1) Calculate the fossa population density in the Midongy-Sud National Park.

2) If slash and burn deforestation cleared 20% of the forest of its deciduous trees. How many km2 were destroyed?

3) If the population density of the fossa decreased by 20% due to the fragmented forests. How many fossa would you expect to find in a 60 km2 area?

1. Population density is 0.26 Fossa/sq km

2. 384.4 sq km of the 1920 sq km was destroyed in the fire

3. 13 Fossas

Explanation:

Fossa are catlike animals and are the largest mammalian carnivores  found in the island area of Madagascar.

It is estimated that there are at least 500 adults Fossa in the Midongy-Sud National Park; although this is unconfirmed & thought to be an overestimation

The Midongy-Sud National Park is a protected area & has a land area of 1922 square kilometers (1922 sq km)

1. Population density is a mathematical measure of the population of living organisms per land area. Mathematically represented as:

Population density = Number of Fossa ÷ Area of Midongy-Sud National Park

Number of Fossa = 500, Area of Midongy-Sud National Park = 1922 sq km

Population density = 500 ÷ 1922

Population density = 0.26 fossa/sq km

2. 20% of the forest area = 0.2 * 1922 = 384.4 sq km

This would imply that if slash and burn deforestation cleared 20% of the Midongy-Sud National Park area, that means 384.4 sq km of the 1920 sq km was destroyed in the fire

3. If Fossa population density reduces by 20%, it implies that only 80% (100% - 20%) of the initial population density is left

Population density = 80% * 0.26 = 0.208 Fossa/sq km

To get the population of Fossa in a 60 sq km, we will multiply the population density obtained above by the land area.

Mathematically,

Population = Population density * Land Area

Population = 0.208 * 60 = 12.48 ≈ 12.5 ≈ 13 Fossas