Answer:
i) [tex]2500(1.11)^n[/tex]
ii) [tex]150(0.6)^n[/tex]
Step-by-step explanation:
1) Here we have to use the growth equation.
[tex]P_1 = P_o(1 +r)^n[/tex]
Where [tex]P_0[/tex] is the initial amount.
[tex]P_1[/tex] is final amount, r - is the rate and n- is the time.
Given: [tex]P_0 = 2500, r = 11% = 0.11[/tex]
Now plug in these given values in the above formula, we get
[tex]P_1 = 2500(1 + 0.11)^n\\P_1 = 2500(1.11)^n\\[/tex]
The answer is [tex]2500(1.11)^n[/tex]
Now let's move to the second equation.
Here the situation is elimination.
So we need to use the following formula.
[tex]P_1 = P_0(1 - r)^n[/tex]
Where [tex]P_0[/tex] is the initial amount.
[tex]P_1[/tex] is final amount, r - is the rate and n- is the time.
Given:
[tex]P_0 = 150, r = 40% = 0.4[/tex]
Now plug in these given values in the above formula, we get
[tex]P_1 = 150(1 - 0.4)^n\\P_1 = 150(0.6)^n\\[/tex]
The answer is [tex]150(0.6)^n[/tex]
