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Sonja [21]
2 years ago
5

An organization advocating for healthcare reform has estimated the average cost of providing healthcare for a senior citizen rec

eiving Medicare to be about $13,000 per year. The article also stated that, with 90% confidence, the margin or error for the estimate is $1,000. Determine the resulting 90% confidence interval for the average cost for healthcare of a senior citizen receiving Medicare.
Mathematics
1 answer:
asambeis [7]2 years ago
7 0

Using the definition of a confidence interval, it is found that the 90% confidence interval for the average cost for healthcare of a senior citizen receiving Medicare is ($12,000, $14,000).

A confidence interval is the <u>sample mean plus/minus the margin of error</u>.

  • In this problem, the sample mean is of $13,000.
  • The margin of error is of $1,000.

Hence:

$13,000 - $1,000 = $13,000

$13,000 + $1,000 = $14,000

The interval is ($12,000, $14,000).

A similar problem is given at brainly.com/question/24869727

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3 years ago
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Answer:

The average rate of change of the function g(x)=x^2+10x+18 over the interval -11 \leq x\leq -1 is -1

Step-by-step explanation:

We are given the function g(x)=x^2+10x+18 over the interval -11 \leq x\leq -1

We need to find average rate of change.

The formula used to find average rate of change is : Average \ rate \ of \ change=\frac{g(b)-g(a)}{b-a}

We have b=-1 and a=-11

Finding g(b) = g(-1)

g(x)=x^2+10x+18\\Putting \ x=-1\\g(-1)=(-1)^2+10(-1)+18\\g(-1)=1-10+18\\g(-1)=9

Finding g(a) = g(-11)

g(x)=x^2+10x+18\\Putting \ x=-11\\g(-11)=(-11)^2+10(-11)+18\\g(-1)=121-110+18\\g(-1)=29

Finding average rate of change

Average \ rate \ of \ change=\frac{g(b)-g(a)}{b-a}\\Average \ rate \ of \ change=\frac{9-29}{-1-(-11)}\\Average \ rate \ of \ change=\frac{-10}{-1+11}\\Average \ rate \ of \ change=\frac{-10}{10}\\Average \ rate \ of \ change=-1

So, the average rate of change of the function g(x)=x^2+10x+18 over the interval -11 \leq x\leq -1 is -1

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3 years ago
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Pretty sure its y > -6.2
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