Please help with correct answer for question number 1 and 2.

The Current Length of the Project is A weeks. Determine the costs to shorten the project. 1 st Week Cost = A/ Hint: Between 10 and 13 and Week Cost = A 3 d Week Cost = A/ 4th Week Cost = A/ Hint: Between 16 and 21 5th Week Cost = A 6th Week Cost = A/ Hint: Between 30 and 34 Note: For each week you are to include only the cost of shortening the activities in that particular week.Question 1 (4 points) ff 'I' '1' i 5 J * 7 1- . r E" - . ,w L. 'U see 5731 I\": ,-A- ' T u 1 la 2 W) ..'f_" K 7__d__.1\""""\\__.m' a 4 ,1. j J: 3 r\" F7 (f/ x_J 7 . ,1} f' 3.; 3 (fix) \\_1 i | ', 7/- \\ Lw /'/ e...\" I\".,. hf :'f T A "_' I i___ I ' 4 f 1 1- \"'5' 1 j '1 .[1 9 j H 12 1- f 13 H- \".r '74 H"! 1'; ' ' \"AN ..-4 _A [4' 3 4' "x a" 4' r' L 4 lri 6 35 \"3 i \\ f." ,5 '- ' / 4 I\" ,. t '\"-' L/ \"_f The Department of Transportation has found that a proposed highway project is not going to be completed before the end of the summer. They have asked the contractor to prepare cost figures to shorten the project by up to 6 weeks. The following table shows the Crashing Costs for each activity. For example. if you needed to shorten Activity 1-2 by 1 week this would cost $12. If you heeded to shorten the same activity by 2 weeks it would cost 816. Note: If an activity lists a cost of Sr". then this activity cannot be any shorter, Activity First Week Third Week 1.2 512 816 S m 2.5 6 8 10 .7 7 8 11 7-11 7 11 m 1-? 12 14 19 3-8 5 --- 8-11 12 1% 14 3-9 14 16 16 12-13 7 11 m 1-4 10 14 18 4-6 12 14 18 6-10 11 12 m 10-13. 14 1% .-. Question 2 (4 points) TIMES (@151 Finish OPTIMISTIC MOST LIKELY PESSIMISTIC t11] Given the Activity on Arrow Diagram and the data found in the table above. calculate each of the following values: Activity A ......... tE = 35 Standard Deviation = 3 Activity B ......... te = A'V Standard Deviation = A\\/ Activity C ......... te = Q/ Standard Deviation = Q/ Activity D ......... tE = A"; Standard Deviation = 1%; Activity E ......... t6 = A\\/ Standard Deviation = 7 Path125 Mean Time : A\\/ Standard Deviation =5 Path1345 Mean Time= ' l 6/ Standard Deviation = l l A'/ Question 3 (2 points) : EQ / 0. 0 Given a simple Project like the one pictured above. Optimistic / Most Likely/ Pessimist Times were assessed and Path Means and Path Standard Deviations were calculated. For Path 1-25 we have a Path Mean = 34 and a Path Standard Deviation = 5. For Path 1-3-4-5 we have a Path Mean = 35 and a Path Standard Deviation = 4. Determine both the Probability that Path 12-5 will finish in 35 Days or Less and the Probability that the Project will finish in 35 Days or Less. EFT. 0977249938 : 0965303611 \\_/ if). 0964069734 : 0942137088 \\1 (7:: 0579259687 : 0289629843 \\_, (:31 0986096601 : 0979973257