Question

The population of a city of 100 people increases at an intrinsic rate of 0.1,but the city has accesses to food and housing for only 200 people what will the current growth rate of the population be?
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Answer 1

Answer:

5/year

Step-by-step explanation:

dN/dT=rN(k-N)/k. k=200 carrying capacity

N=100

r=0.1

dN/dT=0.1*100(200-100)/200

=5

Answer 2

The current growth rate of population be 5.

Given,

Population of city (N) = 100

Intrinsic rate (r)=0.1

capable of carrying (k)=200

To Find,

the current growth rate of the population be

Solution:

$\frac{dN}{dT}$=$\frac{rN(k-N)}{k}$

$\frac{dN}{dT}$=$\frac{0.1*100(200-100)}{200}$

$\frac{dN}{dT}$=5

Hence, the current growth rate of population be 5.

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