1. An electronic games manufacturer producing a new product estimates the annual
ID: 2866782 • Letter: 1
Question
1. An electronic games manufacturer producing a new product estimates the annual sales to be 9000 units. Each year 4% of the units that have been sold will become inoperative. So, 9000 units will be in use after 1 year, [9000 + 0.96(9000)] units will be in use after 2 years, and so on. How many units will be in use after n years?
2. A company buys a machine for $275,000 that depreciates at a rate of 30% per year. Find a formula for the value of the machine after n years. V(n) =?
What is its value(in $) after 4 years? (Round your answer to two decimal places.)
Explanation / Answer
HOPE YOU MEAN THE FOLLOWING
SAY YEAR 1 HE PRODUCED & SOLD 9000 AS GIVEN ..SO IN THE SECOND YEAR HE PRODUCES AGAIN 9000 & SELLS THEM. SINCE,OUT OF THE 9000 SOLD IN YEAR 1 , 4% GOES BAD , WE SHALL HAVE IN ALL 9000 NEW+9000*0.96 IN YEAR 2 . ASSUMING SO ...WE GET ...
So,...U[1]= 9000 units will be in use after 1 year,
U [2] = [9000 + 0.96(9000)] units will be in use after 2 years= 900[1+0.96]
U[3]=[9000+0.96(9000)+(0.96^2)*9000 ] = 9000[1+0.96+0.96^2]
How many units will be in use after n years?
U[N]=9000[1+0.96+0.96^2+..............0.96^(N-1)] = 9000* [ 1 - 0.96^N ] / [1 - 0.96 ] .......
U[N] = 225000*[1 - 0.96^ N] ..........................ANSWER
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A company buys a machine for $275,000 that depreciates at a rate of 30% per year. Find a formula for the value of the machine after n years. V(n) =?
ORIGINAL PRICE = 275000......AFTER 1 YEAR PRICE= 275000*(1-0.3)=275000*(0.7^1)
AFTER N YEARS ...PRICE = 275000*(0.7^N) ...............................
formula for the value of the machine after n years. V(n) =275000*(0.7^N)............................ANSWER
What is its value(in $) after 4 years? (Round your answer to two decimal places.)
PUTTING N=4 WE GET ......V[4] = $ 66027.50 ....................ANSWER
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