Answer: Expected value of the daily cost of operating the machine is 235.264.
Step-by-step explanation:
Since we have given that
E[x]= 0.96 repairs per day
And Var[x] = 0.96 repairs per day.

![E[c]=160+40E[x^2]\\\\E[c]=160+40(Var[x]+(E[x])^2)\\\\E[c]=160+40(0.96+0.96^2)\\\\E[c]=235.264](https://tex.z-dn.net/?f=E%5Bc%5D%3D160%2B40E%5Bx%5E2%5D%5C%5C%5C%5CE%5Bc%5D%3D160%2B40%28Var%5Bx%5D%2B%28E%5Bx%5D%29%5E2%29%5C%5C%5C%5CE%5Bc%5D%3D160%2B40%280.96%2B0.96%5E2%29%5C%5C%5C%5CE%5Bc%5D%3D235.264)
Hence, Expected value of the daily cost of operating the machine is 235.264.
Answer:
Step-by-step explanation:
To find the Taylor series of sinc(x) we will use the taylor series of sin(x). We have that

which is the taylor series expansion based at 0. Then for
, by dividing both sidex by x, we have that

which is the taylor series expansion for the sinc function. Since the series of sine converges for every value of x. Then the taylor series of sinc converges for every value of x, but 0.
Answer:
<h2>It must be shown that both j(k(x)) and k(j(x)) equal x</h2>
Step-by-step explanation:
Given the function j(x) = 11.6
and k(x) =
, to show that both equality functions are true, all we need to show is that both j(k(x)) and k(j(x)) equal x,
For j(k(x));
j(k(x)) = j[(ln x/11.6)]
j[(ln (x/11.6)] = 11.6e^{ln (x/11.6)}
j[(ln x/11.6)] = 11.6(x/11.6) (exponential function will cancel out the natural logarithm)
j[(ln x/11.6)] = 11.6 * x/11.6
j[(ln x/11.6)] = x
Hence j[k(x)] = x
Similarly for k[j(x)];
k[j(x)] = k[11.6e^x]
k[11.6e^x] = ln (11.6e^x/11.6)
k[11.6e^x] = ln(e^x)
exponential function will cancel out the natural logarithm leaving x
k[11.6e^x] = x
Hence k[j(x)] = x
From the calculations above, it can be seen that j[k(x)] = k[j(x)] = x, this shows that the functions j(x) = 11.6
and k(x) =
are inverse functions.
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