Use Algorithm 7 to schedule the largest number of talks in a lecture hall from a proposed set of talks, if the starting and ending times of the talks are 9:00 a. m. and 9:45 a. m.; 9:30 a. m. and 10:00 a. m.; 9:50 a. m. and 10:15 a. m.; 10:00 a. m. and 10:30 a. m.; 10:10 a. m. and 10:25 a. m.; 10:30 a. m. and 10:55 a. m.; 10:15 a. m. and 10:45 a. m.; 10:30 a. m. and 11:00 a. m.; 10:45 a. m. and 11:30 a. m.; 10:55 a. m. and 11:25 a. m.; 11:00 a. m. and 11:15 a. m.

If we arrange the talks from the lowest starting time to the highest ending time we get total of 11 talks.

using algorithm 7 we get answer   (1) - (3) - (6) - (9)   the largest number of talks scheduled.

Step-by-step explanation:

arranging the talks from lowest starting time to the highest ending time.

thus,

we start from the earliest time as

9:00 a.m. and 9:45 a.m which is (1).

After the talk is finished we pick the nearest time for another talk which starts at

9:50 a.m. and 10:15 a.m which is (3).

After this talk we again pick the nearest time for another talk which becomes

10:30 a.m. and 10:55 a.m which is (6).

and lastly

10:45 a.m. and 11:30 a.m which is (9).

Note: we didn't choose other times because we cannot talk at 2 or 3 places at the same time. so we pick another when one talk is finished.

thus the answer is (1) - (3) - (6) - (9) as the largest number of talks scheduled.

If we arrange the talks from the lowest starting time to the highest ending time we get total of 11 talks.

using algorithm 7 we get answer   (1) - (3) - (6) - (9)   the largest number of talks scheduled.

Step-by-step explanation:

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.