Can someone help with these 2 problems?

Review the table on page 3 comparing in-network and out-of-network costs under the Standard Plan. Which statement is TRUE? Insur

Iteru [2.4K]

The true statement is that most insurance plans encourage the use of In-Network doctors by offering lower costs on most services they perform.

<h3>What is the meaning of Health Maintenance Organization?</h3>

An HMO is regarded as an acronym that is used for “Health Maintenance Organization”. It is know to be companies that has been given the power to solely to manage the provision of health care services to the people.

The use of the in-network option is one that tends to save money. The Providers in your network are known to have a form of agreements with your insurance company that tends to make one to save money.

Lean more about insurance plans from

brainly.com/question/1941778

3 0

2 years ago

Bro please help for a brainlist for best answer !!!!!!

tatiyna

The domain of a function are the possible input values of the function.

<em>The domain of the function is (d) Whole numbers</em>

First, we identify the start and end values of x on the graph.

We have:

$Start = 0$

The graph has no end; so:

$End = \infty$

This means that the domain of the graph is $(0,\infty)$

The number of times one can visit the pool cannot be decimal.

Hence, the domain of the function is whole numbers

Read more about domain at:

brainly.com/question/15339465

4 0

2 years ago

Give a 98% confidence interval for one population mean, given asample of 28 data points with sample mean 30.0 and sample standar

klio [65]

We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

$\begin{gathered} 1-\alpha=0.98 \\ \alpha=0.02 \end{gathered}$

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:

$CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack$

Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:

$CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack$

Where (from tables):

$Z_{0.99}=2.33$

Finally, the interval at 98% confidence level is:

$CI(\mu)=\lbrack28.94,31.06\rbrack$

4 0

9 months ago

Help me please!!!!!!!!!!!

Softa [21]

Ratio and portion

taggednow to tagged later comparison

assuming the tagged onws aer in ratio

25-16=9
9/25=0.36=36%

I would assume that the eagle population dropped 36%

4 0

3 years ago

What is the y-coordinate of the point that divides the directed

e-lub [12.9K]

Answer:

(D)5

Step-by-step explanation:

Given the point J(-3,1) and K(8,11).

The line segment that divides the segment from J to K in any given ratio can be determined using the formula.

$P(x,y)=\left(\dfrac{mx_2+nx_1}{m+n} ,\dfrac{my_2+ny_1}{m+n}\right)$

In the given case:

$(x_1,y_1)=(-3,1), (x_2,y_2)=(8,11)$ , m:n=2:3

Since we are to determine the y-coordinate of the point that divides JK into a ratio of 2:3, we have:

$\dfrac{my_2+ny_1}{m+n}=\dfrac{2*11+3*1}{3+2}\\\\=\dfrac{22+3}{5}\\\\=\dfrac{25}{5}\\\\=5$

The y-coordinate of the point that divides the directed line segment from J to K into a ratio of 2:3 is 5.

The correct option is D.

3 0

2 years ago