Biology: Apopulation of naked mole rats begins with 250 individuals in 2015. the population has a carrying capacity of 1,500 individuals. if the population has an intrinsic rate of increase equal to 0.8, how large is the population in 2016? show me step by step, you!

Apopulation of naked mole rats begins with 250

billy1123

30.05.2020

dN/dT = rK (K-N)/K

Here, K is carrying capacity

N = Population size

r = Intrinsic rate

dN/dT = 0.8*250 (1500-250)/1500

dN/dT = 166.66 that is equivalent to 167 animal in a year

Thus, in 2015, there would be 250 individuals and in 2016, there would be 260+167 = 417 animals.

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melisa141b

01.09.2021

$\frac{13.6 g}{cm^3} * \frac{1 Kg}{1000g} * \frac{10^6 cm^3}{m^3} = \frac{13600 Kg}{m^3}$

Explanation:

The process of conversion of 13.6 g/cm3 to kg/m3

1 Kg of weight is equal to 1000 grams of weight

Thus, it can be written as

$1 Kg = 1000g\\1 g = \frac{1 Kg}{1000}$

So the first blank (numerator) will be filled with 1 Kg

Now, 1 meter is equal to 100 centimeter

Thus,

$1 m^3 = (100)^3 cm^3\\1 cm^3 = \frac{1 m^3}{10^6}$

So, the second blank (numerator) will be filled with $10^6\\$

And the third blank (denominator) will be filled with 1 m^3

And the last blank will be 13600

The complete equation will be as follows

$\frac{13.6 g}{cm^3} * \frac{1 Kg}{1000g} * \frac{10^6 cm^3}{m^3} = \frac{13600 Kg}{m^3}$

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