What are some access benefits of the us private health model ?

1 answer:

7 0

1. You Choose Your Doctor
In the private healthcare system you often have more flexibility in choosing a doctor as well as medical facility. For patients that want the same doctor all the time, this can be a very important advantage of this type of system.
2. Shorter Wait Times
If you are having a surgery that is necessary but not life threatening, there are often long wait times. In a private health insurance system the patient will often have shorter wait times because the medical facility is less busy. And even if there are a lot of people waiting, you can often by pass the line by paying a little bit extra for faster medical attention.
3. Improved Facilities
Unfortunately, because the public system is funded with government money it does not have access to as much funds as the private sector one do. This means that public hospitals and health care facilities are often overcrowded, and lacking certain comfort amenities. Private health care facilities, on the other hand, are usually well maintained,
with private wards, private bathrooms, phones, TVs, and better tasting food.

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Which of the following parabolas opens upward and appears narrower than y = −3x^2 + 2x − 1? A. y = 4x^2 − 2x − 1 B. y = −4x^2 +

Kamila [148]

Narrowing it down it's either a or c because the other 2 choices open downward (negative a).
Graphing the 2 parabolas choice c is narrower.
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3 years ago

Life tests on a helicopter rotor bearing give a population mean value of 2500 hours and a population standard deviation of 135 h

stira [4]

Answer:

The percent of the parts are expected to fail before the 2100 hours is 0.15.

Step-by-step explanation:

Given :Life tests on a helicopter rotor bearing give a population mean value of 2500 hours and a population standard deviation of 135 hours.

To Find : If the specification requires that the bearing lasts at least 2100 hours, what percent of the parts are expected to fail before the 2100 hours?.

Solution:

We will use z score formula

$z=\frac{x-\mu}{\sigma}$

Mean value = $\mu = 2500$

Standard deviation = $\sigma = 135$

We are supposed to find If the specification requires that the bearing lasts at least 2100 hours, what percent of the parts are expected to fail before the 2100 hours?

So we are supposed to find P(z<2100)

so, x = 2100

Substitute the values in the formula

$z=\frac{2100-2500}{135}$