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1 year ago

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the administration of community hospital is studying the activity of their cardiovascular telemetry unit, which has 12 beds. dur

ing the last semiannual period (july through december), there were 2,200 inpatient service days. calculate the average daily census for this period? round to a whole number.

1 answer:

hoa [83]1 year ago

6 0

The required the average daily census for this period is 12.

<h3>What is Average?</h3>

A Average can be characterized as the amount of all numbers isolated by the all out number of values. A mean can be characterized as a normal of the arrangement of values in an example of information. At the end of the day, a normal is likewise called the math mean. It is known as a mean to Depict the average.

<h3 /><h3>According to question:</h3>

there were 2,200 inpatient service days.

Number of days between this period = 184 days

So, average daily census = 2200/184

11.95

⇒ 12 up to whole number.

To know more about average visit:

brainly.com/question/2906096

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The corners of a meadow are shown on a coordinate grid. Ethan wants to fence the meadow. What length of fencing is required?

Nuetrik [128]

Answer:

34.6 units

Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$A(-6, 2)) = (x_1, y_1)$

$B(2, 6) = (x_2, y_2)$

$AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}$

$AB = \sqrt{(8)^2 + (4)^2}$

$AB = \sqrt{64 + 16} = \sqrt{80}$

$AB = 8.9$ (nearest tenth)

Distance between B(2, 6) and C(7, 1):

$BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$B(2, 6) = (x_1, y_1)$

$C(7, 1) = (x_2, y_2)$

$BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}$

$BC = \sqrt{(5)^2 + (-5)^2}$

$BC = \sqrt{25 + 25} = \sqrt{50}$

$BC = 7.1$ (nearest tenth)

Distance between C(7, 1) and D(3, -5):

$CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$C(7, 1) = (x_1, y_1)$

$D(3, -5) = (x_2, y_2)$

$CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}$

$CD = \sqrt{(-4)^2 + (-6)^2}$

$CD = \sqrt{16 + 36} = \sqrt{52}$

$CD = 7.2$ (nearest tenth)

Distance between D(3, -5) and A(-6, 2):

$DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$D(3, -5) = (x_1, y_1)$

$A(-6, 2) = (x_2, y_2)$

$DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}$

$DA = \sqrt{(-9)^2 + (7)^2}$

$DA = \sqrt{81 + 49} = \sqrt{130}$

$DA = 11.4$ (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

8 0

3 years ago

Six sophomores and 14 freshmen are competing for two alternate positions on the debate team. Which expression represents the pro

Elanso [62]

Answer:

<em>Choose the first alternative</em>

$\displaystyle P=\frac{_{1}^{6}\textrm{C}\ _{1}^{5}\textrm{C}}{_{2}^{20}\textrm{C}}$

Step-by-step explanation:

<u>Probabilities</u>

The requested probability can be computed as the ratio between the number of ways to choose two sophomores in alternate positions $(N_s)$ and the total number of possible choices $(N_t)$ , i.e.

$\displaystyle P=\frac{N_s}{N_t}$

There are 6 sophomores and 14 freshmen to choose from each separate set. There are 20 students in total

We'll assume the positions of the selections are NOT significative, i.e. student A/student B is the same as student B/student A.

To choose 2 sophomores out of the 6 available, the first position has 6 elements to choose from, the second has now only 5

$_{1}^{6}\textrm{C}\ _{1}^{5}\textrm{C} \text{ ways to do it}$

The total number of possible choices is

$_{2}^{20}\textrm{C} \text{ ways to do it}$

The probability is then

$\boxed{\displaystyle P=\frac{_{1}^{6}\textrm{C}\ _{1}^{5}\textrm{C}}{_{2}^{20}\textrm{C}}}$

Choose the first alternative

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Answer:

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Step-by-step explanation:

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