Consider the following work breakdown structure: Time Estimates (days) Activity Precedes Optimistic Most Likely Pessimistic Start A, B - - - A C, D 44 50 56 B D 45 60 75 C E 42 45 48 D F 31 40 49 E F 27 36 39 F End 58 70 82 What is the probability that the critical path for this project will be completed within 210 days

the probability that the critical path for this project will be completed within 210 days is 0.99725

Step-by-step explanation:

Given that;

Work break down is;  

 Time Estimates (days)

Activity  Precedes  Optimistic  Most Likely  Pessimistic  

Start          A,B              -                    -                   -  

A               C,D             44                50                 56  

B               D                45                60                  75  

C               E                42                45                   48  

D               F                31                 40                   49  

E               F                27                36                   39  

F              End            58                70                   82

we know that;

Expected Time = ( Optimistic Time + ( 4 × most likely time) + pessimistic time) / 6

and

Variance = [( pessimistic time - Optimistic time) / 6 ]²

so

ACTIVITY              EXPECTED TIME                                 VARIANCE

A                   (44 + (4 × 50) + 56) / 6 = 50                    ((56 - 44) / 6)² = 4

B                   (45 + (4 × 60) + 75) / 6 = 60                    ((75 - 45) / 6)² = 25

C                  (42 + (4 × 45) + 48) / 6 = 45                    ((48 - 42) / 6)² = 1

D                   (31 + (4 * 40) + 49) / 6 = 40                     ((49 - 31) / 6)² = 9

E                   (27 + (4 * 36) + 39) / 6 = 35                     ((39 - 27) / 6)² = 4

F                   (58 + (4 * 70) + 82) / 6 = 70                     ((82 - 58) / 6)² = 16

ACTIVITY     DURATION      ES        EF        LS        LF        SLACK

A                   50                      0        50         0         50            0

B                   60                      0        60        30        90           30

C                   45                     50       95        50        95            0

D                   40                     60      100        90       130          30

E                    35                    95       130        95       130           0

F                    70                    130     200       130      200           0

FORWARD PASS:

ES = MAXIMUM EF OF ALL PREDECESSOR ACTIVITIES; 0 IF NO PREDECESSORS ARE PRESENT.

EF = ES + DURATION OF THE ACTIVITY.

CRITICAL PATH = PATH WITH THE LONGEST COMBINED DURATION VALUE AND 0 SLACK.

CRITICAL PATH = ACEF

DURATION OF PROJECT = 200

Now the probability

VARIANCE OF CRITICAL PATH = SIGMA(VARIANCE OF THE CRITICAL PATH) = 25

STANDARD DEVIATION OF CRITICAL PATH = SQRT(VARIANCE OF CRITICAL PATH) = 5

EXPECTED TIME = DURATION OF THE PROJECT = CRITICAL PATH = 200

DUE TIME = 210

Z = (DUE TIME - EXPECTED TIME) / STANDARD DEVIATION OF CRITICAL PATH)

Z = (210 - 200) / 5 = 2

Now form the Normal Distribution Table;

Z = 0.99725

Therefore, the probability that the critical path for this project will be completed within 210 days is 0.99725

step-by-step explanation:

12475909

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!