1.75
4 x 1.75 = 7 so that is the answer
Answer:
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,
- 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.
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.
The answer would be <QRZ
Since you are looking for an angle congruent to <UQR using the alternate interior angles theorem, interior suggests that the angle must be inside the parallel lines, se we can get rid of options <WRT and <TRZ since they are exterior angles
Furthermore, in the alternate interior angles theorem, the two angles must be alternate or opposite of each other which would show that the only possible answer would be <QRZ
hope this helped!