A quantity with an initial value of 6200 decays continuously at a rate of 5.5% per month. What is the value of the quantity after 4 years, to the nearest hundredth?

Answers

Answer 1

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, [tex]q_1 = Q- r \times Q[/tex]

[tex]q_1 = Q(1-r)\cdots(i)[/tex]

The value of quantity after 2 months, [tex]q_2 = q_1- r \times q_1[/tex]

[tex]q_2 = q_1(1-r)[/tex]

From equation (i)

[tex]q_2=Q(1-r)(1-r) \\\\q_2=Q(1-r)^2\cdots(ii)[/tex]

The value of quantity after 3 months, [tex]q_3 = q_2- r \times q_2[/tex]

[tex]q_3 = q_2(1-r)[/tex]

From equation (ii)

[tex]q_3=Q(1-r)^2(1-r)[/tex]

[tex]q_3=Q(1-r)^3[/tex]

Similarly, the value of quantity after n months,

[tex]q_n= Q(1- r)^n[/tex]

As 4 years = 48 months, so puttion n=48 to get the value of quantity after 4 years, we have,

[tex]q_{48}=Q(1-r)^{48}[/tex]

Putting Q=6200 and r=5.5%=0.055, we have

[tex]q_{48}=6200(1-0.055)^{48} \\\\q_{48}=410.32[/tex]

Hence, the value of quantity after 4 years is 410.32.

Answer 2

Answer:

442.44

Step-by-step explanation:


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Step-by-step explanation:

The chemical acidity of a solution is measured in units of pH: pH=-log[H+], where [H+] is the hydrogen ion concentration in the solution. 1.58 x 10⁻⁹ is [H+] if the pH=8.8.

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