Solve problem in photo (8th math)

A. Complete the chart based on the initial conditions:

Nata [24]

Answer:

a.

$Month\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Pop(whole \ ant)\\month1 => x_1=80\times0.94^1=1128\\month2=>x_2=1200\times0.94^2=1060\\month3=>x_3=1200\times0.94^3=996\\month4=>x_4=1200\times0.94^4=936$

b.

$week Number\ \ \ \ \ \ \ \ \ \ \ \ \ \ Mass(g)\\week1 => x_1=80\times1.1^1=88g\\week2=>x_2=80\times1.1^2=96.8g\\week3=>x_3=80\times1.1^3=106.48g\\week4=>x_4=80\times1.1^4=117.128g$

Step-by-step explanation:

a. From the information provided, we can deduce that the population death's follows a Geometric sequence in the form $(a,ar,ar^2,ar^3...)$ where $a-first \ term$ and $r-common \ ratio$

#Since the population is reducing, $r$ can is obtained as $r=1-r=0.94$

#The $n^t^h$ term is obtained using the formula $x_n=ar^(^n^-^1^)$ , given a=1200

The number of ants alive after every month (in first 4 months)

$Month\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Pop(whole \ ant)\\month1 => x_1=80\times0.94^1=1128\\month2=>x_2=1200\times0.94^2=1060\\month3=>x_3=1200\times0.94^3=996\\month4=>x_4=1200\times0.94^4=936$

The ant's alive after 4 months is obtained as the value of $x_5$

$x_n=ar^(^n^-^1^)\\1-x_5=1-1200\times 0.94^4=936.89\\\approx 936$

Hence, 936 ants are alive after 4 months.

b. As with the above question, the kitten population follows a geometric sequence: $(a,ar,ar^2,ar^3...)$ .

#Since it's a growing population , the common ration is the sum of 100% + the growth rate,

$r=1.1$ and $a=80$ and $x_n=ar^(^n^-^1^)$

The population after 4weeks will be:

$week Number\ \ \ \ \ \ \ \ \ \ \ \ \ \ Mass(g)\\week1 => x_1=80\times1.1^1=88g\\week2=>x_2=80\times1.1^2=96.8g\\week3=>x_3=80\times1.1^3=106.48g\\week4=>x_4=80\times1.1^4=117.128g$

8 0

3 years ago

Use >,<, or = to solve the inequality ? 7.5 7.05

mihalych1998 [28]