Answer:
a.
[tex]Month\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Pop(whole \ ant)\\month1 => x_1=80\times0.94^1=1128\\month2=>x_2=1200\times0.94^2=1060\\month3=>x_3=1200\times0.94^3=996\\month4=>x_4=1200\times0.94^4=936[/tex]
b.
[tex]week Number\ \ \ \ \ \ \ \ \ \ \ \ \ \ Mass(g)\\week1 => x_1=80\times1.1^1=88g\\week2=>x_2=80\times1.1^2=96.8g\\week3=>x_3=80\times1.1^3=106.48g\\week4=>x_4=80\times1.1^4=117.128g[/tex]
Step-by-step explanation:
a. From the information provided, we can deduce that the population death's follows a Geometric sequence in the form [tex](a,ar,ar^2,ar^3...)[/tex] where [tex]a-first \ term[/tex] and [tex]r-common \ ratio[/tex]
#Since the population is reducing, [tex]r[/tex] can is obtained as [tex]r=1-r=0.94[/tex]
#The [tex]n^t^h[/tex] term is obtained using the formula [tex]x_n=ar^(^n^-^1^)[/tex], given a=1200
The number of ants alive after every month (in first 4 months)
[tex]Month\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Pop(whole \ ant)\\month1 => x_1=80\times0.94^1=1128\\month2=>x_2=1200\times0.94^2=1060\\month3=>x_3=1200\times0.94^3=996\\month4=>x_4=1200\times0.94^4=936[/tex]
The ant's alive after 4 months is obtained as the value of [tex]x_5[/tex]
[tex]x_n=ar^(^n^-^1^)\\1-x_5=1-1200\times 0.94^4=936.89\\\approx 936[/tex]
Hence, 936 ants are alive after 4 months.
b. As with the above question, the kitten population follows a geometric sequence: [tex](a,ar,ar^2,ar^3...)[/tex].
#Since it's a growing population , the common ration is the sum of 100% + the growth rate,
[tex]r=1.1[/tex] and [tex]a=80[/tex] and [tex]x_n=ar^(^n^-^1^)[/tex]
The population after 4weeks will be:
[tex]week Number\ \ \ \ \ \ \ \ \ \ \ \ \ \ Mass(g)\\week1 => x_1=80\times1.1^1=88g\\week2=>x_2=80\times1.1^2=96.8g\\week3=>x_3=80\times1.1^3=106.48g\\week4=>x_4=80\times1.1^4=117.128g[/tex]
