Answer:
410.32
Step-by-step explanation:
Given that the initial quantity, Q= 6200
Decay rate, r = 5.5% per month
So, the value of quantity after 1 month, ![q_1 = Q- r \times Q](https://tex.z-dn.net/?f=q_1%20%3D%20Q-%20r%20%5Ctimes%20Q)
![q_1 = Q(1-r)\cdots(i)](https://tex.z-dn.net/?f=q_1%20%3D%20Q%281-r%29%5Ccdots%28i%29)
The value of quantity after 2 months, ![q_2 = q_1- r \times q_1](https://tex.z-dn.net/?f=q_2%20%3D%20q_1-%20r%20%5Ctimes%20q_1)
![q_2 = q_1(1-r)](https://tex.z-dn.net/?f=q_2%20%3D%20q_1%281-r%29)
From equation (i)
![q_2=Q(1-r)(1-r) \\\\q_2=Q(1-r)^2\cdots(ii)](https://tex.z-dn.net/?f=q_2%3DQ%281-r%29%281-r%29%20%20%5C%5C%5C%5Cq_2%3DQ%281-r%29%5E2%5Ccdots%28ii%29)
The value of quantity after 3 months, ![q_3 = q_2- r \times q_2](https://tex.z-dn.net/?f=q_3%20%3D%20q_2-%20r%20%5Ctimes%20q_2)
![q_3 = q_2(1-r)](https://tex.z-dn.net/?f=q_3%20%3D%20q_2%281-r%29)
From equation (ii)
![q_3=Q(1-r)^2(1-r)](https://tex.z-dn.net/?f=q_3%3DQ%281-r%29%5E2%281-r%29)
![q_3=Q(1-r)^3](https://tex.z-dn.net/?f=q_3%3DQ%281-r%29%5E3)
Similarly, the value of quantity after n months,
![q_n= Q(1- r)^n](https://tex.z-dn.net/?f=q_n%3D%20Q%281-%20r%29%5En)
As 4 years = 48 months, so puttion n=48 to get the value of quantity after 4 years, we have,
![q_{48}=Q(1-r)^{48}](https://tex.z-dn.net/?f=q_%7B48%7D%3DQ%281-r%29%5E%7B48%7D)
Putting Q=6200 and r=5.5%=0.055, we have
![q_{48}=6200(1-0.055)^{48} \\\\q_{48}=410.32](https://tex.z-dn.net/?f=q_%7B48%7D%3D6200%281-0.055%29%5E%7B48%7D%20%5C%5C%5C%5Cq_%7B48%7D%3D410.32)
Hence, the value of quantity after 4 years is 410.32.