A population has a proportion equal to 0.30. Calculate the following.pdf
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A population has a proportion equal to 0.30. Calculate the following probabilities with n = 100. Find P( ? 0.35). Solution A population has a proportion equal to 0.30. Calculate the following probabilities with n = 100. Find P( ? 0.35). sd of proportion = ?(pq/n) = ?(0.3*0.7/100) = 0.0458 z-score = (0.35-0.30)/0.0458 = 1.09 P( p>0.35) = P(z>1.09) = 0.1379.
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D. The TCP/IP Application layer is virtually identical to the OSI Application layer.
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True or false1 Anabolism results in increased production of biomas.pdf
True or false
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2 Oxygenic phototrophs can use hydrogen sulfide as an electron donor.
3 Oxidation is the process of accepting electrons.
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1. Anabolism: The process of formation of biomolecules during metabolism such as proteins,
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b)There are 44 chromosomes containing 5 ng of DNA.
c)There are 44 chromosomes containing 10 ng of DNA.
d)None of the above.
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Telophase I of meiosis (after cytokinesis). Express your answer as two numbers separated by a
comma (example: 12, 4).
Part C: Write the number of chromosomes and amount of DNA (ng) present per cell at the end of
Anaphase of mitosis.
Part D: Write the number of chromosomes and amount of DNA (ng) present per cell at the end of
Telophase II of meiosis (after cytokinesis).
Solution
The cell cycle of actively dividing eukaryotic cell consists of gap phases (G 1 and G2), S phase
and M phase. In G 1 phase the cell prepares for cell division .In S phase DNA replication occurs
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Thanks for the question
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True or false
1 Anabolism results in increased production of biomass.
2 Oxygenic phototrophs can use hydrogen sulfide as an electron donor.
3 Oxidation is the process of accepting electrons.
4 Water appeared on the earth as a result of bombardment by asteroids.
5 The highest abundance of microbial life in soils appears in the O horizon.
6 Typically, increased oxygenic photosynthetic activity results in a decrease in pH values during
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7 Freshwater represents about 10% of the total water on Earth.
8 Oligotrophic organisms thrive in low nutrient conditions.
9 Establishment of a proton motive force (PMF) is required to generate ATP via substrate level
phosphorylation.
10 Siderphores are used to increase availability of manganese in aerobic environments.
Solution
1. Anabolism: The process of formation of biomolecules during metabolism such as proteins,
polysaccharides, lipids, and nucleic acids etc is known as anabolism, this results in an increase in
the production of biomass.
Therefore, option is: True
2.Oxygenic phototrophs use oxygen as electron donar, while as anoxygenic phototrophs uses
hydrogen sulfide as electron donor.
Therefore, option is: False
3. Oxidation is a process in which a substance looses one or more electrons, while as reduction is
a process of accepting electrons.
Therefore, option is: False
4. The earth got most of the water about 4.6 billion years ago from the asteroid impacts.
Therefore, option is: True.
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The diploid number of the hypothetical animal Geneticus introductus is 2n= 44. Each diploid
nucleus contains 5 ng of DNA in G1.
Part A : What amount of DNA is contained in each nucleus at the end of S phase? What amount
of DNA is contained in each nucleus at the end of phase?
a)There are 22 chromosomes containing 2.5 ng of DNA.
b)There are 44 chromosomes containing 5 ng of DNA.
c)There are 44 chromosomes containing 10 ng of DNA.
d)None of the above.
Part B: Write the number of chromosomes and amount of DNA (ng) present per cell at the end of
Telophase I of meiosis (after cytokinesis). Express your answer as two numbers separated by a
comma (example: 12, 4).
Part C: Write the number of chromosomes and amount of DNA (ng) present per cell at the end of
Anaphase of mitosis.
Part D: Write the number of chromosomes and amount of DNA (ng) present per cell at the end of
Telophase II of meiosis (after cytokinesis).
Solution
The cell cycle of actively dividing eukaryotic cell consists of gap phases (G 1 and G2), S phase
and M phase. In G 1 phase the cell prepares for cell division .In S phase DNA replication occurs
and the DNA content doubles. In G 2 phase, the cell gets ready for mitosis.
In mitosis, two daughter cells with same genetic content as parent are produced. Mitosis occurs
through five stages: prophase, prometaphase, metaphase, anaphase and telophase.
Prophase: This stage extends over half of mitosis. The nuclear membrane disintegrates. The
centrosome of chromosome duplicates to form two daughter centrosomes and migrates to
opposite poles of the cell. The centrosomes organise the production of microtubules, from which
the spindle fibres arise. The chromosomes condense and consist of two identical chromatids (or
sister chromatids) held together at the centromere.
Prometaphase: The chromosomes, move to the middle of the cell. The spindle fibres bind to
centromere of each chromosome at kinetochore.
Metaphase: The chromosomes align along the metaphase plate of the spindle apparatus.
Anaphase: This is the shortest stage of mitosis. The centromeres divide, and the sister chromatids
of each chromosome are separated and move to the opposite poles of the cell. The separated
sister chromatids ordaughter chromosomesreceives a copy of each parental chromosome.
Telophase: This is the final stage of mitosis. The nuclear membrane reforms, the chromosomes
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The Antoine equation often is used to describe vapor pressures: lnP = A B /(T + C) . For
temperatures near the normal boiling point:
a. Derive an expression that gives the behavior of Hvap as a function of temperature which is
thermodynamically consistent with the Antoine equation.
Solution
The Antoine equation is a mathematical expression of the relation between the vapour pressure
and the temperturer of pure substances. The equation was first proposed by Ch. Antoine, a french
researcher, in 1888. The basic form of the equation is:
lnP = A - B / ( T + C )
and it can be transformed into this temperature-explicit form:
T = B ( A - lnP ) - C
where: P is the absolute vapor pressure of a substance
T is the temperature of the substance
A, B and C are substance-specific coefficients (i.e., constants or parameters
log is typically either log10 or loge
A simpler form of the equation with only two coefficients is sometimes used:
lnP = A - ( B \\ T )
which can be transformed to:
T = B ( A - lnP )
The Antoine equation is a semi-empirical equation which expresses vapour pressure as a
function of temperature. A new, rapid and highly accurate method for obtaining its three
constants from experimental data is presented and applied to ethanol, water and 14 anaesthetic
substances where p is the vapour pressure, T is temperature and A, B and C are component-
specific constants.
Usually, the Antoine equation cannot be used to describe the entire saturated vapour pressure
curve from the triple point to the critical point, because it is not flexible enough. Therefore,
multiple parameter sets for a single component are commonly used. A low-pressure parameter
set is used to describe the vapour pressure curve up to the normal boiling point and the second
set of parameters is used for the range from the normal boiling point to the critical point.
The Antoine Equation is a vapor pressure equation and describes the relation between vapor
pressure and temperature for pure water between 0°C and 373.946°C.
For T in Kelvins and P in kPa:
273.150 < T 373.150: A = 16.5699, B = 3984.92, C = 39.724
373.150 < T 647.096: A = 16.7285, B = 4169.84, C = 28.665
Antoine(T) = exp(A B / (T + C))
For temperatures above the critical temperature (373.946°C / 647.096K), where water is a
superfluid, the increase in vapor pressure appears essentially linear† with the same slope (i.e.
dP/dT 235.88 kPa/°C). This behavior appears consistent with the Ideal Gas Law and statements
that supercritical water behaves like an ideal gas. Now if we define Tc = 647.096 K, we can
write
P(T Tc) = Antoine(T)
P(T > Tc) = 235.88 (T Tc) + Antoine(Tc)
Hence, the behaviour of delta hvap as a function of temperature which is termodynamically
consistent with the the antoine equation is derived ..
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The converson of the English to Christianity began a rich period of artistic production in
England, especially in the creation of a. secular mocaics, b. illuminated manuscripts, c.
ecclesiastical treatises, d. gothic cathedrals
Solution
b. illuminated manuscripts
Elaborately decorated manuscripts were prepared after the conversion. Thus b is the correct
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Table 19. Highest Level of Type of organization germ layers \"asymmetry (sponges) Cell Noa
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(flatworm) Hoda Nematoda worms) bila uran 3 and have germ Germ associated with
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Solution
3. Bilateral symmetry
5.Value of body cavity:
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turgescent state of nematode body.
b) Coelom makes the digestive system independent of the body wall. It can act as hydrostatic
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Question 12 (5 points)The reason that mitosis can result in the pr.pdf
Question 12 (5 points)
The reason that mitosis can result in the production of haploid gametes in alternation of
generations is because
Question 12 options:
The sporophyte and gametophyte are both haploid
Eggs and sperm are produced directly by the diploid sporophyte via meiosis
Eggs and sperm are produced by the diploid gametophyte via mitosis
Eggs and sperm are produced by the haploid gametophyte via mitosis
Save
Question 13 (5 points)
In the fern life cycle, how does fertilization occur?
Question 13 options:
The sporangia on the undersides of the leaves produce gametes, which are flung our of the leaf
and find each other on the ground to fuse
The sporophyte produces sperm and eggs in gametangia. The sperm reach the eggs by swimming
through water
The gametophyte produces sperm and eggs in gametangia (antheridia and archegonia). The
sperm reach the eggs by swimming through water
There is no sexual reproduction in ferns
Save
Question 14 (5 points)
The earliest land plants were most likely to have lived in which of the following habitats?
Question 14 options:
arid desert environments
Salt water marshes
Tropical rainforests
fresh water wetlands
Save
The sporophyte and gametophyte are both haploid
Eggs and sperm are produced directly by the diploid sporophyte via meiosis
Eggs and sperm are produced by the diploid gametophyte via mitosis
Eggs and sperm are produced by the haploid gametophyte via mitosis
Solution
Q12 Eggs and sperms are produced by haploid gametophyte by mitosis.
The large, leafy fern is the diploid organism. On the undersurface of its fronds or leaves, its cells
undergo meiosis to create haploid cells. However, these cells do not immediately unite with
others to recreate the diploid state. Instead, they are shed as spores and germinate into small
haploid organisms. Because the diploid organism creates spores, it is called the sporophyte
generation of the life cycle. Upon reaching maturity, the haploid organism creates haploid egg
and sperm cells (gametes) by mitosis . Because the haploid organism creates gametes , it is called
the gametophyte generation of the life cycle. The male gametes (sperm) are then released and
swim to the female egg. Fusion of the gametes creates the new diploid sporophyte, completing
the life cycle.
Q13 The gametophyte produces sperm and eggs in gametangia (antheridia and archegonia). The
sperm reach the eggs by swimming through water
The male organs or antheridia produce and shed numerous motile sperm or antherozoids. These
antherozoids are armed with hair-like or whip-like cilia or flagellae and are able to swim through
water; they do not travel great distances and are only released when free water is available. Only
a continuous film of water is needed. They antherozoids are chemically attracted to the
developing flask-like female organs or archegonia where fertilization of the single egg cell
embedded in each archegonium takes place.
The sporophyte has a double set of chromosomes (i.e. diploid) and i.
peripheral circuit functions built-in to the HC12 microcontroller, t.pdf
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itself or the memories. Access to these is via external pins on the Microcontroller IC package.
What are some of these built-in circuit functions called, and what do they do?
Solution
The peripheral circuit communicate with the Microcontroller using built in cirucit called BUS.
The bus is a set of shared communication lines that serves as the central nervous system of the
computer. Data, address, and control signals are shared by all system components via the bus.
Each component connected to the bus communicates information to and from the bus via its own
bus controller. The data lines,address lines, and control lines allow a specific component to
access data addressed to that component. The data lines are used to communicate words to and
from data.
registers in the various system components such as memory, CPU, and input/output
(I/O) peripherals. The address lines are used to select devices on the bus or specific
data locations within memory. Devices usually have a combinational logic address
decoder circuit that identifies the address code and activates the device..
Mendez Corporation has 10,000 shares of its $100 par value, 7 percen.pdf
Mendez Corporation has 10,000 shares of its $100 par value, 7 percent cumulative
preferred stock outstanding and 50,000 shares of its $1 par value common stock outstanding.
In Mendez’s first four years of operation, its board of directors paid the following
cash dividends: 2011, none; 2012, $120,000; 2013, $140,000; 2014, $140,000.
Determine the dividends per share and total cash dividends paid to the preferred and
common stockholders during each of the four years.
Solution
The dividends per year on preferred stock = 10,000 x $ 100 x 7 %
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Dividend distribution:
Year
Total Dividend Amount
Preferred
Common
2011
-
-
-
2012
$120,000
$120,000
$0
2013
$140,000
$90,000
$50,000
2014
$140,000
$70,000
$70,000
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First dividend is paid to Preferred stock and then excess amount goes to common share holder.
For cumulative preferred stock, any deficiency in dividend, not paid carries over to subsequent
years and is added to that year\'s dividend.
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Preferred Dividends paid in 2012 = $ 120,000
Preferred Dividends shortage on 2012 = $ 20,000
As the Dividend was insufficient for preferred stocks, common stocks holder was not paid any
dividend during 2011 & 2012.
Preferred Dividends paid in 2013 = $ 70,000 + $ 20,000 = $ 90,000
Common Dividends paid in 2013 = $ 140,000 - $ 90,000 = $ 50,000
Preferred Dividends paid in 2014 = $ 70,000
Common Dividends paid in 2014 = $ 140,000 - $ 70,000 = $ 70,000
Computation of Dividends per share:
2011
2012
2013
2014
Preferred Stock (Dividends per share)
$0
$12.00
$9.00
$7.00
Common Stock (Dividends per share)
$0
$0
$1.00
$1.40
Explanation:
Dividends per share = Dividend paid/No. of shares
Year 2012:
Dividends per share for Preferred Stock = $120,000/10,000 = $ 12
Dividend per share for Common Stock = $ 0
Year 2013:
Dividends per share for Preferred Stock = $90,000/10,000 = $ 9
Dividends per share for Common Stock = $ 50,000/50,000 = $ 1
Year 2014:
Dividends per share for Preferred Stock = $70,000 /10,000 = $ 7
Dividends per share for Common Stock = $ 70,000/50,000 = $ 1.40
Year
Total Dividend Amount
Preferred
Common
2011
-
-
-
2012
$120,000
$120,000
$0
2013
$140,000
$90,000
$50,000
2014
$140,000
$70,000
$70,000.
If the dpy-11 and unc-31 genes are linked on the same chromosome, how.pdf
If the dpy-11 and unc-31 genes are linked on the same chromosome, how many phenotypes
would the F_2 progeny show? If the two genes are on different chromosomes and therefore
independently assorting, how many phenotypes would you see in the F_2 worms? List the
phenotype(s). Strawberry plants have varieties that flower when the days are still short, and
other varieties that flower only when the days get longer. The long-day allele is dominant to the
short-day allele. You\'d like to test whether or not this gene is inherited in a Mendelian fashion.
When you cross between heterozygous plants and count the offspring in the F_1 generation, this
is what you find: 85 plants with no flowers on short days and 35 plants with flowers on short
days What would you expect the ratio of phenotypes be? What is your Chi squared value?
Show your work What does this mean?
Solution
2. Linked genes tend to inherit together for the next-generations. No crossing over ororoccurs
independent assortment occurs. No new characters appear. Where as gegenes are on different
chromosomes, they are non linked. Crossing over or independent assortment takes place. New
characters appear along with parental characters. So dpy-11 & unc-31 present on same
chromosome, they are linked and give only parental characters. Only 2 phenotypes appear. If the
said genes present on different chromosomes, they give 4 different phenotypes
3. a. Heterozygous long day representing as Ll. When crossing or hybridization occur between
these genes
Ll x Ll ----------> 3:1 is the phenotypes ratio. Three are long and one is short.
Chi square value can be calculated by using the formula.
(O-E)^2/E
O = observed phenotypes; E = Expected phenotypes.
How does a wave formation influence the resultant waveSolution.pdf
How does a wave formation influence the resultant wave?
Solution
Waves are usually formed by distant wind.Generally larger waves are more powerful than small
waves . Waves can pile up at shore lines and create a rip current.The resultant wave heights Hi
and the wave period Ti are related to the factors that influence the energy transfer between wind
and the ocean..
How are health services paid for Provide a definition for the term .pdf
How are health services paid for? Provide a definition for the term \"third-party payer\", discuss
the different payers currently operating in the market, and assess the importance of each. Please
use a table format to list the information.
Solution
There are two main payers types: single payer and third-party payer. Single payer is when the
person that obtains the health service is the same which pay for it, here relationship is direct taker
and provider of health services.
However, it is more common the presence of a third party payer, which is an external agent that
covers the other health expenses or part of them, they are insurance companies or public social
protection systems.
Single Payer
Third-Party Payer
Type of relation with provider
Direct
Indirect
Capability to compare Prices/Features
Relatively High
Almost Null
Payment proportion
Full
Low percentage
Single Payer
Third-Party Payer
Type of relation with provider
Direct
Indirect
Capability to compare Prices/Features
Relatively High
Almost Null
Payment proportion
Full
Low percentage.
Given the information regarding mitosis and meiosis, clearly and con.pdf
Given the information regarding mitosis and meiosis, clearly and convincingly explain in your
own words how an error in MITOSIS could possibly lead to a gamete containing new genetic
information and thereby giving rise to a new phenotype. Use detail be comprehensive. (7 points)
Solution
The process of mitosis is divided into two phases: Interphase and M-phase.
Interphase is further divided into G1, S and G2 phase. In G1, protein synthesis occurs for S-
phase, in S phase, DNA replicates and in G2 phase, protein synthesis occurs for M-phase to start
properly. Now, during S phase, the ‘2n’ nucleus converts into 4n.
So, the cell entering M-phase is already ‘4n’.
M-phase is the actual mitotic phase, in which the cell divides into two. It contains four phases,
prophase, metaphase, anaphase and telophase. Your question is related with these four phases. In
prophase, the entire DNA converts into chromosomes; the process is known as condensation of
DNA. In metaphase, these chromosomes align on the equator. Now, if alignment at equator is
not correct, then anaphase cannot be started due to metaphase check point.
In anaphase, the aligned chromosomes move towards the opposite poles. This phase is the main
phase, related with formation of gametes. The cell is already ‘4n’. ‘2n’ chromosomes move
towards one end, and the rest of ‘2n’ move towards the other end. Now, suppose an error occurs
in mitosis, and ‘3n’ chromosomes move towards one end, and 1n goes towards the other cell;
then two gametes, ‘3n’ and n will be formed.
This ‘3n’ and ‘n’ gametes are gametes having different genetic information in comparison to
their ‘2n’ counterparts.
Interestingly, Meiosis II will result in gamete formation is mitotic division (equational division).
Haploid cells with \'2c\' amount of DNA leave Meiosis I. Now, in anaphase, suppose one
chromosome\' 2c DNA migrates to one end (due to error) and no DNA migrates to the other end;
then two gametes will be formed; one having all normal chromosomes + one double
chromosome with 2c DNA, and the other having no DNA.
Please note: normally gametes are 1n (haploid) and 1c.
Suppose this gamete with double DNA (2c) fertilizes with a gamete with half DNA (1c), then the
resulting cell will have “all normal chromosomes + one chromosome with ‘3c’ amount of DNA”.
This condition if occurring with chromosome 21 is known as Down\' syndrome.
It is important to note that errors in Mitosis result in non-disjunction; which further results in
many Mendelian disorders; as well as ploidy defects.
** Hope it helps..
First, look to determine what is the earliest era that fossils have .pdf
First, look to determine what is the earliest era that fossils have been found here in Georgia.
Next, see if you can find what the oldest fossil that has been found in Georgia. Give the name of
the fossil, when it was found, and where the fossil was found. Give a brief description of the
organism that was fossilized.
Solution
Trilobites are the oldest fossils found in georgia.They have been found in Georgia’s Appalachian
Plateau which is present the Coosa River in Floyd County.
These fossils are considered to be arthropods,Moreover they had to shed their shells to grow.As a
result the fossils found in these places usually shed shells as bits of individuals or as complete
individuals or as imprints of individuals..
Figure out and write down a definition of a translation that does no.pdf
Figure out and write down a definition of a translation that does not use (x,y) coordinates
Solution
There are many kinds of oprations in mathematics like translation, rotation, reflection, resized
etc. Translation is one of them.
Translation is an operation which is uesd in geometry to move an object in a fixed direction and
for same distance..
explain why right and why wrong Indentify configuration (below) whic.pdf
explain why right and why wrong Indentify configuration (below) which I s not a correct
representation of H and I.
Solution
option C is not correct one.
Explanation:
If we stretch all right hand fingers in the direction of current direction given then the thumb gives
us into the board (into symbol) or out of the board (dot symbol).
In the option C the field should be out of the board, so it is not correct one..
Develop a system flowchart and then write a menu-driven C++ program .pdf
Develop a system flowchart and then write a menu-driven C++ program that uses user-defined
functions arrays, and a random number generator.
Upon program execution, the screen will be cleared and the menu shown below will appear at
the top of the screen and centered. The menu items are explained below.
Help Smallest Largest Quit
H or h ( for Help ) option will invoke a function named help() which will display a help screen.
The help screen(s) should guide the user how to interact with the program, type of data to be
used, and what results would the program produce. Each help screen should remain on the
monitor until the user strikes any key. Once the user completes reading the help() screens,
striking any key will clear the screen and the menu is displayed again.
The main() function will declare an array of 60 elements. The elements are of type double. The
menu options S and L, will ask the user for the actual number of elements for the array. The
program must verify that the actual number of elements does not exceed 60 and is not a negative
number or 0. Your code must check for these.
S or s ( for Smallest ) option will invoke a function named smallest( ) which will prompt the user
for the number of elements for the array to be examined using the function sizeOfArray( ) which
will read and validate the desired array elements and returns it to the calling function. The
program will then use the returned size of the array to fill the array using the number generator
functions shown at the end of this assignment. Your program will ask the user for the range of
values for the elements of the array and uses these numbers in a random generator function to fill
the array. Once the array is filled, the program will call the function findSmallest( ) which will
find and return the smallest number in the array to the calling function. The program will then
call the function frequency(), that will compute and return the frequency of occurrence of the
smallest number. The program will then display the array elements, the smallest number, and its
frequency of occurrence using the function display( ), in the format shown below. The output
shown is for an array of six elements with an array identifier a.
a[0] = xxxx.xx
a[1] = xxxx.xx
a[2] = xxxx.xx
a[3] = xxxx.xx
a[4] = xxxx.xx
a[5] = xxxx.xx
Smallest no. = xxxx.xx Frequency = xx
Note that for the array elements, two digits after the decimal point is required (i.e., 365.24) and
the frequence of occurrence is of type integer.
The function prototypes to be used are as follows:
void smallest(double s[], int size);
where s is the original array declared in the main() function and size is the maximum array size
specified (60 elements in our case).
//gets the desired array size and returns it
int sizeOfArray(void);
//finds the smallest number in the array and returns it via the return statement.
double findSmallest(double s[], int myS);
//where s is the specified array name and myS is the desired size of the array.
Description of nonaxial movement Description of other types of move.pdf
Description of nonaxial movement Description of other types of movement Part B - Uniaxial
Movement Focus your attention on the section called uniaxial movements in parts (b) and (c) in
Focus Figure 8.1 Drag and drop the terms describing a characteristic of uniaxial movement or
the location in which it is found to appropriate blanks in the sentences.
Solution
Answer:
There are four types of synovial joint movements; Uniaxial, Biaxial, Multiaxial and Nonaxial.
Nonaxial movement: It is a gliding movement; no axix around which movement can occur. At a
plane joint (gliding joint), the articulating surfaces of the bones are flat or slightly curved and of
approximately the same size, which allows the bones to slide against each other.
Example: This type of movement found between the carpal bones (intercarpal joints) of the wrist
or tarsal bones (intertarsal joints) of the foot, between the clavicle and acromion of the scapula
(acromioclavicular joint), and between the superior and inferior articular processes of adjacent
vertebrae (zygapophysial joints).
Uniaxial movement: The movement occurs around one axis; only in one plane.
Example: Pivot and Hinge joints
Biaxial movement: Occuring around two axes at right angle to each other.
Example: Saddle and condyloid joint
At a condyloid joint (ellipsoid joint), the shallow depression at the end of one bone articulates
with a rounded structure from an adjacent bone or bones. One example is the radiocarpal joint of
the wrist, between the shallow depression at the distal end of the radius bone and the rounded
scaphoid, lunate, and triquetrum carpal bones.
At a saddle joint, both of the articulating surfaces for the bones have a saddle shape, which is
concave in one direction and convex in the other. The primary example is the first
carpometacarpal joint, between the trapezium (a carpal bone) and the first metacarpal bone at the
base of the thumb.
Multiaxial movement: The movement occurs around several axes.
Example: Ball and Socket joint
The joint with the greatest range of motion is the ball-and-socket joint. At these joints, the
rounded head of one bone (the ball) fits into the concave articulation (the socket) of the adjacent
bone. The hip joint and the glenohumeral (shoulder) joint are the only ball-and-socket joints of
the body..
Centers for Medicare and Medicard Services is a very large purchaser .pdf
Centers for Medicare and Medicard Services is a very large purchaser of health care in the
United States. As part of the American Recovery and Reinvestment Act (ARRA), \'meaningful
use\' is an effort to ensure that health care organizations and providers really do make use of
electronic health record systems in a meaningful way. The following represent key
characteristics of meaningful use except. a. Development of legal health records. b. Use of
electronic prescribing. c. Electronic exchange of information for quality of care. d. Participation
in reporting quality of care measures
Solution
a part of the american recovery and reinvestment act (arra), \'meaningful use\' is an effort to
ensure that he can care organizations and provides do really make use of each health record
systems in a meaningful way. the following represents key characteristics of meaningful use
except
The option is A, Devolepment of legal health records
It does not involve in it,.
An alarm system is to have 4 different devices, each with its sensor..pdf
An alarm system is to have 4 different devices, each with its sensor. One of the sensors is to
have interrupt address 83. Show and explain how that address is created using switches and how
the INTA signal is used to enable a tri-state buffer/driver to place the address on the data lines.
IX. On the 82C55 Parallel Programmable interface, the Control Port has the address ffffh. The
control word 81h will program the lower-C-port as input and the upper-C port as output. (We
don\'t care about the other ports). The address of the C-port is fffdh. Write a program to perform
as follows: Address the Control Port and program the C-port as given above. Read from the
lower-C port. Add 5 to the number that is read. Output the number read to the upper-C port.
Return to step c. (The program will continue until operator intervention).
Solution
In the previous articles we saw about the architecture of 8085 microprocessor. Now we have a
rough idea about how the instructions, data\'s are transferred and processed in 8085
microprocessor. In this article let us discuss in detail about the various signals involved in
transferring data and executing instructions in microprocessor.
The various signals in a microprocessor can be classified as
Power supply and Frequency signals: Signals which aids in supplying power and generating
frequency are associated with this type. Pins like Vcc and ground are classified under this type.
Address signals: Signals associated with the lower order address bus and time multiplexed higher
order address bus comes under this type of signals.
Data Signals: Signals associated with data bus comes under this type.
Control and Status Signals: Signals which are associated with timing and control unit such
HOLD, RW’, WR’ etc. comes under this type of signals.
Interrupt Signals: We know that signals like TRAP, RST 5.5 etc. are interrupt signals. Such
signals come under this category.
Serial I/O signals: These signals are used for giving serial input and output data. Signals like
SID, SOD come under this category.
Acknowledgement Signals: Signals like INTA’, HLDA acts as acknowledgement signal for 8085
microprocessor.
Address Bus:
The pins A8-A15 denote the address bus. They are used for the most significant bit of memory
address.
Address/Data Bus:
AD0-AD7 constitutes the Address/Data bus. They are time multiplexed. These pins are used for
least significant bits of address bus in the first machine clock cycle and used as data bus for
second and third clock cycle.
But what is a clock cycle? What is first clock cycle and second, third so on...
A clock cycle is nothing but the time taken between two adjacent pulses of the oscillator. In
simple words clock cycle refers to the transition between o volts to 5 volts and back to 0 volts.
So the first clock cycle means the first transition of pulse from 0volts to 5 volts and then back to
0 volts.
ALE: Address Latch Enable:
In the previous article we saw how ALE helps in demultiplexing the lower order ad.
A small aquatic crustacean called Ceriodaphnia cornuta has two disti.pdf
A small aquatic crustacean called Ceriodaphnia cornuta has two distinct phenotypes, one with
\"horns\" and one without.The researcher wanted to know whether the horns had any benefit in
discouraging fish from eating the daphnia. So, in a large tank he put many C.cornuta in a 3 to 1
ratio with 3 unhorned to everyone with horns. He then introduced a natural predator of the
daphnia, a small fish. After an hour, he removed the fish, \"sacrificed\" it, and looked at the
daphnia in its belly. Of the 44 daphnia that the fish had eaten, 40 were unhorned and 4 were
horned. Is there any true pattern in which phenotype was eaten? Use chi square.
Solution.
A proof induction has two standard parts. The first establishing tha.pdf
A proof induction has two standard parts. The first establishing that a theorem is true for some
small (usually almost always trivial. Next, an inductive hypothesis is the theorem is assumed to
be true for all cases up to some the theorem is then shown to be true for the next value proves
the theorem (as long as k is finite). As an example, we prove that the Fibonacci numbers, F_4 =
5, ..., F_i = F_i-1 + F_i-2, satisfy F_i
Solution
The induction proof works by verifying whether the example works for all possible values.. Let
us see how it works..
For example :: Consider E(n) 12 + 22 + 32 + ... + n2 = (1/6). n.(n+1).(2n+1)
Now, we need to prove that the above example is true for all possible values..
So, first we consider n = 1.. For n = 1, obviously the sequence is true..
For n=1, LHS = 12
RHS = (1/6).1.(1+1).(2+1) = 6/6 = 1
here LHS = RHS..
Now we have to prove that : if E(k) is true, then E(k+1) is true..
Let us consider E(K) is true..
We have to prove that property is valid for n = k+1
We have to prove: 12 + 22 + 32 + ... + k2 + (k+1)2 = (1/6).(k+1).(k+2)(2k+3)
Left side
= 12 + 22 + 32 + ... + k2 + (k+1)2
= (12 + 22 + 32 + ... + k2) + (k+1)(k+1)
= (1/6). k.(k+1).(2k+1) + (k+1)(k+1)
= (1/6).(k+1). [ k(2k+1) + 6(k+1)]
= (1/6).(k+1).(2 k2 + 7k + 6)
Right side
= (1/6).(k+1).(k+2)(2k+3)
= (1/6).(k+1).(2k2 + 3k + 4k + 6)
Here, LHS = RHS.. I.e., this is true for E(k+1) therefore our assumption is true..
That means the equation is true for all possible values by principle of induction...
1) Compare Figures. What did Da Vinci get right, and what did he dia.pdf
1) Compare Figures. What did Da Vinci get right, and what did he diagram incorrectly? Answer
must be no less than 250 words.
2) How do Prout’s data determine that only HCl is produced in the stomach? Answer must be no
less than 500 words. salivary gland pharanx liver gall bladder duodenum large intestine tongue
salivary gland esophagus stomach pancreas jejunum ileum rectum anus
Solution
1). During the lifetime of Leonardo Da Vinci, he made thousands of pages of notes and
anatomical drawings of human body. He observed human body keenly and drawn them with an
intension to understand the human body’s composition and its function. However, by the time of
his death, his scientific publications were incomplete and unpublished.
But with the observation of Da Vinci’s work, scientists say that they still help us to understand
the human body.
One of the interesting drawings of Da Vinci is, skull. He made a series of drawings of skull. In
one diagram, the front teeth (half position) are cut to show what is behind the teeth. It is difficult
to cut these teeth without damaging the other bones. He also left a drawing of different knives he
used for cutting.
Peter Abraham, Prof. of clinical anatomy at Warwick University in the UK, says that the Da
Vinci is accurate and his drawings are similar to the scientific artists that work now. He
described Da Vinci as a meticulous observer and experimental scientist. Da Vinci have the
ability to draw perfectly what he saw.
The drawings of the upper part of the body, i.e. the liver, stomach, intestine, lungs, heart, the
spinal cord, embryo, etc. But Prof. Abraham says that the uterus is wrong and it indicates the
uterus of a ruminant animal such as cow. He might have used the knowledge of animal
dissection in imagining the human uterus because holding the female corpses is fairly difficult
during those days..
2024.06.01 Introducing a competency framework for languag learning materials ...
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Thesis Statement for students diagnonsed withADHD.ppt
Presentation required for the master in Education.
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The Antoine equation often is used to describe vapor pressures lnP .pdf
The Antoine equation often is used to describe vapor pressures: lnP = A B /(T + C) . For
temperatures near the normal boiling point:
a. Derive an expression that gives the behavior of Hvap as a function of temperature which is
thermodynamically consistent with the Antoine equation.
Solution
The Antoine equation is a mathematical expression of the relation between the vapour pressure
and the temperturer of pure substances. The equation was first proposed by Ch. Antoine, a french
researcher, in 1888. The basic form of the equation is:
lnP = A - B / ( T + C )
and it can be transformed into this temperature-explicit form:
T = B ( A - lnP ) - C
where: P is the absolute vapor pressure of a substance
T is the temperature of the substance
A, B and C are substance-specific coefficients (i.e., constants or parameters
log is typically either log10 or loge
A simpler form of the equation with only two coefficients is sometimes used:
lnP = A - ( B \\ T )
which can be transformed to:
T = B ( A - lnP )
The Antoine equation is a semi-empirical equation which expresses vapour pressure as a
function of temperature. A new, rapid and highly accurate method for obtaining its three
constants from experimental data is presented and applied to ethanol, water and 14 anaesthetic
substances where p is the vapour pressure, T is temperature and A, B and C are component-
specific constants.
Usually, the Antoine equation cannot be used to describe the entire saturated vapour pressure
curve from the triple point to the critical point, because it is not flexible enough. Therefore,
multiple parameter sets for a single component are commonly used. A low-pressure parameter
set is used to describe the vapour pressure curve up to the normal boiling point and the second
set of parameters is used for the range from the normal boiling point to the critical point.
The Antoine Equation is a vapor pressure equation and describes the relation between vapor
pressure and temperature for pure water between 0°C and 373.946°C.
For T in Kelvins and P in kPa:
273.150 < T 373.150: A = 16.5699, B = 3984.92, C = 39.724
373.150 < T 647.096: A = 16.7285, B = 4169.84, C = 28.665
Antoine(T) = exp(A B / (T + C))
For temperatures above the critical temperature (373.946°C / 647.096K), where water is a
superfluid, the increase in vapor pressure appears essentially linear† with the same slope (i.e.
dP/dT 235.88 kPa/°C). This behavior appears consistent with the Ideal Gas Law and statements
that supercritical water behaves like an ideal gas. Now if we define Tc = 647.096 K, we can
write
P(T Tc) = Antoine(T)
P(T > Tc) = 235.88 (T Tc) + Antoine(Tc)
Hence, the behaviour of delta hvap as a function of temperature which is termodynamically
consistent with the the antoine equation is derived ..
The converson of the English to Christianity began a rich period of .pdf
The converson of the English to Christianity began a rich period of artistic production in
England, especially in the creation of a. secular mocaics, b. illuminated manuscripts, c.
ecclesiastical treatises, d. gothic cathedrals
Solution
b. illuminated manuscripts
Elaborately decorated manuscripts were prepared after the conversion. Thus b is the correct
option..
Table 19. Highest Level of Type of organization germ layers .pdf
Table 19. Highest Level of Type of organization germ layers \"asymmetry (sponges) Cell Noa
Cnidaria Pha An tea cnidarians-- jellyfish, hyvha. Radar 4 nerve Platyhelminthes ema toda
(flatworm) Hoda Nematoda worms) bila uran 3 and have germ Germ associated with
gastrulation. Note that deuterostomes product of developmental in which leads threetissue layers
Solution
3. Bilateral symmetry
5.Value of body cavity:
a) Pseudocoelomic fluid is rich in protein and other dissolved substances which is responsible for
turgescent state of nematode body.
b) Coelom makes the digestive system independent of the body wall. It can act as hydrostatic
skeleton. It transports food, waste and other materials..
Question 12 (5 points)The reason that mitosis can result in the pr.pdf
Question 12 (5 points)
The reason that mitosis can result in the production of haploid gametes in alternation of
generations is because
Question 12 options:
The sporophyte and gametophyte are both haploid
Eggs and sperm are produced directly by the diploid sporophyte via meiosis
Eggs and sperm are produced by the diploid gametophyte via mitosis
Eggs and sperm are produced by the haploid gametophyte via mitosis
Save
Question 13 (5 points)
In the fern life cycle, how does fertilization occur?
Question 13 options:
The sporangia on the undersides of the leaves produce gametes, which are flung our of the leaf
and find each other on the ground to fuse
The sporophyte produces sperm and eggs in gametangia. The sperm reach the eggs by swimming
through water
The gametophyte produces sperm and eggs in gametangia (antheridia and archegonia). The
sperm reach the eggs by swimming through water
There is no sexual reproduction in ferns
Save
Question 14 (5 points)
The earliest land plants were most likely to have lived in which of the following habitats?
Question 14 options:
arid desert environments
Salt water marshes
Tropical rainforests
fresh water wetlands
Save
The sporophyte and gametophyte are both haploid
Eggs and sperm are produced directly by the diploid sporophyte via meiosis
Eggs and sperm are produced by the diploid gametophyte via mitosis
Eggs and sperm are produced by the haploid gametophyte via mitosis
Solution
Q12 Eggs and sperms are produced by haploid gametophyte by mitosis.
The large, leafy fern is the diploid organism. On the undersurface of its fronds or leaves, its cells
undergo meiosis to create haploid cells. However, these cells do not immediately unite with
others to recreate the diploid state. Instead, they are shed as spores and germinate into small
haploid organisms. Because the diploid organism creates spores, it is called the sporophyte
generation of the life cycle. Upon reaching maturity, the haploid organism creates haploid egg
and sperm cells (gametes) by mitosis . Because the haploid organism creates gametes , it is called
the gametophyte generation of the life cycle. The male gametes (sperm) are then released and
swim to the female egg. Fusion of the gametes creates the new diploid sporophyte, completing
the life cycle.
Q13 The gametophyte produces sperm and eggs in gametangia (antheridia and archegonia). The
sperm reach the eggs by swimming through water
The male organs or antheridia produce and shed numerous motile sperm or antherozoids. These
antherozoids are armed with hair-like or whip-like cilia or flagellae and are able to swim through
water; they do not travel great distances and are only released when free water is available. Only
a continuous film of water is needed. They antherozoids are chemically attracted to the
developing flask-like female organs or archegonia where fertilization of the single egg cell
embedded in each archegonium takes place.
The sporophyte has a double set of chromosomes (i.e. diploid) and i.
peripheral circuit functions built-in to the HC12 microcontroller, t.pdf
peripheral circuit functions built-in to the HC12 microcontroller, that are not part of the CPU
itself or the memories. Access to these is via external pins on the Microcontroller IC package.
What are some of these built-in circuit functions called, and what do they do?
Solution
The peripheral circuit communicate with the Microcontroller using built in cirucit called BUS.
The bus is a set of shared communication lines that serves as the central nervous system of the
computer. Data, address, and control signals are shared by all system components via the bus.
Each component connected to the bus communicates information to and from the bus via its own
bus controller. The data lines,address lines, and control lines allow a specific component to
access data addressed to that component. The data lines are used to communicate words to and
from data.
registers in the various system components such as memory, CPU, and input/output
(I/O) peripherals. The address lines are used to select devices on the bus or specific
data locations within memory. Devices usually have a combinational logic address
decoder circuit that identifies the address code and activates the device..
Mendez Corporation has 10,000 shares of its $100 par value, 7 percen.pdf
Mendez Corporation has 10,000 shares of its $100 par value, 7 percent cumulative
preferred stock outstanding and 50,000 shares of its $1 par value common stock outstanding.
In Mendez’s first four years of operation, its board of directors paid the following
cash dividends: 2011, none; 2012, $120,000; 2013, $140,000; 2014, $140,000.
Determine the dividends per share and total cash dividends paid to the preferred and
common stockholders during each of the four years.
Solution
The dividends per year on preferred stock = 10,000 x $ 100 x 7 %
= $ 10,000 x $ 100 x 0.07 = $ 70,000
Dividend distribution:
Year
Total Dividend Amount
Preferred
Common
2011
-
-
-
2012
$120,000
$120,000
$0
2013
$140,000
$90,000
$50,000
2014
$140,000
$70,000
$70,000
Explanation:
First dividend is paid to Preferred stock and then excess amount goes to common share holder.
For cumulative preferred stock, any deficiency in dividend, not paid carries over to subsequent
years and is added to that year\'s dividend.
Preferred Dividends unpaid in 2011 = $70,000
Preferred Dividends on 2012 = $70,000 + $ 70,000 = $ 140,000
Preferred Dividends paid in 2012 = $ 120,000
Preferred Dividends shortage on 2012 = $ 20,000
As the Dividend was insufficient for preferred stocks, common stocks holder was not paid any
dividend during 2011 & 2012.
Preferred Dividends paid in 2013 = $ 70,000 + $ 20,000 = $ 90,000
Common Dividends paid in 2013 = $ 140,000 - $ 90,000 = $ 50,000
Preferred Dividends paid in 2014 = $ 70,000
Common Dividends paid in 2014 = $ 140,000 - $ 70,000 = $ 70,000
Computation of Dividends per share:
2011
2012
2013
2014
Preferred Stock (Dividends per share)
$0
$12.00
$9.00
$7.00
Common Stock (Dividends per share)
$0
$0
$1.00
$1.40
Explanation:
Dividends per share = Dividend paid/No. of shares
Year 2012:
Dividends per share for Preferred Stock = $120,000/10,000 = $ 12
Dividend per share for Common Stock = $ 0
Year 2013:
Dividends per share for Preferred Stock = $90,000/10,000 = $ 9
Dividends per share for Common Stock = $ 50,000/50,000 = $ 1
Year 2014:
Dividends per share for Preferred Stock = $70,000 /10,000 = $ 7
Dividends per share for Common Stock = $ 70,000/50,000 = $ 1.40
Year
Total Dividend Amount
Preferred
Common
2011
-
-
-
2012
$120,000
$120,000
$0
2013
$140,000
$90,000
$50,000
2014
$140,000
$70,000
$70,000.
If the dpy-11 and unc-31 genes are linked on the same chromosome, how.pdf
If the dpy-11 and unc-31 genes are linked on the same chromosome, how many phenotypes
would the F_2 progeny show? If the two genes are on different chromosomes and therefore
independently assorting, how many phenotypes would you see in the F_2 worms? List the
phenotype(s). Strawberry plants have varieties that flower when the days are still short, and
other varieties that flower only when the days get longer. The long-day allele is dominant to the
short-day allele. You\'d like to test whether or not this gene is inherited in a Mendelian fashion.
When you cross between heterozygous plants and count the offspring in the F_1 generation, this
is what you find: 85 plants with no flowers on short days and 35 plants with flowers on short
days What would you expect the ratio of phenotypes be? What is your Chi squared value?
Show your work What does this mean?
Solution
2. Linked genes tend to inherit together for the next-generations. No crossing over ororoccurs
independent assortment occurs. No new characters appear. Where as gegenes are on different
chromosomes, they are non linked. Crossing over or independent assortment takes place. New
characters appear along with parental characters. So dpy-11 & unc-31 present on same
chromosome, they are linked and give only parental characters. Only 2 phenotypes appear. If the
said genes present on different chromosomes, they give 4 different phenotypes
3. a. Heterozygous long day representing as Ll. When crossing or hybridization occur between
these genes
Ll x Ll ----------> 3:1 is the phenotypes ratio. Three are long and one is short.
Chi square value can be calculated by using the formula.
(O-E)^2/E
O = observed phenotypes; E = Expected phenotypes.
How does a wave formation influence the resultant waveSolution.pdf
How does a wave formation influence the resultant wave?
Solution
Waves are usually formed by distant wind.Generally larger waves are more powerful than small
waves . Waves can pile up at shore lines and create a rip current.The resultant wave heights Hi
and the wave period Ti are related to the factors that influence the energy transfer between wind
and the ocean..
How are health services paid for Provide a definition for the term .pdf
How are health services paid for? Provide a definition for the term \"third-party payer\", discuss
the different payers currently operating in the market, and assess the importance of each. Please
use a table format to list the information.
Solution
There are two main payers types: single payer and third-party payer. Single payer is when the
person that obtains the health service is the same which pay for it, here relationship is direct taker
and provider of health services.
However, it is more common the presence of a third party payer, which is an external agent that
covers the other health expenses or part of them, they are insurance companies or public social
protection systems.
Single Payer
Third-Party Payer
Type of relation with provider
Direct
Indirect
Capability to compare Prices/Features
Relatively High
Almost Null
Payment proportion
Full
Low percentage
Single Payer
Third-Party Payer
Type of relation with provider
Direct
Indirect
Capability to compare Prices/Features
Relatively High
Almost Null
Payment proportion
Full
Low percentage.
Given the information regarding mitosis and meiosis, clearly and con.pdf
Given the information regarding mitosis and meiosis, clearly and convincingly explain in your
own words how an error in MITOSIS could possibly lead to a gamete containing new genetic
information and thereby giving rise to a new phenotype. Use detail be comprehensive. (7 points)
Solution
The process of mitosis is divided into two phases: Interphase and M-phase.
Interphase is further divided into G1, S and G2 phase. In G1, protein synthesis occurs for S-
phase, in S phase, DNA replicates and in G2 phase, protein synthesis occurs for M-phase to start
properly. Now, during S phase, the ‘2n’ nucleus converts into 4n.
So, the cell entering M-phase is already ‘4n’.
M-phase is the actual mitotic phase, in which the cell divides into two. It contains four phases,
prophase, metaphase, anaphase and telophase. Your question is related with these four phases. In
prophase, the entire DNA converts into chromosomes; the process is known as condensation of
DNA. In metaphase, these chromosomes align on the equator. Now, if alignment at equator is
not correct, then anaphase cannot be started due to metaphase check point.
In anaphase, the aligned chromosomes move towards the opposite poles. This phase is the main
phase, related with formation of gametes. The cell is already ‘4n’. ‘2n’ chromosomes move
towards one end, and the rest of ‘2n’ move towards the other end. Now, suppose an error occurs
in mitosis, and ‘3n’ chromosomes move towards one end, and 1n goes towards the other cell;
then two gametes, ‘3n’ and n will be formed.
This ‘3n’ and ‘n’ gametes are gametes having different genetic information in comparison to
their ‘2n’ counterparts.
Interestingly, Meiosis II will result in gamete formation is mitotic division (equational division).
Haploid cells with \'2c\' amount of DNA leave Meiosis I. Now, in anaphase, suppose one
chromosome\' 2c DNA migrates to one end (due to error) and no DNA migrates to the other end;
then two gametes will be formed; one having all normal chromosomes + one double
chromosome with 2c DNA, and the other having no DNA.
Please note: normally gametes are 1n (haploid) and 1c.
Suppose this gamete with double DNA (2c) fertilizes with a gamete with half DNA (1c), then the
resulting cell will have “all normal chromosomes + one chromosome with ‘3c’ amount of DNA”.
This condition if occurring with chromosome 21 is known as Down\' syndrome.
It is important to note that errors in Mitosis result in non-disjunction; which further results in
many Mendelian disorders; as well as ploidy defects.
** Hope it helps..
First, look to determine what is the earliest era that fossils have .pdf
First, look to determine what is the earliest era that fossils have been found here in Georgia.
Next, see if you can find what the oldest fossil that has been found in Georgia. Give the name of
the fossil, when it was found, and where the fossil was found. Give a brief description of the
organism that was fossilized.
Solution
Trilobites are the oldest fossils found in georgia.They have been found in Georgia’s Appalachian
Plateau which is present the Coosa River in Floyd County.
These fossils are considered to be arthropods,Moreover they had to shed their shells to grow.As a
result the fossils found in these places usually shed shells as bits of individuals or as complete
individuals or as imprints of individuals..
Figure out and write down a definition of a translation that does no.pdf
Figure out and write down a definition of a translation that does not use (x,y) coordinates
Solution
There are many kinds of oprations in mathematics like translation, rotation, reflection, resized
etc. Translation is one of them.
Translation is an operation which is uesd in geometry to move an object in a fixed direction and
for same distance..
explain why right and why wrong Indentify configuration (below) whic.pdf
explain why right and why wrong Indentify configuration (below) which I s not a correct
representation of H and I.
Solution
option C is not correct one.
Explanation:
If we stretch all right hand fingers in the direction of current direction given then the thumb gives
us into the board (into symbol) or out of the board (dot symbol).
In the option C the field should be out of the board, so it is not correct one..
Develop a system flowchart and then write a menu-driven C++ program .pdf
Develop a system flowchart and then write a menu-driven C++ program that uses user-defined
functions arrays, and a random number generator.
Upon program execution, the screen will be cleared and the menu shown below will appear at
the top of the screen and centered. The menu items are explained below.
Help Smallest Largest Quit
H or h ( for Help ) option will invoke a function named help() which will display a help screen.
The help screen(s) should guide the user how to interact with the program, type of data to be
used, and what results would the program produce. Each help screen should remain on the
monitor until the user strikes any key. Once the user completes reading the help() screens,
striking any key will clear the screen and the menu is displayed again.
The main() function will declare an array of 60 elements. The elements are of type double. The
menu options S and L, will ask the user for the actual number of elements for the array. The
program must verify that the actual number of elements does not exceed 60 and is not a negative
number or 0. Your code must check for these.
S or s ( for Smallest ) option will invoke a function named smallest( ) which will prompt the user
for the number of elements for the array to be examined using the function sizeOfArray( ) which
will read and validate the desired array elements and returns it to the calling function. The
program will then use the returned size of the array to fill the array using the number generator
functions shown at the end of this assignment. Your program will ask the user for the range of
values for the elements of the array and uses these numbers in a random generator function to fill
the array. Once the array is filled, the program will call the function findSmallest( ) which will
find and return the smallest number in the array to the calling function. The program will then
call the function frequency(), that will compute and return the frequency of occurrence of the
smallest number. The program will then display the array elements, the smallest number, and its
frequency of occurrence using the function display( ), in the format shown below. The output
shown is for an array of six elements with an array identifier a.
a[0] = xxxx.xx
a[1] = xxxx.xx
a[2] = xxxx.xx
a[3] = xxxx.xx
a[4] = xxxx.xx
a[5] = xxxx.xx
Smallest no. = xxxx.xx Frequency = xx
Note that for the array elements, two digits after the decimal point is required (i.e., 365.24) and
the frequence of occurrence is of type integer.
The function prototypes to be used are as follows:
void smallest(double s[], int size);
where s is the original array declared in the main() function and size is the maximum array size
specified (60 elements in our case).
//gets the desired array size and returns it
int sizeOfArray(void);
//finds the smallest number in the array and returns it via the return statement.
double findSmallest(double s[], int myS);
//where s is the specified array name and myS is the desired size of the array.
Description of nonaxial movement Description of other types of move.pdf
Description of nonaxial movement Description of other types of movement Part B - Uniaxial
Movement Focus your attention on the section called uniaxial movements in parts (b) and (c) in
Focus Figure 8.1 Drag and drop the terms describing a characteristic of uniaxial movement or
the location in which it is found to appropriate blanks in the sentences.
Solution
Answer:
There are four types of synovial joint movements; Uniaxial, Biaxial, Multiaxial and Nonaxial.
Nonaxial movement: It is a gliding movement; no axix around which movement can occur. At a
plane joint (gliding joint), the articulating surfaces of the bones are flat or slightly curved and of
approximately the same size, which allows the bones to slide against each other.
Example: This type of movement found between the carpal bones (intercarpal joints) of the wrist
or tarsal bones (intertarsal joints) of the foot, between the clavicle and acromion of the scapula
(acromioclavicular joint), and between the superior and inferior articular processes of adjacent
vertebrae (zygapophysial joints).
Uniaxial movement: The movement occurs around one axis; only in one plane.
Example: Pivot and Hinge joints
Biaxial movement: Occuring around two axes at right angle to each other.
Example: Saddle and condyloid joint
At a condyloid joint (ellipsoid joint), the shallow depression at the end of one bone articulates
with a rounded structure from an adjacent bone or bones. One example is the radiocarpal joint of
the wrist, between the shallow depression at the distal end of the radius bone and the rounded
scaphoid, lunate, and triquetrum carpal bones.
At a saddle joint, both of the articulating surfaces for the bones have a saddle shape, which is
concave in one direction and convex in the other. The primary example is the first
carpometacarpal joint, between the trapezium (a carpal bone) and the first metacarpal bone at the
base of the thumb.
Multiaxial movement: The movement occurs around several axes.
Example: Ball and Socket joint
The joint with the greatest range of motion is the ball-and-socket joint. At these joints, the
rounded head of one bone (the ball) fits into the concave articulation (the socket) of the adjacent
bone. The hip joint and the glenohumeral (shoulder) joint are the only ball-and-socket joints of
the body..
Centers for Medicare and Medicard Services is a very large purchaser .pdf
Centers for Medicare and Medicard Services is a very large purchaser of health care in the
United States. As part of the American Recovery and Reinvestment Act (ARRA), \'meaningful
use\' is an effort to ensure that health care organizations and providers really do make use of
electronic health record systems in a meaningful way. The following represent key
characteristics of meaningful use except. a. Development of legal health records. b. Use of
electronic prescribing. c. Electronic exchange of information for quality of care. d. Participation
in reporting quality of care measures
Solution
a part of the american recovery and reinvestment act (arra), \'meaningful use\' is an effort to
ensure that he can care organizations and provides do really make use of each health record
systems in a meaningful way. the following represents key characteristics of meaningful use
except
The option is A, Devolepment of legal health records
It does not involve in it,.
An alarm system is to have 4 different devices, each with its sensor..pdf
An alarm system is to have 4 different devices, each with its sensor. One of the sensors is to
have interrupt address 83. Show and explain how that address is created using switches and how
the INTA signal is used to enable a tri-state buffer/driver to place the address on the data lines.
IX. On the 82C55 Parallel Programmable interface, the Control Port has the address ffffh. The
control word 81h will program the lower-C-port as input and the upper-C port as output. (We
don\'t care about the other ports). The address of the C-port is fffdh. Write a program to perform
as follows: Address the Control Port and program the C-port as given above. Read from the
lower-C port. Add 5 to the number that is read. Output the number read to the upper-C port.
Return to step c. (The program will continue until operator intervention).
Solution
In the previous articles we saw about the architecture of 8085 microprocessor. Now we have a
rough idea about how the instructions, data\'s are transferred and processed in 8085
microprocessor. In this article let us discuss in detail about the various signals involved in
transferring data and executing instructions in microprocessor.
The various signals in a microprocessor can be classified as
Power supply and Frequency signals: Signals which aids in supplying power and generating
frequency are associated with this type. Pins like Vcc and ground are classified under this type.
Address signals: Signals associated with the lower order address bus and time multiplexed higher
order address bus comes under this type of signals.
Data Signals: Signals associated with data bus comes under this type.
Control and Status Signals: Signals which are associated with timing and control unit such
HOLD, RW’, WR’ etc. comes under this type of signals.
Interrupt Signals: We know that signals like TRAP, RST 5.5 etc. are interrupt signals. Such
signals come under this category.
Serial I/O signals: These signals are used for giving serial input and output data. Signals like
SID, SOD come under this category.
Acknowledgement Signals: Signals like INTA’, HLDA acts as acknowledgement signal for 8085
microprocessor.
Address Bus:
The pins A8-A15 denote the address bus. They are used for the most significant bit of memory
address.
Address/Data Bus:
AD0-AD7 constitutes the Address/Data bus. They are time multiplexed. These pins are used for
least significant bits of address bus in the first machine clock cycle and used as data bus for
second and third clock cycle.
But what is a clock cycle? What is first clock cycle and second, third so on...
A clock cycle is nothing but the time taken between two adjacent pulses of the oscillator. In
simple words clock cycle refers to the transition between o volts to 5 volts and back to 0 volts.
So the first clock cycle means the first transition of pulse from 0volts to 5 volts and then back to
0 volts.
ALE: Address Latch Enable:
In the previous article we saw how ALE helps in demultiplexing the lower order ad.
A small aquatic crustacean called Ceriodaphnia cornuta has two disti.pdf
A small aquatic crustacean called Ceriodaphnia cornuta has two distinct phenotypes, one with
\"horns\" and one without.The researcher wanted to know whether the horns had any benefit in
discouraging fish from eating the daphnia. So, in a large tank he put many C.cornuta in a 3 to 1
ratio with 3 unhorned to everyone with horns. He then introduced a natural predator of the
daphnia, a small fish. After an hour, he removed the fish, \"sacrificed\" it, and looked at the
daphnia in its belly. Of the 44 daphnia that the fish had eaten, 40 were unhorned and 4 were
horned. Is there any true pattern in which phenotype was eaten? Use chi square.
Solution.
A proof induction has two standard parts. The first establishing tha.pdf
A proof induction has two standard parts. The first establishing that a theorem is true for some
small (usually almost always trivial. Next, an inductive hypothesis is the theorem is assumed to
be true for all cases up to some the theorem is then shown to be true for the next value proves
the theorem (as long as k is finite). As an example, we prove that the Fibonacci numbers, F_4 =
5, ..., F_i = F_i-1 + F_i-2, satisfy F_i
Solution
The induction proof works by verifying whether the example works for all possible values.. Let
us see how it works..
For example :: Consider E(n) 12 + 22 + 32 + ... + n2 = (1/6). n.(n+1).(2n+1)
Now, we need to prove that the above example is true for all possible values..
So, first we consider n = 1.. For n = 1, obviously the sequence is true..
For n=1, LHS = 12
RHS = (1/6).1.(1+1).(2+1) = 6/6 = 1
here LHS = RHS..
Now we have to prove that : if E(k) is true, then E(k+1) is true..
Let us consider E(K) is true..
We have to prove that property is valid for n = k+1
We have to prove: 12 + 22 + 32 + ... + k2 + (k+1)2 = (1/6).(k+1).(k+2)(2k+3)
Left side
= 12 + 22 + 32 + ... + k2 + (k+1)2
= (12 + 22 + 32 + ... + k2) + (k+1)(k+1)
= (1/6). k.(k+1).(2k+1) + (k+1)(k+1)
= (1/6).(k+1). [ k(2k+1) + 6(k+1)]
= (1/6).(k+1).(2 k2 + 7k + 6)
Right side
= (1/6).(k+1).(k+2)(2k+3)
= (1/6).(k+1).(2k2 + 3k + 4k + 6)
Here, LHS = RHS.. I.e., this is true for E(k+1) therefore our assumption is true..
That means the equation is true for all possible values by principle of induction...
1) Compare Figures. What did Da Vinci get right, and what did he dia.pdf
1) Compare Figures. What did Da Vinci get right, and what did he diagram incorrectly? Answer
must be no less than 250 words.
2) How do Prout’s data determine that only HCl is produced in the stomach? Answer must be no
less than 500 words. salivary gland pharanx liver gall bladder duodenum large intestine tongue
salivary gland esophagus stomach pancreas jejunum ileum rectum anus
Solution
1). During the lifetime of Leonardo Da Vinci, he made thousands of pages of notes and
anatomical drawings of human body. He observed human body keenly and drawn them with an
intension to understand the human body’s composition and its function. However, by the time of
his death, his scientific publications were incomplete and unpublished.
But with the observation of Da Vinci’s work, scientists say that they still help us to understand
the human body.
One of the interesting drawings of Da Vinci is, skull. He made a series of drawings of skull. In
one diagram, the front teeth (half position) are cut to show what is behind the teeth. It is difficult
to cut these teeth without damaging the other bones. He also left a drawing of different knives he
used for cutting.
Peter Abraham, Prof. of clinical anatomy at Warwick University in the UK, says that the Da
Vinci is accurate and his drawings are similar to the scientific artists that work now. He
described Da Vinci as a meticulous observer and experimental scientist. Da Vinci have the
ability to draw perfectly what he saw.
The drawings of the upper part of the body, i.e. the liver, stomach, intestine, lungs, heart, the
spinal cord, embryo, etc. But Prof. Abraham says that the uterus is wrong and it indicates the
uterus of a ruminant animal such as cow. He might have used the knowledge of animal
dissection in imagining the human uterus because holding the female corpses is fairly difficult
during those days..
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The French Revolution Class 9 Study Material pdf free download
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Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
● Why and by when do customers need to upgrade to Java 17?
● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
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