SOLUTION: The number of long distance phone calls between two cities in a certain time period varies directly as the populations p1 and p2 of the cities, and inversely as the distance betwee

Question 341947: The number of long distance phone calls between two cities in a certain time period varies directly as the populations p1 and p2 of the cities, and inversely as the distance between them. If 10,000 calls are made between two cities 500 mi apart, having populations of 50,000 and 125,000, find the number of calls between two cities 800 mi apart, having populations of 20,000 and 80,000.

Please help me solve this. It was in the variations portion of my textbook. Not sure where to begin. Thanks!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The number of long distance phone calls between two cities in a certain time period varies directly as the populations p1 and p2 of the cities, and inversely as the distance between them.
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n = k[(p1*p2)/[distance]
If 10,000 calls are made between two cities 500 mi apart, having populations of 50,000 and 125,000,
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Solve for "k": the constant of proportionality
10,000 = k[50000*12000]/500
10,000= k[1200000]
k = 0.0083
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Equation:
n = 0.0083[P1*p2/distance]
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find the number of calls between two cities 800 mi apart, having populations of 20,000 and 80,000.
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n = 0.083[20000*80000/800]
n = 0.083[2000000]
n = 16666 2/3
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Rounding down: n = 16,666 calls
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Cheers,
Stan H.
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