What is the purpose of the first paragraph?

SAT

1 answer:

Yakvenalex [24]3 years ago

8 0

I'm almost positive it's to "describe all of Nawab’s money-making ventures".

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which priority teaching intervention would the nurse include in the care plan for a client who has insomnia

Digiron [165]

The teaching intervention that the nurse would have to include would be to give the patient prescribed insomnia during dinner.

<h3>What is insomnia?</h3>

Insomnia is the term that is used to refer to the fact that a person is having issues with sleeping.

This may be due to underlying issues with health or with mental health. The patient can be helped by giving them prescribed pills that may help them to sleep.

In other to help the person that is suffering from this condition, A nurse would help them through the use of prescribed pills. The issue is that there are usually a lot of side effects when it comes to the use of pills to treat insomnia in a person.

Another way is through the use of sleep therapy to get them to readjust their sleep pattern. If this condition persists, it could be a cause for alarm for the patient.

Read more on insomnia here; brainly.com/question/14525690

#SPJ1

6 0

1 year ago

For a laboratory assignment if the equipment is working the density function of the observed outcome

vivado [14]

Based the density function of the observed outcome, the probability $P(x\leq \frac{1}{3} )$ is equal to 0.1388.

<h3>What is a density function?</h3>

A density function can be defined as a type of function which is used to represent the density of a continuous random variable that lies within a specific range.

Given the following density function:

$f(x)=\left \{ {{2(1-x),\;0 < x < 1} \atop {0,\; otherwise}} \right.$

Next, we would calculate $P(x\leq \frac{1}{3} )$ ;

$P(x\leq \frac{1}{3} )=\int\limits^\frac{1}{3} _0 {f(x)} \, dx \\\\P(x\leq \frac{1}{3} )=2 \int\limits^\frac{1}{3} _0 {(1-x)} \, dx \\\\P(x\leq \frac{1}{3} )=2|x-\frac{x^2}{2} |\limits^\frac{1}{3} _0\\\\P(x\leq \frac{1}{3} )=2[\frac{1}{3} -\frac{1}{2} (\frac{1}{3})^2]\\\\P(x\leq \frac{1}{3} )=2[\frac{1}{3} -\frac{1}{18}]\\\\P(x\leq \frac{1}{3} )=0.1388$