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snow_tiger [21]
3 years ago
9

Evacuate patients/residents according to their ______ condition.

Advanced Placement (AP)
1 answer:
9966 [12]3 years ago
5 0

Answer:

Evacuate patients/residents according to their <u>physical</u> condition.

Explanation:

According to the procedures that should be taken in case of urgent evacuation, patients or residents should be evacuated according to  their physical condition, meaning that certain patients/residents are prioritized. Patients/residents that are ambulatory (able to walk) should be evacuated first,  followed by those that can be transported in a wheelchair, and lastly those that are bedridden, as they are the most difficult to transport (they have to be pushed in their bed, on a stretcher or carried with a blanket).

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the is lslamic empire that developed the most in the period was the ottoman Empire They used certain disagreements that existed in the area around the mediterranean  

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Suppose you roll a regular 6-sided die with the numbers 1 through 6. What is the probability that you roll a 3 and then rolla 47
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3/47

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Pantomiming is the act of miming an object and using it. For example, if you're at an audition, and you see in the script you are supposed to talk on the phone, you would shape your hand as if you had a phone in it. It's like pretending the object is there and interacting with it.

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The regions bounded by the graphs of y=x2 and y=sin2x are shaded in the figure above. What is the sum of the areas of the shaded
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Answer:

The sum of the area of the shaded regions = 0.248685

Explanation:

The sum of the area of the shaded region is given as follows;

The point of intersection of the graphs are;

y = x/2

y = sin²x

∴ At the intersection, x/2 = sin²x

sinx = √(x/2)

Using Microsoft Excel, or Wolfram Alpha, we have that the possible solutions to the above equation are;

x = 0, x ≈ 0.55 or x ≈ 1.85

The area under the line y = x/2, between the points x = 0 and x ≈ 0.55, A₁, is given as follows

1/2 × (0.55)×0.55/2 ≈ 0.075625

The area under the line y = sin²x, between the points x = 0 and x ≈ 0.55, A₂, is given using as follows;

\int\limits {sin^n(x)} \, dx = -\dfrac{1}{n} sin^{n-1}(x) \cdot cos(x) + \dfrac{n-1}{n} \int\limits {sin^{n-2}(x)} \, dx

Therefore;

A_2 = \int\limits^{0.55}_0 {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_0 ^{0.55}

∴ A₂ =1/2 × ((0.55 - sin(0.55)×cos(0.55)) - (0 - sin(0)×cos(0)) ≈ 0.0522

The shaded area, A_{1 shaded} = A₁ - A₂ = 0.075625 - 0.0522 ≈ 0.023425

Similarly, we have, between points 0.55 and 1.85

A₃ = 1/2 × (1.85 - 0.55) × 1/2 × (1.85 - 0.55) + (1.85 - 0.55) × 0.55/2 = 0.78

For y = sin²x, we have;

A_4 = \int\limits^{1.85}_{0.55} {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_{0.55} ^{1.85} \approx 1.00526

The shaded area, A_{2 shaded} = A₄ - A₃ = 1.00526 - 0.78 ≈ 0.22526

The sum of the area of the shaded regions, ∑A = A_{1 shaded} + A_{2 shaded}

∴ A = 0.023425 + 0.22526 = 0.248685

The sum of the area of the shaded regions, ∑A = 0.248685

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