Write a C++ program to get input of a data set of 14 double valu.pdf
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Write a C++ program to get input of a data set of 14 double values (grades) from the standard input device (the keyboard), then calculate and report the sum and the mean of the data set, then the individual DEVIATION of each individual input with respect to the mean, the variance of the dataset, and finally the standard deviation of the data set. This problem requires you to integerate multiple things together: arrays, loops, functions. More specifically, you want to use loops to deal with input, save them to array, or calculate sum and individual deviation, and display results etc. in separate functions. A sample run. Please let me know how many grades you would like to input? 14 Input them one by one: 89 95 72 83 99 54 86 75 92 73 79 75 82 73. Enter a grade: 89 Enter a grade: 95 Enter a grade: 72 Enter a grade: 83 Enter a grade: 99 Enter a grade: 54 Enter a grade: 86 Enter a grade: 75 Enter a grade: 92 Enter a grade: 73 Enter a grade: 79 Enter a grade: 75 Enter a grade: 82 Enter a grade: 73 The sum is: a number here. The mean is: 80.5 GRADE INDIVIDUAL DEVIATION 89 8.5 95 14.5 72 -8.5 83 2.5 99 18.5 54 -26.5 86 5.5 75 -5.5 92 11.5 73 -7.5 79 -1.5 75 -5.5 82 1.5 73 -7.5 The variance is: 124.679 standard deviation: 11.166 Write a C++ program to get input of a data set of 14 double values (grades) from the standard input device (the keyboard), then calculate and report the sum and the mean of the data set, then the individual DEVIATION of each individual input with respect to the mean, the variance of the dataset, and finally the standard deviation of the data set. This problem requires you to integerate multiple things together: arrays, loops, functions. More specifically, you want to use loops to deal with input, save them to array, or calculate sum and individual deviation, and display results etc. in separate functions. A sample run. Please let me know how many grades you would like to input? 14 Input them one by one: 89 95 72 83 99 54 86 75 92 73 79 75 82 73. Enter a grade: 89 Enter a grade: 95 Enter a grade: 72 Enter a grade: 83 Enter a grade: 99 Enter a grade: 54 Enter a grade: 86 Enter a grade: 75 Enter a grade: 92 Enter a grade: 73 Enter a grade: 79 Enter a grade: 75 Enter a grade: 82 Enter a grade: 73 The sum is: a number here. The mean is: 80.5 GRADE INDIVIDUAL DEVIATION 89 8.5 95 14.5 72 -8.5 83 2.5 99 18.5 54 -26.5 86 5.5 75 -5.5 92 11.5 73 -7.5 79 -1.5 75 -5.5 82 1.5 73 -7.5 The variance is: 124.679 standard deviation: 11.166 Write a C++ program to get input of a data set of 14 double values (grades) from the standard input device (the keyboard), then calculate and report the sum and the mean of the data set, then the individual DEVIATION of each individual input with respect to the mean, the variance of the dataset, and finally the standard deviation of the data set. This problem requires you to integerate multiple things together: arrays, loops, functions. More specifically, you want to use loops to deal with input, save them to.
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Main
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//Increment counter
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2
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Program Design:
Main
// This program will calculate the average of 10 integer numbers
// Declare variables
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// Loop through 10 numbers
// Prompt for positive integer
// Get input
// test input value for gt 0 if (value > 0)
//Increment counter
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// Prompt for positive integer
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//Calculate average
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End
Test Plan:
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Test Case
Input
Expected Output
1
1 1 1 0 1
2 0 1 3 2
Average = 1.2
2
100 100 100 100 -100
Input a positive value
100 200 -200 200 200
Input a positive value
200 200
average is 120.0
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Pseudocode: Main
// This program will calculate the average of 10 positive integers.
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Input value
If (value >=0)
sum = sum + value count=count+1
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// Calculate average avg = sum/count
// Print results
End //End of Main
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The following is the C Code that will compile in execute in the online.
System Verilog 2009 & 2012 enhancements
This document summarizes enhancements made to System Verilog in 2009 and 2012. The 2009 enhancements included final blocks, bit selects of expressions, edge detection for DDR logic, fork-join improvements, and display enhancements. The 2012 enhancements extended enums, added scale factors to real constants and mixed-signal assertions, introduced aspect-oriented programming features, and removed X-optimism using new keywords. It also proposed signed operators and discussed some high-level problems not yet addressed.
control statements of clangauge (ii unit)
This document outlines control structures in programming, including selection structures like if/else and repetition structures like while and for loops. It provides examples of algorithms using pseudocode and C++ code that employ counter-controlled and sentinel-controlled repetition. Key concepts covered include flowcharts, logical operators, and avoiding logic errors. Nested control structures and the switch statement are also discussed.
03b loops
The document discusses loops and repetition structures in C++. It explains while loops, their syntax, and how they work. A while loop repeats actions while a condition is true. Each iteration, the condition is evaluated. If true, the loop body executes and control returns to the condition. If false, execution continues after the loop. Pseudocode and C++ code examples are provided to illustrate how a while loop can be used to repeatedly purchase items from a shopping list until it is empty. Flowcharts further demonstrate the flow of control in a while loop.
Copy of dti2143/dam31303 chap 1 problem solving and program design
The document discusses program design methods including pseudocode, flowcharts, and their conventions. It provides examples of pseudocode for different problems and their corresponding flowcharts. Key points covered are the basic structure of pseudocode using keywords, and the common symbols used in flowcharts to represent different elements like processes, decisions, and flow of steps.
Csphtp1 04
The document outlines control structures in programming including algorithms, pseudocode, selection structures like if/else, repetition structures like while loops, and case studies applying these concepts. Key topics are counter-controlled and sentinel-controlled repetition for loops, and nesting control structures with if statements inside loops. Examples in C# demonstrate accumulating values in loops and nested if/else statements.
Testing techniques
The document discusses various software testing techniques including black box testing, white box testing, and grey box testing. It provides details on specific techniques such as equivalence partitioning, boundary value analysis, statement coverage, condition coverage, function coverage, and cyclomatic complexity. The objective is to understand these techniques so they can be used effectively to test applications and find defects.
LectureNotes-05-DSA
Arrays allow storing and accessing multiple values under a single variable name. They are indexed collections that can hold values of built-in or user-defined types. Multidimensional arrays can model matrices by using multiple indices to access elements. Parameter arrays allow methods to accept a variable number of arguments by specifying the last parameter as a type followed by ellipses.
Lec 8 03_sept [compatibility mode]
This document discusses Java data types, variables, and methods. It covers primitive data types like int, float, and boolean. It describes rules for identifiers and conventions for naming variables and classes. The document defines parameter types, implicit and explicit parameters, and how to call methods. It explains constructing objects with the new operator and accessor and mutator methods. The summary concludes with a discussion of static methods and how to call them without an object.
Similar to Write a C++ program to get input of a data set of 14 double valu.pdf (16)
CMIS 102 HANDS-ON LAB WEEK 6 OVERVIEW THIS HANDS-ON LAB ALLOWS YOU TO FOLLOW ...
CMIS 102 HANDS-ON LAB WEEK 6 OVERVIEW THIS HANDS-ON LAB ALLOWS YOU TO FOLLOW ...
OverviewThis hands-on lab allows you to follow and experiment w.docx
OverviewThis hands-on lab allows you to follow and experiment w.docx
Copy of dti2143/dam31303 chap 1 problem solving and program design
Copy of dti2143/dam31303 chap 1 problem solving and program design
More from arshiartpalace
Which plants contain lignified vascular tissue (Select all that appl.pdf
Sporophytes of seedless vascular plants contain lignified vascular tissue. While some bryophytes have internal pipelines to transport fluid, only vascular plants have lignin-strengthened vascular tissue made of xylem and phloem. The sporophyte is the dominant phase in the life cycle of seedless vascular plants that contains this lignified vascular tissue.
Which of the neurotransmitters are metabotropic and which are ionotr.pdf
Which of the neurotransmitters are metabotropic and which are ionotropic?
Solution
METABOTROPIC NEUROTRANSMITTERS:
Cholinergic Muscarinic Receptors
Metabotropic Glutamate Receptors
Dopamine Receptors
Adrenergic Receptors
GABABETA Receptors
Serotonin Receptors
Histamine Receptors
Adenosine Receptors
Opioid Receptors
IONOTROPIC NEUROTRANSMITTER:
glutamatergic (EPSPs) and GABAergic (IPSP).
What is the range of y= absolute value f(x) if f(x) = -x^2 And can .pdf
What is the range of y= absolute value f(x) if f(x) = -x^2? And can you explain how you got the
answer? Thanks
Solution
f(x)=-x^2
before commenting about the range of this firstly check its domain
now we can see that there is not a single point where the function is not defined
hence the domain is all real
now as x2 is always positive whether x is +ve or -ve
hence
-x2 is always negaive which is our function
hence
f(x)=-x2 is alwayys negative
hence its range is (-infinnity, 0].
G. Will a pumping test that is run for seven days give more informati.pdf
G. Will a pumping test that is run for seven days give more information than a pumping test run
for one day? If so, what type of information? Why?
Solution
Answer:-
Yes , Pumping test that is run for seven days give more information than pumping test run for
one day.
Pumping test give information about:-
(i) How much ground water can be extracted from a well based on long term yield and well
efficiency.
(ii) The hydraulic properties of an aquifer.
(iii) Suitable depth of pump
(iv) Information on water quality and its variability with time.
Pumping test is done for obtaining knowledge about ground water. By the help of pumping test
we can know quality of water..
What challenges, strengths, or weaknesses do you see Baxter Internat.pdf
What challenges, strengths, or weaknesses do you see Baxter International Inc. would have in
2016 with a liquidity ratio of 2.4 and a profitability ratio of 0.31. In 2017 the company liquidity
ratio was 2.6 and their profitability ratio was 0.41. Please be articulate in your rationale for both
years and no hand written answers.
What challenges, strengths, or weaknesses do you see Exxon Mobil Corp. would have in 2016
with a liquidity ratio of 0.9 and a profitability ratio of 0.02. In 2017 the company liquidity ratio
was 0.8 and their profitability ratio was 0.05. Please be articulate in your rationale for both years
and no hand written answers.
Solution
Answer) The liquidity ratio study the firm\'s short term solvency and its ability to pay off the
current liabilities. This ratio provides the information about the firms liquidity and the primary
concern is the firm’s ability to pay its current liabilities.
Liquidity Ratio (2016) - The Baxter International Inc.\'s liquidity ratio is 2.4 means more than 2
which denotes good liquidity position as the firm has sufficient cash to meet current obligations
and the greater safety of funds of the short-term creditors. It is the strength for the firm as the
firm is in strong financial position which represents margin of safety for current liabilities.
Profitability Ratio (2016): Profitability means the ability of a company to earn a profit. Firm’s
profitability is very important both for stockholders and creditors because revenue in the form of
dividends is being derived from profits, as well as profits are one source of funds for covering
debts. Profitability ratio analysis is a good way to measure Company’s performance. The Baxter
International Inc.\'s profitability ratio is 0.31 0r 30% it is a good indicator for the firm. Generally,
20% to 30% gross profit margin is considered to be adequate for a business concern to meet its
fixed obligations, dividend and all other appropriations.
Liquidity Ratio (2017) - The Baxter International Inc.\'s liquidity ratio is 2.6 means more than 2
which denotes good liquidity position as the firm has sufficient cash to meet current obligations
and the greater safety of funds of the short-term creditors..It is the strength for the firm as the
firm is in strong financial position which represents margin of safety for current liabilities.
Profitability Ratio (2017) : The Baxter International Inc.\'s profitability ratio is 0.41 0r 40% it is
a good indicator for the firm. It has increased as compared to last which indicates the company is
earning good profit.
Liquidity Ratio (2016) - The Exxon Mobil Corp. \'s liquidity ratio is 0.9 means less than 1 which
denotes poor liquidity position as the firm doesn\'t have sufficient cash to meet current
obligations. It shows the weakness of company as the company isn\'t doing very well--they
can\'t pay back all the money they owe with the cash they have on hand and will have to start
selling long-term assets, or look at refinancing .
Three historical figures were identified as working with vita statist.pdf
Three historical figures were identified as working with vita statistics. Who were these
individuals? William Farr, Lemuel Shattuck, George Soper John Graunt, William Farr, Lemuel
Shattuck Benjamin Jesty, Robert Koch, John Snow Louis Pasteur, John Graunt, George Soper
Solution
option B.
John Graunt, William Farr, Lemuel Shattuck.
John Graunt developed early human statistical and census methods.
William Farr is considered as the founder of medical statistics.
Lemuel Shattuck was one of the members in American statistical Association..
What are postretirement benefits other than pensionsSolutionB.pdf
What are postretirement benefits other than pensions?
Solution
Benefits, other than pension distributions, paid to employees during their retirement years. Most
post-retirement benefits include life insurance and medical plans. Although these benefits are
mostly employer-paid, retired employees often share in the cost of these benefits through co-
payments, payment of deductibles and making employee contributions to the plan when required..
There are parts of a true sustained competitive advantage Ther.pdf
There are parts of a true sustained competitive advantage?
There are parts of a true sustained competitive advantage?
Solution
There are four component of a true sustained competitive advantage:
1) Valuable
2) Rare
3))Imperfectly imitable resources
4)Non-substitute
_____________________________________________________________________________
_____.
Frequency .java Word frequency counter package frequ.pdf
Frequency .java
/**
* Word frequency counter
*/
package frequency;
import java.util.Iterator;
/**
*
* @author UMD CS
*/
public class Frequency implements Iterable{
private Node first;
private int N;
Frequency(){
N = 0;
first = null;
}
@Override
public Iterator iterator() {
return new ListIterator();
}
/**
*
* List iterator
*
*/
private class ListIterator implements Iterator{
private Node current;
private int index ;
ListIterator(){
current = first;
index = 0;
}
@Override
public boolean hasNext() {
return current != null;
}
public String next() {
if(!hasNext()){
return null;
}
String word = current.key;
int count = current.count;
String r = \"(\"+word + \",\" + Integer.toString(count)+\")\";
current = current.next;
return r;
}
@Override
public void remove() {
}
}
/**
*
* Node class
*
*/
private class Node {
private String key;
private int count;
private Node next;
Node(String item){
key = item;
count = 1;
next = null;
}
@Override
public String toString(){
return \"(\"+key +\",\"+count+\")\";
}
}
/*
* Inserts a word into the linked list. If the word exists, increment the
* count by q.
*/
public void insert(String word){
if(word.equals(\"\")){
return;
}
//TODO
/*
* implement here
*/
}
/**
*
* @param str input string
* This method splits a string into words and pass the words to insert method
*
*/
public void insertWords(String str){
String delims = \"[ .,?!\'\\\"()}{;/<>&=#-:\\\\ _]+\";
String[] words = str.split(delims);
for(String s: words){
s = s.toLowerCase();
insert(s);
}
}
/**
* prints the word frequency list
*/
public void print(){
Node c = first;
while(c != null){
System.out.print(\"(\"+c.key + \",\" + c.count+\")\");
c = c.next;
}
System.out.print(\"\ \");
}
}
--------------------------------------------------------------------------------------------------------------------
------------------------------------------------------
WordFrequency.java
package frequency;
import utils.In;
import utils.Stopwatch;
/**
*
* @author UMD CS
*/
public class WordFrequency {
/**
* @param input source
* @param
* This method receives input resource and return the word frequency string
* DO NOT CHANGE THIS METHOD.
*/
public static String freq(String inputSource, int maxLines){
In in;
// Frequency class can only count the frequency of strings. DO NOT CHANGE THIS.
Frequency freq = new Frequency();
int MAX = 100;
String inputSourceName = inputSource;
try {
in = new In(inputSourceName);
while (!in.isEmpty()) {
String s = in.readLine();
//System.out.println(s);
freq.insertWords(s);
}
}
catch (Exception e) {
e.printStackTrace();
}
StringBuilder strFrequency = new StringBuilder();
int cnt = 0;
for(String s: freq){
strFrequency.append(s);
strFrequency.append(\",\");
cnt++;
if(cnt >= maxLines){break;}
}
return strFrequency.toString();
}
/**
*
*/
public static void main(String[] args) {
In in;
Frequency freq = new Frequency();
int MAX = 100;
// String inputSourceName =
\"http://www.cs.umd.edu/class/summer2015/cmsc132/projects/P3_WordFrequency/test1.html\";
String .
The alternative hypothesis typically Corresponds to the presumed def.pdf
The alternative hypothesis typically Corresponds to the presumed default state of nature
Contests the status quo for which a corrective action may be required States the probability of
rejecting the null hypoyhesis when it is false States the probability of rejecting the null
hypoyhesis when it is true
Solution
states the porbability of rejuecting the null hypothesis when it is false.
Suppose the continuous random variable X has probability density func.pdf
Suppose the continuous random variable X has probability density function f(x) = x^2/3 for -2
LT xLT 1, 0 else where. Find P{-2 LTX LT -1). Find E(X). Find V(X). Find the cumulative
distribution function (cdf) of X, F(x). Be sure to define the cdf for all x (-infinity, infinity).
Show F(x) using clear notation. The most common mistake for the cdf is forgetting the lower
bound of the support set, which is -2 here.
Solution.
Drag and drop the phrases to the appropriate status during the cockin.pdf
Drag and drop the phrases to the appropriate status during the cocking of the myosin head phase.
Answers may be used or not at all. Status of ATP Seamus of the myosin head. Sketch of actin
and myosin.
Solution
1. status of ATP - hydrolysed to ADP and Pi .
2. status of the myosin head - returnes to low energy position
3. status of actin and myosin - actin and myosin detached.
Some bacteria have one or more flagella. What is true about these st.pdf
Bacterial flagella allow bacteria to move around in their environment. They are built from multiple proteins that originally had different functions. Flagella are found concentrated on one part of the bacterial cell.
our company has an opening for a junior database administrator. As a.pdf
our company has an opening for a junior database administrator. As a senior database
administrator, your manager has developed the following questions to ask potential new hires
and he wants you to provide the answers to the following questions so they can be forwarded to
Human Resources: Explain Data Manipulation Language (DML). What is the difference between
Data Definition Language (DDL) and Data Control Languages (DCL)? Provide an example of an
SQL query that creates and executes four individual tables; Student, Class, InstructorID, and
Room Number with not null and default constraints.
Solution
Data Manupulation Language(DML):
-->It permits the users to manipulate data in database.
-->manupulation involves in inserting data into database tables,retrieving existing data,
deleting data from eisting tables and modifying existing data.
-->It is mostly incorporated in SQL databases.
-->SELECT,UPDATE,INSERT INTO,DELETE commands comes under DML.
------------------------------------------------------------------------------------------------------
DDL: Data definition language
-->It is used to define the database structure or schema.
1)CREATE-It is used to create objects in database
2)ALTER - It is used to alters the structure of the database
3)DROP - It is used to delete objects from the database
4)TRUNCATE -It is used to remove all records from a table, including all spaces allocated
for the records are removed
5)RENAME - It is used to rename an object
DCL:data control language
-->It is used to control the data in database
1)GRANT - It gives users access privileges to database.
2)REVOKE - It withdraw access privileges given with the GRANT command
----------------------------------------------------------------------------------------------------------
1)CREATE TABLE Student{
Student_name varchar(50) NOT NULL DEFAULT \'Mouni\',
Student_Id int NOT NULL DEFAULT 10
}
CREATE TABLE Class{
Class_name varchar(15) NOT NULL DEFAULT \'CSE\'
InstructorId int NOT NULL DEFAULT \'100\'
}
CREATE TABLE InstructorID{
InstructorID int NOT NULL DEFAULT \'100\'
Instructor_name varchar(15) NOT NULL DEFAULT \'Thomson\'
}
CREATE TABLE Room{
Room_number int NOT NULL DEFAULT \'12A\'
Room_naem varchar(10) DEFAULT \'omega\'
}.
Part ABriefly, explain the current hypotheses about how brown alg.pdf
Part A:
Briefly, explain the current hypotheses about how brown algae, red algae, chlorophytes and
charophytes (couple of groups of green algae), and land plants are related. What are the
characteristics of the land plants that distinguish it from the most closely related algal group, and
(if applicable) how do each of these characteristics benefit plants in a terrestrial environment?
Part B:
What are the differences between plant cells, animal cells, and bacterial cells. And what are the
characteristics of different plant cell types.
Solution
(1)Evolution of plants has facilitated the survival of life on earth and these land plants has their
monophyletic lineage embedded with that of green algae and these green algae are believed to be
the oldest eukaryotic lineages.
Hypothesis states that land plants are related to brown algae, red algae, chlorophytes and
charophytes. There are so many evidences to prove that these land plants share similar
characteristics with that of brown algae, red algae, chlorophytes and charophytes. Plants and
algae are eukaryotic and both posses the chlorophyll, which allows them to make their own food,
both plants and algae have similar cell walls, stores energy in the form of starch, follows two
stage life cycle.
Sporopollenin, chlorophyll a, cellulose, chlorophyll b are common to both charophytes and land
plants, but plants has lignin and is absent in case of charophytes. Some green algae exhibit
alternation of generations like plants. All land plants exhibit alternation of generations. But no
charophytes exhibit alternation of generations.
Charophytes are very similar to plants as they both have homologous chloroplasts (chlorophyll b,
beta carotene is present, thylakoid are arranged as grana, DNA as genetic material), with respect
to biochemical similarity, both has cellulose cell wall, similar enzymes in peroxisomes, when it
comes to mitosis and cytokinesis, follow similar mechanisms like disappearance of nuclear
envelope and till cytokinesis the spindle will remain. Based on the above features the
Charophytes are considered as the closest living algal relatives of land plants
Most of the varieties of algal species live in extremely damp terrestrial environment, whereas
most of the plants live on land and some plants can live in water also, but algae cannot survive
on dry soil and they are very adapted to it. In case of alga, the surrounding water supports it and
the whole alga is involved in helping it in the process of photosynthesis, absorption of water and
minerals, holdfast will provides the anchoring. But in case of plants, each and every organ is
specialized for different functions, like leaves for photosynthesis, cuticle to prevent water loss,
stomata for gaseous exchange, stem to support, root to provide anchorage. But alternation of
generation is absent in charophytes, but it is found in all plants providing them extra benefit to
survive.
2)
Animal Cell
Plant cell
Bacteria
These are eukaryotes
These ar.
Please show the whole code... Im very confused. Matlab code for ea.pdf
Please show the whole code... I\'m very confused. Matlab code for earth orbit Using a software
package such as MATLAB, create a plot of an Earth orbit with a = 31, 800 km and e = 0.6 for 0
lessthanorequalto theta lessthanorequalto 2 pi. The origin of the plot should be Earth. Perigee
should be in the direction of the positive x axis (therta = 0). On this plot, you should label (by
hand) the semimajor axis, semiminor axis, radius of perigee, radius of apogee, and the
semiparameter (semilatus rectum) of the ellipse.
Solution
---------------------------------------------------------------------------------------------------------------------
-------------------------------------------------
Elliptic orbit:
%periapsis
rp = 500e3 + getradius(\'earth\');
%eccentricity
ec = .7;
%semimajor axis
a = getsemimajoraxis(rp,ec);
%apoapsis
ra = 2*a-rp;
%true anomaly
theta = [0:.1:360];
orbit = ellipse(theta,a,ec,\'earth\');
figure(1), plot(orbit.x,orbit.y); axis equal
Re = getradius(\'earth\');
x = Re*cosd(theta);
y = Re*sind(theta);
hold on
plot(x,y,\'color\',[0 1 1])
hold off
figure(2), plot(theta,orbit.V/1000)
xlabel(\'\\theta\'), ylabel(\'V (km/s)\')
%current time
c = clock;
%reset hour and second
c(4) = 12; %hour
c(5) = 0; %minute
c(6) = orbit.tau; %seconds
%ellapsed time
datetime(c)
%returns 14-Feb-2015 21:34:54, so it\'s about 9.5 hours later still on
%Monday
2
ans =
22-Feb-2015 21:34:54
-----------Elliptic orbit
%periapsis
rp = 500e3 + getradius(\'earth\');
%eccentricity
ec = .7;
%semimajor axis
a = getsemimajoraxis(rp,ec);
%apoapsis
ra = 2*a-rp;
%true anomaly
theta = [0:.1:360];
orbit = ellipse(theta,a,ec,\'earth\');
figure(1), plot(orbit.x,orbit.y); axis equal
Re = getradius(\'earth\');
x = Re*cosd(theta);
y = Re*sind(theta);
hold on
plot(x,y,\'color\',[0 1 1])
hold off
figure(2), plot(theta,orbit.V/1000)
xlabel(\'\\theta\'), ylabel(\'V (km/s)\')
%current time
c = clock;
%reset hour and second
c(4) = 12; %hour
c(5) = 0; %minute
c(6) = orbit.tau; %seconds
%ellapsed time
datetime(c)
%returns 14-Feb-2015 21:34:54, so it\'s about 9.5 hours later still on
%Monday
2
ans =
22-Feb-2015 21:34:54
3
The circular orbits at perigee and apogee are
7.6167 and 3.1997 km/s, respectively.
-----Parabolic orbit:
Sphere of influence is actually a chapter 5 concept, but is the distance at which the gravitational
forces
between two objects is equal. Thus,
While an approximation, if we just estimate this as the point between the sun and the earth
where
these forces balance, then if the orbital radius of Earth is , then
, and thus solution for r_E in gives
r_soi = getsphereofinfluence(\'earth\',\'sun\');
display([\'Radius from center of earth to sphere of influence is approximately \' num2str(r_soi)]);
%perifocal distance
rp = 500e3 + getradius(\'earth\');
%eccentricty
ec = 1; %always so for a a parabola
theta = -180:1:180;
orbit = parabola(theta,rp,\'earth\');
figure(3), plot(orbit.x,orbit.y); axis equal
Re = getradius(\'earth\');
x = Re*cosd(theta);
y = Re*sind(theta);
.
Monomeric single pass trans-membrane proteins generally span the bil.pdf
Monomeric single pass trans-membrane proteins generally span the bilayer membrane with an
alphahelical domain. Which of the three sequences below is most likely to be a trans-membrane
segment.
A. I T L I Y F G N M S S V T Q T I L L I S
B. I L L I F F G V M A L V I V V I L L I A
C. L L K K F F R D M A A V H E T I L E E S
EXPLAIN YOUR ANSWER DO NOT JUST ANSWER THE
QUESTION!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Solution
A Trans-membrane Protein (TP) is a type of integral membrane protein. A Trans-membrane
Protein is called so because the polypeptide chain is exposed on both sides of the biological
membrane.
It spans the entire biological membrane or plasma membrane and is attached to it permanently.
They function as gateways to permit the transport of specific substances across the plasma
membrane. For this, these proteins (TP) undergo a conformational change that is a temporary
change in their structure.
Trans-membrane Segment: It usually denotes a single trans membrane alpha helix of a trans
membrane protein, also known as an integral protein. This segment is thermodynamically stable
in the membrane.
We all know that the secondary structure of proteins refers to the spirally coiled polypeptide
chain called alpha helical form or alpha helix.
Of the three 20 amino acid trans membrane sequences;
Sequence B is most likely to be a trans- membrane segment because it is composed primarily of
hydrophobic amino acids, therefore can be stably integrated into lipid bilayer.
Sequence A: It contains many polar amino acids therefore it cannot be integrated into the
hydrophobic interior of lipid bilayer of plasma membrane.
Sequence C: It contains many charged amino acids which again is disfavored into the
hydrophobic interior of lipid bilayer..
Match the following to one or more layers of the Internet protocol st.pdf
Match the following to one or more layers of the Internet protocol stack: Provide user services
such as e-mail and web browsing Ensure reliable transmission of packets
Solution
Ans: b - Transport Layer
Transport layer: Responsible for reliable transmission of data and service specification between
hosts. The major responsibility of this layer is data integrity--that data transmitted between hosts
is reliable and timely..
Kent and Emily have a baby boy Dustin by in vitro fertilization. Dust.pdf
Kent and Emily have a baby boy Dustin by in vitro fertilization. Dustin does not resemble either
one of his parents, so they suspect that the clinic made a mistake during the procedure and
implanted the wrong embryo. At Dustin\'s first well-baby checkup they insist that they get his
blood type. Baby Dustin is type B-negative. Kent is A-positive and Emily is AB-positive. Given
this information answer the following questions. (a) What are Dustin\'s possible genotypes?
__________ (b) What are Kent\'s possible genotypes? __________ (c) What are Emily\'s
possible genotypes? __________ (d) Is it possible that Dustin is Kent and Emily\'s son?
__________ (e) Explain your answer in (d) __________ Carrie is AB-positive and her father is
AB-negative. Carrie marries Ryan, who is AB-negative. (f) What is Carrie\'s genotype?
__________ (g)What is Ryan\'s genotype? __________ (h) What are the phenotypic ratios of
blood types (for both the ABO gene and Rh gene) that would be expected among their children?
__________ (Show your work in a Punnett square below!)
Solution
Answer:
a). Dustin’s possible genotypes: IA IB : Rh+ Rh+ or IA IB : Rh+ Rh-
b). Kent’s possible genotypes:
IA IA : Rh+ Rh+ or IA IA : Rh+ Rh- or IA i : Rh+ Rh+ or IA i : Rh+ Rh-
c). Emily’s possible genotypes: IB IB : Rh- Rh- or IB i : Rh- Rh-
d). Yes Dustin belongs to Kenta and Emily’s son.
e).
The gamete, IA: Rh+ of Kent’s and the gamete, IB: Rh- of Emily’s fuse to produce IA IB : Rh+
Rh- (AB-positive).
Rh+ is dominant over Rh-
IA & IB are codominant.
f). Carrie’s genotype (AB-positive): IAIB: Rh+Rh-
g). Ryan’s genotype (AB-negative): IAIB: Rh-Rh-
IAIB: Rh+Rh- x IAIB: Rh-Rh-
IARh-
IBRh-
IARh+
IAIA Rh+Rh-
(A+)
IAIBRh+Rh- \\
(AB+)
IA Rh-
IAIA Rh-Rh-
(A-)
IAIB Rh-Rh-
(AB-)
IARh-
IBRh-
IARh+
IAIA Rh+Rh-
(A+)
IAIBRh+Rh- \\
(AB+)
IA Rh-
IAIA Rh-Rh-
(A-)
IAIB Rh-Rh-
(AB-).
Isabel Lopez started Biz Consulting, a new business, and completed th.pdf
Isabel Lopez started Biz Consulting, a new business, and completed the following transactions
during its first year of operations a. I. Lopez invests $69,000 cash and office equipment valued at
$37,000 in the company in exchange for its common stock b. The company purchased a
$309,000 building to use as an office. Biz paid $41,000 in cash and signed a note payable
promising to pay the $268,000 balance over the next ten years. c. The company purchased office
equipment for $5,400 cash. d. The company purchased $3,500 of office supplies and $1,700 of
office equipment on credit. e. The company paid a local newspaper $960 cash for printing an
announcement of the office\'s opening. f. The company completed a financial plan for a client
and billed that client $4,200 for the service g. The company designed a financial plan for another
client and immediately collected an $8.100 cash fe h. The company paid $1,600 cash in
dividends to the owner . The company received $3.200 cash as partial payment from the client
described in transaction f payment of $850 cash on the equipment purchased in transaction d k.
The company paid $2.400 cash for the office secretary\'s wages for this period. Required: 2.
Enter the amount of each transaction on individual items of the accounting equation. (Enter
reductions to account balances with a minus sign.) Assets Office Office Accounts
Solution
Answer
Assets
Liabilities
Equity
Cash
Accounts receivables
Office Supplies
Office
Equip
ments
Buildings
Accounts Payable
Notes Payable
Common Stock
Dividends
Revenues
Expenses
a
69000
37000
106000
b
-41000
309000
268000
28000
0
0
37000
309000
0
268000
106000
0
0
0
c
-5400
5400
22600
0
0
42400
309000
0
268000
106000
0
0
0
d
3500
1700
5200
22600
0
3500
44100
309000
5200
268000
106000
0
0
0
e
-960
960
21640
0
3500
44100
309000
5200
268000
106000
0
0
960
f
4200
4200
21640
4200
3500
44100
309000
5200
268000
106000
0
4200
960
g
8100
8100
29740
4200
3500
44100
309000
5200
268000
106000
0
12300
960
h
-1600
1600
28140
4200
3500
44100
309000
5200
268000
106000
1600
12300
960
i
3200
-3200
31340
1000
3500
44100
309000
5200
268000
106000
1600
12300
960
j
-850
-850
30490
1000
3500
44100
309000
4350
268000
106000
1600
12300
960
k
-2400
2400
Bal
28090
1000
3500
44100
309000
4350
268000
106000
1600
12300
3360
Assets
Liabilities
Equity
Cash
Accounts receivables
Office Supplies
Office
Equip
ments
Buildings
Accounts Payable
Notes Payable
Common Stock
Dividends
Revenues
Expenses
a
69000
37000
106000
b
-41000
309000
268000
28000
0
0
37000
309000
0
268000
106000
0
0
0
c
-5400
5400
22600
0
0
42400
309000
0
268000
106000
0
0
0
d
3500
1700
5200
22600
0
3500
44100
309000
5200
268000
106000
0
0
0
e
-960
960
21640
0
3500
44100
309000
5200
268000
106000
0
0
960
f
4200
4200
21640
4200
3500
44100
309000
5200
268000
106000
0
4200
960
g
8100
8100
29740
4200
3500
44100
309000
5200
268000
106000
0
12300
960
h
-1600
1600
28140
4200
3500
44100
309000
5200
268000
106000
1600
12300
960
i
3200
-3200
31340
1000
3500
44100
.
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- 2. Enter a grade: 54 Enter a grade: 86 Enter a grade: 75 Enter a grade: 92 Enter a grade: 73 Enter a grade: 79 Enter a grade: 75 Enter a grade: 82 Enter a grade: 73 The sum is: a number here. The mean is: 80.5 GRADE INDIVIDUAL DEVIATION
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- 6. Enter a grade: 73 The sum is: a number here. The mean is: 80.5 GRADE INDIVIDUAL DEVIATION 89 8.5 95 14.5 72 -8.5 83 2.5 99 18.5 54 -26.5 86 5.5 75 -5.5
- 7. 92 11.5 73 -7.5 79 -1.5 75 -5.5 82 1.5 73 -7.5 The variance is: 124.679 standard deviation: 11.166 Write a C++ program to get input of a data set of 14 double values (grades) from the standard input device (the keyboard), then calculate and report the sum and the mean of the data set, then the individual DEVIATION of each individual input with respect to the mean, the variance of the dataset, and finally the standard deviation of the data set. This problem requires you to integerate multiple things together: arrays, loops, functions. More specifically, you want to use loops to deal with input, save them to array, or calculate sum and individual deviation, and display results etc. in separate functions. A sample run.
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- 10. 54 -26.5 86 5.5 75 -5.5 92 11.5 73 -7.5 79 -1.5 75 -5.5 82 1.5 73 -7.5 The variance is: 124.679 standard deviation: 11.166 Solution Edit & Run
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