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Computers and Technology, 21.06.2019 21:00, kitkatwolf
Analyze the following code. int x = 1; while (0 < x) & & (x < 100) system. out. println(x++); a. the loop runs forever. b. the code does not compile because the loop body is not in the braces. c. the code does not compile because (0 < x) & & (x < 100) is not enclosed in a pair of parentheses. d. the numbers 1 to 99 are displayed. e. the numbers 2 to 100 are displayed.
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Computers and Technology, 22.06.2019 10:00, kimmmmmmy333
According to alisa miller foreign news bureaus
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Computers and Technology, 23.06.2019 03:10, nxusasmangaliso8780
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; }
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Computers and Technology, 23.06.2019 09:00, amberpublow7
Which best describes the role or restriction enzymes in the analysis of edna a. to break dna into fragments that vary in size so they can be sorted and analyzed b. to amplify small amounts of dna and generate large amounts of dna for analysis c. to purify samples of dna obtained from the environment so they can be analyzed d. to sort different sizes of dna fragments into a banding pattern that can be analyzed
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