2. Suppose you put $100 into a savings account paying 2.5 interest each year com
ID: 2911144 • Letter: 2
Question
2. Suppose you put $100 into a savings account paying 2.5 interest each year compounded annually) (o) Wihat percean wllr 3 yqun. (c) One student solved the problem this way: 100 x 1.025 x 1.025 x 1.025 107.69. 107.69 100 an why 7.69. So the answer is 7.69%. Explain this student's work. What does 7.69% represent in this problem 3. The pesticide DDT was used in the US and later banned. The half-life of DDT is about 15 years. sample is 100 grams. (b) According to your model, how much of the initial sample will remain after 60 years? (e) How many years will it take for the sample to decay to 1 gram?Explanation / Answer
2)
(a)
amount in account after t years =100*(1+(2.5/100))t
amount in account after 3 years =100*(1+(2.5/100))3
amount in account after 3 years =100*(1+0.025)3
amount in account after 3 years =100*1.0253
amount in account after 3 years =107.6890625
percent you earn = 100*(final amount -initial amount)/initial amount
percent you earn = 100*(107.6890625 -100)/100
percent you earn = 7.6890625
percent you earn = 7.69 approximately
(b)
answer is not 7.5% because it doesnot take into account the interest earned on the interest of first and second year.
7.5% is just the simple interest.
here we are asked to calculate compound interest when compounded anually.
(c)
initial amount in account =100
amount in account after first year =100*1.025=102.5
amount in account after second year =102.5*1.025=105.0625
amount in account after third year =105.0625*1.025=107.6890625
total interest earned = 107.6890625-100
total interest earned = 7.6890625 dollars
which is approximately 7.69 % of initial amount of $100
7.69% represent the percentage growth in bank balance of the initial amount of $100
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3)
(a)
A(t)=(A(0))ekt
where A(0) =100 gram is initial amount
at half life ,amount will be half of initial amount
=>A(t)=(1/2)*100
=>100ek*15=(1/2)(100)
=>ek*15=(1/2)
=>15k=ln(1/2)
=>k=(1/15)ln(1/2)
=>k=ln(1/2)(1/15) or ek =(1/2)(1/15)
=>k=-0.046209812 approximately
A(t)=100ekt
A(t)=100e-0.046209812t or A(t)=100*(1/2)(t/15)
(b)
after 60 years, t =60
A(60)=100*(1/2)(60/15)
A(60)=100*(1/2)4
A(60) =6.25
6.25 grams of DDT remain after 60 years
which is 6.25% of initial sample
(c)
A(t)=1 gram
100e-0.046209812t =1
=>e0.046209812t =100
=>0.046209812t =ln(100)
=>t=(1/0.046209812)ln(100)
=>t=99.65784
it takes nearly 100 years for the sample to decay to 1 gram
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