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3 years ago

How long should the mpi be retained? permanently 25 years 50 years 10 years?

2 answers:

7 0

Master Patient Index will be kept by the hospital or medical institution permanently. It is also called Enterprise or enterprise-wide master patient index and it is a database that is used to healthcare organization to maintain its consistency, accurate information of the patients and essential medical documents or data of the patients managed within the departments. It is important to have an accurate identification of individuals that are in the Master Patient (person) Index or MPI in an organization. The MPI is also the registry of every patient in a healthcare it is used to identify, match, de-duplicate and cleanses the record of the individual. It is helpful to the healthcare organization to have this because they can find faster the data or information of the patients they searching.

Marrrta [24]3 years ago

5 0

Answer:

the correct answer is 10years.

Explanation:

<em>The Master Patient Index Module </em>is essential for the accurate sharing of clinical data across healthcare organizations. It is a central component of any future-oriented eHealth( it is the support that the cost-effective and safe use of information and communications technologies offers to health and related fields). Using a powerful matching algorithm, it compares patient data from disparate systems, links related records under an Enterprise Unique Identifier and creates a Golden Record.

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Copper was the first metal to be produced from its ore because it is the easiest to smelt, that is, to refine by heating in the

Ganezh [65]

Answer:

57.48%

Explanation:

Calculate the mass of 1 mole of malachite:

MM Cu = 63.55

MM O = 16.00

MM H = 1.01

MM C = 12.01

$(Cu_{2}(OH)_{2}CO_{3})$

A mole of malachite has:

2 moles of Cu

5 moles of O

2 moles of H

1 mole of C

MW Malachite = 2*MM(CU) + 5*MM(O) + 2*MM(H) + 1 *MM(C)

MW Malachite = 2*63.55 + 5*16.00 + 2*1.01 + 1*12.01

MW Malachite = 221.13

Mass of Cu in a mole of Malachite = 2*MM(CU) = 127.1

Now divide the mass of Cu by the mass of Malachite

$%Cu = \frac{127.1}{221.13} =0.5748=57.48%$

7 0

3 years ago

An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, h

vaieri [72.5K]

Answer:

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

Explanation:

Length of the base = l

Width of the base = w

Height of the pyramid = h

Volume of the pyramid = $V=\frac{1}{3}lwh$

We have:

Rate at which water is filled in cube = $\frac{dV}{dt}= 45 cm^3/s$

Square based pyramid:

l = 6 cm, w = 6 cm, h = 13 cm

Volume of the square based pyramid = V

$V=\frac{1}{3}\times l^2\times h$

$\frac{l}{h}=\frac{6}{13}$

$l=\frac{6h}{13}$

$V=\frac{1}{3}\times (\frac{6h}{13})^2\times h$

$V=\frac{12}{169}h^3$

Differentiating V with respect to dt:

$\frac{dV}{dt}=\frac{d(\frac{12}{169}h^3)}{dt}$

$\frac{dV}{dt}=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$45 cm^3/s=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times h^2}$

Putting, h = 4 cm

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times (4 cm)^2}$

$=13.20 cm/s$

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

3 0

3 years ago

In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in

RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 $m^{3}$

Density of water = 1000 $kg/m^{3}$

Therefore, mass of water = Density × Volume

= $1000 kg/m^{3} \times 0.25 m^{3}$

= 250 kg

Initial Temperature of water ( $T_{1}$ ) = $20^{o}C$

Final temperature of water = $140^{o}C$

Heat of vaporization of water ( $dH_{v}$ ) at $140^{o}C$ is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point ( $140^{o}C$ ) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

= 62.5 (kg) × 2133 (kJ/kg)

= $133.3 \times 10^{3}$ kJ

All this heat was supplied in 60 min = 60(min) × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = $\frac{133.3 \times 10^{3}kJ}{3600 s}$ = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from $20^{o}C$ to $140^{o}C$ of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

= 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

= 125520 kJ

Time required = Heat required / Power rating

= $\frac{125520}{37}$

= 3392 sec

Time required to raise the temperature from $20^{o}C$ to $140^{o}C$ of 0.25 $m^{3}$ water is calculated as follows.

$\frac{3392 sec}{60 sec/min}$

= 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0

3 years ago

Liquid acetone is ______________.

MaRussiya [10]

A is the answer maybe

5 0

3 years ago

Which statement describes a chemical property of sodium

Alenkasestr [34]

Answer:

Option b is show the chemical property of sodium....

4 0

3 years ago