Question

Please help me!

Answer 1

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

• decreases yearly

⇔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