1) y = 5000(1 +.06) 4 2) 5000 = a(1 +.06) 4 3) 5000 = 2000(1 +.06) x 4) 5000 = 2000(1 + r) 4 6312.4...

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Page 1: 1) y = 5000(1 +.06) 4 2) 5000 = A(1 +.06) 4 3) 5000 = 2000(1 +.06) x 4) 5000 = 2000(1 + r) 4 6312.4 5000/(1.06) 4  3960.47 5000/2000  1.06 x Log(2.5)
Page 2: 1) y = 5000(1 +.06) 4 2) 5000 = A(1 +.06) 4 3) 5000 = 2000(1 +.06) x 4) 5000 = 2000(1 + r) 4 6312.4 5000/(1.06) 4  3960.47 5000/2000  1.06 x Log(2.5)

1) y = 5000(1 + .06)4

2) 5000 = A(1 + .06)4

3) 5000 = 2000(1 + .06)x

4) 5000 = 2000(1 + r)4

6312.4

5000/(1.06)4 3960.47

5000/2000 1.06x

Log(2.5) = x log 1.06Log(2.5)/log 1.06 = x 15.73

5/2 = (1 + r)4

)1(2/54 r1.257 = (1 + r) so r = .257 or 25.7%

Variable by itself

Variable times something

Variable is exponent

Variable has exponent

Type in calculator

Divide by everything

Logs

Take nth root

Page 3: 1) y = 5000(1 +.06) 4 2) 5000 = A(1 +.06) 4 3) 5000 = 2000(1 +.06) x 4) 5000 = 2000(1 + r) 4 6312.4 5000/(1.06) 4  3960.47 5000/2000  1.06 x Log(2.5)

The formula to find money compounded is given by the following formula: A = P(1 + r/n)nt where n is the number of times per year, r is the rate, and t is the years, P is the principal. Quarterly: n = 4; bi-weekly: n = 26; semi-monthy: n= 241) Find how much 500$ will be in 10 years if you invest into an account that compounds it monthly at 5%? 2) How long will it take you to double your money invested at 5.35% compounded it quarterly.

3) Find out how much money you need to invest if there will be $1600 after 5 years if compounded weekly at 4%?

4) What is the rate if $3500 becomes 5000$ compounded monthly for 7 years?

Page 4: 1) y = 5000(1 +.06) 4 2) 5000 = A(1 +.06) 4 3) 5000 = 2000(1 +.06) x 4) 5000 = 2000(1 + r) 4 6312.4 5000/(1.06) 4  3960.47 5000/2000  1.06 x Log(2.5)

1) Find how much 500$ will be in 10 years if you invest into an account that compounds it monthly at 5%?

2) How long will it take you to double your money invested at 5.35% compounded it quarterly.

3) Find out how much money you need to invest if there will be after 5 years if 1600$ is compounded weekly at 4%?

4) What is the rate if $3500 becomes 5000$ compounded monthly for 7 years?

Page 5: 1) y = 5000(1 +.06) 4 2) 5000 = A(1 +.06) 4 3) 5000 = 2000(1 +.06) x 4) 5000 = 2000(1 + r) 4 6312.4 5000/(1.06) 4  3960.47 5000/2000  1.06 x Log(2.5)

Remember that compounded continuously is A=Pert

1) How much should you invest now so that you have $160,000 in 18 years if money is compounded 4% continuously?

2) What is the rate if it takes 12 years to double your money compounded continously?

Page 6: 1) y = 5000(1 +.06) 4 2) 5000 = A(1 +.06) 4 3) 5000 = 2000(1 +.06) x 4) 5000 = 2000(1 + r) 4 6312.4 5000/(1.06) 4  3960.47 5000/2000  1.06 x Log(2.5)

Ex) A Snickers candy bar was $.25 in 1980. It was $.85 in 2008. What is the rate of inflation? (0, .25) and (28, .85). Exp regression and you get y=.25(1.047)x. So it’s 4.7% growthOr .85= .25(1 + r)28

3.4 = (1 + r)28

28th root of 3.4 is 1.044 so 4.7%

5) In 1930, the cost of toothpaste was $.05. In 2008, the cost was $1.25. A) What is the rate of inflation?B) Predict when the toothpaste will be $1.60

6) In 1971 the first Starbucks Opened. Now there are 19,555 stores worldwide. A) What is the rate of growthB) Predict how many Starbucks there will be in 2020 if the trend

continues

Page 7: 1) y = 5000(1 +.06) 4 2) 5000 = A(1 +.06) 4 3) 5000 = 2000(1 +.06) x 4) 5000 = 2000(1 + r) 4 6312.4 5000/(1.06) 4  3960.47 5000/2000  1.06 x Log(2.5)

Slides about the APPS….

Page 8: 1) y = 5000(1 +.06) 4 2) 5000 = A(1 +.06) 4 3) 5000 = 2000(1 +.06) x 4) 5000 = 2000(1 + r) 4 6312.4 5000/(1.06) 4  3960.47 5000/2000  1.06 x Log(2.5)

Half Life: y = a(1/2)t/k where y is the final amount, a is the initial amount, t is the time in years and k is the amount of years for it to be ½ of its value. 18)What is the exponential equation for an unknown

substance that has a ½ life of 1000 years?

19) A person’s body’s half life is about 20 years. How long has the body been in the ground if 1/6 of the body is remaining?

20) The ½ life of Carbon-14 is 5760 years. Suppose that a piece of pottery has been in the ground for 5000 years. What % of Carbon-14 is remaining?

21) The ½ life of Carbon-14 is 5760 years. If 1/8 of a sample survives, how long was it in the ground?