Bronquios, Bronquiolos, Alvéolos. Inspiración, expiración.

Pulmones, aires

Respiración, Inhalación.​

This problem has been solved! see the answeran evil king has a cellar containing n bottles of expensive wine, and his guards have just caught a spy trying to poison the king’s wine. fortunately, the guards caught the spy after he succeeded in poisoning only one bottle. unfortunately, they don’t know which one. to make matters worse, the poison the spy used was very deadly; just one drop diluted even a billion to one will still kill someone. even so, the poison works slowly; it takes a full month for the person to die. design a scheme that allows the evil king to determine exactly which one of his wine bottles was poisoned in just one month’s time while expending at most o(logn) of his taste testers. you are not allowed to use n testers, one for each bottle, and see which one tester dies after 30 days.(no pseudocode needed. just answer in words, how many testers you use, how you use them and why you correctly identify the poisoned bottle in 30 days)

Which of the following would you find on a network

Fill in the following program so that it will correctly calculate the price of the orange juice the user is buying based on the buy one get one sale.#include //main functionint main() { int cartons; float price, total; //prompt user for input information printf("what is the cost of one container of oj in dollars? \n"); scanf(" [ select ] ["%d", "%c", "%f", "%lf"] ", & price); printf("how many containers are you buying? \n"); scanf(" [ select ] ["%d", "%c", "%f", "%lf"] ", & cartons); if ( [ select ] ["cartons / 2", "cartons % 1", "cartons % 2", "cartons % price", "cartons / price", "cartons / total"] [ select ] ["=", "==", "! =", "< =", "> =", "< "] 0) total = [ select ] ["price * cartons", "cartons * price / 2 + price", "(cartons / 2) * price", "cartons / (2.0 * price)", "(cartons / 2.0) * price + price", "((cartons / 2) * price) + price"] ; else total = ((cartons / 2) * price) + price; printf("the total cost is $%.2f.\n", total); return 0; }