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The population of the LessShy neighborhood has been growing steadily since 1984. In 1986, the population..

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Hey Geoffrey -- looks like 8 yrs produced 4800 extras ... a "straight-liner" of 600 per annum ... drop back 2 years from 1986 to 1984 by removing 1200 ... 36000 for 1984 ... y= 600x + 36000 ,,, chk: 10 yrs out to 1994 ought to have 6000 more or 42000 ... Best regards, sir :)

Hi Geoffrey,

My name's Ken, and I'd like to show you how I'd go about solving this problem.

 

It seems like we want to see how the town population depends on the year. This means that town population is our dependent variable (y) and year is our independent variable (x).

 

The rate (m) at which y changes with respect to x can be found by taking the average:

m = (y2 - y1) / (x2 - x1)

Plugging in our two years as x1, x2 and our two populations as y1, y2 we get

m = (42000 - 37200) people / (1994 - 1986) year = 4800 people / 8 years = 600 people / year

m = 600 people / year

I would then use the point slope formula (y - y1) = m * ( x - x1) to find an equation for the line.

(y - 37200) = 600 * (x - 1986)

y = 600 x - 1986 * 600 + 37200

y = 600 x - 1154400

 

You can then plug in x = 1994 and check to see if we get the correct result of y = 42000

 

I hope that I've helped you solve this problem!

(1986,37200) and (1994,42000) are your points.  (Y2-Y1)/(X2-X1) = slope, m

m=600 so, Y=600X+b.   x= # of years past 1984 so for point (1986,37200) X =2 years.

Plugging in 2 for X and 37200 for Y gives you 37200=600*2+b or 37200-(600*2) = b

b = 36000.   Y=600X+36000

To check your work, plug in a point.  42000 = (1994-1984)*600+36000.

10*600=6000.   6000+36000= 42000 so the equation is correct.

2 formulas are created:

1) 37200 = 2m + b

2) 42000 = 10m + b

Solving the first equation for m:

37200 – b = 2m

(37200 – b)/2 = m

Then inputting the value of m into the second equation:

42000 = 10( (37200 – b)/2 ) +b

Solving this for b:

42000 = 5( (37200 – b) ) + b

42000 = 186000 – 5b + b

42000 = 186000 – 4b

4b = 186000 – 42000

4b = 144000

b = 36000

Inputting the numeric value of b into the first equation to find the numeric value of m:

m = (37200 – b)/2

m = (37200 – 36000)/2

m = 1200/2

m = 600

Let’s double check these values in both formulas:

1) 37200 = 2m + b

37200 = 2(600) + 36000

37200 = 1200 + 36000

37200 = 37200

2) 42000 = 10m + b

42000 = 10(600) + 36000

42000 = 6000 + 36000

42000 = 42000

Check and Check.

Therefore we get the formula below in the format of y = mx + b

y = 600x + 36000

Use the information .." In 1986, the population was 37200 people. In 1994, it was 42000 people." to find the growth rate in this period, i.e.,

(42,000 -37,200)/8 = 4800/8=600

where 8 is the number of years from 1986 to 1994. This number, 600, is your m in the equation of a line like  y=mx+b.

So, how do you find b?

We now that in 1986 the population is 37,200 and that x = 1986-1984 =2. Hence:

37,200 = 600*2+ b 

from which we find b = 36,000. In this way we have found the equation we were after, i.e., :

y = 600*x + 36,000

where x is the number of years past 1984. So x=1 represents 1985, x=2 represents 1986, and so on.

Note that for x=10, i.e. in 1994, your linear equation gives a population of 600*10+36,000 = 42,000 which is exactly the number you were given for the population in 1994.

 

Since x is the number of years since 1984, you have two known points:

1986: (1986-1984,42000) = (2,37200)
1994: (1994-1984,42000) = (10,42000)

Use the slope formula to calculate the slope of the line:
m = (y2-y1)/(x2-x1)
m = (42000-37200)/(10-2) = 4800 / 8 = 600

Use m and the point-slope equation for a line:
y-y1 = m (x-x1),   where (x1, y1) is a known point.  You can use either point.
y - 37200 = 600(x - 2) = 600x - 1200
y = 600x - 1200 + 37200
y = 600x + 36000

check:
1986: 37200 = 600 * 2 + 36000 = 1200 + 36000 = 37200
1994: 42000 = 600 * 10 + 36000 = 6000 + 36000 = 42000