This exponential growth/decay (in this case decay because r<1) of the form:
f=ir^t, f=final value, i=initial value, r=common ratio or "rate", t=time.
Since the population decreases by 4.5% each year the common ratio is:
r=(100-4.5)/100=0.955 so we can say
P(t)=8500(0.955^t)
....
7000=8500(0.955^t)
14/17=(955/1000)^t taking the natural log of both sides
ln(14/17)=t ln(955/1000)
t=ln(14/17)/ln(955/1000)
t≈4.22 years (to nearest hundredth of a year)
Since t is the years since 2010, the population will fall to 7000 in the year (2010+4.22=2014.22, more than four years will have elapsed) 2015.
Answer:y = (19 - 8x)/3
Step-by-step explanation:
add 8x to both sides ,
-3y = -19+8x
dividing both sides by -3
y = (19 - 8x)/3
19.27x6=115.62x1.065=123.1353
88.22x1.065=93.9543
321.77x1.065=342.68505
total purchase price=559.77
Suppose we have had S visitors in n months since EIS (Entry Into Service) of the site, then :
<span>S(n) = 8100 + 8100 * 3 + 8100 * 3^2 + 8100 * 3^3 + ... +8100 * 3^n </span>
<span>S(n) = 8100 * [ 1 + 3 + 3^2 + 3^3 + ... + 3^n ] </span>
<span>and we know that : </span>
<span>1 + 3 + 3^2 + 3^3 + ... + 3^n = (3^(n+1) - 1)/(3 - 1) </span>
<span>= (3^(n+1) - 1)/2 </span>
<span>therefore </span>
<span>S(n) = 8100 * (3^(n+1) - 1)/2 </span>
<span>finally : </span>
<span>S(n) = 4050 * (3^(n+1) - 1) </span>
<span>and this allows you to obtain the month n, if you know S(n) : </span>
<span>3^(n+1) = S(n) /4050 + 1 </span>
<span>therefore : </span>
<span>n = ln[S(n) /4050 + 1] /ln3 - 1 </span>
<span>hope it' ll help !!</span>
