Classification of organisms is based upon a number of physical and p.pdf
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Classification of organisms is based upon a number of physical and physiological features. The structure and organization of reproductive organs remains in important feature usually used to distinguish organisms. Some of the features can be discussed as below: 1. Fruiting body: A very broad classification can be made on the basis whether the fruiting body is present or not. Presence of a fleshy or modified cover around the seed is a characteristic feature of angiosperms whereas a naked seed located directly above a vegetative structure of a plant is a feature of gymnosperms. 2. Spores sacs: Spore releasing sacs are characteristic features found on the lower scales of the leaf blades of lower and higher ferns. Their arrangement and mode of release determines the nature, frequency and ease of sexual reproduction in these lower plants. 3. Pollens: Higher plants, including both angiosperms and gymnosperms utilize pollen grains for sexual reproduction. The mode of transfer of pollen determines the geographical range of pollination in plants. Sulphur shower is a critical example for sexual reproduction in pines (gymnosperms) which distinguishes them from other trees. 4. Sporocarp: A sporocarp or a fruiting body is a bulging of the vegetative extension from the main structure of a lower fungi which contains a group of cells which either follow meiotic division or mitotic division and serve as the reproductive structures. Their growth and nature can be monitored microscopically and the fungus can be thus classified appropriately. Some algae also contain a unique localization and arrangement of haploid cells called spores which can be present in singles or doublets and thus represent a very unique discrimination from other organisms. These, these set of information provide an insight into usability of features of a reproductive structure for classification of organisms. Solution Classification of organisms is based upon a number of physical and physiological features. The structure and organization of reproductive organs remains in important feature usually used to distinguish organisms. Some of the features can be discussed as below: 1. Fruiting body: A very broad classification can be made on the basis whether the fruiting body is present or not. Presence of a fleshy or modified cover around the seed is a characteristic feature of angiosperms whereas a naked seed located directly above a vegetative structure of a plant is a feature of gymnosperms. 2. Spores sacs: Spore releasing sacs are characteristic features found on the lower scales of the leaf blades of lower and higher ferns. Their arrangement and mode of release determines the nature, frequency and ease of sexual reproduction in these lower plants. 3. Pollens: Higher plants, including both angiosperms and gymnosperms utilize pollen grains for sexual reproduction. The mode of transfer of pollen determines the geographical range of pollination in plants. Sulphur shower is a critical example for.
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Division : pteridophyta
Class: lucopsida
Order: Selaginellales
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A beautiful presentation on selaginella plant .
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Division : pteridophyta
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Order: Selaginellales
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Which is a common trait among all land plantsVascular tissues tra.pdf
Which is a common trait among all land plants?
Vascular tissues transport water and nutrients, and provide support
Seeds facilitate survival and dispersal of the species
Flowers facilitate cross-pollination
The embryo is protected within maternal tissues
The haploid phase of the life cycle is the dominant phase
Flowers facilitate cross-pollination
The embryo is protected within maternal tissues
The haploid phase of the life cycle is the dominant phase
Solution
First we see common traits of land plants and then see listed traits in details one by one:
1. Ability to withstand desiccation. Extant land plants have a cuticle and guard cells.
2. Ability to withstand the effects of more intense radiation, particularly DNA-damaging
radiation. Extant land plants have several compounds in their vacuoles that absorb UV. Since the
vacuole of a plant occupies most of a mature cell, this helps protect the DNA in other organelles.
3. Ability to protect their spores from desiccation. Early land plants have spores that are encased
in a sporopollenin wall. Sporopollenin is a very resistant polymer, resistant to UV and almost
everything including desiccation, squashing, etc. To remove sporopollenin from spores, one boils
them in a mixture of acetic and hydrochloric acid.
4. Ability to move solutions from the ground to portions of the plant that are not in contact with
the ground, and from the photosynthetic portions of the plant to non-photosynthetic portions.
Some land plants do this better than others.
5. Ability to support themselves. Aquatic plants float; terrestrial plants cannot do so. Most
terrestrial plants have lignin in some of their conducting cells. There is some debate as to
whether this was selected for by the advantages of growing tall or the need to protect against
embolism in the conducting cells. Since both are important, it seems most realistic to accept that
both contributed to the success of plants with the ability to manufacture lignin, the tracheophytes
or vascular plants.
6. Ability to acquire the carbon dioxide required for photosynthesis from the atmosphere. This
ability is associated with stomatal cells, specialized cells that surround openings (stomates) in the
outer cell layer of land plants.
Now we see Options given one by one:
Vascular tissues transport water and nutrients, and provide support
Transport of Water and Minerals:
The xylem tissue transports water and minerals. It consists of interconnected vessels and
tracheids organized into continuous conducting tubes stretching from the roots to the leaves.
These tubes carry water and minerals to all parts of the plant.
Plants absorb water from the soil through the root and transport it to the stem, leaves and
flowers. Roots have root hairs that are unicellular, thin-walled outgrowths of the epiblema (skin
of the root).
The root hairs are in close contact with the thin film of water surrounding the soil particles.
There are mineral salts such as nitrates, chlorides, sulphates, phos.
Copyright © The McGraw-Hill Companies, Inc. Permission required .docx
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chapter 16
Protists
& Fungi
16.6 General Biology of Protists
There are four kingdoms in the Domain Eukarya.
Protista, Fungi, Plantae and Animalia
All organisms in Eukarya have organelles including a nucleus.
Protists vary in size from microscopic to macroscopic (very large)
Most protists are unicellular, but they are very complex
Most organisms in Protista reproduce by asexual reproduction
17-2
Organizing Organisms
Organisms can be categorized by how they obtain nutrients
Heterotrophs – organisms that consume food and digest it inside their bodies.
Autotrophs – organisms that make their own food using photosynthesis. While we think of plants as doing photosynthesis, some protists and some bacteria can also do photosynthesis. Cyanobacteria make a significant amount of atmospheric oxygen.
Decomposers – digest food outside their body and absorb the nutrients across the cell wall.
17-3
Types of Protists
Protists can be separated into 3 categories based on their nutrition
Protozoa consume food (heterotrophs) and digest it internally. Also they typically can move.
Algae that can do photosynthesis (autotrophs). Underscoring the debate over how to categorize protists, some sources list algae as plants, but your book lists them as protists.
Slime molds that absorb nutrients that were digested outside their body.
17-4
Protozoa
Amoeba proteus
Pseudopods - extensions that form when cytoplasm goes in a particular direction
Feed by phagocytosis, surrounding prey with pseudopods and digesting it
Paramecium caudatum
The most structurally complex and specialized of all protozoans
The majority are free-living, some are parasitic
17-5
Green algae are
ancestral to plants
Green algae (Approximately 7,500 species)
Not always green
Inhabit a variety of environments
Oceans, freshwater, snowbanks, bark of trees, backs of turtles
Phytoplankton are microscopic autotrophs that live in water. They produce about half of the oxygen available in the atmosphere and form the basis of
Algae can form colonies of loosely organized independent cells or end-to-end chains of cells that form filaments.
17-6
Algae
Volvox, a colonial green alga
A Volvox colony is a hollow sphere with thousands of cells in a single layer on the outside
The cells are usually connected by strands of cytoplasm
Smaller daughter colonies are contained within the ball
Spirogyra, multicellular, filamentous algae
Forms long filaments of cells end-to-end
Can reproduce asexually and sexually
Has a chloroplast that performs photosynthesis
Stores glucose in a pyrenoid
17-7
16.8 A Fungus Is Not a Plant
Fungi (domain Eukarya)
Fungi are heterotrophs that release digestive enzymes into the external environment and digest their food outside the body, then absorb nutrients across their cell walls. They play an important role as decomposers.
Fungal cells are different from plant c.
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ALGAE Advance Microbiology.pptx
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Fungus
Fungus, plural fungi, any of about 99,000 known species of organisms of the kingdom Fungi, which includes the yeasts, rusts, smuts, mildews, molds, and mushrooms. There are also many funguslike organisms, including slime molds and oomycetes (water molds), that do not belong to kingdom Fungi but are often called fungi. Many of these funguslike organisms are included in the kingdom Chromista. Fungi are among the most widely distributed organisms on Earth and are of great environmental and medical importance. Many fungi are free-living in soil or water; others form parasitic or symbiotic relationships with plants or animals.
A fungus (/ˈfʌŋɡəs/; plural: fungi[3] or funguses[4]) is any member of the group of eukaryotic organisms that includes microorganisms such as yeasts and molds, as well as the more familiar mushrooms. These organisms are classified as a kingdom, Fungi, which is separate from the other eukaryotic life kingdoms of plants and animals.
A characteristic that places fungi in a different kingdom from plants, bacteria, and some protists is chitin in their cell walls. Similar to animals, fungi are heterotrophs; they acquire their food by absorbing dissolved molecules, typically by secreting digestive enzymes into their environment. Fungi do not photosynthesise. Growth is their means of mobility, except for spores (a few of which are flagellated), which may travel through the air or water. Fungi are the principal decomposers in ecological systems. These and other differences place fungi in a single group of related organisms, named the Eumycota (true fungi or Eumycetes), which share a common ancestor (form a monophyletic group), an interpretation that is also strongly supported by molecular phylogenetics. This fungal group is distinct from the structurally similar myxomycetes (slime molds) and oomycetes (water molds). The discipline of biology devoted to the study of fungi is known as mycology (from the Greek μύκης, mukēs, meaning "fungus"). In the past, mycology was regarded as a branch of botany, although it is now known fungi are genetically more closely related to animals than to plants.
Classification Of Plant
I give this slide for your learning from the pictures. I put many pictures in the slide. It is easy way to learn from this slide. This slide tells us about the plants and their classification . It is copyright from the textbook of 9th standard. This is biology.
Biological classification Part-IV
The topic of discussion is Kingdom plantae & Kingdom Animalia, The detailed characteristic of Thallophytes, Bryophytes & Pteridophytes along with Phanerogams such as Gymnosperms & Angiosperms have been discussed. The general characteristic of Virus (TMV) and Bacteriophage along with their life cycle are also discussed. Finally the ecologically important LIchens and Mycorrhizae are also discussed.
You have to leave solids and liquids out of the equation. This lea.pdf
You have to leave solids and liquids out of the equation.
This leaves only product B for the product, nothing for the reactants.
Initially there was no product B.
At equilibrium: you determine the molarity by multiplying 2.5% by (0.5M/10L)= .00125 =
which rounds to answer E) 1.3e-3
it was a trick question.
Solution
You have to leave solids and liquids out of the equation.
This leaves only product B for the product, nothing for the reactants.
Initially there was no product B.
At equilibrium: you determine the molarity by multiplying 2.5% by (0.5M/10L)= .00125 =
which rounds to answer E) 1.3e-3
it was a trick question..
Tungsten is a body centered cubic latticeSolutionTungsten is a.pdf
Tungsten is a body centered cubic lattice
Solution
Tungsten is a body centered cubic lattice.
More Related Content
Similar to Classification of organisms is based upon a number of physical and p.pdf
Which is a common trait among all land plantsVascular tissues tra.pdf
Which is a common trait among all land plants?
Vascular tissues transport water and nutrients, and provide support
Seeds facilitate survival and dispersal of the species
Flowers facilitate cross-pollination
The embryo is protected within maternal tissues
The haploid phase of the life cycle is the dominant phase
Flowers facilitate cross-pollination
The embryo is protected within maternal tissues
The haploid phase of the life cycle is the dominant phase
Solution
First we see common traits of land plants and then see listed traits in details one by one:
1. Ability to withstand desiccation. Extant land plants have a cuticle and guard cells.
2. Ability to withstand the effects of more intense radiation, particularly DNA-damaging
radiation. Extant land plants have several compounds in their vacuoles that absorb UV. Since the
vacuole of a plant occupies most of a mature cell, this helps protect the DNA in other organelles.
3. Ability to protect their spores from desiccation. Early land plants have spores that are encased
in a sporopollenin wall. Sporopollenin is a very resistant polymer, resistant to UV and almost
everything including desiccation, squashing, etc. To remove sporopollenin from spores, one boils
them in a mixture of acetic and hydrochloric acid.
4. Ability to move solutions from the ground to portions of the plant that are not in contact with
the ground, and from the photosynthetic portions of the plant to non-photosynthetic portions.
Some land plants do this better than others.
5. Ability to support themselves. Aquatic plants float; terrestrial plants cannot do so. Most
terrestrial plants have lignin in some of their conducting cells. There is some debate as to
whether this was selected for by the advantages of growing tall or the need to protect against
embolism in the conducting cells. Since both are important, it seems most realistic to accept that
both contributed to the success of plants with the ability to manufacture lignin, the tracheophytes
or vascular plants.
6. Ability to acquire the carbon dioxide required for photosynthesis from the atmosphere. This
ability is associated with stomatal cells, specialized cells that surround openings (stomates) in the
outer cell layer of land plants.
Now we see Options given one by one:
Vascular tissues transport water and nutrients, and provide support
Transport of Water and Minerals:
The xylem tissue transports water and minerals. It consists of interconnected vessels and
tracheids organized into continuous conducting tubes stretching from the roots to the leaves.
These tubes carry water and minerals to all parts of the plant.
Plants absorb water from the soil through the root and transport it to the stem, leaves and
flowers. Roots have root hairs that are unicellular, thin-walled outgrowths of the epiblema (skin
of the root).
The root hairs are in close contact with the thin film of water surrounding the soil particles.
There are mineral salts such as nitrates, chlorides, sulphates, phos.
Copyright © The McGraw-Hill Companies, Inc. Permission required .docx
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chapter 16
Protists
& Fungi
16.6 General Biology of Protists
There are four kingdoms in the Domain Eukarya.
Protista, Fungi, Plantae and Animalia
All organisms in Eukarya have organelles including a nucleus.
Protists vary in size from microscopic to macroscopic (very large)
Most protists are unicellular, but they are very complex
Most organisms in Protista reproduce by asexual reproduction
17-2
Organizing Organisms
Organisms can be categorized by how they obtain nutrients
Heterotrophs – organisms that consume food and digest it inside their bodies.
Autotrophs – organisms that make their own food using photosynthesis. While we think of plants as doing photosynthesis, some protists and some bacteria can also do photosynthesis. Cyanobacteria make a significant amount of atmospheric oxygen.
Decomposers – digest food outside their body and absorb the nutrients across the cell wall.
17-3
Types of Protists
Protists can be separated into 3 categories based on their nutrition
Protozoa consume food (heterotrophs) and digest it internally. Also they typically can move.
Algae that can do photosynthesis (autotrophs). Underscoring the debate over how to categorize protists, some sources list algae as plants, but your book lists them as protists.
Slime molds that absorb nutrients that were digested outside their body.
17-4
Protozoa
Amoeba proteus
Pseudopods - extensions that form when cytoplasm goes in a particular direction
Feed by phagocytosis, surrounding prey with pseudopods and digesting it
Paramecium caudatum
The most structurally complex and specialized of all protozoans
The majority are free-living, some are parasitic
17-5
Green algae are
ancestral to plants
Green algae (Approximately 7,500 species)
Not always green
Inhabit a variety of environments
Oceans, freshwater, snowbanks, bark of trees, backs of turtles
Phytoplankton are microscopic autotrophs that live in water. They produce about half of the oxygen available in the atmosphere and form the basis of
Algae can form colonies of loosely organized independent cells or end-to-end chains of cells that form filaments.
17-6
Algae
Volvox, a colonial green alga
A Volvox colony is a hollow sphere with thousands of cells in a single layer on the outside
The cells are usually connected by strands of cytoplasm
Smaller daughter colonies are contained within the ball
Spirogyra, multicellular, filamentous algae
Forms long filaments of cells end-to-end
Can reproduce asexually and sexually
Has a chloroplast that performs photosynthesis
Stores glucose in a pyrenoid
17-7
16.8 A Fungus Is Not a Plant
Fungi (domain Eukarya)
Fungi are heterotrophs that release digestive enzymes into the external environment and digest their food outside the body, then absorb nutrients across their cell walls. They play an important role as decomposers.
Fungal cells are different from plant c.
Reproduction in organisms, Class XII
A Detail Description of Chapter Reproduction in Organisms of NCERT Class XII. The matter can be used by the Students of class X also.
Fungi
This is a very detailed slide on the topic 'Fungi'. I hope this slide is beneficial to everyone. Also don't forget to 'Like' if u like this slide! Thank you!
ALGAE Advance Microbiology.pptx
Microbiology - Algae
Algae is an informal term for a large and diverse group of photosynthetic eukaryotic organisms. It is a polyphyletic grouping that includes species from multiple distinct clades.
Algae are sometimes considered plants and sometimes considered "protists" (a grab-bag category of generally distantly related organisms that are grouped on the basis of not being animals, plants, fungi, bacteria, or archaeans).
Std 12 chapter 2 REPRODUCTION IN FLOWERING PLANTS
BIOLOGY STD 12
HELPFUL FOR NEET/ GSET/NET EXAMINATION PREPARATION
STRUCTURE OF FLOWERS, PARTS OF FLOWER, POLLINATION
Plant Tissues
This powerpoint presentation is mainly about the type of plant tissues, major parts of plant with their modifications.
Fungus
Fungus, plural fungi, any of about 99,000 known species of organisms of the kingdom Fungi, which includes the yeasts, rusts, smuts, mildews, molds, and mushrooms. There are also many funguslike organisms, including slime molds and oomycetes (water molds), that do not belong to kingdom Fungi but are often called fungi. Many of these funguslike organisms are included in the kingdom Chromista. Fungi are among the most widely distributed organisms on Earth and are of great environmental and medical importance. Many fungi are free-living in soil or water; others form parasitic or symbiotic relationships with plants or animals.
A fungus (/ˈfʌŋɡəs/; plural: fungi[3] or funguses[4]) is any member of the group of eukaryotic organisms that includes microorganisms such as yeasts and molds, as well as the more familiar mushrooms. These organisms are classified as a kingdom, Fungi, which is separate from the other eukaryotic life kingdoms of plants and animals.
A characteristic that places fungi in a different kingdom from plants, bacteria, and some protists is chitin in their cell walls. Similar to animals, fungi are heterotrophs; they acquire their food by absorbing dissolved molecules, typically by secreting digestive enzymes into their environment. Fungi do not photosynthesise. Growth is their means of mobility, except for spores (a few of which are flagellated), which may travel through the air or water. Fungi are the principal decomposers in ecological systems. These and other differences place fungi in a single group of related organisms, named the Eumycota (true fungi or Eumycetes), which share a common ancestor (form a monophyletic group), an interpretation that is also strongly supported by molecular phylogenetics. This fungal group is distinct from the structurally similar myxomycetes (slime molds) and oomycetes (water molds). The discipline of biology devoted to the study of fungi is known as mycology (from the Greek μύκης, mukēs, meaning "fungus"). In the past, mycology was regarded as a branch of botany, although it is now known fungi are genetically more closely related to animals than to plants.
Classification Of Plant
I give this slide for your learning from the pictures. I put many pictures in the slide. It is easy way to learn from this slide. This slide tells us about the plants and their classification . It is copyright from the textbook of 9th standard. This is biology.
Biological classification Part-IV
The topic of discussion is Kingdom plantae & Kingdom Animalia, The detailed characteristic of Thallophytes, Bryophytes & Pteridophytes along with Phanerogams such as Gymnosperms & Angiosperms have been discussed. The general characteristic of Virus (TMV) and Bacteriophage along with their life cycle are also discussed. Finally the ecologically important LIchens and Mycorrhizae are also discussed.
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Copyright © The McGraw-Hill Companies, Inc. Permission required .docx
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You have to leave solids and liquids out of the equation. This lea.pdf
You have to leave solids and liquids out of the equation.
This leaves only product B for the product, nothing for the reactants.
Initially there was no product B.
At equilibrium: you determine the molarity by multiplying 2.5% by (0.5M/10L)= .00125 =
which rounds to answer E) 1.3e-3
it was a trick question.
Solution
You have to leave solids and liquids out of the equation.
This leaves only product B for the product, nothing for the reactants.
Initially there was no product B.
At equilibrium: you determine the molarity by multiplying 2.5% by (0.5M/10L)= .00125 =
which rounds to answer E) 1.3e-3
it was a trick question..
Tungsten is a body centered cubic latticeSolutionTungsten is a.pdf
Tungsten is a body centered cubic lattice
Solution
Tungsten is a body centered cubic lattice.
This is the final code which meets your requirement.Than youCard.j.pdf
This is the final code which meets your requirement.Than you
Card.java
import java.util.Random;
public class Card {
//declaring instance variables
private String suit;
private int face_value;
public static int count=52;
//Creating an instance of Random class reference
Random rand;
//Default Constructor
public Card() {
super();
//Passing random class object to the Random class reference
rand = new Random();
}
//Getters and setters
public String getSuit() {
return suit;
}
public void setSuit(String suit) {
this.suit = suit;
}
public int getFace_value() {
return face_value;
}
public void setFace_value(int face_value) {
this.face_value = face_value;
}
public int getCount() {
return count;
}
//This shuffle() method will shuffle the deck
public void shuffle()
{
//Picking a card from the deck randomly and find its face value and suit
this.face_value = 1 + rand.nextInt((9) + 1);
int suit_random_value = 1 + rand.nextInt((3) + 1);
if(suit_random_value==1)
{
this.suit=\"Clubs\";
}
else if(suit_random_value==2)
{
this.suit=\"hearts\";
}
else if(suit_random_value==3)
{
this.suit=\"Spades\";
}
else if(suit_random_value==4)
{
this.suit=\"diamonds\";
}
this.count=count-1;
}
}
________________________________________________________________________
DriverClass.java
public class DriverClass {
public static void main(String[] args) {
//creating an array which can hold 52 Card class Objects
Card c[]=new Card[52];
//Creating 52 card class objects and storing them into an array
for(int i=0;i<52;i++)
{
//Creating Card class Object
c[i]=new Card();
}
//Displaying each card face value and its suit.
for(int i=0;i<52;i++)
{
//Shuffling the deck before every card picking from the deck
c[i].shuffle();
//Displaying randomly picked card face value and its suit
System.out.println(c[i].getFace_value()+\" of \"+c[i].getSuit()+\" Cards remaining
\"+c[i].getCount());
}
}
}
_________________________________________________________________________
Output:
8 of diamonds
Cards remaining 51
4 of Clubs
Cards remaining 50
9 of Clubs
Cards remaining 49
7 of Spades
Cards remaining 48
9 of diamonds
Cards remaining 47
7 of diamonds
Cards remaining 46
4 of diamonds
Cards remaining 45
9 of Clubs
Cards remaining 44
5 of hearts
Cards remaining 43
7 of Spades
Cards remaining 42
10 of Clubs
Cards remaining 41
7 of diamonds
Cards remaining 40
8 of diamonds
Cards remaining 39
6 of hearts
Cards remaining 38
4 of hearts
Cards remaining 37
6 of Spades
Cards remaining 36
4 of Clubs
Cards remaining 35
3 of diamonds
Cards remaining 34
9 of diamonds
Cards remaining 33
3 of diamonds
Cards remaining 32
5 of Spades
Cards remaining 31
4 of Clubs
Cards remaining 30
3 of Spades
Cards remaining 29
6 of diamonds
Cards remaining 28
3 of hearts
Cards remaining 27
9 of hearts
Cards remaining 26
7 of hearts
Cards remaining 25
3 of hearts
Cards remaining 24
10 of Clubs
Cards remaining 23
2 of hearts
Cards remaining 22
9 of Clubs
Cards remaining 21
8 of diamonds
Cards remaining 20
8 of Clubs
Cards remaining 19
2 of h.
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disciplines. Certain topics have \"statistical\" in their name but relate to manipulations of
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astronomical data.
· Biostatistics is a branch of biology that studies biological phenomena and observations by
means of statistical analysis, and includes medical statistics.
· Business analytics is a rapidly developing business process that applies statistical methods
to data sets (often very large) to develop new insights and understanding of business
performance & opportunities
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process to the state of the system via application of mathematical or statistical methods.
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can be applied to any kind of dynamic population, that is, one that changes over time or space.
· Econometrics is a branch of economics that applies statistical methods to the empirical
study of economic theories and relationships.
· Environmental statistics is the application of statistical methods to environmental science.
Weather, climate, air and water quality are included, as are studies of plant and animal
populations.
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such as petroleum geology, hydrogeology, hydrology, meteorology,oceanography, geochemistry,
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mathematics and formal science that uses methods such as mathematical modeling, statistics, and
algorithms to arrive at optimal or near optimal solutions to complex problems.
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(Cydonia oblonga) belong to the rose family (Rosaceae), and include literally hundreds of
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variations throughout the centuries that some authors lump them all into one species, Malus
domestica. They all originated in western Asia (or Eurasia) and are characterized by fleshy fruits
called pomes. In the pome, a thick, fleshy hypanthium layer (also called the floral cup or calyx
tube) surrounds (and is fused with) the seed-bearing ovary or core. The sepals, petals and
stamens arise from the rim of the hypanthium. Since the ovary is situated below the attachment
of the sepals, petals and stamens, it is termed \"inferior\" in technical plant taxonomy books. The
fleshy hypanthium of a rose (Rosa) surrounds a cluster of small one-seeded achenes. Since the
achenes represent separate ripened ovaries all derived from a single flower, the entire structure
(called a rose hip) can be considered an aggregate fruit or etaerio. Rose hips are eaten raw and
are ground up as a supplemental source of vitamin C (ascorbic acid).
Bing cherries (Prunus avium) showing the long stalk (pedicel) and fleshy drupe containng a hard,
stony, seed-bearing endocarp. Sweet cherries such as these are usually considered to belong to P.
avium, while sour cherries belong to the P. cerasus group. There are literally hundreds of
varieties of cherries. Maraschino cherries are made from sweet cherries which have been
bleached, deseeded, and soaked in a sugar solution to which red food coloring and flavoring have
been added. Maraschino cherries are commonly covered with chocolate, placed as a decorative
topping in ice cream sundies, and in mixed drinks.
The raspberry or hindberry is the edible fruit of a multitude of plant species in the genus Rubus,
most of which are in the subgenusIdaeobatus; the name also applies to these plants themselves.
Raspberries are perennial, with woody stems.
Raspberries are grown for the fresh fruit market and for commercial processing into individually
quick frozen (IQF) fruit, purée, juice, or as dried fruit used in a variety of grocery products.
Traditionally, raspberries were a mid-summer crop, but with new technology, cultivars, and
transportation, they can now be obtained year-round. Raspberries need ample sun and water for
optimal development. Raspberries thrive in well-drained soil with a pH of between 6 and 7 with
ample organic matter to assist in retaining water. While moisture is essential, wet and heavy soils
or excess irrigation can bring on Phytophthora root rot which is one of the most serious pest
problems facing red raspberry. As a cultivated plant in moist temperate regions, it is easy to grow
and has a tendency to spread unless pruned. Escaped raspberries frequently appear as garden
weeds, spread by seeds found in bird droppings.
An indiv.
Polygon .java package chegg2;public class Polygon { Var.pdf
Polygon .java
package chegg2;
public class Polygon {
// Variables
private int numSides;
private double sideLength;
private double xCord;
private double yCord;
// Default Constructor
public Polygon () {
numSides = 3;
sideLength = 5.0;
xCord = 0.0;
yCord = 0.0;
}
// Constructor
public Polygon (int pSides, double numLength, double numXCord, double numYCord) {
numSides = pSides;
sideLength = numLength;
xCord = numXCord;
yCord = numYCord;
}
// Setter
public void setSides(int mSides) {
numSides = mSides;
System.out.println(\"setSides() results: \"+this.numSides);
}
// Getter
public int getSides() {
return numSides;
}
}
TestPolygon .java
package chegg2;
import java.util.Scanner;
public class TestPolygon {
public static void main(String[] args) {
Polygon poly[] = new Polygon[2];
Polygon poly2 = new Polygon();
poly[0] = poly2;
Scanner input = new Scanner(System.in);
for(int i=1; i<2; i++){
Polygon obj = new Polygon();
System.out.println(\"New Polygons\");
System.out.println(\"--------------------------------------------------\");
System.out.print(\"Enter Number of Sides:\");
obj.setSides(input.nextInt());
poly[i] = obj;
} input.close();
System.out.println(\"Polygons\");
System.out.println(\"--------------------------------------------------\");
for(int i=0; i<2; i++) {
poly[i].setSides(poly[i].getSides());
}
}
}
Output:
run:
New Polygons
--------------------------------------------------
Enter Number of Sides:9
setSides() results: 9
Polygons
--------------------------------------------------
setSides() results: 3
setSides() results: 9
BUILD SUCCESSFUL (total time: 1 second)
Solution
Polygon .java
package chegg2;
public class Polygon {
// Variables
private int numSides;
private double sideLength;
private double xCord;
private double yCord;
// Default Constructor
public Polygon () {
numSides = 3;
sideLength = 5.0;
xCord = 0.0;
yCord = 0.0;
}
// Constructor
public Polygon (int pSides, double numLength, double numXCord, double numYCord) {
numSides = pSides;
sideLength = numLength;
xCord = numXCord;
yCord = numYCord;
}
// Setter
public void setSides(int mSides) {
numSides = mSides;
System.out.println(\"setSides() results: \"+this.numSides);
}
// Getter
public int getSides() {
return numSides;
}
}
TestPolygon .java
package chegg2;
import java.util.Scanner;
public class TestPolygon {
public static void main(String[] args) {
Polygon poly[] = new Polygon[2];
Polygon poly2 = new Polygon();
poly[0] = poly2;
Scanner input = new Scanner(System.in);
for(int i=1; i<2; i++){
Polygon obj = new Polygon();
System.out.println(\"New Polygons\");
System.out.println(\"--------------------------------------------------\");
System.out.print(\"Enter Number of Sides:\");
obj.setSides(input.nextInt());
poly[i] = obj;
} input.close();
System.out.println(\"Polygons\");
System.out.println(\"--------------------------------------------------\");
for(int i=0; i<2; i++) {
poly[i].setSides(poly[i].getSides());
}
}
}
Output:
run:
New Polygons
--------------------------------------------.
probability that the trajectory falls between 1.2 and 1.4 =(1.4-1.2).pdf
probability that the trajectory falls between 1.2 and 1.4 =(1.4-1.2)/(2.5-1) = 0.134
Solution
probability that the trajectory falls between 1.2 and 1.4 =(1.4-1.2)/(2.5-1) = 0.134.
OutputFibonacci till 20 0 1 1 2 3 5 8 13 21 .pdf
Output:
Fibonacci till 20:
0
1
1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
Program:
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main {
public static void main (String[] args) throws java.lang.Exception {
final int LIMIT = 20;
int[] fib = new int[LIMIT];
// init fib 0 and first val
fib[0] = 0;
fib[1] = 1;
for (int i = 2; i < LIMIT; ++i) {
fib[i] = fib[i - 1] + fib[i - 2];
}
System.out.println(\"Fibonacci till \" + LIMIT + \":\");
for (int i = 0; i < LIMIT; ++i) {
System.out.println(fib[i]);
}
}
}
Iterative solution is more efficient than recursive. The iterative solution do not compute the same
values again and again as recursive solution. Also because we are not using recursion the os
stack is not used hence computation is faster.
Solution
Output:
Fibonacci till 20:
0
1
1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
Program:
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main {
public static void main (String[] args) throws java.lang.Exception {
final int LIMIT = 20;
int[] fib = new int[LIMIT];
// init fib 0 and first val
fib[0] = 0;
fib[1] = 1;
for (int i = 2; i < LIMIT; ++i) {
fib[i] = fib[i - 1] + fib[i - 2];
}
System.out.println(\"Fibonacci till \" + LIMIT + \":\");
for (int i = 0; i < LIMIT; ++i) {
System.out.println(fib[i]);
}
}
}
Iterative solution is more efficient than recursive. The iterative solution do not compute the same
values again and again as recursive solution. Also because we are not using recursion the os
stack is not used hence computation is faster..
Matter is made up of electrically charge particle but the Constituen.pdf
Matter is made up of electrically charge particle but the Constituents of objects have opposite
charges, adding up to electric neutrality overall. Think of it in this way that two bring two
nucleus very close together we need a lot of energy because when we talk about nucleus it\'s only
the +ve(protons) charge we are talking about. That is why the fusion of two atoms is very tough
task. But in case of atoms or molecules that are the smallest entity of matter that can individually
exist; it always contain equal number of proton and electron and hence are electrically neutral on
a whole. Atom being neutral on whole there no force on other atoms
Solution
Matter is made up of electrically charge particle but the Constituents of objects have opposite
charges, adding up to electric neutrality overall. Think of it in this way that two bring two
nucleus very close together we need a lot of energy because when we talk about nucleus it\'s only
the +ve(protons) charge we are talking about. That is why the fusion of two atoms is very tough
task. But in case of atoms or molecules that are the smallest entity of matter that can individually
exist; it always contain equal number of proton and electron and hence are electrically neutral on
a whole. Atom being neutral on whole there no force on other atoms.
it is a polynomial of degree 3Solutionit is a polynomial of de.pdf
it is a polynomial of degree 3
Solution
it is a polynomial of degree 3.
JUnit is a Regression Testing Framework used by developers to implem.pdf
JUnit is a Regression Testing Framework used by developers to implement unit testing in Java,
and accelerate programming speed and increase the quality of code. JUnit Framework can be
easily integrated with either of the following
Features of JUnit Test Framework
JUnit test framework provides the following important features
Fixtures
Fixtures is a fixed state of a set of objects used as a baseline for running tests. The purpose of a
test fixture is to ensure that there is a well-known and fixed environment in which tests are run so
that results are repeatable
Test Suites
A test suite bundles a few unit test cases and runs them together. In JUnit, both @RunWith and
@Suite annotation are used to run the suite test.
Test Runners
Test runner is used for executing the test cases. Here is an example that assumes the test class
TestJunit already exists.
JUnit Classes
JUnit classes are important classes, used in writing and testing JUnits. Some of the important
classes are
Assert Contains a set of assert methods.
TestCase Contains a test case that defines the fixture to run multiple tests.
TestResult Contains methods to collect the results of executing a test case.
Solution
JUnit is a Regression Testing Framework used by developers to implement unit testing in Java,
and accelerate programming speed and increase the quality of code. JUnit Framework can be
easily integrated with either of the following
Features of JUnit Test Framework
JUnit test framework provides the following important features
Fixtures
Fixtures is a fixed state of a set of objects used as a baseline for running tests. The purpose of a
test fixture is to ensure that there is a well-known and fixed environment in which tests are run so
that results are repeatable
Test Suites
A test suite bundles a few unit test cases and runs them together. In JUnit, both @RunWith and
@Suite annotation are used to run the suite test.
Test Runners
Test runner is used for executing the test cases. Here is an example that assumes the test class
TestJunit already exists.
JUnit Classes
JUnit classes are important classes, used in writing and testing JUnits. Some of the important
classes are
Assert Contains a set of assert methods.
TestCase Contains a test case that defines the fixture to run multiple tests.
TestResult Contains methods to collect the results of executing a test case..
Intitially take a graph sheet and plot the points and the whole figu.pdf
Intitially take a graph sheet and plot the points and the whole figure as it is.
Next draw the velocity diagram.Due to technical issues, I am not able to upload the graphical
representation.
Apply corialis acceleration component at link 6.
Solution
Intitially take a graph sheet and plot the points and the whole figure as it is.
Next draw the velocity diagram.Due to technical issues, I am not able to upload the graphical
representation.
Apply corialis acceleration component at link 6..
Incapsula Enterprise is the best mitigation service provider with th.pdf
Incapsula Enterprise is the best mitigation service provider with the gold award winner and the
other two with silver and bronze award are F5 Silverline DDOS Protection and Arbor Cloud
respectively
With DDoS attacks growing in complexity and size daily, you need a DDoS protection service
with a robust network and variety of mitigation techniques to thwart any attacks directed at your
site. We found that Incapsula Enterprise, F5 Silverline DDoS Protection and Arbor Cloud
offered the best protection.
Incapsula’s growing global network and scrubbing capacities equip it with the size needed to
prevent large-scale volumetric attacks,27 data centers are connected which are located all over
the world, the service is prepared for even the most advanced DDoS attacks. In addition to
enormous network capacity and a high number of data centers, Incapsula hosts a variety of
mitigation techniques that intercept application layer, volumetric and protocol attacks.
With the recent acquisition of Defense.net, F5 has adjusted its service to offer cloud-based
mitigation technique, on-site and hybrid deployment methods,both hardware and the cloud is
used, F5 Silverline Protection is equipped for sophisticated DDoS attacks. This service surpasses
Incapsula in scrubbing capacity, it provides 2TB of bandwidth to filter, clean and remove
infected traffic. While the service does not offer unlimited mitigation, this protection service can
be used on-demand or always on, customizing it to fit the needs of any company.
Third one comes Arbor Cloud which offers a large network and scrubbing capacity and have
1.14TB of bandwidth, this server prepare the system for specialised volumetric attacks.variety of
protection techniques are provided by this service including IP blocking, rate limiting, automatic
bot discernment and more. While the cloud-based mitigation techniques are impressive, Arbor is
known for its hardware. The service is in a unique position, it\'s hardware is the challenge for it\'s
competitors
similar mitigation techniques is provided by all DDOS mitigation service providers to keep an
eye on any attacks and constant monitoring, they differ in price, deployment methods and
network size.It is not necessary that bigger mean better, the services that offer larger networks
and more mitigation technologies allow for more customization that you employ the best
protection service for your IT infrastructure.
Solution
Incapsula Enterprise is the best mitigation service provider with the gold award winner and the
other two with silver and bronze award are F5 Silverline DDOS Protection and Arbor Cloud
respectively
With DDoS attacks growing in complexity and size daily, you need a DDoS protection service
with a robust network and variety of mitigation techniques to thwart any attacks directed at your
site. We found that Incapsula Enterprise, F5 Silverline DDoS Protection and Arbor Cloud
offered the best protection.
Incapsula’s growing global network and scrubbing capacit.
Go with the most obvious one first...Carboxylic Acids will have a.pdf
Go with the most obvious one first...
Carboxylic Acids: will have a really broad and large absorption range in the 3500-2500cm-1 area
(aka. the \"valley of the acid\") as well as a carbonyl peak in the 1700cm-1 area. That would be
spectrum 4.
Alcohols:will also have a strong and broad absorption in the 3500-3000cm-1 region, but nowhere
near as broad as the acid. It just looks like a really fat absorption line, rather than a well...
Anyway, the characteristic spectrum for the alcohol is spectrum 5.
Amines:lucky for you, the amine given in this question is a primary amine. There\'s going to be
two N-H stretching peaks in the 3300cm-1 region. There\'s also no carbonyl on the structure
given, so there should be no C=O stretching at 1700cm-1.This is spectrum 3. Don\'t worry about
the small absorption in the 1500-1700cm-1 area.
Now for the fine details...
The ester and aldehyde spectra are pretty similar, but aldehydes absorb at around 2750cm-1. It\'s
not totally clear on the spectra in the question but it looks like there\'s a 2750cm-1 absorption in
spectrum 2, so I\'d say that spectrum 2 is the aldehyde, and spectrum 1 is the ester.
Summary:
Spectrum 1 = structure B
Spectrum 2 = structure E
Spectrum 3 = structure C
Spectrum 4 = structure D
Spectrum 5 = structure A
Solution
Go with the most obvious one first...
Carboxylic Acids: will have a really broad and large absorption range in the 3500-2500cm-1 area
(aka. the \"valley of the acid\") as well as a carbonyl peak in the 1700cm-1 area. That would be
spectrum 4.
Alcohols:will also have a strong and broad absorption in the 3500-3000cm-1 region, but nowhere
near as broad as the acid. It just looks like a really fat absorption line, rather than a well...
Anyway, the characteristic spectrum for the alcohol is spectrum 5.
Amines:lucky for you, the amine given in this question is a primary amine. There\'s going to be
two N-H stretching peaks in the 3300cm-1 region. There\'s also no carbonyl on the structure
given, so there should be no C=O stretching at 1700cm-1.This is spectrum 3. Don\'t worry about
the small absorption in the 1500-1700cm-1 area.
Now for the fine details...
The ester and aldehyde spectra are pretty similar, but aldehydes absorb at around 2750cm-1. It\'s
not totally clear on the spectra in the question but it looks like there\'s a 2750cm-1 absorption in
spectrum 2, so I\'d say that spectrum 2 is the aldehyde, and spectrum 1 is the ester.
Summary:
Spectrum 1 = structure B
Spectrum 2 = structure E
Spectrum 3 = structure C
Spectrum 4 = structure D
Spectrum 5 = structure A.
Flexible benefit plans gives employees a choice between qualified be.pdf
Flexible benefit plans gives employees a choice between qualified benefits (non taxable) and
cash. An example of qualified benefits is medical plans. Flexible benefit plans includes health
insurance and retirement benefits.
The 2 main types of flexible benefit plans are cafeteria plans and flexible spending accounts.
Under the cafeteria plan, employees have the option of choosing from several different benefit
packages. Employees can select the option of receiving some or all of the employer\'s nontaxable
benefits or receiving cash or other taxable benefits.
The benefits offered under cafeteria plans are health and group insurance, medical
reimbursement schemes for non-insured expenses and vacation days.
Flexible spending accounts (FSA) is another prevalant flexible benefit plan. FSA is a tax
deferred savings account. These accounts are formed by employers and helps the employee in
meeting certain medical expenses that are not a part of employer\'s insurance plan.
Solution
Flexible benefit plans gives employees a choice between qualified benefits (non taxable) and
cash. An example of qualified benefits is medical plans. Flexible benefit plans includes health
insurance and retirement benefits.
The 2 main types of flexible benefit plans are cafeteria plans and flexible spending accounts.
Under the cafeteria plan, employees have the option of choosing from several different benefit
packages. Employees can select the option of receiving some or all of the employer\'s nontaxable
benefits or receiving cash or other taxable benefits.
The benefits offered under cafeteria plans are health and group insurance, medical
reimbursement schemes for non-insured expenses and vacation days.
Flexible spending accounts (FSA) is another prevalant flexible benefit plan. FSA is a tax
deferred savings account. These accounts are formed by employers and helps the employee in
meeting certain medical expenses that are not a part of employer\'s insurance plan..
EnvironmentLet assume the environment is a grid of colored tiles(.pdf
Environment:
Let assume the environment is a grid of colored
tiles(WHITE,BLUE,GREEN,BLACK,YELLOW).Robot is standing on a tile.Every tile has
a(X,Y) Position
We use a function getTileColor(x,y) to know color of current tile(x,y position).
ALGORITHM:
TRESUREHUNT(x,y)
1. color=getColor(x,y)
2. IF color=YELLOW
3. Return \"tiles\"+x+y+\" Is a tresure\"
4. ELSE IF color=WHITE
5. TRESUREHUNT(x,y-1) // move to front tile by decrementing y by 1
6 ELSE IF color=BLUE
7. TRESUREHUNT(x-1,y) // move to left bydecrementing x by 1
8. ELSE IF color=GREEN
9. TRESUREHUNT(x+1,y) // move to right tile by incrementing x by 1
10. ELSE IF color=BLACK
11. TRESUREHUNT(x,y+2) // move to front tile by incrementint y by 2
12. ELSE
13. Return \"Unsucessfull\"
12.END iF
C++ CODE
//for simplicity i represent entite grid in a 2;D array .Each cell is viewed as a tile.It contains
arbitary color values,0-BLACK
//1-BLUE,2;GREEN,14;YELLOW,15-WHITE
//accordingly the x,y value is changed. if current cell(tile)=15(WHITE) robot moves to front tile.
// so x=x-1 and y=y
#include
#include
void tresurehunt(int x,int y);
int grid[][7]={ {1, 1, 1, 14,2, 0, 15},
{0, 1, 1, 1, 0, 0, 2},
{2, 2 ,1, 0, 15,14,1},
{15,15,14,0, 1, 2, 1},
{15,0, 0, 1, 2, 2, 14},
{14,14,14,15,0, 2, 2},
{15,15,15,1, 1, 2, 1}};
int main()
{
tresurehunt(4,4);
return 0;
}
void tresurehunt(int x,int y)
{
//int color=getpixel(x,y);
int color=grid[x][y];
if(color==YELLOW)
{
cout<<\"Tresure is in:\"<
Solution
Environment:
Let assume the environment is a grid of colored
tiles(WHITE,BLUE,GREEN,BLACK,YELLOW).Robot is standing on a tile.Every tile has
a(X,Y) Position
We use a function getTileColor(x,y) to know color of current tile(x,y position).
ALGORITHM:
TRESUREHUNT(x,y)
1. color=getColor(x,y)
2. IF color=YELLOW
3. Return \"tiles\"+x+y+\" Is a tresure\"
4. ELSE IF color=WHITE
5. TRESUREHUNT(x,y-1) // move to front tile by decrementing y by 1
6 ELSE IF color=BLUE
7. TRESUREHUNT(x-1,y) // move to left bydecrementing x by 1
8. ELSE IF color=GREEN
9. TRESUREHUNT(x+1,y) // move to right tile by incrementing x by 1
10. ELSE IF color=BLACK
11. TRESUREHUNT(x,y+2) // move to front tile by incrementint y by 2
12. ELSE
13. Return \"Unsucessfull\"
12.END iF
C++ CODE
//for simplicity i represent entite grid in a 2;D array .Each cell is viewed as a tile.It contains
arbitary color values,0-BLACK
//1-BLUE,2;GREEN,14;YELLOW,15-WHITE
//accordingly the x,y value is changed. if current cell(tile)=15(WHITE) robot moves to front tile.
// so x=x-1 and y=y
#include
#include
void tresurehunt(int x,int y);
int grid[][7]={ {1, 1, 1, 14,2, 0, 15},
{0, 1, 1, 1, 0, 0, 2},
{2, 2 ,1, 0, 15,14,1},
{15,15,14,0, 1, 2, 1},
{15,0, 0, 1, 2, 2, 14},
{14,14,14,15,0, 2, 2},
{15,15,15,1, 1, 2, 1}};
int main()
{
tresurehunt(4,4);
return 0;
}
void tresurehunt(int x,int y)
{
//int color=getpixel(x,y);
int color=grid[x][y];
if(color==YELLOW)
{
cout<<\"Tresure is in:\"<.
Assets=Liabilities+Paid in capital+Retained earnings1..pdf
Assets
=
Liabilities
+
Paid in capital
+
Retained earnings
1.
260000
=
260000
Total
260000
=
260000
2.
32000 + (8000)
=
24000
Total
284000
=
24000
+
260000
+
0
3.
80000
=
80000
Total
364000
=
104000
+
260000
+
0
4.
(60000) + 100000
=
40000
Total
404000
=
104000
+
260000
+
40000
5.
(3750)
=
(3750)
Total
400250
=
104000
+
260000
+
36250
6.
(5650)
=
(5650)
394600
=
98350
+
260000
+
36250
7.
(60000)
=
(60000)
334600
=
38350
+
260000
+
36250
8
45000 + (45000)
=
Total
334600
=
38350
+
260000
+
36250
9.
(800)
=
(800)
Total
333800
=
38350
+
260000
+
35450
Assets
=
Liabilities
+
Paid in capital
+
Retained earnings
1.
260000
=
260000
Total
260000
=
260000
2.
32000 + (8000)
=
24000
Total
284000
=
24000
+
260000
+
0
3.
80000
=
80000
Total
364000
=
104000
+
260000
+
0
4.
(60000) + 100000
=
40000
Total
404000
=
104000
+
260000
+
40000
5.
(3750)
=
(3750)
Total
400250
=
104000
+
260000
+
36250
6.
(5650)
=
(5650)
394600
=
98350
+
260000
+
36250
7.
(60000)
=
(60000)
334600
=
38350
+
260000
+
36250
8
45000 + (45000)
=
Total
334600
=
38350
+
260000
+
36250
9.
(800)
=
(800)
Total
333800
=
38350
+
260000
+
35450
Solution
Assets
=
Liabilities
+
Paid in capital
+
Retained earnings
1.
260000
=
260000
Total
260000
=
260000
2.
32000 + (8000)
=
24000
Total
284000
=
24000
+
260000
+
0
3.
80000
=
80000
Total
364000
=
104000
+
260000
+
0
4.
(60000) + 100000
=
40000
Total
404000
=
104000
+
260000
+
40000
5.
(3750)
=
(3750)
Total
400250
=
104000
+
260000
+
36250
6.
(5650)
=
(5650)
394600
=
98350
+
260000
+
36250
7.
(60000)
=
(60000)
334600
=
38350
+
260000
+
36250
8
45000 + (45000)
=
Total
334600
=
38350
+
260000
+
36250
9.
(800)
=
(800)
Total
333800
=
38350
+
260000
+
35450
Assets
=
Liabilities
+
Paid in capital
+
Retained earnings
1.
260000
=
260000
Total
260000
=
260000
2.
32000 + (8000)
=
24000
Total
284000
=
24000
+
260000
+
0
3.
80000
=
80000
Total
364000
=
104000
+
260000
+
0
4.
(60000) + 100000
=
40000
Total
404000
=
104000
+
260000
+
40000
5.
(3750)
=
(3750)
Total
400250
=
104000
+
260000
+
36250
6.
(5650)
=
(5650)
394600
=
98350
+
260000
+
36250
7.
(60000)
=
(60000)
334600
=
38350
+
260000
+
36250
8
45000 + (45000)
=
Total
334600
=
38350
+
260000
+
36250
9.
(800)
=
(800)
Total
333800
=
38350
+
260000
+
35450.
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Hypothesis Test for population variance
If we have a sample from an underlying normal distribution and variance ?
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Hypothesis Test for population variance
If we have a sample from an underlying normal distribution and variance ?.
ANSWERFundamental configurations of Windows printer Deployments.pdf
ANSWER:
Fundamental configurations of Windows printer Deployments:
The Following sections describe four fundamental configuraions that are the basis of most
Windows Printer Deployments
1.Direct Printing
2.Locally Attached printer Sharing
3.Network-attached Printing
4.Network-Attached printer sharing
you can scale these configurations up to accommodate a network of virtually any size.
1.DIRECT PRINTING:
The simplest print architecture consists of one print device connected to one computer, also
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2.LOCALLY ATTACHED PRINTER SHARING:
In addition to printing from an application running on that computer, you can also share the
printer (and the print device) with other users on the same network. In this arrangement, the
computer with the locally attached print device functions as a print server. Figure 2-14 shows the
other computers on the network, which are known as the print clients.
In the default Windows Server 2012 R2 printer-sharing configuration, each client uses its own
printer and printer driver. As before, the application running on the client computer sends the
print job to the printer and the printer driver renders the job, based on the capabilities of the print
device.
The main advantage of this printing arrangement is that multiple users, located anywhere on the
network, can send jobs to a single print device connected to a computer functioning as a print
server. The downside is that processing the print jobs for many users can impose a significant
burden on the print server. Although any Windows computer can function as a print server, you
should use a workstation for this purpose only when you have no more than a handful of print
clients to support or you have a very light printing volume.
3.NETWORK-ATTACHED PRINTING:
The printing solutions discussed thus far involve print devices connected directly to a computer
using a USB or other port. Print devices do not necessarily have to be attached to computers,
however. You can connect a print device directly to the network instead. Many print device
models are equipped with network interface adapters, enabling you to attach a standard network
cable. Some print devices have expansion slots into which you can install a network printing
adapter you have purchased separately. Finally, for print devices with no networking capabilities,
standalone network print servers are available, which connect to the network and enable you to
attach one or more print devices. Print devices so equipped have their own IP addresses and
typically have an embedded web-based configuration interface.
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make is to decide which computer will function as th.
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