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dem82 [27]
4 years ago
9

Hypothesis: Family centered care is a significant predictor of children's health-related quality of life, independent of illness

severity. This hypothesis is a ____________ hypothesis.
Mathematics
1 answer:
Nostrana [21]4 years ago
8 0

Answer:

This hypothesis is a Directional Hypothesis.

Step-by-step explanation:

Family centered care is a significant predictor of children's health-related quality of life, independent of illness severity.

This hypothesis is a Directional Hypothesis.

Directional hypothesis predicts the way in which the independent variable will affect the dependent variable.

Here, independent variable is - family centered care

Dependent variable is - health related quality of life

We can see that when family centered care is of great quality, there is no severe illness for children.

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Answer:

1 solution

Step-by-step explanation:

4x + x + 3 = 4x (Subtract 4x from both sides)

x + 3 = 0 (Subtract 3 from both sides)

x = -3

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Solve for x x + 47 = -23
givi [52]
X = -23 - 47 
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Indicate the equation of the given line in standard form.The line containing the median of the trapezoid whose vertices are R(-1
maxonik [38]
RS => y - 5 = (8 - 5)/(1 - (-1)) (x - (-1))
y - 5 = 3/2 (x + 1) => slope = 3/2
ST => y - 8 = (-2 - 8)/(7 - 1) (x - 1)
y - 8 = -10/6 (x - 1) = -5/3 (x - 1) => slope = -5/3
TU => y - (-2) = (0 - (-2))/(2 - 7) (x - 7)
y + 2 = 2/5(x - 7) => slope = 2/5
UR => y = 5/(-1 - 2) (x - 2)
y = -5/3 (x - 2) => slope = -5/3

The median is the line joining the midpoints of the non-parallel sides.
Midpoint of RS = ((-1 + 1)/2, (5 + 8)/2) = (0, 13/2)
Midpoint of TU = ((7 + 2)/2, -2/2) = (9/2, -1)

Equation of the line joining (0, 13/2) and (9/2, -1) is given by y - 13/2 = (-1 - 13/2)/(9/2) x
y - 13/2 = (-15/2)/(9/2) x
y - 13/2 = -15/9x
18y - 117 = -30x
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7 0
3 years ago
Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt.
Oliga [24]

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes =  (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

6 0
3 years ago
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