At noon, ship A is 130 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 20 km/h. How fast is
the distance between the ships changing at 4:00 PM?
2 answers:
Answer:
36.1 km/h
Step-by-step explanation:
rate at which the distance is changing is given by relative velocity(vR)
Look at the diagram. THe diagram shows relative motion. From relative motion diagram, velocity vector diagram is drawn.
From velocity vector diagram, it can be seen that relative velocity or vR can be calculated using pythagoras theorem.
vR²= vA² + vB²
vR²= 30²+ 20²
vR= 36.1 km/h
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Answer:
n = 66.564
Step-by-step explanation:
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Where:
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S = standard deviation.
E = error range.
2. We will find each of these three data and replace them in the formula.
"z" theoretically is a value that measures how many standard deviations an element has to the mean. For each confidence level there is an associated z value. In the question, this level is 99%, which is equivalent to a z value of 2.58. To find this figure it is not necessary to follow any mathematical procedure, it is enough to make use of a z-score table, which shows the values for any confidence interval.
The standard deviation is already provided by the question, it is S = 100.
Finally, "E" is the acceptable limit of sampling error. In the example, we can find this data. Let us note that in the end it says that the director wishes to estimate the mean number of admissions to within 1 admission, this means that she is willing to tolerate a miscalculation of just 1 admission.
Once this data is identified, we replace in the formula:
3. The corresponding mathematical operations are developed:
n= 66.564