The age of the Earth is estimated from the amount of uranium isotope .pdf
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The age of the Earth is estimated from the amount of uranium isotope 238 remaining. Radioisotopes like U^238 slowly decay over time turning into either another isotope or another element when they release radiation. Estimates are that only half of the U^238 that was originally present in the Earth still exists. Since the half life (time it takes a radioisotope to lose 1/2 its material) is 4 5 times 10^9 that puts the age of the Earth at 4 5 billion years U^238 decays into Th^234, which decays into Pa^234, which decays into U^234. The earliest fossil evidence of bacteria is 3 5 billion years old. Vascular plants began to appear 400 million years ago. Modern humans are estimated to have appeared about 200,000 years ago. At which of these times would the most original U^238 remain? The time bacteria first became fossilized The time vascular plants appeared The time modem humans appeared At which of these times would the least original U^238 remain? The time bacteria first became fossilized The time vascular plants appeared The time modem humans appeared Fossils found in rock suggest that: Compare these time spans by changing the scale from YEARS to SECOND: let 1 year = 1 sec. Thus, in this scale the average biology student is 20 seconds old. At this scale, how many days ago did humans appear? 1.7 hours 1 day 1.7 days 2 days 2.7 days How many years ago did the dinosaurs vanish? 2 years 2.1 years 2.2 days 2.3 days 2.4 days How many years ago did flowering plants begin to dominate? 1.5 years 2 years 2.5 years 3 years 3.5 years How many years ago did plants start to live on land? 10 years 11 years 12 years 13 years 14 years How many years ago did cells first fossilize? 91 years 111 years 121 years 131 years 141 years Solution a) Most U238 existed when the bacteria first bacame fossilised. The earliest fossil evidence of bacteria is 3.5 billion years ago b) Least U238 remains at time when first modern human appeared 200,000 years ago 2a) 180,000 years = 180,000 sec = 2.08 days or 2 days. Ans iv b) 2.06 years c)2.5 years d) 12.6 years e)111 years.
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Question 2 options:
dehydrogenations
anaerobic respiration
fermentation
digestion
oxidative phosphorylation
heterotrophs
aerobic respiration
ATP synthase
autotrophs
substrate-level phosphorylation
glycolysis
High-energy electrons from chemical bonds are transferred to a final electron acceptor, which is
oxygen
An enzyme that couples the reentry of protons into the matrix with the phosphorylation of ADP
to ATP
ATP synthetase makes ATP with energy from a proton gradient generated by an electron
transport chain.
Organisms that live on the organic compounds produced by other organisms.
The use of inorganic molecules as final electron acceptors for an electron transport chain.
A series of 10 reactions used to catabolize glucose and produce ATP by substrate levels
phosphorylation.
Reactions in which electrons lost are accompanied by protons.
ATP is formed by transfer of a phosphate group directly to ADP from a phosphate-bearing
intermediate.
The enzymatic breakdown of large molecules into smaller ones.
The electrons generated by glycolysis are donated to organic molecules and NAD+ is recycled
Organisms that harvest energy from the sun and convert it to chemical energy by photosynthesis.
dehydrogenations
anaerobic respiration
fermentation
digestion
oxidative phosphorylation
heterotrophs
aerobic respiration
ATP synthase
autotrophs
substrate-level phosphorylation
glycolysis1.
High-energy electrons from chemical bonds are transferred to a final electron acceptor, which is
oxygen2.
An enzyme that couples the reentry of protons into the matrix with the phosphorylation of ADP
to ATP3.
ATP synthetase makes ATP with energy from a proton gradient generated by an electron
transport chain.4.
Organisms that live on the organic compounds produced by other organisms.5.
The use of inorganic molecules as final electron acceptors for an electron transport chain.6.
A series of 10 reactions used to catabolize glucose and produce ATP by substrate levels
phosphorylation.7.
Reactions in which electrons lost are accompanied by protons.8.
ATP is formed by transfer of a phosphate group directly to ADP from a phosphate-bearing
intermediate.9.
The enzymatic breakdown of large molecules into smaller ones.10.
The electrons generated by glycolysis are donated to organic molecules and NAD+ is
recycled11.
Organisms that harvest energy from the sun and convert it to chemical energy by
photosynthesis.
Solution
Answer :
Dehydrogenations : Reactions in which electrons lost are accompanied by protons.
Anaerobic respiration : The use of inorganic molecules as final electron acceptors for an electron
transport chain.
Fermentation : The electrons generated by glycolysis are donated to organic molecules and
NAD+ is recycled
Digestion : The enzymatic breakdown of large molecules into smaller ones.
oxidative phosphorylation : ATP synthetase makes ATP with energy from a proton gradient
generated by an electron transport chain.
Heterotrophs : Organi.
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This document discusses human evolution and provides learning objectives related to dating methods, major anatomical features of primates, and trends in hominid evolution. It describes how radioactive dating of rocks and fossils can determine ages and discusses isotopes like carbon-14, potassium-40, and uranium that are used. Key anatomical primate features include grasping limbs and stereoscopic vision. Major hominid lineages discussed include Ardipithecus, Australopithecus, and various Homo species. The incompleteness of the fossil record results in uncertainties about the exact relationships between hominid species.
the geologic timescale general biology 2.pptx
The document provides information about the Geologic Time Scale, which is a system that chronologically relates geological strata and fossils to time periods in Earth's history. It describes the four main divisions of the Geologic Time Scale (eon, era, period, epoch) and discusses some of the major time periods within each, including key events and life forms that existed during these periods, from the earliest parts of Earth's history to the present day. A brief quiz is included at the end to test the reader's understanding.
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- Evolution is defined as the gradual development of simple matters into more complex forms over time. Darwin was the first to put together evidence to explain evolution through his theory of natural selection.
- Primary claims of evolution include that the Earth is over 4 billion years old, life emerged around 3.8 billion years ago, extinction is a major feature of biological evolution, and humans evolved from pre-existing species.
- Major claims of evolutionary theory include that natural selection is the main mechanism of evolution, whereby individuals with advantageous traits are more likely to survive and reproduce, passing on those traits; all living organisms share a common ancestry; and DNA evidence provides powerful support for the interrelatedness of all life.
5.00 History of time
Fossils, Molds, Casts provides information about fossils and the geological time scale. It aims to describe how fossils record Earth's history and evidence of evolution. Key points include that fossils are preserved plants and animals, the law of superposition establishes layer age relationships, and absolute dating determines fossil ages in years using radioisotope methods. The geological time scale is a calendar used by scientists to outline Earth and life history, broken into eras like the Paleozoic, Mesozoic, and Cenozoic. Mass extinctions have occurred throughout Earth's history.
Geologic time scale_1
The geologic time scale is a system used to describe Earth's history by dividing it into units of time based on changes in lifeforms and rock layers. It begins over 4.6 billion years ago with the formation of Earth and extends to the present. Major divisions include eons like the Precambrian and Phanerozoic, eras like the Paleozoic and Cenozoic, and periods like the Jurassic and Quaternary. Life began as single-celled organisms and evolved over billions of years to include complex plants and animals, with major extinction events periodically wiping out many species. The current geologic period is the Quaternary, which began over 2 million years ago and includes the evolution
Similar to The age of the Earth is estimated from the amount of uranium isotope .pdf (20)
More from arihantsherwani
Suppose Michael creates an RSA cryptosystem with a very large modulu.pdf
Suppose Michael creates an RSA cryptosystem with a very large mod ulus n for which the
factorization of n cannot be found in a reasonable amount of time. Suppose that Nikita sends
messages to Michael by representing each alphabetic character as an integer between 0 and 26
(A corresponds to 1, B to 2, etc., and a space to 0), then encrypts each number separately using
Michael\'s RSA cryptosystem. Is this method secure? Explain your answer.
Solution
This method is not secure, it is just equal to a \"Ceaser Cipher\", that is a permutation of letters of
alphabet, which is well-known to be easily broken using a frequency analysis..
Match the terms to the definitions provided.Question 2 optionsd.pdf
Match the terms to the definitions provided.
Question 2 options:
dehydrogenations
anaerobic respiration
fermentation
digestion
oxidative phosphorylation
heterotrophs
aerobic respiration
ATP synthase
autotrophs
substrate-level phosphorylation
glycolysis
High-energy electrons from chemical bonds are transferred to a final electron acceptor, which is
oxygen
An enzyme that couples the reentry of protons into the matrix with the phosphorylation of ADP
to ATP
ATP synthetase makes ATP with energy from a proton gradient generated by an electron
transport chain.
Organisms that live on the organic compounds produced by other organisms.
The use of inorganic molecules as final electron acceptors for an electron transport chain.
A series of 10 reactions used to catabolize glucose and produce ATP by substrate levels
phosphorylation.
Reactions in which electrons lost are accompanied by protons.
ATP is formed by transfer of a phosphate group directly to ADP from a phosphate-bearing
intermediate.
The enzymatic breakdown of large molecules into smaller ones.
The electrons generated by glycolysis are donated to organic molecules and NAD+ is recycled
Organisms that harvest energy from the sun and convert it to chemical energy by photosynthesis.
dehydrogenations
anaerobic respiration
fermentation
digestion
oxidative phosphorylation
heterotrophs
aerobic respiration
ATP synthase
autotrophs
substrate-level phosphorylation
glycolysis1.
High-energy electrons from chemical bonds are transferred to a final electron acceptor, which is
oxygen2.
An enzyme that couples the reentry of protons into the matrix with the phosphorylation of ADP
to ATP3.
ATP synthetase makes ATP with energy from a proton gradient generated by an electron
transport chain.4.
Organisms that live on the organic compounds produced by other organisms.5.
The use of inorganic molecules as final electron acceptors for an electron transport chain.6.
A series of 10 reactions used to catabolize glucose and produce ATP by substrate levels
phosphorylation.7.
Reactions in which electrons lost are accompanied by protons.8.
ATP is formed by transfer of a phosphate group directly to ADP from a phosphate-bearing
intermediate.9.
The enzymatic breakdown of large molecules into smaller ones.10.
The electrons generated by glycolysis are donated to organic molecules and NAD+ is
recycled11.
Organisms that harvest energy from the sun and convert it to chemical energy by
photosynthesis.
Solution
Answer :
Dehydrogenations : Reactions in which electrons lost are accompanied by protons.
Anaerobic respiration : The use of inorganic molecules as final electron acceptors for an electron
transport chain.
Fermentation : The electrons generated by glycolysis are donated to organic molecules and
NAD+ is recycled
Digestion : The enzymatic breakdown of large molecules into smaller ones.
oxidative phosphorylation : ATP synthetase makes ATP with energy from a proton gradient
generated by an electron transport chain.
Heterotrophs : Organi.
Simple linear regression and multiple regression are both based on t.pdf
Simple linear regression and multiple regression are both based on the use of quantitative
variables to predict a single dependent variable. It is the number of independent variables that
changes.
Describe a regression relationship using multiple independent variables.
Solution
Consider Y: weight
Let X1: gender
X2: age
X3: height
X4: nature of job
X5: heredity factors
This is one example where multiple regression is used..
Sand, Mell, and Rand are partners who share incomes and losses in a .pdf
Sand, Mell, and Rand are partners who share incomes and losses in a 1:4:5 ratio. After lengthy
disagreements among the partners and several unprofitable periods, the partners decided to
liquidate the partnership. Before the liquidation, the partnership balance sheet showed the
following:
Cash $10,000
Total “other assets,” $106,000
Total liabilities, $88,000
Sand, Capital, $1,200
Mell, Capital, $11,700
Rand, Capital, $15,100
The “other assets” were sold for $ 85,000. Proceeds from the sale of other assets were used to
payoff existing liabilities.
Determine the following:
1)The gain (or loss) realized on the sale of the assets and recording of liabilities’ payment.
2)The balances in the partners’ capital accounts after the distribution of this gain or loss to the
capital accounts.
3)Assume that if any capital deficits exist, they are not made up and the deficient partner pays
down his or her deficit to zero. How much cash will each of the partners receive in the final
liquidation?
Solution
Balance Sheet Assets Amount Liabilities Amount Cash $ 10,000.00 Total
liabilities $ 88,000.00 Other Assets $ 106,000.00 Partner\'s Capital A/c Sand $
1,200.00 Mell $ 11,700.00 RAND $ 15,100.00 Total $ 116,000.00 Total $
116,000.00 Journal Entries Particular Amount(DR) Amount(CR) 1) Cash
A/c $ 85,000.00 Loss on realization $ 21,000.00 To Other Assets $
106,000.00 (Being amount realized on sale of assets) Sand=($21000*1/10) $
2,100.00 Mell=($21000*4/10) $ 8,400.00 Rand=($21000*5/10) $ 10,500.00
To Loss on realization $ 21,000.00 (Being loss on realization distribute to
partner\'s in their profit sharing ratio) Total Liabilities $ 88,000.00 To Cash
$ 88,000.00 (Being amount paid to total liabilities) 2) Sand\'s Capital
A/c Particular Amount(Dr) Particular Amount(Cr) To Loss on realization $
2,100.00 By Balance B/d $ 1,200.00 By Cash A/c $ 900.00 Total $ 2,100.00
Total $ 2,100.00 Mell\'s Capital A/c Particular Amount(Dr) Particular
Amount(Cr) To Loss on realization $ 8,400.00 By Balance B/d $ 11,700.00 To
Cash $ 3,300.00 Total $ 11,700.00 Total $ 11,700.00 Rand\'s Capital A/c
Particular Amount(Dr) Particular Amount(Cr) To Loss on realization $ 10,500.00 By
Balance B/d $ 15,100.00 To Cash $ 4,600.00 Total $ 15,100.00 Total $
15,100.00 3) Cash A/c Particular Amount(Dr) Particular Amount(Cr) To
Balance b/d $ 10,000.00 By Total Liabilities $ 88,000.00 To Other assets $ 85,000.00
By Mell\'s Capital A/c $ 3,300.00 To Sand\'s Capital A/c $ 900.00 By Rand\'s
Capital A/c $ 4,600.00 Total $ 95,900.00 Total $ 95,900.00.
PLEASE ANSWER ALL THE QUESTIONES BELOW AS DETAILED AS POSSIBLE. THAN.pdf
PLEASE ANSWER ALL THE QUESTIONES BELOW AS DETAILED AS POSSIBLE.
THANK YOU!
Why do all (-)RNA virus have an RNA polymerase in its virion?
How does vesicular stomatitis virus regulate its mRNA transcription?
How does vesicular stomatitis virus regulate Protein synthesis?
State numerous factors for Ebola virus being an emerging deadly pathogen for us.
Why can’t influenza virus replicate in a cell where the nucleus has been removed?
PLEASE ANSWER ALL THE QUESTIONES ABOVE AS DETAILED AS POSSIBLE.
THANK YOU!
Solution
1) Negative-sense RNA has a nucleotide sequence complementary to the mRNA that it encodes.
Like DNA, this RNA cannot be translated into protein directly. Instead, it must first be
transcribed into a positive-sense RNA that acts as an mRNA.
In order to get transcribed into a positive-sense RNA these viruses need a RNA Polymerase
enzyme. As soon as the virion enters the host cell the virus uses its own RNA replicase, known
as RNA dependent RNA Polymerase to form positive RNA template strands through
complementary base pairing. This positive sense RNA strand acts as mRNA and can be
translated into structural capsomere proteins and viral RNA polyemrase enzymes for further
synthesis of virions.
2) The RNA-dependent RNA polymerase of vesicular stomatitis virus, a nonsegmented negative-
strand RNA virus, directs two discrete RNA synthetic processes, transcription and replication.
The two short extragenic regions at the genomic termini, the 3 leader (Le) and the complement
of the 5 trailer (TrC), contain essential signals for these processes.
The viral genome comprises a 47-nt leader region (Le), five transcriptional units for the N, P, M
(matrix), G (glycoprotein), and L mRNAs, and a 59-nucleotide trailer region (Tr), arranged in the
order 3 Le-N-P-M-G-L-Tr 5. The 3 Le is thought to act as a promoter for transcription of the
viral mRNAs.
The promoter for mRNA transcription was shown to include specific sequence elements within
Le at positions 19 to 29 and 34 to 46, a separate element at nucleotides 47 to 50, the
nontranscribed leader-N gene junction.
Transcription process
3) Translation in Vesicular Stomatitis Virus
3) Ebola, commonly known as Ebola hemorrhagic fever, is a rare and deadly disease caused by
infection with one of the Ebola virus strains. Ebola can cause disease in humans and nonhuman
primates (monkeys, gorillas, and chimpanzees). Ebola viruses belong to a family of viruses
called Filoviridae, and there are five known species of Ebola virus
5) Influenza A virus belongs to the family of Orthomyxoviridae. Orthomyxoviridae viruses are
one of two RNA viruses that replicate in the nucleus (the other being retroviridae). The reason is
that the machinery of orthomyxo viruses cannot make their own mRNAs. The influenza virus
has a negative sense RNA genome. In the host nucleus, the virus does primary transcription to
produce necessary proteins for replication..
Livingstone is a marble freak and loves to paly at school. During th.pdf
Livingstone is a marble freak and loves to paly at school. During the first break, he doubled the
number of marbles he had. During the second break, he lost 4 marbles. During the third break, he
increased his stock of marbles by one-third. During lunch, however, he lost half of his marbles.
After school, livingston played again and tripled what he had. He went home with 72 marbles.
How much did he start with? work backword to answer this question.
Solution
consider he had x marblles
During the first break, he doubled the number of marbles he had. = 2x
During the second break, he lost 4 marbles =2x-4
During the third break, he increased his stock of marbles by one-third =2x-4 +1/3 (2x-4)
During lunch, however, he lost half of his marbles. = (2x-4 +1/3 (2x-4))/2
After school, livingston played again and tripled what he had = 3(2x-4 +1/3 (2x-4))/2
He went home with 72 marbles
3(2x-4 +1/3 (2x-4))/2 = 72
(3/2)(4/3)(2x-4) = 72
4x-8 =72
4x = 80
x=20
he start with 20 marbles.
Invertebrates colonized terrestrial environments independently of ve.pdf
Invertebrates colonized terrestrial environments independently of vertebrates. However, they
faced similar challenges adapting to land. Compare the adaptations of flatworms, mollusks and
arthropods in regards to: i) water conservation, 2) movement/body support and 3) reproduction
that allowed them to successfully colonize terrestrial environments.
Solution
Free-living flatworms are carnivores and scavengers that eat organic matter,which they acquire
through their pharynx.Tapeworms attach to their host’s intestine and absorb food through the
body wall.Flatworms move by contracting muscles against a hydrostatic skeleton.Flatworms
reproduce asexually by pinching in half;they can also reproduce sexually inside a host by
producing eggs that are dispersed.The flat body shape of a flatworm maximizes the surface area
in contact with the environment.
Mollusks have a mantle that secretes a shell in most species,a muscular foot that provides
movement,a visceral mass containing the internal organs,and a radula.Mollusks feed by filtering
organic particles out of the water;by scraping algae off rocks;or by hunting prey.Mollusks move
by contracting muscles against a hydrostatic skeleton.Many mollusks protect themselves with a
hard shells,others can evade predators by changing color and darting away,often in a cloud of
ink.
Arthropods use their jointed appendages for movement,defense,prey capture,sensing the
environment,ornaments,weapons,and copulation.Arthropods feed on almost every imaginable
inorganic substance,either by chewing or sucking.On land,arthropods exchange gases via tubules
called tracheae or via book lungs;arthropods that live in water respire with gills.Arthropods have
specialized organs that remove nitrogenous wastes,which are deposited into the end of the
digestive tract and ejected though the anus.Arthropods have specialized sense organs that are
sensitive to light,sound,scents,chemicals, potential mates,food proximity,touch,air currents,and
body position.Arthropods move by contracting muscles attached internally to the
exoskeleton.Although there are asexually reproducing species, most arthropods reproduce
sexually,using internal fertilization.The exoskeleton offers some protection against predators, but
many arthropods defend themselves by fleeing,biting,stinging,pinching,making noises,or giving
off foul odors.Camouflage is another defense,as are wing spots that may startle
predators.Aquatic arthropods respire with gills.Terrestrial forms rely on diffusion through tiny
tubes called trachea.Trachea are cuticle-lined air ducts that branch throughout the body,and open
in tiny holes called spiracles,located along the abdomen.Insects can open and close these
spiracles,to conserve water that would otherwise be lost to evaporation from the open tubes.Their
reliance on diffusion for respiration is one of the reasons that insects are small.Being small,an
arthropod has a larger relative surface area than a human, so it loses more water.The exoskele.
Is there a similarity transformation that maps Delta ABC to Delta DBF.pdf
Is there a similarity transformation that maps Delta ABC to Delta DBF? If so, identify the
similarity transformation and write a similarity statement. Explain your answer.
Solution
yes there is a similarity in both the triangle
since all the three sides are in same proportion to the corresponding sides in the other triangle,
therefore these triangle are similar
AC/DF=CB/FE=AB/DE=1/2
SO SSS similarity transformation.
Felix and Janet Green live in Swarthmore, PA. Janets father, Larry,.pdf
Felix and Janet Green live in Swarthmore, PA. Janet\'s father, Larry, lives in Sweden. For each
of the following transactions that occur in their lives, identify whether it is included in the
calculate investment (I), government purchases (G), exports (X), or imports (M). Check all that
apply.
Solution
1. It is part of investment (I)
2. Yes, It is part of imports (M)
3. Yes, It is part of consumption (C)
4. Yes, It is part of consumption (C)
5. Yes, It is part of government spendig (G).
Do the leaves of all plants have an abscission layer If not, what h.pdf
Do the leaves of all plants have an abscission layer? If not, what happens to the leaves on such
plants?
Solution
Plants exhibit the phenomenon of abscission when it no longer needs its parts like leaves, flowers
and fruits. Most of the plants undergo abscission to conserve nutrients and energy. Almost all
plants and trees of deciduous, coniferous and rain forests shed leaves by forming abscission
layer. Abscission layer formation shed the plant part with deposition of secondary structures
made up of parenchyma, suberin and lignin as a protective layer.
Abscission in plants take place under 3 major conditions- 1. to conserve water and
photosynthetic energy; 2. to resist the pests or as part defense mechanism and 3. to improve
photosynthetic efficiency of plants. Generally, the leaves that became older are shed through
abscission. This is because, the older leaves lose chlorophylls and may develop other pigments
like carotenes, xanthophylls and anthocyanins abundantly. If abscission does not take place, the
older leaves remaining on plant may not carry out photosynthesis efficiently as a chlorophyllous
leaf does. Due to this plant may exhibit poor growth and nutrient deficiency and may even die
due to inadequate nutrient synthesis..
Discuss the outcomes and challenges surrounding shortages in various.pdf
Discuss the outcomes and challenges surrounding shortages in various markets. Below are a few
examples from NBC learn. Pick one of these to discuss: Bees Tamiflu Workers Rice
Solution
The report notes that the record population can be viewed as a success because it means people
are living longer — average life expectancy has increased from about 48 years in the early 1950s
to about 68 in the first decade of the 21st century — and more children are surviving worldwide.
But not everyone has benefited from a higher quality of life.
In some of the poorest countries women are having more babies, stymieing development and
perpetuating poverty; in some of the wealthier countries low fertility rates and a shortage of
workers are raising concerns about the sustainability of economic growth and social programs.
World population by the billions
1 billion - 1804
2 billion - 1927
3 billion - 1959
4 billion - 1974
5 billion - 1987
6 billion - 1999*
7 billion - 2011
*Year when the milestone was observed by the United Nations.
Source: United Nations Population Fund
\"This report makes the case that with planning and the right investments in people now — to
empower them to make choices that are not only good for themselves but for our global
commons — our world of 7 billion can have thriving, sustainable cities, productive labor forces
that can fuel economic growth, youth populations that contribute to the well-being of economies
and societies, and a generation of older people who are healthy and actively engaged in the social
and economic affairs of their communities,\" writes Babatunde Osotimehin, executive director of
the UNFPA..
Create a C program that implements The Game of Life cellular auto.pdf
Create a C program that implements \"The Game of Life\" cellular automata (CA) simulation. A
few comments: You will want 2 two-dimensional arrays. One that holds the current state of the
\"universe\" and second that is generated from the first using the Game of Life rules, which are
given below. After the computations for filling the second array are complete, the contents of the
second array can be copied back to the first. You can make the array as big as you want, but you
should probably use 20 times 20 as a minimum. As discussed in class, the outside boundary of
the \"universe\" can be a bit tricky. To keep it simple for this exercise, you can assume that the
outside boundary cells are always \"dead\". You can use whatever character (or number) that
you want to represent the \"living\" cells. The dead cells should probably be spaces. The rules
for whether a particular cell lives or dies are determined by the 8 surrounding cells. If a particular
cell is alive, then: If only 0 or 1 of the surrounding cells are alive, the cell dies. (It\'s lonely.) If
4 or more of the surrounding cells are alive, the cell dies. (Too much competition.) If 2 or 3 of
the surrounds cells are alive, the cell stays alive. If a particular cell is dead, then it comes to life
if exactly 3 of the surround cells are alive. You will have to pick some starting configuration for
the universe. An easy thing to do is fill the array randomly, and then let things go and see what
evolves. Write your program as described above and demo the operation to your TA. Then try
some modifications. First, try using a different starting configurations. Look here for some
ideas:. Gosper\'s Glider Gun is a famous one. Or try a simple spaceship or oscillator. Then try
changing the rules determining life and death. A common variation is to include a \"birth\" when
6 neighboring cells are alive in addition to the \"3-cell\" birth. Finally, there will be a short quiz
covering 2D arrays. This exercise should be very familiar.
Solution
#include #include #include #include #include #include #include #define ROWS 60 #define
COLS 60 #define OFF 0 #define ON 1 #define BLACK 8, 0, 0, 0, 0 #define scr_width 600
#define scr_height 600 #define cell_width (scr_width / ROWS) #define cell_height (scr_height /
COLS) We must hold two copies of the board so that we can analyze board and make *
changes to temp when killing/creating cells so that we don\'t cause changes * to the board to
affect the following cells. */ char board[ROWS][COLS]; char temp[ROWS][COLS];
SDL_Rect cells[ROWS][COLS]; /* Stores positions of each cell for blits. */ void
randomize_board(void); void initialize_grid(SDL_Surface* screen); void
blit_board(SDL_Surface* bcell, SDL_Surface* screen); int num_neighbours(int x, int y); void
update_board(void); int clear_board(SDL_Surface* screen, Uint32 color); int
clear_cell(SDL_Surface* screen, int x, int y, Uint32 color); void initialize_cells_array(void); int
main(void) { SDL_Init(SDL_INIT_VIDEO).
Copy a String in Reverse Order Write a program with a loop and ind.pdf
Copy a String in Reverse Order
Write a program with a loop and indirect addressing that copies a string from source to target,
reversing the character order in the process. Use the following variables:
source BYTE \"This is the source string\",0
target BYTE SIZEOF source DUP(\'#\')
and here is what I have got so far
;reverse
INCLUDE IRVINE32.INC
.DATA
SOURCE byte \"this is the source string\", 0
target BYTE SIZEOF source DUP(\'#\')
.code
main PROC
Call Clrscr
mov esi,0
mov edi,SIZEOF source-TYPE source
mov ecx,LENGTHOF source
L1: mov al, source[edi]
mov target[esi]
sub edi TYPE source
loop L1
exit
main ENDP
END main
Solution
program with a loop and indirect addressing that copies a string from source to target
MOV SI, source ; Get source address
MOV DI, (target + SIZEOF source) ; Get the ending address for target
LOOP:
MOV AL, [SI] ; Get a byte
MOV [DI], AL ; Store a byte
INC SI ; Move forward one in source
DEC DI ; Move back one in target
CMP AL, 0 ; Have we reached the zero termination?
JNZ LOOP
RET.
Compare and contrast prokaryotic cells and eukaryotic cells. Compare.pdf
Compare and contrast prokaryotic cells and eukaryotic cells. Compare and contrast passive
transport and active transport. Compare and contrast animal cells and plant cells. Summarize
the endosymbiosis theory including the evidence that supports it.
Solution
1. Similarities between prokaryotic cells and eukaryotic cells:
Differences between Prokaryotic cells and Eukaryotic cells:
2. Similarities between active and passive Transport:
Differentiate between Active and Passive Transport:Prokaryotic CellsEukaryotic cellsCells
don\'t have nucleus.Cells have nucleus.They don\'t have organelles.They have organelles.They
all are unicellular.They may be unicellular or multicellular..
Case Study 1Hotel worker Danny Ruiz was living with his wife and f.pdf
Case Study 1
Hotel worker Danny Ruiz was living with his wife and four children in a cramped New York
apartment when he saw a television ad promising the family a way out. “Why rent when you can
own your own home?” Penn- sylvania builder Gene Percudani asked. The company even offered
to pay his rent for a year, while he saved for a down payment. So the Ruiz family fled the cityhe
Pocono Mountains, where they bought a three- bedroom Cape Cod home for $171,000.
However, when they tried to refinance less than two years later, the home was valued at just
$125,000. “I just about flipped,” said Mr. Ruiz. Later Mrs. Ruiz remarked about her husband,
“He went nuts.” Percudani, a 51-year-old native of Queens, New York, built a thriv- ing
homebuilding business in this market, running folksy television ads offering New Yorkers new
homes in Pennsylvania. If they joined Percudani’s program, called “Why Rent,” homeowners
would find financing through another of his companies, Chapel Creek Mort- gage, which
brokered loans from J. P. Morgan Chase and the company’s Chase Manhattan Mortgage unit.
For years, the “Why Rent” program appealed to workers with modest salaries, such as Eberht
Rios, a truck driver for UPS. Rios bought a home in the Poco- nos for $140,000. This year, when
he tried to refinance, he was told the home was valued at only $100,000. One local appraiser,
Dominick Stranieri, signed off on most of the “Why Rent” deals that state officials now say were
overpriced, including the Rios and Ruiz homes. Percudani’s firm picked Stranieri as his
appraiser because of his quick work and low fee of $250, instead of the typical $300 to $400. In
exchange for a steady stream of work, Mr. Stranieri accepted without ques- tion valuations from
Percudani’s company. Other common methods of creating revenues include investors and others
buying distressed proper- ties and then, using inflated appraisals, selling them for a big profit. In
order to secure the efforts of a “dirty appraiser,” those involved with the fraud would pay up to
$1,500 under the table on top of the appraiser’s stan- dard fee of $400. Another unique twist to
the plot is that few of the people involved in making mortgage loans have a long- term interest in
them. Traditionally, bankers made loans directly and held them, giving the lenders a strong
incentive to find fair appraisals to protect their interest. Today, however, many appraisers are
picked by independent mortgage brokers, who are paid per transaction and have little stake in the
long-term health of the loans. Many lenders have also lost a long-term interest in their loans,
because they sell them off to investors. Appraisers increasingly fear that if they don’t go along
with higher valuations sought by bro- kers, their business will dry up. Do you think a county
appraiser would do a lot bet- ter than a private practitioner? Joel Marcus, a New York–based
attorney recently had his property valued at $2.2 million by a county appraiser, up fro.
Assignment Submission Assignment, you submit answers by question by .pdf
Assignment Submission Assignment, you submit answers by question by question parts. The
number of submissions remaining for each question part only changes if you submit or change
the answer. Assignment Scoring Assignment submission is used for your score. The complex
number z = -1 + i in polar form is z =. The complex number z = 2(cos(pi/6) + i sin(pi/6)) in
rectangular form is z = ___________. The complex number graphed below can be expressed in
rectangular form as z = or in polar form as z = _________.
Solution
a)given z =-1+ 1i
r =[(-1)2+12]
r=2
=tan-1(1/-1)
=-tan-1(1/1)
=-(/4)
=(3/4)
z=2 [cos(3/4) +i sin(3/4)]
====================================
b)z=2 [cos(/6) +i sin(/6)]
z=2 [(3 /2) +i(1/2)]
z=3 +i1
z=3 +i
c) z =(1+ 1i)
z=2 [cos(/4) +i sin(/4)].
A linear dynamical system can be created for two masses connected by.pdf
A linear dynamical system can be created for two masses connected by springs between one
another, and connected to opposing walls. The state vector at time t is a 4?×?1 vector consisting
of the displacement and velocity of each of the two masses. The transition matrix for this
dynamical system is given by the following matrix.
[?i
i
? 30
? ? 30]
Solution.
A 7 year old female with a history of previous UTI’s suddenly develo.pdf
A 7 year old female with a history of previous UTI’s suddenly develops painful, dark-orange
colored urine, with fever and back pain. Her urinalysis reveals dipstick tests that are positive for
protein (2+), blood (1+), leukocyte esterase (4+), and nitrate (4+). The microscopic examination
yields 20-25 WBC/HPF, 5-10 RBC/HPF, 5-10 WBC Casts/LPF, and numerous bacteria. The
patient has been taking Tylenol and vitamin C supplements daily for the past 24 hours, since the
condition began.
Based on the results 1) what diagnosis/condition would you surmise at this point? 2) what
dipstick/microscopic test support this? 3) is the patient’s history significant? 4) can the
medication or vitamin supplement have any impact on the test results?
Solution
Women often experience urinary tract infections that are often marked by pain in the abdomen or
lower back. The urine is dark-orange colored due to blood mixing with dark yellow urine. The
infection may be due to bacteria E. coli, (MayoClinic).
Leukocyte esterase test dipstick that indicates; the presence of white blood cells either as whole
cells or as lysed cells; supports this assumption and the microscopic examination yielding 20-25
WBC/HPF also backs this assumption.
Yes with all the medical history and the details of the tests done; one can proceed with the
treatment.
Yes according to researches; Vitamin C spiking of urine has shown to demonstrate false-negative
results at various concentrations, (Ko DH, 2015) in case of Dipstick Test Results..
5. Toxco, Inc. emits a noxious gaseous substance as a waste product o.pdf
5. Toxco, Inc. emits a noxious gaseous substance as a waste product of its operations. As part of
its manufacturing process, it also generates liquid solvent wastes such as toluene. Analyze which
if any, of these materials are regulated under RCRA. If there is any additional information that
would be needed to answer this question, identify what that is. (15 points)
Solution
Following the discovery of an illegal meth lab by law enforcement, officers collect
samples of the chemicals used in that lab as evidence for prosecution. After all forensic samples
are gathered, a Hazardous Materials (Hazmat) Team is summoned to clean up the remaining
chemicals and other by-products or wastes. Estimates indicate that for each pound of meth
produced, between five and six pounds of highly toxic waste is generated.1
The by-products or
wastes removed by the Hazmat Teams are generally treated as RCRA hazardous waste without
further analysis, and are sent to a RCRA permitted site for treatment and disposal.
State and Federal officials also have found that during the operation of the meth lab, the
residue from meth production may enter porous building materials such as wallboard, carpets
and furnishings.2
In addition to the potential contamination of the building and its contents, there
may be further contamination from the improper disposal of these by-products or wastes by the
producers of the meth, which is typically poured into the indoor plumbing drains that flow either
into the city sewer or individual sewage treatment systems (e.g., septic systems), or modified to
allow drainage directly onto the soil near the meth lab. The other disposal method typically seen
at functioning meth labs is a burn or burial pit. The primary environmental concern from
drainage of these materials into the soil or from disposal into a burn or burial pit is
contamination
of ground water by the chemicals, including volatile organic chemicals used in the production of
meth, as well as vapor intrusion which involves soil vapor gas from ground water plumes
infiltrating, and accumulating in, nearby buildings (residences and nonresidential sites) at
dangerous levels.3
In most States, the property owner is responsible for decontamination of the property. In
some States, the health department or another agency may evaluate the property for risks from
long-term exposure to residual chemicals, including the potential environmental impacts of
chemical spills into the environment or improper waste disposal.
V. How Does RCRA Regulate Hazardous Waste?
The determination of whether a particular waste is a Subtitle C RCRA hazardous waste is
a multi-step process that depends on a number of factors, including the chemical character and
composition of the waste. The regulations for making the hazardous waste determination are
found in 40 CFR part 261.4
The first step is to determine whether the waste in question is a solid waste as defined
under RCRA. The RCRA regulations require this determinatio.
21. Two fair coins are tossed. Let E be the event not more than one h.pdf
21. Two fair coins are tossed. Let E be the event not more than one head and F the event at least
one of each face. Define the sample space S = {HH, HT, TH, TT) and assign probability ¼ to
each simple event. a. Determine the P(E) b. Find the P(F) c. Find the P(E intersection F) d. Are E
and F independent or dependent? 23. Three fair coins are tossed. Let E be the event not more
than one head and F the event at least one of each face. Define the sample space S = {HHH,
HHT, ..., TTT} and assign probability 1/8 to each simple event. a. Determine the P(E) b. Find the
P(F) c. Find the P(E intersection F) d. Are E and F independent or dependent?
Solution
21.
a)
E be the Event = { Not more than one Head } = {(H,T) (T,H) (T,T) }
b)
F be the Event = { atleast one of each face } = { (H,T) (T,H)} =
Sample Space = { (H,T) (T,H) (T,T) (H,H) }
c.
P( E n F) = {(H,T) (T,H)}
d.
Dependent
Note : We are suppose to answer only 1 question in time.
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