In 2021 California passed a law to provide all public schoo.pdf

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In 2021, California passed a law to provide all public school children with free lunch and breakfast. Public health professionals have asked the state to make sure the meals are healthy and nutritious, to reduce the chance the program might contribute to childhood obesity. What are public health professionals asking California to consider? Balance the interests of the state and schools Maximize the benefits over harms Manage a tragedy of the commons Interpret the general welfare clause broadly.

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In 2019, scientists conducted a research study about spicy food as a potential risk factor for gastroesophageal reflux disease (GERD). Scientists enrolled 500 patients with GERD. An additional 1,000 persons without GERD were also enrolled, matched on gender to patients with GERD. Participants self-reported their recent spicy foods consumption. What study design did the scientists use to answer their research question? a. Case-control study b. Cross-sectional study c. Prospective cohort study X d. Randomized clinical trial X.

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In a city of half a million, there are initially 400 cases of a particularly virulent strain of flu. The Centers for Disease Control and Prevention in Atlanta claims that the cumulative number of infections of this flu strain will increase by 40% per week if there are no limiting factors. Make a logistic model of the potential cumulative number of cases of flu as a function of weeks from initial outbreak, and determine how long it will be before 100,000 people are infected..

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In a 10-year graph with an explanation of Idaho Panhandle: Hypothetically speaking A forested region with regular annual prescribed fire(RX), has a creek running through it. In a semi-arid region, during the months of winter, which are typically wet each year and dry each year in the summer. What would the results be in year 7, the month of July of the watershed when a wildfire hits (2 months before full suppression) during the summertime? In terms of the watershed, what would happen with runoff, local ecology, streamflow, changes in sediment, animal wildlife activities, etcetera)..

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In 2-D, let x=[x1,x2]. Write down the explicit cubic feature mapping (x) as a vector; i.e., (x)=[f1(x1,x 2),,fN(x1,x2)] Hint Hint: (x) should be a 10-dimensional vector. (x)= You have used 0 of 20 attempts The cubic_features function in features.py is already implemented for you. That function can handle input with an arbitrary dimension and compute the corresponding features for the cubic Kernel. Note that here we don't leverage the kernel properties that allow us to do a more efficient computation with the kernel function (without computing the features themselves). Instead, here we do compute the cubic features explicitly and apply the PCA on the output features..

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import yfinance as yf (yf.Tickers(tickers=['AAPL', 'TSLA']).history(period='max', auto_adjust=False, progress=False).assign(Date = lambda x:x.index.tz_localize(None)).set_index('Date').rename_axis(columns=['Variable', 'Ticker']).to_csv('yahoo.csv')) Read the daily returns data for AAPL and TSLA saved in file yahoo.csv Calculate the decimal daily returns for AAPL and TSLA from July 2010 through February 2023 (inclusive) and assign to data frame returns..

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In 2020, Natural Selection, a nationwide computer dating service has $508 million assets and $204 million of liabilities. Earnings before interests and taxes were $124 million, interest expense was $28.5million, the tax rate was 40%, principal payment requirements were $23.9 million, and annual dividends were 20% per share on $20 million share outstanding. Calculate the following for natural selection: liabilities-to-equity ratio times-interest-earned ratio times burden covered.

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In 2017, Americans spent a record-high $9.1 billion on Halloween-related purchases (the balance website). Sample data showing the amount, in dollars, 16 adults spent on a Halloween costume are as follows. a. What is the estimate of the population mean amount adults spend on a Halloween costume (to 2 decimals)? $ b. What is the sample standard deviation (to 2 decimals)? $ c. Provide a 95% confidence interval estimate of the population standard deviation for the amount adults spend on a Halloween costume (to 2 decimals). Use Table 11.1 $$.

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import java.util.LinkedList; import java.util.Queue; import java.util.Stack; class Sudoku_Solver_question { public static int type = 0; public static int height = 0; // Node exploration holds the sudoku board in every iteration private static class Exploration { char[][] board; int curCell = 0; // constructor method public Exploration(char[][] board, int curCell) { this.board = cloneBoard(board); this.curCell = curCell; } // clones the board for every state private char[][] cloneBoard(char[][] board) { char[][] newBoard = new char[board.length][board[0].length]; copyIntoBoard(board, newBoard); return newBoard; } } public static boolean BFS(char[][] board) { // TODO write BFS search algorithm } public static boolean DFS(char[][] board) { // TODO write DFS search algorithm } /** * parameter: index * return: index * Returns the index of a row based on type(6x6 / 9x9) of the board */ private static int getRowForIndex(int idx) { return idx / type; } /** * parameter: index * return: index * Returns the index of a column based on type(6x6 / 9x9) of the board */ private static int getColumnForIndex(int idx) { return idx % type; } /** * Parameters: source board & destination board * return: null * Copies the source board to destination board in every state */ protected static void copyIntoBoard(char[][] src, char[][] dst) { for (int i = 0; i < src.length; i++) { System.arraycopy(src[i], 0, dst[i], 0, src[i].length); } } /** * Parameters: Board & box index number * returns: true / false * Validates the box, returns true if there is no duplicate number in the box else false */ private static boolean isValidSquare(char[][] board, int s) { boolean[] isPresent = new boolean[board.length+1]; for (int r = (s / 3) * height; r < ((s / 3) + 1) * height; r++) { for (int c = (s % 3) * height; c < ((s % 3) + 1) * height; c++) { char value = board[r][c]; if (value == '.') { continue; } int intVal = Character.getNumericValue(value); if (isPresent[intVal]) { return false; } isPresent[intVal] = true; } } return true; } /** * Parameters: Board & column index number * returns: true / false * Validates the box, returns true if there is no duplicate number in the column else false */ private static boolean isValidColumn(char[][] board, int c) { boolean[] isPresent = new boolean[board.length+1]; for (int i = 0; i < board.length; i++) { char value = board[i][c]; if (value == '.') { continue; } int intVal = Character.getNumericValue(value); if (isPresent[intVal]) { return false; } isPresent[intVal] = true; } return true; } /** * Parameters: Board & row index number * returns: true / false * Validates the box, returns true if there is no duplicate number in the row else false */ private static boolean isValidRow(char[][] board, int r) { boolean[] isPresent = new boolean[board.length+1]; for (int i = 0; i < board.length; i++) { char value = board[r][i]; if (value == '.') { continue; } int intVal = Character.getNumericValue(value); if (isPresent[intVal]) { return false; } isPre.

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In 2010 the Pew Research Center reported that teenagers send on average 50 text messages every day. A researcher suspects this number is out of date, and wishes to determine if teens are sending fewer texts than they used to. A random sample of 25 teens were found to be sending an average of 47.75 texts per day, with a standard deviation of 15. Testing at 10% significance and stating the null and alternative hypothesis, what can the researcher conclude?.

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Implementing CRM Across 3CA Ian was marketing and sales director for 3CA, responsible for increasing sales across all market sectors in Australia. 3CA was an international organisation with offices in Europe, Australia, New Zealand, Singapore, and Malaysia. Each country office operated largely independently, but they were encouraged by the global executive to co-operate where clear synergies were identifiable. Nine months previously, the country CEOs for Australia and New Zealand met at a conference. The New Zealand CEO expressed serious interest in the customer relationship management (CRM) system that Ian and his team had recently designed and implemented in Australia. Implementation of the system was believed to have contributed to an 810% improvement in customer retention over a 12-month period, and the system had been recognised as outstanding by marketing industry associations. Ian was regularly called on to talk about his work at industry events. The Australian CEO volunteered Ian to work with his counterpart in New Zealand to explore the potential benefits of implementing the system in New Zealand. Ian approached the task with great enthusiasm, but found his counterpart, a woman named Anna Williams, to be less enthused. She believed that their existing systems were fit for purpose and that to embark on a new systems implementation project would be a distraction to the business. She was nevertheless instructed by her CEO to participate in a joint study. In the early stages of the project Ian quickly realised that the Australian system would be relatively easy to install and would probably deliver similar benefits in New Zealand to those delivered in Australia. The biggest piece of work would be helping sales and marketing staff to understand how best to utilise the system, rather than the installation of the system itself. Anna did not put much effort into the project, but nor did she present any obstacles, allowing Ian to talk to all of her staff, most of whom became excited at the prospect of being able to work with the new system. At the end of six months Ian prepared a preliminary report outlining benefits, drawbacks, and risks attached to installing the new system. The report made an overwhelming case for moving ahead with implementation. Anna did not contribute much to the writing of the report, though Ian provided her with every opportunity to do so. Shortly before the report was completed, Anna went on maternity leave and was unavailable to sign the report. Ian nevertheless submitted the report to his COO who, he assured Ian, forwarded the report on to the Australian CEO. Three months later, Ian had received no feedback from the COO or CEO. He asked the COO several times for an update but had been told that the company had other immediate priorities. He would receive feedback on the report after the executive team had time to address it. Meanwhile Ian was being pressed by his staff and Annas staff in New Zealand. They all recog.

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Implement special cases here: (define (delete t v) (let ((path (find t v))) (cond (( empty? t ) ) ((empty? (bst-root t)) ) ((not ( = v (node-value (car path))) ) (( > (node-count (car path)) 1) decrement count) ((equal? (car path) (bst-root t)) ;node were trying to delete (cond ; nest conditiona; ( (let-AND right (car path) empty) ; case 0.1 root <- null ((let-root empty, right empty) ; case 1 root <- left car path ; case 1.1 ((left-empty, right not empty) ; case 1.2 root <- right car path (else (delete-node path)) (else (delete-node path)) ) ) ) ) ) ) ) ) ) Special cases Top-Level Function: Delete Value from BST: (delete t v) The first parameter t is a reference to a bst, and the second parameter v is a value to find and delete. Like the insert function, delete will directly resolve a few special cases. For the general cases, it will call (find t v) to obtain a list of nodes that represent a path from the root to the node to be deleted, and will then call a helper function (delete-node path) to handle the rest of the work. In the case of delete the helper function will further rely on secondary helper functions to resolve cases and subcases. Deletion logically breaks down into cases based on how many child nodes the node to delete has. Each of those cases has subcases. The large number of cases implies the need for a correspondingly larger set of test cases: a test case to trigger each deletion case. Delete Cases Before we look at the detailed code for delete in Racket, lets look at the general approach. As for some of the other functions, it will be convenient to create a top-level function (delete t v) and several helper functions that manage specific detailed cases. There will be some initial special cases that should be managed separately, before launching any lower level helper function. Special cases for (delete t v): Tree t is empty: end the function because there is nothing to do. At this point, call (find t v) to create the path from root to node to delete if it is present. Assume the list returned is bound to a symbol named path. The node of interest will be the first node in the list: (car path) By examining node (car path), you can determine if the value to delete is present or not. Value v is not in the tree: end the function because there is nothing to do. Value v is in the tree: o Is the count field > 1? Dont delete the node, simply decrement its count field. o Is the count field = 1? In this case, we must move forward to actually delete the node. Is the node to delete the root of the tree? Does the node have 0 or 1 child nodes? This corresponds to delete cases 0.1, 1.1, and 1.2. Delete the node and update the root of the tree. o If we reach this pointvalue v is in the tree, count field is 1, node is not the root if it has 0 or 1 child nodes then the general delete case applies, call (delete-node path) The remaining delete cases will be covered by the (delete-node path) helper function. Some general considerations for node n (.

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