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vodomira [7]
3 years ago
11

Use the following information to compute the ALOS and median LOS and range for Community Nursing Center. The discharge date is J

une 2, 20XX (a non-leap year). Round the ALOS to once decimal point.
Community Nursing Center
Discharge Report
June 2, 20XX
Patients Admission Date LOS
1 January 2, 20XX 179
2 January 10, 20XX 171
3 February 8, 20XX 142
4 February 10, 20XX 140
5 February 26, 20XX 124
6 March 1, 20XX 118
7 March 6, 20XX 113
8 March 12, 20XX 107
9 March 15, 20XX 104
10 April 1, 20XX 88
11 April 15, 20XX 73
12 May 3, 20XX 56
13 May 5, 20XX 54
14 May 6, 20XX 53
15 May 18, 20XX 41
Mathematics
1 answer:
zaharov [31]3 years ago
5 0

Answer:

ALOS = 104.2 days.

Median LOS = 107 days.

Step-by-step explanation:

ALOS = Average length of stay = (Total length of stay)/(number of patients)

Median = variable in middle when the variables are arranged in ascending or descending order.

a) ALOS

Total length of stay = 179+171+142+140+124+118+113+107+104+88+73+56+54+53+41 = 1563

Number of patients = 15

ALOS = (1563/15) = 104.2

b) Median LOS

As the LOS are, they are arranged in descending form, the LOS in the middle (that is, at the 8th position) is that of the patient admitted on March 12, 20XX; 107 days.

Hope this Helps!!!

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Z^4-5(1+2i)z^2+24-10i=0
mixer [17]

Using the quadratic formula, we solve for z^2.

z^4 - 5(1+2i) z^2 + 24 - 10i = 0 \implies z^2 = \dfrac{5+10i \pm \sqrt{-171+140i}}2

Taking square roots on both sides, we end up with

z = \pm \sqrt{\dfrac{5+10i \pm \sqrt{-171+140i}}2}

Compute the square roots of -171 + 140i.

|-171+140i| = \sqrt{(-171)^2 + 140^2} = 221

\arg(-171+140i) = \pi - \tan^{-1}\left(\dfrac{140}{171}\right)

By de Moivre's theorem,

\sqrt{-171 + 140i} = \sqrt{221} \exp\left(i \left(\dfrac\pi2 - \dfrac12 \tan^{-1}\left(\dfrac{140}{171}\right)\right)\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= \sqrt{221} i \left(\dfrac{14}{\sqrt{221}} + \dfrac5{\sqrt{221}}i\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= 5+14i

and the other root is its negative, -5 - 14i. We use the fact that (140, 171, 221) is a Pythagorean triple to quickly find

t = \tan^{-1}\left(\dfrac{140}{171}\right) \implies \cos(t) = \dfrac{171}{221}

as well as the fact that

0

\sin\left(\dfrac t2\right) = \sqrt{\dfrac{1-\cos(t)}2} = \dfrac5{\sqrt{221}}

(whose signs are positive because of the domain of \frac t2).

This leaves us with

z = \pm \sqrt{\dfrac{5+10i \pm (5 + 14i)}2} \implies z = \pm \sqrt{5 + 12i} \text{ or } z = \pm \sqrt{-2i}

Compute the square roots of 5 + 12i.

|5 + 12i| = \sqrt{5^2 + 12^2} = 13

\arg(5+12i) = \tan^{-1}\left(\dfrac{12}5\right)

By de Moivre,

\sqrt{5 + 12i} = \sqrt{13} \exp\left(i \dfrac12 \tan^{-1}\left(\dfrac{12}5\right)\right) \\\\ ~~~~~~~~~~~~~= \sqrt{13} \left(\dfrac3{\sqrt{13}} + \dfrac2{\sqrt{13}}i\right) \\\\ ~~~~~~~~~~~~~= 3+2i

and its negative, -3 - 2i. We use similar reasoning as before:

t = \tan^{-1}\left(\dfrac{12}5\right) \implies \cos(t) = \dfrac5{13}

1 < \tan(t) < \infty \implies \dfrac\pi4 < t < \dfrac\pi2 \implies \dfrac\pi8 < \dfrac t2 < \dfrac\pi4

\cos\left(\dfrac t2\right) = \dfrac3{\sqrt{13}}

\sin\left(\dfrac t2\right) = \dfrac2{\sqrt{13}}

Lastly, compute the roots of -2i.

|-2i| = 2

\arg(-2i) = -\dfrac\pi2

\implies \sqrt{-2i} = \sqrt2 \, \exp\left(-i\dfrac\pi4\right) = \sqrt2 \left(\dfrac1{\sqrt2} - \dfrac1{\sqrt2}i\right) = 1 - i

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So our simplified solutions to the quartic are

\boxed{z = 3+2i} \text{ or } \boxed{z = -3-2i} \text{ or } \boxed{z = 1-i} \text{ or } \boxed{z = -1+i}

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