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# College Algebra #6

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In this exercise we consider data from the Statistical Abstract of the United States on the fraction of women married for the first time in 1960 whose marriage reached a given anniversary number. The data show that the fraction of women who reached their fifth anniversary was 0.928. After that, for each one-year increase in the anniversary number, the fraction reaching that number drops by about 2%. These data describe constant percentage change, so it is reasonable to model the fraction M as an exponential function of the number n of anniversaries since the fifth.

(a) What is the yearly decay factor for the exponential model?

M = ________

(b) Find an exponential model for M as a function of n. (Let n = 0 represent the fifth anniversary.) \

(c) According to your model, what fraction of women married for the first time in 1960 celebrated their 45th anniversary? (Take n = 40.) Round your answer to three decimal places.

__________

idenny  Mar 17, 2017

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(a) Call the number who married in 1960 = M

And.....5 years later......the number who remained married = .928M

And we have that one year later 98% of these remain married....so.......  0.98 (.928M)  = .90944M

So

.90944M  =  .928M*e^(k * 1)   divide by M

.90944 = .928*e^(k)    divide by .928

.90944/.928  = e^(k)     take the ln of both sides

ln (.90944/.928) =  ln e^(k)

k = ln (.90944/.928)  = -.02  = the decay factor

(b)  The function is .928M * e^(-.02n)     where n is the time in years since the 5th anniversary

(c) 45 years later we have

.928M * e ^(-.02 * 40)  ≈ .417M   [about 41.% remain married ]

CPhill  Mar 17, 2017
#1
+88871
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(a) Call the number who married in 1960 = M

And.....5 years later......the number who remained married = .928M

And we have that one year later 98% of these remain married....so.......  0.98 (.928M)  = .90944M

So

.90944M  =  .928M*e^(k * 1)   divide by M

.90944 = .928*e^(k)    divide by .928

.90944/.928  = e^(k)     take the ln of both sides

ln (.90944/.928) =  ln e^(k)

k = ln (.90944/.928)  = -.02  = the decay factor

(b)  The function is .928M * e^(-.02n)     where n is the time in years since the 5th anniversary

(c) 45 years later we have

.928M * e ^(-.02 * 40)  ≈ .417M   [about 41.% remain married ]

CPhill  Mar 17, 2017
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I dont understand how you got "98%"

Can you show me an equation on how you got that.?

idenny  Mar 17, 2017
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C. was wrong,

but I did .928*.98^40= .414 which was correct

idenny  Mar 17, 2017
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If something drops 2% from one year to the next....it's only 98% of what it was the year before....does that make sense???

CPhill  Mar 17, 2017
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ah so it went from 100% to 98% because of the 2% drop

and I see.

idenny  Mar 17, 2017
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I actually rounded the decay factor....if you use -.0202, you get

.928M * e ^(-.0202 * 40)   =  .413655  which is close to .414

CPhill  Mar 17, 2017