And now find the number of instructions executed in this time for each processor. make sure you use enough digits of significance to uncover any difference; don’t let rounding make it look like they have run the same number if in fact they have not. p1: 4.589 * 10^10 instr p2: 4.663 * 10^10 instr for the next 2 parts, we discover we can reduce the execution time for each processor by 18% from part (b) above but it leads to an increase of 11% in the cpi for each processor. (4 pts) what is the new execution time for both processors? always show your formulas.

In deadlock avoidance using banker’s algorithm, what would be the consequence(s) of: (i) a process declaring its maximum need as maximum possible for each resource. in other words, if a resource a has 5 instances, then each process declares its maximum need as 5. (ii) a process declaring its minimum needs as maximum needs. for example, a process may need 2-5 instances of resource a. but it declares its maximum need as 2.

For a typical middle-income family, what is the estimated cost of raising a child to the age of 18? $145,500 $245,340 $304,340 $455,500

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; }