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This document provides data on outdoor temperatures recorded throughout the day. It then lists tasks to analyze the data using a quadratic regression model. The tasks include plotting the temperature vs. time data, determining the quadratic polynomial best fit curve, calculating the r-squared value to assess the goodness of fit, using the model to estimate temperatures at given times, finding the time of maximum temperature, and estimating the times when the temperature equals specific target values.

Formation

pls show work
Quadratic Numbers
Input Time of Day (hour)
Target Outdoor Temperature
10.5
49.0
Quadratic Regression (QR)
Data: On a particular day in April, the outdoor temperature was recorded at 8 times of the day,
and the following table was compiled.
Time of day
(hour)
x
Temperature
(degrees F.)
y
7
35
9
50
11
56
13
59
14
61
17
62
20
59
23
44
REMARKS: The times are the hours since midnight. For instance, 7 means 7 am, and 13 means
1 pm.
The temperature is low in the morning, reaches a peak in the afternoon, and then decreases.

Tasks for Quadratic Regression Model (QR)
(QR-1) Plot the points (x, y) to obtain a scatterplot. Note that the trend is definitely non-linear.
Use an appropriate scale on the horizontal and vertical axes and be sure to label carefully.
(QR-2) Find the quadratic polynomial of best fit and graph it on the scatterplot. State the formula
for the quadratic polynomial.
(QR-3) Find and state the value of r2, the coefficient of determination. Discuss your findings. (r2
is calculated using a different formula than for linear regression. However, just as in the linear
case, the closer r2 is to 1, the better the fit. Just work with r2, not r.) Is a parabola a good curve to
fit to this data?
(QR-4) Use the quadratic polynomial to make an outdoor temperature estimate. Temperature is
listed above. Be sure to use the quadratic regression model to make the estimate (not the values
in the data table). State your results clearly -- the time of day and the corresponding outdoor
temperature estimate.
(QR-5) Using algebraic techniques we have learned, find the maximum temperature predicted by
the quadratic model and find the time when it occurred. Report the time to the nearest quarter
hour (i.e., __:00 or __:15 or __:30 or __:45). (For instance, a time of 18.25 hours is reported as
6:15 pm.) Report the maximum temperature to the nearest tenth of a degree. Show work.
(QR-6) Use the quadratic polynomial together with algebra to estimate the time(s) of day when
the outdoor temperature is a specific target temperature. (Target numbers SHOWN ABOVE)
Report the time(s) to the nearest quarter hour. Be sure to use the quadratic model to make the
time estimates (not values in the data table). Show work. State results clearly -- the target
temperature and the associated time(s). Show work.
pls show work
Quadratic Numbers
Input Time of Day (hour)
Target Outdoor Temperature
10.5
49.0
Solution
extend the time and I will answer!

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Poor shooting percentages in recent men\'s DI basketball Final Four games have caused many to question decisions made by the NCAA concerning the layout of the court used for the Final Four. While domed football stadiums have long been used for the Final Four, beginning with the 2009 Final Four the NCAA has required that: i) the court be raised, ii) the entire stadium seating capacity be used (that is, large curtains cannot be erected to seal off portions of the stadium), and iii) temporary seats be installed on the football field around the court to accommodate more students from the participating schools. Players and coaches complain that the resulting dome shooting environment (depth perception, sight lines, etc.) is vastly different from that in the arenas in which players are accustomed to playing. This Excel file NCAA Men\'s Final Four Shooting Percentages shows the 72 team shooting percentages in all Final Four games from 2000 to 2011. Use the data from 2000-2008 and the data from 2009-2011 to calculate a 90% confidence interval for %u03BCpre2009 - %u03BC2009+, the difference in the mean team Final Four shooting percentages before and after the venue changes made by the NCAA. Note: Calculate sample means to 3 decimal place and sample standard deviations to 4 decimal places. % lower bound % upper bound Field Goal Shooting PercentageTeamYearGame 55.9Mich St2000Final 55.9Florida2000Final 55.9Mich St2000Semi 55.9Wisconsin2000Semi 55.9Florida2000Semi 55.9N Carolina2000Semi 55.9Arizona2001SemiST DEVST Error 55.9Mich St2001Semi5.72423E-146.74607E-15 55.9Duke2001Semi 55.9Maryland2001Semi 55.9Duke2001Final 55.9Arizona2001Final 55.9Indiana2002Final 55.9Oklahoma2002Semi 55.9Kansas2002Semi 55.9Maryland2002Final 55.9Maryland2002Semi 55.9Indiana2002Semi 55.9Marquette2003Semi 55.9Texas2003Semi 55.9Kansas2003Final 55.9Syracuse2003Final 55.9Kansas2003Semi 55.9Syracuse2003Final 55.9Georgia Tech2004Final 55.9Duke2004Semi 55.9UConn2004Final 55.9Oklahoma St2004Semi 55.9Georgia Tech2004Semi 55.9UConn2004Semi 55.9Michigan St2005Semi 55.9Illinois2005Final 55.9Louisville2005Semi 55.9Illinois2005Semi 55.9N Carolina2005Semi 55.9N Carolina2005Final 55.9LSU2006Semi 55.9UCLA2006Final 55.9Geo Mason2006Semi 55.9UCLA2006Semi 55.9Florida2006Semi 55.9Florida2006Final 55.9UCLA2007Semi 55.9Ohio St2007Semi 55.9Ohio St2007Final 55.9Georgetown2007Semi 55.9Florida2007Final 55.9N Carolina2008Semi 55.9UCLA2008Semi 55.9Memphis2008Final 55.9Memphis2008Semi 55.9Kansas2008Final 55.9Kansas2008Semi 55.9Florida2008Semi 55.9Villanova2009Semi 55.9Michigan St2009Final 55.9N Carolina2009Semi 55.9UConn2009Semi 55.9Michigan St2009Semi 55.9N Carolina2009Final 55.9Butler2010Semi 55.9Butler2010Final 55.9W Virginia2010Semi 55.9Michigan St2010Semi 55.9Duke2010Final 55.9Duke2010Semi 55.9Butler2011Final 55.9Kentucky2011Semi 55.9Uconn2011Final 55.9Butler2011Semi 55.9VCU2011Semi 55.9UConn2011Semi Solution 2000-2008: sample mean,x1=43.400 sample standard deviation,s1=6.4241 sample size,n1=54 20.

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Pope Urban II\'s Speech at Clermont (1095) in Lualdi, How did urban ii appeal to his audience? Why do you think this language motivated people in the audience to join the Crusade? What did the Crusade offer them? Solution Urban II makes perhaps the most influential speech of the Middle Ages, giving rise to the Crusades by calling all Christians in Europe to war against Muslims in order to reclaim the Holy Land, with a cry of “Deus vult!” or “God wills it!” Born Odo of Lagery in 1042, Urban was a protege of the great reformer Pope Gregory VII. Like Gregory, he made internal reform his main focus, railing against simony (the selling of church offices) and other clerical abuses prevalent during the Middle Ages. Urban showed himself to be an adept and powerful cleric, and when he was elected pope in 1088, he applied his statecraft to weakening support for his rivals, notably Clement III. By the end of the 11th century, the Holy Land—the area now commonly referred to as the Middle East—had become a point of conflict for European Christians. Since the 6th century, Christians frequently made pilgrimages to the birthplace of their religion, but when the Seljuk Turks took control of Jerusalem, Christians were barred from the Holy City. When the Turks then threatened to invade the Byzantine Empire and take Constantinople, Byzantine Emperor Alexius I made a special appeal to Urban for help. This was not the first appeal of its kind, but it came at an important time for Urban. Wanting to reinforce the power of the papacy, Urban seized the opportunity to unite Christian Europe under him as he fought to take back the Holy Land from the Turks. At the Council of Clermont, in France, at which several hundred clerics and noblemen gathered, Urban delivered a rousing speech summoning rich and poor alike to stop their in-fighting and embark on a righteous war to help their fellow Christians in the East and take back Jerusalem. Urban denigrated the Muslims, exaggerating stories of their anti-Christian acts, and promised absolution and remission of sins for all who died in the service of Christ.

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Political analysts estimate the probability that Politician A will run as the Democrat candidate for president in 2016 is 45%, and the probability that Politician B will run as the Republican candidate is 20%. If their political decisions are independent, then what is the probability that only Politican A runs for president? Solution the probabilty of required event = P(politixian A is democrat candidate)*P(politician B is not a Republican candidate) this is true since their political decisions are independant. HENCE ANSWER = 0.45*0.8=0.36 hence 36% chances.

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Please Write in CPP.Thank you Create a class called BinaryTreeDeque. This binary tree implements all of the functionality of the BinaryTree class, but instead of storing its values in a tree structure, the element values are stored in a deque, using pointers to maintain the relationship between the binary tree and its node values. Additionally: deque * getValues() - returns a copy of the deque the current array of values (which should be in insertion order). Solution #include #include #include #define SIZE 50 class Node { public: int data; Node *rt,*lt; }; class Que { int front,rear; int size; Node *array; public: Node* newNode(int data) { Node *temp; temp->data=data; temp->lt=temp->rt=NULL; return(temp); } Que* createQue(int size) { Que *que=new Que; que->front=que->rear=-1; que->size=size; /* que->array=new Node; for(int i=0;iarray[i]=NULL;*/ return que; } int isEmpty(Que *que) { return que->front==-1; } int isFull(Que *que) { return que->rear==que->size-1; } int hasOnlyOneItem(Que *que) { return que->front==que->rear; } void Enque(Node *root,Que *que) { if(isFull(que)) que->array[++que->rear]=*root; if(isEmpty(que)) ++que->front; } Node *Deque(Que *que) { if(isEmpty(que)) return NULL; Node temp=que->array[que->front]; if(hasOnlyOneItem(que)) que->front=que->rear=-1; else ++que->front; return &temp; } Node* getFront(Que *que) { return &que->array[que->front]; } int hasBothChild(Node *temp) { return temp && temp->lt && temp->rt; } void insert(Node **root,int data,Que *que) { Node *temp=new Node; temp->data=data; if(!*root) *root=temp; else { Node *front=getFront(que); if(!front->lt) front->lt=temp; else if(!front->rt) front->rt=temp; if(hasBothChild(front)) Deque(que); } Enque(temp,que); } void levelOrder(Node *root) { Que *que=createQue(SIZE); Enque(root,que); while(!isEmpty(que)) { Node *temp=Deque(que); cout<data; if(temp->lt) Enque(temp->lt,que); if(temp->rt) Enque(temp->rt,que); } } }; void main() { Node *root=NULL; Que q1; Que *que=q1.createQue(SIZE); for(int i=1;i<=12;i++) q1.insert(&root,i,que); q1.levelOrder(root); }.

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Please write in C++ and should be able to compile and debug.Thank you Create a class called BinaryTreeDeque. This binary tree implements all of the functionality of the BinaryTree class, but instead of storing its values in a tree structure, the element values are stored in a deque, using pointers to maintain the relationship between the binary tree and its node values. Additionally: deque * getValues() - returns a copy of the deque the current array of values (which should be in insertion order). Solution Following are steps to insert a new node in Complete Binary Tree. 1. If the tree is empty, initialize the root with new node. 2. Else, get the front node of the queue. …….If the left child of this front node doesn’t exist, set the left child as the new node. …….else if the right child of this front node doesn’t exist, set the right child as the new node. 3. If the front node has both the left child and right child, Dequeue() it. 4. Enqueue() the new node. Code :- ------------------------- binarytreedequeue.h #ifndef BINARYTREEDEQUE_H #define BINARYTREEDEQUE_H // Program for linked implementation of complete binary tree #include #include // For Queue Size #define SIZE 50 // A tree node struct node { int data; struct node *right,*left; }; // A queue node struct Queue { int front, rear; int size; struct node* *array; }; class BinaryTreeDeque { public: BinaryTreeDeque(); struct node* newNode(int data); struct Queue* createQueue(int size); int isEmpty(struct Queue* queue); int isFull(struct Queue* queue); int hasOnlyOneItem(struct Queue* queue); void Enqueue(struct node *root, struct Queue* queue); struct node* Dequeue(struct Queue* queue); struct node* getFront(struct Queue* queue); int hasBothChild(struct node* temp); void insert(struct node **root, int data, struct Queue* queue); void display(struct Queue* queue); node * getValues(struct Queue* queue); }; #endif // BINARYTREEDEQUE_H binarydequeue.cpp --------------------------------------- #include \"binarytreedeque.h\" BinaryTreeDeque::BinaryTreeDeque() { } // A utility function to create a new tree node struct node* BinaryTreeDeque::newNode(int data) { struct node* temp = (struct node*) malloc(sizeof( struct node )); temp->data = data; temp->left = temp->right = NULL; return temp; } // A utility function to create a new Queue struct Queue* BinaryTreeDeque::createQueue(int size) { struct Queue* queue = (struct Queue*) malloc(sizeof( struct Queue )); queue->front = queue->rear = -1; queue->size = size; queue->array = (struct node**) malloc(queue->size * sizeof( struct node* )); int i; for (i = 0; i < size; ++i) queue->array[i] = NULL; return queue; } // Standard Queue Functions int BinaryTreeDeque::isEmpty(struct Queue* queue) { return queue->front == -1; } int BinaryTreeDeque::isFull(struct Queue* queue) { return queue->rear == queue->size - 1; } int BinaryTreeDeque::hasOnlyOneItem(struct Queue* queue) { return queue->front == queue->rear; } void BinaryTreeDeque::Enqueue(struct node *root, s.

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Please solve and explain: 1. You toss a coin 270 times. Find the probability that exactly 135 of the 270 tosses will be heads. 2. About 4% of the scientists in the National Academy of Sciences believe in PSI (parapsychology such as psychokinesis, telepathy, and astral projection). At a convention 144 members show up. Let X be the number of scientists at the convention who believe in PSI. You may assume for the following questions that X is binomially distributed with parameters n=144 and p=0.04. (a) How many scientists at the convention do you expect do believe in PSI? (b) What are the chances that none of the scientists at the convention believes in PSI? (c) What are the chances that at least 4 scientists at the convention believe in PSI? (d) What is the standard deviation of X? Solution a)scientists at the convention do you expect do believe in PSI=mean=140*4/100=6(approx of 5.6); b)134; c)134.4 d)5.76.

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Please solve problem. Thanks. Suppose that you toss a coin and roll a die. The sample space is shown in the figure below. (a) What is the probability of obtaining tails and a five? (Enter the probability as a fraction.) (b) What is the probability of obtaining tails or a five? (Enter the probability as a fraction.) (c) What is the probability of obtaining heads and a two? (Enter the probability as a fraction.) Solution 1. sample space , S = 12 so P(obtaining tails and a five) = 1/12 2 . P(obtaining tails or a five) = 1/2 + 1/6 = 2/3 3. P(obtaining heads and a two) = 1/2 * 1/6 = 1/12.

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