tonne to Kilogram converter calculator?
Metric tonne to Kilogram converter is a free and online calculator that converts Metric tonnes to Kilograms.
### How do I use Metric tonne to Kilogram converter?
1. Either copy and paste or type the amount of metric tonnes to convert in the metric tonnes input.
2. Click on the Convert button.
3. It will convert metric tonnes into kilograms and output it in the kilograms input.
### Which browsers are supported?
All mayor web browsers are supported, including Internet Explorer, Microsoft Edge, Firefox, Chrome, Safari and Opera.
### Which devices does Metric tonne to Kilogram converter work on?
Metric tonne to Kilogram converter calculator works in any device that supports any of the browsers mentioned before. It can be a smartphone, desktop computer, notebook, tablet, etc. | 1,096 | 3,687 | {"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-2021-04 | latest | en | 0.506728 |
http://www.vbforums.com/showthread.php?849283-Approaching-a-MOO-1-style-game-Ways-to-do-a-map-and-place-stars&s=fed8a6d876c47c11869085492ec48256&p=5186605 | 1,529,733,817,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864943.28/warc/CC-MAIN-20180623054721-20180623074721-00286.warc.gz | 526,343,918 | 22,698 | # Thread: Approaching a MOO 1 style game - Ways to do a map and place stars
1. ## Approaching a MOO 1 style game - Ways to do a map and place stars
Anyway if you aren't familiar with Masters of Orion 1, it uses a 2D map of space with stars spread randomly around. Although it shows stars in all practical terms the stars are actually planets that may or may not be habitable.
The game map can be several sizes (24 to 100+ stars). When performing your turn you see only a portion of the entire game map (basically zoomed in so you can see details) then when the game is processing your commands and figuring out what the Computer players are doing it displays the entire map with ship movement etc being displayed. You can also get the zoomed out view during your turn by hitting the "Map" button. Clicking on stars/ships displays its details in a sidebar.
Everything on the map is clickable and when you do click on a star, ship, monster, whatever, it switches back to the zoomed in view with the clicked item centered and highlighted (or close enough to centered if the object was on the edges of the map).
Navigating around the map is done by simply clicking near the edges and the view slides in that direction.
Ships in the game travel in a straight line between stars and distances are calculated as the crow flies, not by any kind of grid cell counts though there is obviously some form of grid use when it comes to star placements.
So I'm looking for suggestions on how to achieve this kind of 2D graphics functionality (not necessarily code, just point to examples/write ups/whatever you think would be useful). I'm flying blind as I've never even attempted something in the 2D map realm, its all a mystery at the moment.
As an example, how to place stars and ships and make them selectable?. Should I have some sort in internal grid (math only) for placements? Something like a grid where the click position is divided by the number of cells which gives an index to look up data (either X/Y or a single dimension calculated by an X index and adding Y * XMax (e.g. one row has 200 positions, clicking in the area of the first row gives X + 0, second row gives X + 200, third X + 400, you get the idea). Or am I barking up the wrong tree entirely?
Next is the map graphic - sticking a picture in a scrolling picture box is probably not going to cut it (though it might, this is just my assumption).
I found an old bit of source for a similar game and it used Gorgon library (which I believe is the same idea as XNA/Monoplay). I get the feeling I should probably look into some form of library for 2D games. Anyone have a favorite?
I'll stop babbling now. Looking forward to any suggestions people may have!
2. ## Re: Approaching a MOO 1 style game - Ways to do a map and place stars
I pulled up an image on Google by searching for "Masters of Orion 1" and it looks pretty straight forward. To me, this looks like a PictureBox that has a static background image of space and dynamic drawings of the stars. In terms of making the items selectable, what I'd do is create a class to store the information that would be pulled up when clicked and have at least one property indicate where the object is in relation to the map (a Point property). Then whenever the user clicks on the Image, you'd get the location of where the user just clicked and check if it is within the bounds of one of your stars/ships.
Really, something this simple doesn't warrant any kind of external library. This could all be done very easily using GDI+.
3. ## Re: Approaching a MOO 1 style game - Ways to do a map and place stars
I messed around with XNA/Monogame and while the 2d stuff seems exceptionally easy it sure gobbles up CPU time. An empty game (no logic in update, nothing do draw) uses up 15% of my total CPU time on a 4 ghz 8 core processor. While I understand why it just seems a crime to waste all that CPU time for what is a completely event driven game. I guess if I wanted things to animate while the game is doing nothing it would be handy but i can probably achieve the same thing with a timer and GDI.
I have been playing around with using pictureboxes and rotating images for depicting fleets and that's been a bit of a learning curve, but not bad. Going to see if having a few hundred picture boxes has any serious impact on performance. Someone also suggested using panels since I don't need most of the picture box functionality. I'll try both and see what happens. If there's too much overhead I'll try drawing the "sprites" directly though from what I've read (not extensive yet) it looks like I have to handle masking the background and re-drawing it when sprites move etc. I like the idea of using a container though since then I can have click events and not have to figure out what was clicked by coordinates (not that its hard in any way).
Not that I mind all the experimentation. I picked MOO because I know the game so I can concentrate on learning 2D graphics rather than designing a game from scratch. Later perhaps I'll try working on my own game.
Thanks for the suggestions. It confirms what I was thinking.
4. ## Re: Approaching a MOO 1 style game - Ways to do a map and place stars
By the way, I have a tutorial on how to use GDI+ for beginners in the utility bank here: http://www.vbforums.com/showthread.p...-for-Beginners
It sounds like you have the basics down, but if I were you I'd still run through the tutorial to avoid some common pitfalls.
5. ## Re: Approaching a MOO 1 style game - Ways to do a map and place stars
Yeah I was perusing that area.
I was going to get fancy and have fleets point in the direction of movement but then realized that mostly negates the usefulness of one of the technologies which shows you the destination of a fleet. So now I just have to figure out why a manually placed picturebox transparency works but the ones I add manually don't. I've set the parent to the background picturebox etc... More experimentation required.
6. ## Re: Approaching a MOO 1 style game - Ways to do a map and place stars
Just to add to this ... I had some trouble before with transparencies and it turned out, that even though I saved my images with a transparent color selected, it would not be recognized when loaded. even though it might not be your problem, keep in mind that in some cases, the difference between an Indexed image and a regular RGB, can sometimes be confused between your application and whatever raster editor you are using.
Just a side note on some of my past troubles.
7. ## Re: Approaching a MOO 1 style game - Ways to do a map and place stars
Actually yeah, I read something somewhere about specifying the pixelformat when loading. Been busy so I haven't had a chance to look but that's probably it. I guess when you choose a pic from the properties menu it loads it as 32bpp but from the program it may not.
8. ## Re: Approaching a MOO 1 style game - Ways to do a map and place stars
This is a go to website for me when it concerns visual basic. Chances are, that there is a code snippet there that might help you. also, check out the authors books, They are high quality references and tutorials
http://vb-helper.com/index_vbnet.html
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Featured | 1,676 | 7,391 | {"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-2018-26 | latest | en | 0.949135 |
https://www.exactlywhatistime.com/days-from-date/july-11/28-days | 1,708,996,350,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474669.36/warc/CC-MAIN-20240226225941-20240227015941-00568.warc.gz | 774,748,792 | 6,207 | # What date is 28 days added from Thursday July 11, 2024?
## Thursday August 08, 2024
Adding 28 days from Thursday July 11, 2024 is Thursday August 08, 2024 which is day number 221 of 2024. This page is designed to help you the steps to count 28, but understand how to convert and add time correctly.
• Specific Date: Thursday July 11, 2024
• Days from Thursday July 11, 2024: Thursday August 08, 2024
• Day of the year: 221
• Day of the week: Thursday
• Month: August
• Year: 2024
## Calculating 28 days from Thursday July 11, 2024 by hand
Attempting to add 28 days from Thursday July 11, 2024 by hand can be quite difficult and time-consuming. A more convenient method is to use a calendar, whether it's a physical one or a digital application, to count the days from the given date. However, our days from specific date calculatoris the easiest and most efficient way to solve this problem.
If you want to modify the question on this page, you have two options: you can either change the URL in your browser's address bar or go to our days from specific date calculator to enter a new query. Keep in mind that doing these types of calculations in your head can be quite challenging, so our calculator was developed to assist you in this task and make it much simpler.
## Thursday August 08, 2024 Stats
• Day of the week: Thursday
• Month: August
• Day of the year: 221
## Counting 28 days forward from Thursday July 11, 2024
Counting forward from today, Thursday August 08, 2024 is 28 from now using our current calendar. 28 days is equivalent to:
28 days is also 672 hours. Thursday August 08, 2024 is 60% of the year completed.
## Within 28 days there are 672 hours, 40320 minutes, or 2419200 seconds
Thursday Thursday August 08, 2024 is the 221 day of the year. At that time, we will be 60% through 2024.
## In 28 days, the Average Person Spent...
• 6014.4 hours Sleeping
• 799.68 hours Eating and drinking
• 1310.4 hours Household activities
• 389.76 hours Housework
• 430.08 hours Food preparation and cleanup
• 134.4 hours Lawn and garden care
• 2352.0 hours Working and work-related activities
• 2163.84 hours Working
• 3541.44 hours Leisure and sports
• 1921.92 hours Watching television
## Famous Sporting and Music Events on August 08
• 1810 Urdu poet Mirza Ghalib marries Maaroof, daughter of Nawab Ilahi Baksh, and moves to Delhi.
• 1900 1st International Lawn Tennis Challenge (precursor to Davis Cup) begins at Longwood Cricket Club in Massachusetts, won 3-0 by US over British Isles | 675 | 2,517 | {"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} | 3.78125 | 4 | CC-MAIN-2024-10 | latest | en | 0.924824 |
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I tried solving the following question but am not sure I did so correctly and simply wished to verify it here.
In 50.3 gr of a compound whose formula is $$Na_2X_4O_7$$ (where X is an unknown atom) are 1.75 moles of oxygen atoms.
a) What is the number of Na atoms in the sample?
I figured the number of moles would simply be 1.75*2/7=0.5, hence number of atoms would be $$0.5*N_A$$.
b) What is the molar mass of X?
I figured that since there are 1.75 moles of oxygen at 16 gr/mole and 0.5 moles of Na at approx. 23 gr/mole, the MW of oxygen and Na together would be 1.75*16+0.5*23=39.5 gr, hence the MW of X would be: (50.3-39.5)/4=2.7 gr/mole.
I am now told that when the compound comes in contact with 500ml of hot water (in excess), the following reaction takes place:
$$Na_2X_4O_{7(s)}+H_2O ->XH_{3(g)}+Na^+_{(aq)}+OH^-_{(aq)}+O_{2(g)}$$
Moreover, in a certain process 0.84 moles of $$XH_{3(g)}$$ were formed.
1) How many grams of gaseous oxygen were formed in the process?
I first balanced the reaction thus:
$$Na_2X_4O_{7(s)}+7H_2O ->4XH_{3(g)}+2Na^+_{(aq)}+2OH^-_{(aq)}+6O_{2(g)}$$
I figured that if 0.84 moles of $$XH_3$$ were formed then 0.84*1.5 would be the number of oxygen moles formed, weighing 20.16 gr.
2) How many grams of $$Na_2X_4O_{7(s)}$$ reacted in the process?
I figured that 0.84/4 moles of $$Na_2X_4O_{7(s)}$$ must have reacted in the process, weighing (0.21/3.25) * 50.3=3.25 gr.
3) What are the volumes of the oxygen and the $$XH_{3(g)}$$ formed in the process at room conditions?
I figured $$V_{O_2}=n_{O_2}*0.082*298$$ and $$V_{XH_3}=n_{XH_3}*0.082*298$$, where $$n_{O_2}=1.26 moles$$ and $$n_{XH_3}=0.84 moles$$
4) What is the concentration of $$OH^-$$ ions in the solution obtained (500ml)?
I figured the answer would simply be 0.42/0.5=0.84M, as 0.42 moles of $$OH^-$$ were formed.
I'd appreciate any comments on my attempt above.
I tried solving the following question but am not sure I did so correctly and simply wished to verify it here.
In 50.3 gr of a compound whose formula is $$Na_2X_4O_7$$ are 1.75 moles of oxygen atoms.
a) What is the number of Na atoms in the sample?
I figured the number would simply be 1.75*2/7=0.5, hence number of atoms would be $$0.5*N_A$$.
b) What is the molar mass of X?
I figured that since there are 1.75 moles of oxygen at 16 gr/mole and 0.5 moles of Na at approx. 23 gr/mole, the MW of oxygen and Na together would be 1.75*16+0.5*23=39.5 gr, hence the MW of X would be: (50.3-39.5)/4=2.7 gr/mole.
I am now told that when the compound comes in contact with 500ml of hot water (in excess), the following reaction takes place:
$$Na_2X_4O_{7(s)}+H_2O ->XH_{3(g)}+Na^+_{(aq)}+OH^-_{(aq)}+O_{2(g)}$$
Moreover, in a certain process 0.84 moles of $$XH_{3(g)}$$ were formed.
1) How many grams of gaseous oxygen were formed in the process?
I first balanced the reaction thus:
$$Na_2X_4O_{7(s)}+7H_2O ->4XH_{3(g)}+2Na^+_{(aq)}+2OH^-_{(aq)}+6O_{2(g)}$$
I figured that if 0.84 moles of $$XH_3$$ were formed then 0.84*1.5 would be the number of oxygen moles formed, weighing 20.16 gr.
2) How many grams of $$Na_2X_4O_{7(s)}$$ reacted in the process?
I figured that 0.84/4 moles of $$Na_2X_4O_{7(s)}$$ must have reacted in the process, weighing (0.21/3.25) * 50.3=3.25 gr.
3) What are the volumes of the oxygen and the $$XH_{3(g)}$$ formed in the process at room conditions?
I figured $$V_{O_2}=n_{O_2}*0.082*298$$ and $$V_{XH_3}=n_{XH_3}*0.082*298$$, where $$n_{O_2}=1.26 moles$$ and $$n_{XH_3}=0.84 moles$$
4) What is the concentration of $$OH^-$$ ions in the solution obtained (500ml)?
I figured the answer would simply be 0.42/0.5=0.84M, as 0.42 moles of $$OH^-$$ were formed.
I'd appreciate any comments on my attempt above.
I tried solving the following question but am not sure I did so correctly and simply wished to verify it here.
In 50.3 gr of a compound whose formula is $$Na_2X_4O_7$$ (where X is an unknown atom) are 1.75 moles of oxygen atoms.
a) What is the number of Na atoms in the sample?
I figured the number of moles would simply be 1.75*2/7=0.5, hence number of atoms would be $$0.5*N_A$$.
b) What is the molar mass of X?
I figured that since there are 1.75 moles of oxygen at 16 gr/mole and 0.5 moles of Na at approx. 23 gr/mole, the MW of oxygen and Na together would be 1.75*16+0.5*23=39.5 gr, hence the MW of X would be: (50.3-39.5)/4=2.7 gr/mole.
I am now told that when the compound comes in contact with 500ml of hot water (in excess), the following reaction takes place:
$$Na_2X_4O_{7(s)}+H_2O ->XH_{3(g)}+Na^+_{(aq)}+OH^-_{(aq)}+O_{2(g)}$$
Moreover, in a certain process 0.84 moles of $$XH_{3(g)}$$ were formed.
1) How many grams of gaseous oxygen were formed in the process?
I first balanced the reaction thus:
$$Na_2X_4O_{7(s)}+7H_2O ->4XH_{3(g)}+2Na^+_{(aq)}+2OH^-_{(aq)}+6O_{2(g)}$$
I figured that if 0.84 moles of $$XH_3$$ were formed then 0.84*1.5 would be the number of oxygen moles formed, weighing 20.16 gr.
2) How many grams of $$Na_2X_4O_{7(s)}$$ reacted in the process?
I figured that 0.84/4 moles of $$Na_2X_4O_{7(s)}$$ must have reacted in the process, weighing (0.21/3.25) * 50.3=3.25 gr.
3) What are the volumes of the oxygen and the $$XH_{3(g)}$$ formed in the process at room conditions?
I figured $$V_{O_2}=n_{O_2}*0.082*298$$ and $$V_{XH_3}=n_{XH_3}*0.082*298$$, where $$n_{O_2}=1.26 moles$$ and $$n_{XH_3}=0.84 moles$$
4) What is the concentration of $$OH^-$$ ions in the solution obtained (500ml)?
I figured the answer would simply be 0.42/0.5=0.84M, as 0.42 moles of $$OH^-$$ were formed.
I'd appreciate any comments on my attempt above.
1
# Stoichiometry question
I tried solving the following question but am not sure I did so correctly and simply wished to verify it here.
In 50.3 gr of a compound whose formula is $$Na_2X_4O_7$$ are 1.75 moles of oxygen atoms.
a) What is the number of Na atoms in the sample?
I figured the number would simply be 1.75*2/7=0.5, hence number of atoms would be $$0.5*N_A$$.
b) What is the molar mass of X?
I figured that since there are 1.75 moles of oxygen at 16 gr/mole and 0.5 moles of Na at approx. 23 gr/mole, the MW of oxygen and Na together would be 1.75*16+0.5*23=39.5 gr, hence the MW of X would be: (50.3-39.5)/4=2.7 gr/mole.
I am now told that when the compound comes in contact with 500ml of hot water (in excess), the following reaction takes place:
$$Na_2X_4O_{7(s)}+H_2O ->XH_{3(g)}+Na^+_{(aq)}+OH^-_{(aq)}+O_{2(g)}$$
Moreover, in a certain process 0.84 moles of $$XH_{3(g)}$$ were formed.
1) How many grams of gaseous oxygen were formed in the process?
I first balanced the reaction thus:
$$Na_2X_4O_{7(s)}+7H_2O ->4XH_{3(g)}+2Na^+_{(aq)}+2OH^-_{(aq)}+6O_{2(g)}$$
I figured that if 0.84 moles of $$XH_3$$ were formed then 0.84*1.5 would be the number of oxygen moles formed, weighing 20.16 gr.
2) How many grams of $$Na_2X_4O_{7(s)}$$ reacted in the process?
I figured that 0.84/4 moles of $$Na_2X_4O_{7(s)}$$ must have reacted in the process, weighing (0.21/3.25) * 50.3=3.25 gr.
3) What are the volumes of the oxygen and the $$XH_{3(g)}$$ formed in the process at room conditions?
I figured $$V_{O_2}=n_{O_2}*0.082*298$$ and $$V_{XH_3}=n_{XH_3}*0.082*298$$, where $$n_{O_2}=1.26 moles$$ and $$n_{XH_3}=0.84 moles$$
4) What is the concentration of $$OH^-$$ ions in the solution obtained (500ml)?
I figured the answer would simply be 0.42/0.5=0.84M, as 0.42 moles of $$OH^-$$ were formed.
I'd appreciate any comments on my attempt above. | 2,697 | 7,598 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 60, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-05 | latest | en | 0.936394 |
https://www.coursehero.com/file/6183805/PHYS-422-Lecture-10-Superposition-II/ | 1,516,314,850,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887621.26/warc/CC-MAIN-20180118210638-20180118230638-00256.warc.gz | 897,717,745 | 24,335 | PHYS 422 Lecture 10 Superposition II
# PHYS 422 Lecture 10 Superposition II - The Superposition of...
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Unformatted text preview: The Superposition of Waves II Standing Wave In general: ( ) ( ) ( ) t x g C t x f C t x v v + + − = 2 1 , ψ two waves traveling in opposite direction Consider 2 waves, incident and reflected: ( ) I I I t kx E E ξ ω + − = sin ( ) R R R t kx E E ξ ω + + = sin n n ( ) ( ) [ ] R I I t kx t kx E E ξ ω ξ ω + + + + − = sin sin ( ) ( ) 2 cos 2 sin 2 sin sin β α β α β α − + = + ⎞ ⎛ − ⎞ ⎛ + + = cos sin I R R I kx ξ ξ ξ ξ ⎟ ⎠ ⎜ ⎝ ⎟ ⎠ ⎜ ⎝ 2 2 I ω an select rigin and that: s n Can select x origin and t= 0 so that: ( ) ( ) t kx E E I ω cos sin 2 = (Typically E =0 on the surface of a metal mirror) Standing Wave s n ( ) ( ) t kx E E I ω cos sin 2 = λ Animation courtesy of Dr. Dan Russell, Kettering University Standing Wave and Resonance If the number of λ /2 is integer in example above, the string can oscillate forever (if there are no losses) - resonance. Standing Electromagnetic Wave 890 tto Wiener experiment 1890 - Otto Wiener experiment Where is the energy when E is zero?...
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PHYS 422 Lecture 10 Superposition II - The Superposition of...
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Ask a homework question - tutors are online | 538 | 1,823 | {"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} | 3.296875 | 3 | CC-MAIN-2018-05 | latest | en | 0.718074 |
https://brilliant.org/problems/constant-product/ | 1,529,515,635,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863830.1/warc/CC-MAIN-20180620163310-20180620183310-00000.warc.gz | 572,950,681 | 11,374 | # Constant Product?
If the number of ordered quintuplets $$(x_1,x_2,x_3,x_4,x_5)$$ such that $\displaystyle\prod_{i=1}^{5} x_i=1458000$ is of the form $$2^a \times 3^b \times 5^c \times 7^d$$, then find $$a+b+c+d$$.
Details and Assumptions:
• $$x_1,x_2,x_3,x_4,x_5 \in \mathbb{N}$$
• $$a,b,c,d \in \mathbb{N}$$
× | 141 | 315 | {"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} | 2.796875 | 3 | CC-MAIN-2018-26 | latest | en | 0.388845 |
https://oeis.org/A264932 | 1,716,499,181,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00848.warc.gz | 371,731,844 | 4,112 | The OEIS mourns the passing of Jim Simons and is grateful to the Simons Foundation for its support of research in many branches of science, including the OEIS.
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A264932 a(1) = 1, a(2) = 3, a(n)= H_4(a(n-1),a(n-2)). 1
1, 3, 3, 7625597484987 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS See A054871 for definitions and key links. Possible other sequence with the same 4 entries: a(1)=1; a(2)= 3; a(n)=a(n-1)^a(n-2)^a(n-2); LINKS Table of n, a(n) for n=1..4. MATHEMATICA RecurrenceTable[{a[1]==1, a[2]==3, a[n]==a[n-1]^a[n-2]^a[n-2]}, a, {n, 4}] (* Vincenzo Librandi, Dec 17 2015 *) PROG (Magma) I:=[1, 3]; [n le 2 select I[n] else Self(n-1)^Self(n-2)^Self(n-2): n in [1..5]]; // Vincenzo Librandi, Dec 17 2015 CROSSREFS Sequence in context: A076253 A081174 A292832 * A171604 A132139 A045951 Adjacent sequences: A264929 A264930 A264931 * A264933 A264934 A264935 KEYWORD nonn AUTHOR Natan Arie Consigli, Dec 16 2015 STATUS approved
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Last modified May 23 16:36 EDT 2024. Contains 372765 sequences. (Running on oeis4.) | 491 | 1,440 | {"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-2024-22 | latest | en | 0.642186 |
https://www.youthkiawaaz.com/web-stories/most-repeated-questions-in-cbse-maths-class-10/ | 1,719,220,900,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865348.3/warc/CC-MAIN-20240624084108-20240624114108-00377.warc.gz | 938,273,464 | 12,527 | ## Most Repeated Questions In CBSE Maths Class 10?
By Ishika
13 March, 2024
In CBSE Class 10 Mathematics, students often encounter a variety of question types that test their understanding of mathematical concepts and problem-solving skills.
# "
## Here are the most repeated types of questions in CBSE maths class 10:
### 1. Direct Questions:
These are straightforward questions that require recalling definitions, formulas, or theorems. They usually involve definitions, simple calculations, or identifying properties. For example, “Define a quadratic equation and give an example.”
These questions present real-life situations that require mathematical analysis and problem-solving skills. They may involve calculating distances, areas, volumes, or solving equations based on given conditions. For example, “A train travels from City A to City B at a speed of 60 km/h and returns at a speed of 80 km/h. Find the average speed for the whole journey.”
## 2. Word Problems:
These questions require students to prove geometric theorems or algebraic identities using deductive reasoning and logical arguments. For example, “Prove that the opposite sides of a parallelogram are equal.”
### 4. Graph-based Questions:
These questions involve interpreting graphs, drawing graphs based on given data, or solving problems using graphical representations. For example, “Plot the points (2, 3), (4, 5), and (6, 7) on the coordinate plane and connect them to form a triangle. Determine its area.”
By understanding and practicing these types of questions, students can enhance their proficiency in CBSE Class 10 Mathematics and excel in their exams. | 338 | 1,650 | {"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-2024-26 | latest | en | 0.910914 |
https://maplesoft.com/support/help/maple/view.aspx?path=PolyhedralSets%2FZPolyhedralSets%2FIsEmpty | 1,726,804,010,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652130.6/warc/CC-MAIN-20240920022257-20240920052257-00498.warc.gz | 341,532,171 | 24,369 | IsEmpty - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
PolyhedralSets[ZPolyhedralSets]
IsEmpty
check whether a given ZPolyhedralSet is empty or not
Calling Sequence IsEmpty(zpoly)
Parameters
zpoly -
Description
• IsEmpty(zpoly) returns true if zpoly is empty, and otherwise returns false.
Examples
> $\mathrm{with}\left(\mathrm{PolyhedralSets}\right):$
> $\mathrm{with}\left(\mathrm{ZPolyhedralSets}\right):$
Create a Z-polyhedron in the two-dimensional space with a system of three linear inequalities.
> $\mathrm{ineqs}≔\left[x+\frac{2y}{3}\le 4,-x+y\le \frac{7}{8},\frac{1}{11}x+\frac{24}{11}\le y\right]$
${\mathrm{ineqs}}{≔}\left[{x}{+}\frac{{2}{}{y}}{{3}}{\le }{4}{,}{-}{x}{+}{y}{\le }\frac{{7}}{{8}}{,}\frac{{x}}{{11}}{+}\frac{{24}}{{11}}{\le }{y}\right]$ (1)
> $\mathrm{zp}≔\mathrm{ZPolyhedralSet}\left(\mathrm{ineqs},\left[x,y\right]\right)$
Check whether it is empty.
> $\mathrm{ZPolyhedralSets}:-\mathrm{IsEmpty}\left(\mathrm{zp}\right)$
${\mathrm{true}}$ (2)
Double-check that latter answer.
> $\mathrm{EnumerateIntegerPoints}\left(\mathrm{zp}\right)$
$\left[\right]$ (3)
Create another Z-polyhedron in the three-dimensional space with a system of linear inequalities.
> $\mathrm{ineqs}≔\left[0\le -16+2y+z,0\le -72+4x+4y+3z,0\le 2y-z,0\le -24+4x+4y-3z,0\le -4x+4y+3z,0\le 48-4x+4y-3z,0\le 48-4x-4y+3z,0\le 8-2y+z,0\le -24+4x-4y+3z,0\le 24-2y-z,0\le 24+4x-4y-3z,0\le 96-4x-4y-3z\right]$
${\mathrm{ineqs}}{≔}\left[{0}{\le }{-}{16}{+}{2}{}{y}{+}{z}{,}{0}{\le }{-}{72}{+}{4}{}{x}{+}{4}{}{y}{+}{3}{}{z}{,}{0}{\le }{2}{}{y}{-}{z}{,}{0}{\le }{-}{24}{+}{4}{}{x}{+}{4}{}{y}{-}{3}{}{z}{,}{0}{\le }{-}{4}{}{x}{+}{4}{}{y}{+}{3}{}{z}{,}{0}{\le }{48}{-}{4}{}{x}{+}{4}{}{y}{-}{3}{}{z}{,}{0}{\le }{48}{-}{4}{}{x}{-}{4}{}{y}{+}{3}{}{z}{,}{0}{\le }{8}{-}{2}{}{y}{+}{z}{,}{0}{\le }{-}{24}{+}{4}{}{x}{-}{4}{}{y}{+}{3}{}{z}{,}{0}{\le }{24}{-}{2}{}{y}{-}{z}{,}{0}{\le }{24}{+}{4}{}{x}{-}{4}{}{y}{-}{3}{}{z}{,}{0}{\le }{96}{-}{4}{}{x}{-}{4}{}{y}{-}{3}{}{z}\right]$ (4)
> $L≔\mathrm{Lattice}\left(\mathrm{Matrix}\left(\left[\left[1,0,2\right],\left[0,-1,1\right],\left[0,0,2\right]\right]\right),\mathrm{Vector}\left(\left[0,0,1\right]\right)\right)$
${L}{≔}{\mathrm{Lattice}}{}\left(\left[\begin{array}{ccc}{1}& {0}& {2}\\ {0}& {-1}& {1}\\ {0}& {0}& {2}\end{array}\right]{,}\left[\begin{array}{c}{0}\\ {0}\\ {1}\end{array}\right]\right)$ (5)
> $\mathrm{vars}≔\left[x,y,z\right]$
${\mathrm{vars}}{≔}\left[{x}{,}{y}{,}{z}\right]$ (6)
> $\mathrm{zp}≔\mathrm{ZPolyhedralSet}\left(\mathrm{ineqs},\mathrm{vars},':-\mathrm{lattice}'=L\right)$
${\mathrm{zp}}{≔}\left\{\begin{array}{lll}{\text{Relations}}& {:}& \left\{\begin{array}{:}{0}{\le }{2}{}{y}{-}{z}\\ {0}{\le }{-}{16}{+}{2}{}{y}{+}{z}\\ {0}{\le }{8}{-}{2}{}{y}{+}{z}\\ {0}{\le }{24}{-}{2}{}{y}{-}{z}\\ {0}{\le }{-}{4}{}{x}{+}{4}{}{y}{+}{3}{}{z}\\ {0}{\le }{-}{72}{+}{4}{}{x}{+}{4}{}{y}{+}{3}{}{z}\\ {0}{\le }{-}{24}{+}{4}{}{x}{-}{4}{}{y}{+}{3}{}{z}\\ {0}{\le }{-}{24}{+}{4}{}{x}{+}{4}{}{y}{-}{3}{}{z}\\ {0}{\le }{24}{+}{4}{}{x}{-}{4}{}{y}{-}{3}{}{z}\\ {0}{\le }{48}{-}{4}{}{x}{-}{4}{}{y}{+}{3}{}{z}\\ {0}{\le }{48}{-}{4}{}{x}{+}{4}{}{y}{-}{3}{}{z}\\ {0}{\le }{96}{-}{4}{}{x}{-}{4}{}{y}{-}{3}{}{z}\end{array}\right\\\ {\text{Variables}}& {:}& \left[{x}{,}{y}{,}{z}\right]\\ {\text{Parameters}}& {:}& \left[\right]\\ {\text{ParameterConstraints}}& {:}& \left\{\begin{array}{}\end{array}\right\\\ {\text{Lattice}}& {:}& {\text{ZSpan}}\left(\left[\begin{array}{ccc}{1}& {0}& {2}\\ {0}& {-1}& {1}\\ {0}& {0}& {2}\end{array}\right]{,}{,}{,}\left[\begin{array}{c}{0}\\ {0}\\ {1}\end{array}\right]\right)\end{array}\right\$ (7)
Check whether it is empty.
> $\mathrm{ZPolyhedralSets}:-\mathrm{IsEmpty}\left(\mathrm{zp}\right)$
${\mathrm{false}}$ (8)
Double-check that latter answer.
> $\mathrm{Point_from_pzp}≔\mathrm{SamplePoint}\left(\mathrm{zp}\right)$
${\mathrm{Point_from_pzp}}{≔}\left[{x}{=}{10}{,}{y}{=}{7}{,}{z}{=}{9}\right]$ (9)
References
Rachid Seghir, Vincent Loechner, and Benoı̂t Meister. "Integer affine transformations of parametric Z-polytopes and applications to loop nest optimization." Proceedings of TACO, Vol. 9(2):8:1–8:27, 2012.
Rui-Juan Jing and Marc Moreno Maza. "Computing the Integer Points of a Polyhedron, I: Algorithm." Proceedings of CASC 2017: 225-241, Springer.
Rui-Juan Jing and Marc Moreno Maza. "Computing the Integer Points of a Polyhedron, II: Complexity Estimates." Proceedings of CASC 2017: 242-256, Springer.
Compatibility
• The PolyhedralSets:-ZPolyhedralSets:-IsEmpty command was introduced in Maple 2023.
• For more information on Maple 2023 changes, see Updates in Maple 2023. | 2,054 | 4,673 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 21, "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} | 3.34375 | 3 | CC-MAIN-2024-38 | latest | en | 0.380902 |
https://www.examsegg.com/upseat-maths-practice-paper.html | 1,642,378,412,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300253.51/warc/CC-MAIN-20220117000754-20220117030754-00564.warc.gz | 835,322,234 | 20,026 | Engineering Entrance Sample Papers
# UPSEE Maths Practice Paper
#### Uttar Pradesh State Entrance Exam (UPSEE) Maths Practice Paper with Answers [PDF]:
Ques: Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. The relation R is
(a) Reflexive
(b) Transitive
(c) Not symmetric
(d) A function
Ans:- (c)
Ques: x + y + z + 2 = 0 together with x + y + z + 3 = 0 represents in space
(a) A line
(b) A point
(c) A plane
(d) None of these
Ans:- (d)
Ques: n – 1C3 + n – 1C4 > nC3 then the value of n is
(a) 7
(b) < 7
(c) > 7
(d) None of these
Ans:- (c)
Ques: Three forces P, Q and R act along the sides BC, AC and BA of an equilateral triangle ABC. If their resultant is a force parallel to BC through the centroid of the triangle ABC, then
(a) P = Q = R
(b) P = 2Q = 2R
(c) 2P = Q + 2R
(d) 2P = 2Q = R
Ans:- (b)
Ques: If a matrix A is such that 4A3 + 2A2 + 7A + I = O, then A–1 equals
(a) (4A2 + 2A + 7I)
(b) –(4A2 + 2A + 7I)
(c) –(4A2 – 2A + 7I)
(d) (4A2 + 2A – 7I)
Ans:- (b)
Related: EAMCET Chemistry Sample Paper
Ques: A particle possess two velocities simultaneously at an angle of tan–1 12/5 to each other. Their resultant is 15 m/s. If one velocity is 13m/s, then the other will be
(a) 5 m/s
(b) 4 m/s
(c) 12 m/s
(d) 13m/s
Ans:- (b)
Ques: If sin x + sin2 x = 1, then the value of expression cos12 x + 3 cos10 x + 3 cos8 x + cos6 x – 1 is equal to
(a) 0
(b) 1
(c) –1
(d) 2
Ans:- (a)
Ques: A box contains 100 tickets numbered 1, 2 …… 100. Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than 10. The minimum number on them is 5 with probability
(a) 1/8
(b) 13/15
(c) 1/7
(d) None of these
Ans:- (b)
Ques: If y = 2x is a chord of the circle x2 + y2 – 10x = 0, then the equation of the circle of which this chord is a diameter, is
(a) x2 + y2 – 2x + 4y = 0
(b) x2 + y2 + 2x + 4y = 0
(c) x2 + y2 – 2x + 4y = 0
(c) x2 + y2 – 2x – 4y = 0
Ans:- (d)
Ques: If there are n harmonic means between 1 and 1/31 and the ratio of 7th and (n – 1)th harmonic means is 9 : 5
then the and value of n will be
(a) 12
(b) 13
(c) 14
(d) 15
Ans:- (c)
Ques: If mean = (3 median – mode) k, then the value of k is
(a) 1
(b) 2
(c) 1/2
(d) 3/2
Ans:- (c)
Ques: If | a x b | = 4 and | a.b |= 2, then |a|2 |b|2 =
(a) 2
(b) 6
(c) 8
(d) 20
Ans:- (d)
Related: BITS Maths Sample Paper
Ques: If the lines of regression are 3x + 12y = 19 and 3y + 9x = 46 then rxy will be
(a) 0.289
(b) – 0.289
(c) 0.209
(d) None of these
Ans:- (b)
Ques: The value of the nearest root of the equation x3 + x – 1 = 0 after third iteration by Newton-Raphson method near x = 1 is
(a) 0.51
(b) 0.42
(c) 0.67
(d) 0.55
Ans:- (c)
Ques: Let G denote the set of all n x n non-singular matrices with rational numbers as entries. Then under multiplication
(a) G is a subgroup
(b) G is a finite abelian group
(c) G is an infinite, non-abelian group
(d) G is infinite, abelian
Ans:- (c)
Ques: In Boolean Algebra, the zero element ‘0’
(a) Has two values
(b) Is unique
(c) As atleast two values
(d) None of these
Ans:- (b)
Related: VITEEE Sample Paper
Ques: What is the decimal equivalent of the octal number 1217
(a) 640
(b) 620
(c) 650
(d) 655
Ans:- (d) | 1,294 | 3,204 | {"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} | 3.859375 | 4 | CC-MAIN-2022-05 | latest | en | 0.795889 |
http://teachersnetwork.org/teachnet-lab/ps101/bglasgold/rocks/lesson_4.htm | 1,670,140,891,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710968.29/warc/CC-MAIN-20221204072040-20221204102040-00305.warc.gz | 53,026,661 | 4,230 | LESSON 4: HOW CAN WE TELL THE HARDNESS OF ROCKS?
AIMS: 1. What is Moh's Scale of Hardness?
2. What is the scratch test?
MOTIVATION: Have a sample of the following rocks on display for the children to touch: The first two rocks (talc and feldspar) were taken from
http://ivyhall.district96.k12.il.us/4th/kkhp/RocksandMinerals/PinkFeldspar.html
and the third picture, quartz, is from http://casdn.neu.edu/~geology/department/staff/colgan/iceland/minerals/quartz.htm
The fourth sample, mica, is from http://cobweb.net/~bug2/rock4.htm
The fifth sample, granite, is from http://ivyhall.district96.k12.il.us/4th/kkhp/RocksandMinerals/rocks.html
### PROCEDURE:
1. Have students use their fingernails to see if they can make a scratch mark on any of the rock samples. Have them log on to http://ivyhall.district96.k12.il.us/4th/kkhp/RocksandMinerals/RockQuiz.html and read how they can judge the hardness of the various objects they will use to test their rocks.
### The following Moh's Scale of Hardness was taken from
and the picture samples are from http://ivyhall.district96.k12.il.us/4th/kkhp/RocksandMinerals/RockQuiz.html
## topaz
#### spinel
(available in rock shops)
## diamond
### 2. Students will perform a scratch test on the five rock samples. See the following lab activity:
LAB ACTIVITY:
Problem: How hard is a rock?
Hypothesis: We think if we use our fingernail, penny, and an iron nail, then we can determine how hard a rock is.
Materials: penny, nail, crayons, 5 rock samples, fingernail
Procedure: 1. Label each rock 1, 2, 3, 4, 5. 2. Try to scratch each rock. Use your fingernail first, then the penny, and last, the nail. 3. Record what happened. Put an X on the chart of the rock was scratched.
Observations: Use the chart to record your observations.
Rock Fingernail Penny Nail #1 #2 #3 #4 #5
1. How many rocks could you scratch with your fingernail?_________ Penny?________ Nail?_________
Conclusion:
1. Which rock was the hardest? How do you know? ________________________________
_________________________________________________________________________
2. Which rock is the softest? How do you know?___________________________________
_________________________________________________________________________
_________________________________________________________________________ | 612 | 2,357 | {"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} | 3.890625 | 4 | CC-MAIN-2022-49 | latest | en | 0.823211 |
https://www.numbersaplenty.com/15021216 | 1,686,300,558,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655446.86/warc/CC-MAIN-20230609064417-20230609094417-00544.warc.gz | 970,186,920 | 3,598 | Search a number
15021216 = 253277451
BaseRepresentation
bin111001010011…
…010010100000
31001021011020100
4321103102200
512321134331
61253542400
7241451430
oct71232240
931234210
1015021216
11852a731
125044a00
13315c1c2
141dd02c0
1514baae6
hexe534a0
15021216 has 72 divisors (see below), whose sum is σ = 48825504. Its totient is φ = 4291200.
The previous prime is 15021197. The next prime is 15021239. The reversal of 15021216 is 61212051.
It is a tau number, because it is divible by the number of its divisors (72).
It is a Harshad number since it is a multiple of its sum of digits (18).
It is an unprimeable number.
It is a polite number, since it can be written in 11 ways as a sum of consecutive naturals, for example, 1710 + ... + 5741.
It is an arithmetic number, because the mean of its divisors is an integer number (678132).
Almost surely, 215021216 is an apocalyptic number.
15021216 is a gapful number since it is divisible by the number (16) formed by its first and last digit.
It is an amenable number.
15021216 is an abundant number, since it is smaller than the sum of its proper divisors (33804288).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
15021216 is a wasteful number, since it uses less digits than its factorization.
15021216 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 7474 (or 7463 counting only the distinct ones).
The product of its (nonzero) digits is 120, while the sum is 18.
The square root of 15021216 is about 3875.7213522130. The cubic root of 15021216 is about 246.7374263355.
Adding to 15021216 its reverse (61212051), we get a palindrome (76233267).
The spelling of 15021216 in words is "fifteen million, twenty-one thousand, two hundred sixteen". | 530 | 1,802 | {"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} | 3.265625 | 3 | CC-MAIN-2023-23 | latest | en | 0.882315 |
https://www.educationindex.com/essay/Confidence-Intervals-F3W88JAJ2Z | 1,701,247,998,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100057.69/warc/CC-MAIN-20231129073519-20231129103519-00154.warc.gz | 878,694,133 | 10,703 | # Confidence Intervals
2086 words 9 pages
Confidence Intervals
Consider the following question: someone takes a sample from a population and finds both the sample mean and the sample standard deviation. What can he learn from this sample mean about the population mean?
This is an important problem and is addressed by the Central Limit Theorem. For now, let us not bother about what this theorem states but we will look at how it could help us in answering our question.
The Central Limit Theorem tells us that if we take very many samples the means of all these samples will lie in an interval around the population mean. Some sample means will be larger than the population mean, some will be smaller. The Central Limit Theorem goes on to state that 95% of the sample means will lie
…show more content…
Example 2:
The Roman Arches is an Italian restaurant. The manager wants to estimate the average amount a customer spends for lunch, Monday through Friday. A random sample of 115 customers' lunch tabs gives a mean of \$12.75 with a standard deviation of \$3.37. What is the 99% confidence interval for the mean of the lunch tabs for all customers?
Solution:
[pic]. Thus the manager can be 99% confident that the mean of all customers' lunch tabs is somewhere between \$11.94 and \$13.56.
Comment: If the population standard deviation is known, then we could use the formula for those cases where the sample size is less than 30. But since for most practical situations we will not know the population standard deviation, we will simply use the formula for those cases where the sample size is greater than 30.
Confidence Intervals when Sample Proportions are known
In situations where a sample proportion, denoted by[pic], is known, and we want to get information about the population proportion p, we will use the formula [pic].
Example 3:
A representative of a consumer organization took a random sample of 250 egg cartons from the dairy section of a very large supermarket and found that 80 cartons had at least one broken egg. Find a 90% confidence interval for the proportion of cartons in the population that had at least one broken egg in them.
Solution:
Use[pic][pic]. Thus we are 90% confident that the proportion of egg cartons in the population containing
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• ###### Biophysical Ecology and Pattern Recognition
1675 words | 7 pages | 657 | 2,856 | {"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} | 4.40625 | 4 | CC-MAIN-2023-50 | latest | en | 0.933576 |
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