Answer:
It would take approximately 305 s to go to 99% completion
Explanation:
Given that:
y = 50% = 0.5
n = 1.7
t = 100 s
We need to first find the parameter k from the equation below.
[tex]exp(-kt^n)=1-y[/tex]
taking the natural logarithm of both sides:
[tex]-kt^n=ln(1-y)\\kt^n=-ln(1-y)\\k=-\frac{ln(1-y)}{t^n}[/tex]
Substituting values:
[tex]k=-\frac{ln(1-y)}{t^n}= -\frac{ln(1-0.5)}{100^1.7} = 2.76*10^{-4}[/tex]
Also
[tex]t^n=-\frac{ln(1-y)}{k}\\t=\sqrt[n]{-\frac{ln(1-y)}{k}}[/tex]
Substituting values and y = 99% = 0.99
[tex]t=\sqrt[n]{-\frac{ln(1-y)}{k}}=\sqrt[1.7]{-\frac{ln(1-0.99)}{2.76*10^{-4}}}=304.6s[/tex]
∴ t ≅ 305 s
It would take approximately 305 s to go to 99% completion
Given:
The equation will be:
→ [tex]exp(-kt^n) = 1-y[/tex]
By taking "log" both sides, we get
→ [tex]-kt^n = ln(1-y)[/tex]
[tex]k = -\frac{ln(1-y)}{l^n}[/tex]
By substituting the values, we get
[tex]= 2.76\times 10^{-4}[/tex]
Now,
→ [tex]t^n = - \frac{ln(1-y)}{k}[/tex]
[tex]t = n\sqrt{-\frac{ln(1-y)}{k} }[/tex]
By substituting the value of "y", we get
[tex]= \sqrt[1.7]{-\frac{ln(1-0.99)}{2.76\times 10^{-4}} }[/tex]
[tex]= 304.6 \ s[/tex]
Thus the above answer is right.
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