Answer:
The correct answer is E.
Step-by-step explanation:
First I look for the relationships between the coefficients. The first term corresponds to the base value.
[tex]\frac{8*d}{h^{2}} = \frac{8}{1} * \frac{d}{h^{2} }[/tex]
The second term corresponds to the value obtained when the density of the underlying material is doubled and the daily usage of the equipment is halved.
[tex]\frac{2*8*d}{(0,5*h)^{2}} = \frac{16*d}{0,25 * h^{2}} = \frac{16}{0,25} * \frac{d}{h^{2} }[/tex]
With the I get the ratio of coefficients of [tex]\frac{d}{h^{2}}[/tex] :
[tex]\frac{8}{1}[/tex] = 8
[tex]\frac{16}{0,25}[/tex] = 64
Now I calculate the percentage increase in the useful life of the equipment as:
% = [tex]\frac{64}{8}[/tex] * %100 = %800
