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loris [4]
3 years ago
12

The assistant director of HIM is evaluating software that would use electronic logging of the location of incomplete and delinqu

ent records as they move through the completion process. What departmental function is this most useful for?
Computers and Technology
1 answer:
Mazyrski [523]3 years ago
3 0

Answer:

The answer is "Chart tracking".

Explanation:

It is also known as monitoring systems, which is required for the optimization of health records. In t he very first steps on both the road for electronic health records is indeed a properly designed tracking system. It often recognized as line charts outside the area of quality assurance, which represent the performance of the system across time.  

  • Patterns, cycles, and wide deviations could be noticed in the future investigated.  
  • Actions are computed against such a period on both the x-axis in a running graph, shown on the y-axis, and other choices can't be correct because it is not defined in the question.
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Answer:

<u><em>DEAR HERE IS YOUR ANSWER:</em></u>

<u><em>T</em></u>he Patch Tool is part of the healing brush set of tools. These are the go-to tools for retouching and repairing your images. The Patch Tool is primarily used to repair larger areas of an image, or get rid of any distractions or blemishes.

  • The patch tool was introduced into Photoshop at the same time as the Healing Brush
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Explanation:

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Find the root using bisection method with initials 1 and 2 for function 0.005(e^(2x))cos(x) in matlab and error 1e-10?
frutty [35]

Answer:

The root is:

c=1.5708

Explanation:

Use this script in Matlab:

-------------------------------------------------------------------------------------

function  [c, err, yc] = bisect (f, a, b, delta)

% f the function introduce as n anonymous function

%       - a y b are the initial and the final value respectively

%       - delta is the tolerance or error.

%           - c is the root

%       - yc = f(c)

%        - err is the stimated error for  c

ya = feval(f, a);

yb = feval(f, b);

if  ya*yb > 0, return, end

max1 = 1 + round((log(b-a) - log(delta)) / log(2));

for  k = 1:max1

c = (a + b) / 2;

yc = feval(f, c);

if  yc == 0

 a = c;

 b = c;

elseif  yb*yc > 0

 b = c;

 yb = yc;

else

 a = c;

 ya = yc;

end

if  b-a < delta, break, end

end

c = (a + b) / 2;

err = abs(b - a);

yc = feval(f, c);

-------------------------------------------------------------------------------------

Enter the function in matlab like this:

f= @(x) 0.005*(exp(2*x)*cos(x))

You should get this result:

f =

 function_handle with value:

   @(x)0.005*(exp(2*x)*cos(x))

Now run the code like this:

[c, err, yc] = bisect (f, 1, 2, 1e-10)

You should get this result:

c =

   1.5708

err =

  5.8208e-11

yc =

 -3.0708e-12

In addition, you can use the plot function to verify your results:

fplot(f,[1,2])

grid on

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