T ime
V alue
M oney
-----------------------
press APPS ---> ENTER ----> ENTER to get to TVM solver menu
N= number of times interest is compounded(#)
I% = interest rate as APR(%)
PV = present value($)
PMT = payment per period ($)
FV = future value ($)
P/Y = payments per year (#)
C/Y = compoundings per year (#)
ex. I want to buy an 04' corvette.
It costs $33 995 plus PST on book value
($33 995 x 1.07 = $36374.65)
I have $2 000.00 to make a down payment.
I need to take out a $34 374.65 loan (obvisously) at an interest rate of 7%
and is to be paid off in 5 years with monthly payments.
I want to calculate my monthly payment.
so I enter this on the TVM Solver
N = 5 x 12=60
I% = 7
PV = 34 374.65
PMT = (press ALPHA ---> ENTER) answer should be -680.66
FV= 0
P/Y = 12
C/Y= 12
---highlight END
Wednesday, October 7, 2009
Tuesday, October 6, 2009
Question 15
15. On Murky Island there is a bear population of 400. Every year, 12% of the bear population dies. Each spring an average of 56 bear cubs are born.
a)determine the bear pop ulation over the next 5 years. Show your work.
year 0 - 400 bears x 88% survival rate(12% mortality rate) + 56 bears born year 1 - 408
year 2 - 415
year 3 - 421
year4 - 426
year5 - 431
b) What will the bear population be after it stabilizes?
The bear population will be at 465, conservatively when it stabilzes.
c) Construct a population graph of the sequence extending the x-axis to include the first 50 years. Lbel the graph.
d) Devise a plan to stabilize the bear population between 400 and 450 bears (inclusive) by the 50 th year using the following parameters:
-he intial bear population is 400
-he birth rate remains at 56 bear cubs per year
the death rate remains at 12% of the population each year
you must cath and transport bears off the island each year
you must stop recreational fishing which results in an increase of up to a maximum of 10 more bears each year.
State the stblized bear population. Show your work to explain how you arrived at your answer.
400 bears x 88% survival rate + 56 bears born + 10 due to fishing restrictions - 16 bears are transported off the island = (400*.88+50).
The stabilized bear population would be at about 415 animals using this formula (400*.88+50)Monday, October 5, 2009
question number 11
I've gotten a little bit behind on my blog posts ,so i am posting the solutions to two sequences questions from the handout (#11 and 15).
11. Scientists have determined that 8 parts per million (ppm) of oil in water is an acceptable level. After an oil spill, the level of oil in a spill zone rises to 900ppm. Its is known that each week the spilled oil is reduced by 40%.
a) How many weeks will it take for the oil to be at an acceptable level?
week 0- 900ppm of oil in water x 60% retention rate of oil in water (40% depletion rate)
week 1 - 540 x .6
week 2 - 324 x .6
week 3 - 194.4
week 4 - 116.64
week 5 - 69.984
week 6 - 41.99
week 7 - 25.194
week 8 - 15.117
week 9 - 9.07
week 10 - 5.442 ppm of oil
It would take 10 weeks for the oil to be at an acceptable level.
* I came about this answer by entering the formula 900* .6 on the homescreen of my calculator and pressing enter consecutively
b)Chemicals could be added to the spill zone at the beginning of each week to speed up the process. There are two treatments available:
Treatment 1 oil is reduced by an additional 50 ppm at a cost of $1 million per treatment
Treatment2 oil is reduced by an additional 10 ppm at a cost of $550 000 per treatment
Compare treatment 1 and 2 in order to recommend a treatment, if any. Analyze the cost and time required for each treatment. Support your answer by providing cost and time.
Treatment 1 Treatment 2
week 0 - 900 x 54.4% ( 45.6 removal rate) week 0 - 900 x 58.9% (41.1 % removal rate)
week 1 - 489.6 week 1 - 530
week 2- 266.342 week 2 - 312.229
week 3 - 144.89 week 3 - 183.903
week4 - 78.82 week 4 - 108.319
week 5 - 42.878 week 5 - 63.8
week 6 - 23.326 week 6 - 37.578
week 7 - 12.689 week 7 - 22.664
week 8 - 6.903 week 8- 13.037
week 9 - 7.679
I would recommend neither treatment, on account that one would have to double any of each treatment to have any greater effect on the amount of time that it takes for the oil to dissipate from the water, and doubling either of the treatments would have a tremendously significant dollar value.
11. Scientists have determined that 8 parts per million (ppm) of oil in water is an acceptable level. After an oil spill, the level of oil in a spill zone rises to 900ppm. Its is known that each week the spilled oil is reduced by 40%.
a) How many weeks will it take for the oil to be at an acceptable level?
week 0- 900ppm of oil in water x 60% retention rate of oil in water (40% depletion rate)
week 1 - 540 x .6
week 2 - 324 x .6
week 3 - 194.4
week 4 - 116.64
week 5 - 69.984
week 6 - 41.99
week 7 - 25.194
week 8 - 15.117
week 9 - 9.07
week 10 - 5.442 ppm of oil
It would take 10 weeks for the oil to be at an acceptable level.
* I came about this answer by entering the formula 900* .6 on the homescreen of my calculator and pressing enter consecutively
b)Chemicals could be added to the spill zone at the beginning of each week to speed up the process. There are two treatments available:
Treatment 1 oil is reduced by an additional 50 ppm at a cost of $1 million per treatment
Treatment2 oil is reduced by an additional 10 ppm at a cost of $550 000 per treatment
Compare treatment 1 and 2 in order to recommend a treatment, if any. Analyze the cost and time required for each treatment. Support your answer by providing cost and time.
Treatment 1 Treatment 2
week 0 - 900 x 54.4% ( 45.6 removal rate) week 0 - 900 x 58.9% (41.1 % removal rate)
week 1 - 489.6 week 1 - 530
week 2- 266.342 week 2 - 312.229
week 3 - 144.89 week 3 - 183.903
week4 - 78.82 week 4 - 108.319
week 5 - 42.878 week 5 - 63.8
week 6 - 23.326 week 6 - 37.578
week 7 - 12.689 week 7 - 22.664
week 8 - 6.903 week 8- 13.037
week 9 - 7.679
I would recommend neither treatment, on account that one would have to double any of each treatment to have any greater effect on the amount of time that it takes for the oil to dissipate from the water, and doubling either of the treatments would have a tremendously significant dollar value.
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