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Biology Exams With Solutions

Statistics

AP Biology Exams

• University of Georgia has an AP Biology site, with many example essay questions discussed.
• CSUN has a Source Book for Teaching Science with a compilation of AP Biology Essay Questions and the associated scoring guidelines.
• 4tests.com lets you take an online exam resembling AP Biology.
• The College Board information on the AP Biology Exam.
• College Board sample exams, 1999-2010.

Introductory Biology Exams with Solutions

• Bryn Mawr Bio 103 exams with solutions.
• George Mason University Biology 104, Introductory Biology II, Exams and Answers, 2006-2010, from Arndt F. Laemmerzahl.
• University of Georgia Bio 102 sample tests. From Dr. Peggy Brickman.
• Kansas State Biology 198 sample exam with solutions.
• Missouri State Biology 121 sample exams with solutions.
• MIT Introductory Biology sample exams with solutions from 2004. Also see 2005 and 2006 .
• Ramapo College Biology 1 sample quizzes with solutions.
• Riverside Community College Biology 1 sample problems with solutions.
• San Jose State Biology 109 old exams with solutions. From Dr. Sally Veregge.
• Union County Community College Bio 101 sample exams with solutions. From Dr. Igor V. Oksov.
• UIC BIOS 101 sample problems.
• University of Maine BIO 110 has sample exams with solutions. Multiple choice.
• University of Tennessee, Martin General Biology sample exams. No solutions. Also some review materials. From Dr. T. David Pitts.

Molecular Biology Exams with Solutions

• Columbia C2005 Introduction to Cellular and Molecular Biology. Sample Exam with solutions.
• Davidson College sample Molecular Biology exams, 1997-2010.
• Lewis and Clark University Molecular Biology L211, Old exams with answer keys. Text: Molecular Biology of the Gene by Watson.
• Michigan State BMB 526 Molecular Biology and Genetics.
• UC Berkeley Molecular and Cell Biology 1A sample problems with solutions. Also see Practice Practical Exam and Answers .
• UC Berkeley Extension X107 Principles of Molecular Biology Tests with Answer Keys. From Bob Bruner. Text: Molecular Biology by Weaver.
• UCSD BIMM 100 Molecular Biology, Sample Exams and Exam Keys.

Cell Biology Exams with Solutions

• Carnegie Mellon Cell Biology 03-240 Exams and Answers.
• University of Iowa Cell Biology BSc-303 exams with solutions.
• Texas A
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5 Free Solutions To Create A Question-Answer Website

Question answer websites (like Stack Overflow ) are great platforms for enabling communities to share knowledge and help each other.

Although they are kinda similar-in-function with the forum applications (check 9 Widely-Used And Free Forum Softwares ), the main focus is finding the (best) answers .

Here are 5 free and open source or hosted platforms to create your own question-answer website:

Question2Answer is an application that is built with PHP-MySQL that has a points-based reputation system.

Users can comment into any questions and they can be tagged.

It can betotallystyled with CSS and has an admin interface to manage the whole application.

QSQA enables anyone to instantly create a full-featured question-answer community where users can earn point/badges for their contribution.

It is written in Python and powered by the Django application framework.

Answerbase is a powerful and hosted service for creating Q A websites without any installation.

The front-end can be customized completely with the help of ready-to-use templates, ability to update every image/text and inserting custom CSS styles.

A feature-rich admin panel eases the management of the system, users and content.

Once the website is created, it gets a mobile-friendly version automatically and there is an API provided for integration with 3rd party products.

This question-answer application is written in Ruby, MongoMapper and MongoDB.

It is multilingual and, optionally, users can create their own websites in a free hosted platform rather than installing it.

Qwench is a simple-yet-functional application.

It comes with a points/rewards system, an entry can be posted as a knowledge-base article (for corporate sites) and it can be customized via CSS.

Qhub is a free hosted platform for creating a q a website where the look-feel can be customized and members can be managed easily.

You can optionally use a URL under Qhub or your own domain name.

And Others:

• Stacked . it is an ASP.NET alternative but not improved for a while.
• CNPROG . a pretty powerful one which is Django-powered (QSQA is a for of this application)
• AskBot . a fork of CNPROG.

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Mathematical Induction – Problems With Solutions

The principle of mathematical induction is used to prove that a given proposition (formula, equality, inequality�) is true for all positive integer numbers greater than or equal to some integer N.

Let us denote the proposition in question by P (n), where n is a positive integer. The proof involves two steps:

Step 1: We first establish that the proposition P (n) is true for the lowest possible value of the positive integer n.

Step 2: We assume that P (k) is true and establish that P (k+1) is also true

Use mathematical induction to prove that

1 + 2 + 3 +. + n = n (n + 1) / 2

for all positive integers n.

Solution to Problem 1:

• Let the statement P (n) be

1 + 2 + 3 +. + n = n (n + 1) / 2

• STEP 1: We first show that p (1) is true.

Right Side = 1 (1 + 1) / 2 = 1

• Both sides of the statement are equal hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 + 2 + 3 +. + k = k (k + 1) / 2

• and show that p (k + 1) is true by adding k + 1 to both sides of the above statement

1 + 2 + 3 +. + k + (k + 1) = k (k + 1) / 2 + (k + 1)

• The last statement may be written as

1 + 2 + 3 +. + k + (k + 1) = (k + 1)(k + 2) / 2

• Which is the statement p(k + 1).
• 1 2 + 2 2 + 3 2 +. + n 2 = n (n + 1) (2n + 1)/ 6

For all positive integers n.

Solution to Problem 2:

• Statement P (n) is defined by

1 2 + 2 2 + 3 2 +. + n 2 = n (n + 1) (2n + 1)/ 2

• STEP 1: We first show that p (1) is true.

Left Side = 1 2 = 1

Right Side = 1 (1 + 1) (2*1 + 1)/ 6 = 1

• Both sides of the statement are equal hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 2 + 2 2 + 3 2 +. + k 2 = k (k + 1) (2k + 1)/ 6

• and show that p (k + 1) is true by adding (k + 1) 2 to both sides of the above statement

1 2 + 2 2 + 3 2 +. + k 2 + (k + 1) 2 = k (k + 1) (2k + 1)/ 6 + (k + 1) 2

• Set common denominator and factor k + 1 on the right side

= (k + 1) [ k (2k + 1)+ 6 (k + 1) ] /6

• Expand k (2k + 1)+ 6 (k + 1)

= (k + 1) [ 2k 2 + 7k + 6 ] /6

• Now factor 2k 2 + 7k + 6.

= (k + 1) [ (k + 2) (2k + 3) ] /6

• We have started from the statement P(k) and have shown that

1 2 + 2 2 + 3 2 +. + k 2 + (k + 1) 2 = (k + 1) [ (k + 2) (2k + 3) ] /6

• Which is the statement P(k + 1).
• Use mathematical induction to prove that

1 3 + 2 3 + 3 3 +. + n 3 = n 2 (n + 1) 2 / 4

for all positive integers n.

Solution to Problem 3:

• Statement P (n) is defined by

1 3 + 2 3 + 3 3 +. + n 3 = n 2 (n + 1) 2 / 4

• STEP 1: We first show that p (1) is true.

Left Side = 1 3 = 1

Right Side = 1 2 (1 + 1) 2 / 4 = 1

• hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 3 + 2 3 + 3 3 +. + k 3 = k 2 (k + 1) 2 / 4

• add (k + 1) 3 to both sides

1 3 + 2 3 + 3 3 +. + k 3 + (k + 1) 3 = k 2 (k + 1) 2 / 4 + (k + 1) 3

• factor (k + 1) 2 on the right side

= (k + 1) 2 [ k 2 / 4 + (k + 1) ]

• set to common denominator and group

= (k + 1) 2 [ k 2 + 4 k + 4 ] / 4

= (k + 1) 2 [ (k + 2) 2 ] / 4

• We have started from the statement P(k) and have shown that

1 3 + 2 3 + 3 3 +. + k 3 + (k + 1) 3 = (k + 1) 2 [ (k + 2) 2 ] / 4

• Which is the statement P(k + 1).
• Prove that for any positive integer number n. n 3 + 2 n is divisible by 3

Solution to Problem 4:

• Statement P (n) is defined by

n 3 + 2 n is divisible by 3

• STEP 1: We first show that p (1) is true. Let n = 1 and calculate n 3 + 2n

3 is divisible by 3

• hence p (1) is true.
• STEP 2: We now assume that p (k) is true

k 3 + 2 k is divisible by 3

is equivalent to

k 3 + 2 k = 3 M. where M is a positive integer.

• We now consider the algebraic expression (k + 1) 3 + 2 (k + 1); expand it and group like terms

(k + 1) 3 + 2 (k + 1) = k 3 + 3 k 2 + 5 k + 3

= [ k 3 + 2 k] + [3 k 2 + 3 k + 3]

= 3 M + 3 [ k 2 + k + 1 ] = 3 [ M + k 2 + k + 1 ]

• Hence (k + 1) 3 + 2 (k + 1) is also divisible by 3 and therefore statement P(k + 1) is true.
• Prove that 3 n n 2 for n = 1, n = 2 and use the mathematical induction to prove that 3 n n 2 for n a positive integer greater than 2.

Solution to Problem 5:

• Statement P (n) is defined by
• STEP 1: We first show that p (1) is true. Let n = 1 and calculate 3 1 and 1 2 and compare them
• 3 is greater than 1 and hence p (1) is true.
• Let us also show that P(2) is true.
• Hence P(2) is also true.
• STEP 2: We now assume that p (k) is true
• Multiply both sides of the above inequality by 3

3 * 3 k 3 * k 2

• The left side is equal to 3 k + 1. For k , 2, we can write

k 2 2 k and k 2 1

• We now combine the above inequalities by adding the left hand sides and the right hand sides of the two inequalities

2 k 2 2 k + 1

• We now add k 2 to both sides of the above inequality to obtain the inequality

3 k 2 k 2 + 2 k + 1

• Factor the right side we can write

3 * k 2 (k + 1) 2

• If 3 * 3 k 3 * k 2 and 3 * k 2 (k + 1) 2 then

3 * 3 k (k + 1) 2

• Rewrite the left side as 3 k + 1

3 k + 1 (k + 1) 2

• Which proves tha P(k + 1) is true
• Prove that n. 2 n for n a positive integer greater than or equal to 4. (Note: n! is n factorial and is given by 1 * 2 *. * (n-1)*n.)

Solution to Problem 6:

• Statement P (n) is defined by
• STEP 1: We first show that p (4) is true. Let n = 4 and calculate 4. and 2 n and compare them
• 24 is greater than 16 and hence p (4) is true.
• STEP 2: We now assume that p (k) is true
• Multiply both sides of the above inequality by k + 1

k! (k + 1) 2 k (k + 1)

• The left side is equal to (k + 1). For k , 4, we can write
• Multiply both sides of the above inequality by 2 k to obtain

2 k (k + 1) 2 * 2 k

• The above inequality may be written

2 k (k + 1) 2 k + 1

• We have proved that (k + 1)! 2 k (k + 1) and 2 k (k + 1) 2 k + 1 we can now write
• We have assumed that statement P(k) is true and proved that statment P(k+1) is also true.
• Use mathematical induction to prove De Moivre’s theorem

[ R (cos t + i sin t) ] n = R n (cos nt + i sin nt)

for n a positive integer.

Solution to Problem 7:

• STEP 1: For n = 1

[ R (cos t + i sin t) ] 1 = R 1 (cos 1*t + i sin 1*t)

• It can easily be seen that the two sides are equal.
• STEP 2: We now assume that the theorem is true for n = k, hence

[ R (cos t + i sin t) ] k = R k (cos kt + i sin kt)

• Multiply both sides of the above equation by R (cos t + i sin t)

[ R (cos t + i sin t) ] k R (cos t + i sin t) = R k (cos kt + i sin kt) R (cos t + i sin t)

• Rewrite the above as follows

[ R (cos t + i sin t) ] k + 1 = R k + 1 [ (cos kt cos t – sin kt sin t) + i (sin kt cos t + cos kt sin t) ]

• Trigonometric identities can be used to write the trigonometric expressions (cos kt cos t – sin kt sin t) and (sin kt cos t + cos kt sin t) as follows

(cos kt cos t – sin kt sin t) = cos(kt + t) = cos(k + 1)t

(sin kt cos t + cos kt sin t) = sin(kt + t) = sin(k + 1)t

• Substitute the above into the last equation to obtain

[ R (cos t + i sin t) ] k + 1 = R k + 1 [ cos (k + 1)t + sin(k + 1)t ]

• It has been established that the theorem is true for n = 1 and that if it assumed true for n = k it is true for n = k + 1.

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It’s Easy To Get Started

Our 27 Gifted Faculty have written and hosted more than 105 video training modules in six categories: Leadership, Youth Development Play, Mental Health Behavior, Physical Emotional Safety, Supervision, and Creative Literacy. This virtual library, called “Leadership Essentials,” has awesome content for new staff, returning staff, program staff, and supervisory staff. Watch some full-length samples and you’ll instantly see why day camps, overnight camps, parks recs, and summer schools all choose EOT to complement their on-site training.

To view this video please enable JavaScript, and consider upgrading to a web browser that supports HTML5 video

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Previews From Our Faculty

Our 27 Gifted Faculty have written and hosted more than 105 video training modules in six categories: Leadership, Youth Development Play, Mental Health Behavior, Physical Emotional Safety, Supervision, and Creative Literacy. This virtual library, called “Leadership Essentials,” has awesome content for new staff, returning staff, program staff, and supervisory staff. Watch some full-length samples and you’ll instantly see why day camps, overnight camps, parks recs, and summer schools all choose EOT to complement their on-site training.

To view this video please enable JavaScript, and consider upgrading to a web browser that supports HTML5 video

To view this video please enable JavaScript, and consider upgrading to a web browser that supports HTML5 video

To view this video please enable JavaScript, and consider upgrading to a web browser that supports HTML5 video

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Algebra: Fraction Problems

In these lessons, we will learn how to solve fraction word problems that deal with fractions and algebra. Remember to read the question carefully to determine the numerator and denominator of the fraction.

We will also learn how to solve word problems that involve comparing fractions, adding mixed numbers, subtracting mixed numbers, multiplying fractions and dividing fractions.

Fraction Word Problems using Algebra

2/3 of a number is 14. What is the number?

Step 1: Assign variables.

Step 2: Solve the equation

The numerator of a fraction is 3 less than the denominator. When both the numerator and denominator are increased by 4, the fraction is increased by fraction.

Let the numerator be x,
then the denominator is x + 3,
and the fraction is $$\frac<>$$
When the numerator and denominator are increased by 4, the fraction is $$\frac<><>$$
$$\frac<><> – \frac<> = \frac<<12>><<77>>$$
77(x + 4)(x + 3) – 77x(x+7) = 12(x + 7)(x + 3)
77x 2 + 539x + 924 – 77x 2 – 539x = 12x 2 + 120x + 252
12x 2 + 120x – 672 = 0
x 2 + 10x – 56 = 0
(x – 4)(x + 14) = 0
x = 4 (negative answer not applicable in this case)

How to solve Fraction Word Problems using Algebra?
Examples:
(1) The denominator of a fraction is 5 more than the numerator. If 1 is subtracted from the numerator, the resulting fraction is 1/3. Find the original fraction.
(2) If 3 is subtracted from the numerator of a fraction, the value of the resulting fraction is 1/2. If 13 is added to the denominator of the original fraction, the value of the new fraction is 1/3. Find the original fraction.
(3) A fraction has a value of 3/4. When 14 is added to the numerator, the resulting fraction has a value equal to the reciprocal of the original fraction, Find the original fraction.

• Show Step-by-step Solutions

Algebra Word Problems with Fractional Equations
Solving a fraction equation that appears in a word problem
Example:
One third of a number is 6 more than one fourth of the number. Find the number.

• Show Step-by-step Solutions

Fraction and Decimal Word Problems
How to solve algebra word problems with fractions and decimals?
Examples:
(1) If 1/2 of the cards had been sold and there were 172 cards left, how many cards were printed?
(2) Only 1/3 of the university students wanted to become teachers. If 3,360 did not wan to become teachers, how many university were there?
(3) Rodney guessed the total was 34.71, but this was 8.9 times the total. What was the total?

• Show Step-by-step Solutions

Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.

You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

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Fraction Word Problems: Examples

Here are some examples of fraction word problems.
The first example is a one-step word problem.
The second example shows how blocks can be used to help illustrate the problem.
The third example is a two-step word problem.
More examples and solutions using the bar modeling method to solve fraction word problems are shown on the videos. The bar modeling method also called tape diagrams are use in Singapore Math and Common Core.

Martha spent of her allowance on food and shopping. What fraction of her allowance had she left?

She had of her allowance left.

of a group of children were girls. If there were 24 girls, how many children were there in the group?

1 unit = 24 3 = 8

5 units = 5 8 = 40

There were 40 children in the group.

Sam had 120 teddy bears in his toy store. He sold of them at $12 each. How much did he receive? Step 1: Find the number of teddy bears sold. He sold 80 teddy bears. Step 2: Find how much money he received. 80 12 = 960 He received$960.

Bar Modeling with Fractions
Examples:
1) Grace thought that a plane journey would take 7/10 hr but the actual journey took 1/5 hr longer. How long did the actual journey take?
2) Timothy took 2/3 hr to paint a portrait. This was 1/3 hr shorter than the time he took to paint scenery. How long did he take to paint scenery?

• Show Step-by-step Solutions

Word Problem on Subtracting Fractions From Whole Numbers
Example: A craft store has a 9 yd spool of ribbon. In the morning, a customer buys 1/5 yd of ribbon from the spool. In the afternoon another customer buys 7/10 yd of ribbon from the spool. How much ribbon is left?

• Show Step-by-step Solutions
How to solve real world problems involving multiplication of fractions by using visual fraction models or equations to represent the problem?
Examples:
1. At the animal shelter 4/6 of the animals are cats. Of the cats 1/2 are male. What fraction of the animals at the shelter are male cats?
2. A taco recipe called for 2/3 cup of cheese per taco. If Andrew wanted to make 3 tacos, how much cheese would he need?

• Show Step-by-step Solutions

How to solve word problems using tape diagrams and fraction-by-fraction multiplication?
Examples:
1. Ethan is icing 30 cupcakes. He spreads mint icing on 1/5 of the cupcakes and chocolate on 1/2 of the remaining cupcakes. The rest will get vanilla frosting. How many cupcakes have vanilla frosting?
2. Maddox puts 1/4 of her lawn-mowing money in savings and uses 1/2 of the remaining money to pay back her suster. If she has \$15 left, how much did she have at first?

• Show Step-by-step Solutions

Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.

You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

NCERT Solutions for Class 9 #answering #service #pricing

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Chapter-wise, detailed solutions to the questions of the NCERT Books are provided with the objective of helping students compare their answers with the sample answers.

Key features of class 9 NCERT Solutions:

• Homework made easy with chapter-wise detailed solutions to the questions of NCERT class 9 textbooks
• Properly organized solutions in chapter-wise and page-wise format help find you find your solutions quickly
• Beautifully Animated Videos help you understand solutions of difficult questions of class 9
• Detailed subjective answers for questions which are easy to understand and learn
• Carefully designed class 9 NCERT solutions by experts match the curriculum exactly
• NCERT solutions for class 9 are available for all chapters of all the subjects

Select a Subject
to start

Taking the burden off the shoulders of the students of collecting good study materials while preparing for the exam, at Meritnation.com, we are offering class 9 NCERT solutions. Prepared by our experts, the class 9 NCERT Solutions are not only precise and to the point, but are also easy to understand and learn.

Special care has been take to ensure that the exact format of questions and answers are maintained while preparing the class 9 NCERT solutions. Both subjective and MCQ type questions with proper answers and solutions in the NCERT solution enables the student to prepare in a comprehensive manner.

With the class 9 NCERT solutions, students can study any particular chapter from any subject such as Science, Math, Social Science, English, Hindi, etc. Not only that, students also get the opportunity to take chapter test or revise according to their requirement.

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Physics Exams With Solutions

AP Physics

• College Board Physics B Past AP exams and sample problems from the College Board.
• College Board AP Physics B Practice Exams Multiple choice and free response, from old exams.
• Farragut High School Mr. Milligan’s AP Physics C with solutions.
• College Board AP Physics C Practice Questions Mechanics and electricity and magnetism.
• College Board AP Physics C Mechanics Practice Questions Free response only.

General Physics

• University of Wisconsin Sample midterms and finals from Physics 207.
• University of Gaziantep Advanced engineering and physics exams from General Physics II.
• University of Wisconsin Oshkosh Physics 109 tests with answers, class covers Kinematics, Dynamics, and Waves.
• MIT Open Courseware Covers a wide array of physics subjects, from mechanics to astrophysics.
• Rutgers University Practice Final for Physics 204, only the final works.
• Louisiana State University Sample exams for Physics 2001, an introductory physics course that covers basic and rotational mechanics.
• Portland State University Sample tests with answers from Physics 203, a more difficult general physics course.
• MIT Open Courseware Covers a wide array of physics subjects, from mechanics to astrophysics.
• Purdue University Exams for General Physics 221.

Mechanics Exams

• University of Hawaii Lots of midterms and finals from Physics 151, a class that covers mechanics and thermodynamics.
• University of Rochestor Tests with answers from Physics 141, a mechanics course that also covers special relativity.
• CSU Fresno Multiple choice sample exams with answers and quizzes from Physics 2A.
• Princeton University Physics 203, difficult mechanics exams with solutions. Some advanced concepts, eigenvectors and rotational frequency.
• Michigan Technological University Sample mechanics exams from Physics 2100.
• Johns Hokpkins University Sample mechanics midterms and final from Classical Mechanics II, covers Lagrangian and Newtonian mechanics.

Electricity and Magnetism Exams

• University of Hawaii Physics 272 exams and resources on E
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Physics Problems And SolutionsFor Real World Applications

This website contains a variety of physics problems, complete with solutions, taken from real world applications. The intent is to stimulate awareness and understanding of the physical world for people who are interested in physics.

Maybe you want to know how the physics you were taught in school applies to the “real world”. This website will answer that question.

One of the main areas of focus here will be classical mechanics; which is basically, the study of forces and motion. It is one of the most interesting areas of physics. And so many things from our everyday lives, such as sports and amusement parks, have a strong relationship to classical mechanics.

If something moves in a manner more complicated than a straight line, chances are there’s some good physics involved. The topics covered in this website will show examples of this.

It’s inescapable; the physics of everyday things is all around us. It’s just a matter of expressing them in the well-known language of physics. However, this will not be a trivial or superficial treatment. The explanations and solutions will go right to the heart of the problem, but will be clear and concise so that you can appreciate the wonders of the physical world, and understand how they actually work.

The solutions for the problems will not just consist of “plug ‘n chug” formula usage, and number crunching. Intuitive explanations will be included so you have a “feel” for what’s happening.

Feel free to look at the different examples of physics problems. What’s your interest? Sports? Amusement parks? Battle machines and weapons? It’s all here. There is plenty for you to get started.

For those who want to learn theory, there is an education section which is subdivided into different areas of physics. This section was created to assist students and teachers in their courses, and for those who just want to learn some theory.

In addition, there is a kids section to help kids learn science, with an emphasis on physics. This is a useful resource for parents and teachers who want to engage their kids in the sciences, and help plant the “seeds” for future interest, and a potential career, in the sciences.

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Street Lighting Control

Street lighting, one of the most essential services provided by municipal authorities, is a cost and management challenge. New technological advances can deliver up to 50% reductions in energy usage. Smart control, including proactive maintenance, can slash annual operating costs significantly.

T-Light TM – Telematics Wireless s solution for Smart Lighting enables utilities and maintenance companies to benefit from an improved level of cost-effective, reliable and timely service, by controlling and managing street light operations. It helps provide a safer, urban environment for people, and reduces carbon footprints, contributing to environmental sustainability.

The T-Light family offers Smart Lighting Control and Monitoring Systems differentiated by their coverage, features, configurations, and prices.

This variety, flexibility, and modularity enable customers with a range of light control system requirements to identify solutions ranging from local to wide area metropolitain networks.

The T-Light Family Includes:

• T-Light Galaxy a wide area network utilizing single Base Station covers an area of up to 20 km radius and monitoring directly up to 50,000 luminaries
• T-Light Pro a robust multi hop RF Mesh network with self configuring, self healing capabilities, non line of sight links. Provides local or wide area coverage .
• T-Light Basic RF Mesh network for self-contained local area deployments (e.g. campuses, dealerships, parking lots, gas stations etc).

These systems feature various configuration choices. They also offer seamless scalability as the customer s needs change and expand.

The Smart Street Lighting Controlling system establishes a wireless network between all participating light poles and provides the operator with a web-accessed efficient, automatic or manual, control of the light pole s operation. The system may also apply to a smart lighting system, where the light poles already have, or can have, a wired connection.

The complete and reliable smart light solution features the following elements:

• T-Light LCU, Light Control Unit. installed on top or inside the luminaire (external NEMA or internal configuration), enabling the transmission of information, and the reception of control commands for the luminaire s LED / HID fixtures. Includes built-in energy meter and possesses autocommissioning functonality .
• The information from and to the T-Light LCU is routed via the T-Light DCU, Data Communication Unit. and via the Internet, using GPRS/3G or Ethernet connections directly to the BackOffice application.
• T-Light CMS, Control and Management System. is a web-enabled BackOffice application, accessible at any location in the world – simply by using a standard browser, such as Internet Explorer or Google Chrome.