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tatuchka [14]
2 years ago
5

Solve problem in photo (8th math)

Mathematics
1 answer:
murzikaleks [220]2 years ago
4 0
A hrhehrhrjrhrrhndndjd
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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]

7.5 > 7.05
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