Answer:
Step-by-step explanation:
After one year
A=p(1+r/n)^nt
=2000(1+0.03/12)^12*1
=2000(1+0.0025)^12
=2000(1.0025)^12
=2000(1.0304)
=$2060.8
After two-years
A=p(1+r/n)^nt
=2060.8(1+0.03/12)^12*2
=2060.8(1+0.0025)^24
=2060.8(1.0025)^24
=2060.8(1.0618)
=$2188.157
After three years
A=p(1+r/n)^nt
=2188.157(1+0.03/12)^12*3
=2188.157(1+0.0025)^36
=2188.157(1.0025)^36
=2188.157(1.0941)
=$2394.063
Answer:
Option C. $6,012
Step-by-step explanation:
we know that
The formula to calculate the depreciated value is equal to
where
V is the the depreciated value
P is the original value
r is the rate of depreciation in decimal
t is Number of Time Periods
in this problem we have
t = 7 years
P = $8,000
r = 0.04
substitute in the formula above
Hope this helps :)
<span>All the information we have are the probabilities, and what we need is the lowest number: so let's choose the smallest probability among the numbers: 0.0065%, B 0.0037%,C 0.0108%,D 0.0029%, E 0.0145%. The smallest of the numbers is 0.0029% -it starts with two 00s and the number that follows, 2, is smaller than all there others - so the smallest probability is in option D - and the model would be the corresponding model (but we're missing some information here) </span>
The answer to this question is 2
