+10 An automobile is depreciating at 10% per year, every year. A $22,000 car depreciating at this rate can be modeled by the equation V(t) = 22,000(0.90)t. What is an equivalent equation for this vehicle at a daily depreciation and what is it worth (rounded to the nearest thousand dollar) 5 years after purchase?
a) V(t) = 22,000(0.90)365t, $13,000
b) V(t) = 22,000(0.9997)365t, $13,000
c) V(t) = 22,000(1.00026)365t, $35,000
d)V(t) = 22,000(0.9997)t, $22,000
+10 The intensity, Io, of a light source is reduced to I after passing through d meters of fog according to the formula I = Ioe−0.12d. If 1,200 lumens of light are present initially, what is an equivalent expression written as a percentage rate of lumens lost?
Hint: Find the value of e−0.12 on your calculator.
a) I = 1,200ed I = 1,200(0.887)d
b) I = 1,200e0.88d I = 1,200(0.88)d
c)I = 1,200ed I = 1,200(0.887)d
d)I = 1,200e0.88d I = 1,200(0.88)d
+130466 First question.....
We need to find the effective rate of daily depreciation, thusly...
(.90)^t = (r)^(365t) take the log of both sides
log(.90)^t = log(r)^(365t)
(t) log(.90) = (365t)log(r)
(1/365) log(.90) = log(r)
10^[(1/365) log(.90) ] = r = ..9997
So....the correct expression is :
22000(,9997)^(365t) and the value after 5 years = about $13000
So (b) is the correct answer
