WHERE IS THE MATH PROBLEM
The average decrease in value can be solve by first solving the value at year 1 and year 2
and i think the equation is <span>f(x)=10,000(0.73)^x</span>
at x = 1
<span>f(x) = 10,000(0.73)^x
</span><span>f(1) = 10,000(0.73)^(1)
f(1) = 7300
at x = 2
</span><span>f(x) = 10,000(0.73)^x
</span><span>f(2) = 10,000(0.73)^(2)
f(2) = 5329
so the average decrease = ( 7300 - 5329) = $ 1971 per year</span><span>
</span>
So basically your gonna regaurd 63.5 as 60 +3.5, 60 is 5*12 and the remaining 3.5 is less than the divisor 5, so it would be the remainder
anwser is 63.5/5 treat as 60/5+3.5/5
so anwser is 60/5+3.5/5=0.7=12.7
