Question

Help please. i really need to pass​

Answer 1

Number of year, when final value is $61,412.50 = 3 year

Number of year, when final value is $20,000 = 10 year (Approx.)

Explanation:

Given:

Depreciation model

y = 100,000[0.85]ˣ

Where x = Number of year

Find:

Number of year, when final value is $61,412.50

Number of year, when final value is $20,000

Computation:

Depreciation model

y = 100,000[0.85]ˣ

61,412.50 = 100,000[0.85]ˣ

0.614125 = [0.85]ˣ

x = 3

Number of year, when final value is $61,412.50 = 3 year

Depreciation model

y = 100,000[0.85]ˣ

20,000 = 100,000[0.85]ˣ

0.2 = [0.85]ˣ

x = 10 (Approx.)

Number of year, when final value is $20,000 = 10 year (Approx.)