Answer:
the first one
Step-by-step explanation:
A graph showing the Earliest Start Times (EST) for project tasks is computed left to right based on the predecessor task durations. For dependent tasks, the earliest start time will be the latest of the finish times of predecessor tasks.
The first graph appears to appropriately represent the table values, using edges to represent task duration, and bubble numbers to represent start times.
The second graph does not appropriately account for duration of predecessor tasks.
The third graph seems to incorrectly compute task completion times (even if you assume that the edge/bubble number swap is acceptable).
9514 1404 393
Answer:
0.06164
Step-by-step explanation:
The effective annual rate obtained by compounding nominal annual rate r monthly is ...
eff rate = (1 +r/12)^12 -1
Then the value of r is ...
r = 12×((eff rate) +1)^(1/12) -1)
For the given effective rate, that is ...
r = 12×(1.06341^(1/12) -1) ≈ 0.06164 . . . . nominal annual interest rate
You can draw a diagram of the situation, representing it by a right triuangle.
In the right triangle the hypotenuse is the distance indicated by the radio signal, 1503 m, and the angle is 41°.
The opposed leg to 41° is the height of the ballon.
Then you can use the sine trigonometric function.
sine (41°) = opposed leg / hypotenuse = x / 1503 m => x = 1503m * sine(41°)
=> x = 986m.
Answer: 986m
"He starts both trains at the same time. Train A returns to its starting point every 12 seconds and Train B returns to its starting point every 9 seconds". Basically, what you need to do is find the least common multiple. The least common multiple of 12 and 9 is 36, so the least amount of time, in seconds, that both trains will arrive at the starting points at the same time is 36 seconds.
