(a) The original pairs give r=0.547; the altered pairs give r=0.984. Sorting makes a moderate positive association appear very strong.

Keep the original pairings for the first calculation and use y-values 15,20,23,27,28,31 for the second.

(b) The site with x=2 actually has y=23, not 15; the site with x=12 has y=27, not 31. Correlation depends on the pairs, not just the separate lists. Sort complete paired records so each site's x and y remain together.

The marginal values are unchanged, but the measured relationship has been replaced.

(c) The observations show association and do not establish causation. Site size or staffing could influence both variables. Randomly allocate comparable sites to different training-hour schedules and compare outcomes while controlling other conditions.

Accept another plausible confounder and a defensible controlled comparison.

(a) frac12(((300+k) / (4))+((316+k) / (4)))=112, so 616+2k=896 and k=140. The component means are 110 and 114, whose average is 112.

Set up the given centred average before solving for the original count.

(b) At t=5, average 118 and 119 to obtain 118.5. At t=6, average 119 and 123 to obtain 121. A centred value at t=7 would also require quarter 9, which is unavailable.

Adjacent four-quarter means are centred halfway between their own time positions.

(c) The centred values rise from 112 at quarter 3 to 121 at quarter 6, supporting an upward underlying movement. Only two annual cycles are observed, so repeated seasonal effects cannot yet be distinguished confidently from irregular changes.

Use numerical smoothed evidence and the limited number of repeated seasons.

(a) A_(n+1)=1.0055A_n+450, A_0=3000. Interest is applied before the end-of-month deposit.

The multiplier acts on the existing balance only.

(b) A_(36)approx$21515.92. Monthly deposits total 36(450)=$16200. Including the initial $3000, total contributions are $19200, so interest earned is 21515.92-19200=$2315.92.

Iterate without premature rounding, then separate all contributed principal from growth.

(c) A_(42)approx$24973.18<$25000<A_(43)approx$25560.53, so the first month is 43.

Compare consecutive recurrence-table entries around the target.

(d) The model is C_(n+1)=1.0055(C_n+450). It gives C_(36)approx$21614.15, about $98.24 more because each deposit earns one additional month of interest.

Move the deposit inside the interest multiplication.

(a) At A, inflow and outflow are both 9; at B, both are 6; at C, both are 7. At D, inflow is 4+4=8 but outflow is 7. One unit per hour would accumulate at D, contrary to the no-storage condition.

Capacity compliance alone does not ensure conservation.

(b) Set AC=6, AD=3, CT=8. Then A sends 6+3=9, C sends 6+2=8, and D sends 3+4=7. B remains balanced at 6. The three changed flows are within capacities 6, 4 and 8; all other capacities remain respected. The total flow is 15.

Divert one unit from A-D to A-C-T without altering either source flow.

(c) The maximum flow is 15. The cut separating T from all other vertices contains Cto T and Dto T, with capacity 8+7=15. This equals the feasible flow, proving optimality.

Match a feasible flow to an upper bound supplied by a cut.

(a) A finishes at day 4 and B at day 5. C runs from day 4 to day 10; D runs from day 5 to day 8. E runs from day 10 to day 14, so the project finishes at day 14.

A forward scan gives the precedence-only schedule.

(b) C before D: C runs 4-10, D runs 10-13, and E finishes at 17. D before C: D runs 5-8, C runs 8-14, and E finishes at 18. The minimum is 17 days. Continuous, non-overlapping activities C and D must occur in one of these two orders; idle time cannot improve either finish.

Apply both predecessor readiness and crew availability.

(c) Yes. Let the first crew perform C from 4 to 10, and the second perform D from 6 to 9. A is complete before C starts; both A and B are complete before D starts. E can run from 10 to 14. The dependent chain A-C-E takes 4+6+4=14 days, so no schedule can finish earlier.

Use the stated availability interval without shortening any activity.

(a) The total is 2+6+1+4=13 technician-hours. All jobs finish after max(2,6,1,4)=6 hours. Concurrent jobs contribute separately to total labour, but elapsed completion time is determined by the longest assigned job.

Distinguish a sum from a maximum.

(b) Use A2,B3,C1,D4, taking 2,3,5,4 hours, so all finish after 5 hours. Job 1 takes at least 5 hours with any technician; this lower bound proves that 5 hours is optimal.

Construct a feasible one-to-one assignment and use the unavoidable duration of job 1.

(c) No. Under the five-hour deadline, A must do job 2, C must do job 1, and D must do job 4, leaving job 3 for B. This forces total labour 2+3+5+4=14, exceeding the 13-hour limit.

A minimum-total assignment need not minimise elapsed completion time.

(a) A: 6.0411%; B: 5.8000%; C: 5.8439%.

Use the compounding frequency for each nominal rate.

(b) A, then C, then B. Nominal rates use different compounding frequencies and are not directly comparable.

Compare rates on a common effective basis.

(c) A: $22489.42; B: $22387.28; C: $22405.85.

Use each plan's actual period rate and number of periods.

(d) Plan B then grows $19920 to about $22297.73. Plan A remains best at $22489.42.

Apply the fee before compounding and compare final values.

(a) For L_(all), the absolute errors are 1.85, 2.70 and 3.55, giving a mean of 2.70. For L_(recent), they are 0.75, 0.30 and 0.15, giving a mean of 0.40. Select L_(recent).

The comparison uses observations outside both fitting windows, not in-sample fit.

(b) L_(recent)(18)=32.4+7.45(18)=166.50.

Substitute month 18 into the selected linear trend.

(c) 166.50(1.08)=179.82. Capacity 170 is insufficient by 9.82, so another arrangement is required.

Restore the seasonal effect before comparing with capacity. Allow correct follow-through from part (b).

(d) The recent line was fitted to only six observations, and month 18 lies beyond both the fitting and checking periods. For example, compare each new actual demand with its reseasonalised forecast and review / refit after absolute error exceeds 10 units for three consecutive months. Repeated large errors would suggest the model no longer supports the capacity decision. Other defensible rules are acceptable; the numbers 10 and three are examples only.

Credit any two valid limitations and any clearly defined, justified rule. A numerical threshold is not mandatory.

(a) The total is 4+3+9+2+6+4=28 km. Each of A, B, C and D has degree 3, so all four are odd. A closed Eulerian trail requires every degree to be even.

Count each physical road once and apply the closed-trail condition.

(b) Pair AB with CD: 4+4=8. Pair AC with BD: 3+6=9. Pair AD with BC: 7+2=9, since A-C-D is shorter than direct AD. The minimum extra distance is 8 km.

Compare path distances rather than automatically taking the direct road.

(c) One route is A-B-C-A-D-C-D-B-A. It uses AB and CD twice and every other road once, for 28+8=36 km. Other valid Eulerian routes in this augmented network are acceptable.

Demonstrate that the lower bound from the pairing comparison is attainable.

(d) Only B and C need their degrees changed to even. Repeat BC, the shortest B-C path of 2 km; A and D remain odd endpoints of a semi-Eulerian trail. The minimum total becomes 28+2=30 km.

An open trail has exactly its two endpoints odd.