Park A
Y=150(6/5)ˣ
Y=150(1+0.20)ˣ
Park B
Y=150(4/5)ˣ
Y=150(1-0.20)ˣ
Neither of these
Y=6/5.(150)ˣ
Y=4/5.(150)ˣ
Step-by-step explanation:
As given in the question, the population of bear
- increases year after year
⇔It means the given actions are compounding in fashion
We know that for Compound Increase/Decrease
Y=X(1+R/100)ⁿ
Where
Y= final number after intended “n” time
X=Initial number at the starting
R= Rate of compounding
ⁿ= period into consideration
FOR PARK A
Since the bear population increases by 20% per annum
Y= 150(1+20/100)ˣ
⇒Y=150(1+0.20)ˣ
similarly
Y=150(1+1/5)ˣ (reducing the fraction 20/100)
⇒Y=150(6/5)ˣ
Where Y= population of the bear after x years
X= years
Initial population of bear =150(given)
FOR PARK B
Since the bear population decreases by 20% per annum
Y= 150(1-20/100)ˣ
⇒Y=150(1-0.20)ˣ
similarly
Y=150(1-1/5)ˣ (reducing the fraction 20/100)
⇒Y=150(4/5)ˣ
The remaining of the function would group under none of these
