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vovikov84 [41]
3 years ago
7

Help please and thank you

Mathematics
1 answer:
Sidana [21]3 years ago
6 0

Answer:

It's discrete and non-linear

Step-by-step explanation:

it is not continuous and linear because it does not form a line going in one direction the entire time

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Costs are rising for all kinds of medical care. The mean monthly rent at assisted-living facilities was reported to have increas
polet [3.4K]

Answer:

a) The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

Step-by-step explanation:

a) Develop a 90% confidence interval estimate of the population mean monthly rent.

Our sample size is 120.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

df = 120-1 = 119

Then, we need to subtract one by the confidence level \alpha and divide by 2. So:

\frac{1-0.90}{2} = \frac{0.10}{2} = 0.05

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 119 and 0.05 in the t-distribution table, we have T = 1.6578.

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

s = \frac{650}{\sqrt{120}} = 59.34

Now, we multiply T and s

M = T*s = 59.34*1.6578 = 98.37

The lower end of the interval is the mean subtracted by M. So it is 3486 - 98.37 = $3387.63.

The upper end of the interval is the mean added to M. So it is 3486 + 98.37 = $3584.37.

The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) Develop a 95% confidence interval estimate of the population mean monthly rent.

Now we have that \alpha = 0.95

So

\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025

With 119 and 0.025 in the t-distribution table, we have T = 1.9801.

M = T*s = 59.34*1.9801 = 117.50

The lower end of the interval is the mean subtracted by M. So it is 3486 - 117.50 = $3368.5.

The upper end of the interval is the mean added to M. So it is 3486 + 117.50 = $3603.5.

The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) Develop a 99% confidence interval estimate of the population mean monthly rent.

Now we have that \alpha = 0.99

So

\frac{1-0.95}{2} = \frac{0.05}{2} = 0.005

With 119 and 0.025 in the t-distribution table, we have T = 2.6178.

M = T*s = 59.34*2.6178 = 155.34

The lower end of the interval is the mean subtracted by M. So it is 3486 - 155.34 = $3330.66.

The upper end of the interval is the mean added to M. So it is 3486 + 155.34 = $3641.34.

The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.

The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

4 0
3 years ago
Given the definitions of f(x) and g(x) below, find the value of (g o f)(0).
Mamont248 [21]

Answer:

its 3

so its

(4x+4)^2-2(4x+4)-5

so first you have to do

(4x+4)(4x+4)

the answer is 16x^2+32x+16

and then you do -2(4x+4) and then you add them so then its

16x^2+32x+16-8x-8-5

and then you just add like terms so its

16x^2+24x+3

and then you insert 0 for x

so the answer is 3

Step-by-step explanation:

7 0
3 years ago
In a tutoring session, 2/3 of an hour was spent reviewing math problems. Adelina attended 3/4 of the tutoring session. How much
fgiga [73]

Answer:

Time spend by Adelina at the tutoring session is 30 minutes.

Step-by-step explanation:

Given : In a tutoring session, \frac{2}{3} of an hour was spent reviewing math problems and Adelina attended \frac{3}{4} of the tutoring session.

We have to find the time Adelina spend at the tutoring session.

We know, 1 hour = 60 minutes.

Total time taken in tutoring session = \frac{2}{3}\times 60=40 minutes.

also, Adelina attended \frac{3}{4} of the tutoring session that is \frac{3}{4} of 40 minutes.

that is \frac{3}{4}\times 40=30minutes.

Thus, time spend by Adelina at the tutoring session is 30 minutes.


 

4 0
3 years ago
Question below here you can rear if not aks me please and thank you ​
Vanyuwa [196]
180-48.2-75=56.8 thank you
5 0
2 years ago
HELP⚠️Which equation represents the line that passes through the points (-3, 7) and (9, -1)?
Alborosie

Answer:

-2/3x+5

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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