Which statement is true.?

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Answer 1

Answer:

Answer:

Step-by-step explanation:

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For many years, businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to less inflation in health care prices and employees paying for a larger portion of health care benefits. A recent survey showed that 62% of employers are likely to require higher employee contributions for health care coverage this year relative to last year. Suppose the survey was based on a sample of 800 companies likely to require higher employee contributions for health care coverage this year relative to last year. At 95% confidence, compute the margin of error for the proportion of companies likely to require higher employee contributions for health care coverage. (Round your answer to four decimal places.) Compute a 95% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage. (Round your answers to four decimal places.)

Answers

Answer:

95% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage.

(0.5868 , 0.6532)

Step-by-step explanation:

Step(i):-

Given the survey was based on a sample of 800 companies

Given size 'n' = 800

A recent survey showed that 62% of employers are likely to require higher employee contributions for health care coverage this year relative to last year

sample proportion

p⁻ = 0.62

Step(ii):-

The margin of error for the proportion of companies likely to require higher employee contributions for health care coverage.

$M.E= Z_(0.05) \sqrt{(p^(-) (1-p^(-)) )/(n) }$

$M.E= 1.96\sqrt{(0.62 (1-0.62 )/(800) }$

M.E = 0.017 X 1.96

M.E = 0.03

Step(iii):-

95% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage.

$(p^(-) - Z_(0.05) \sqrt{(p^(-) (1-p^(-)) )/(n) } , p^(-) +Z_(0.05) \sqrt{(p^(-) (1-p^(-)) )/(n) })$

$(0.62 - 1.96\sqrt{(0.62 (1-0.62 )/(800) } ,0.62+1.96\sqrt{(0.62 (1-0.62 )/(800) }$

( 0.62 - 0.0332 , 0.62+0.0332)

(0.5868 , 0.6532)

Final answer:

The margin of error for the proportion of companies likely to require higher employee contributions for health care coverage is approximately 0.0245. The 95% confidence interval for the proportion of companies likely to require higher employee contributions is (0.5955, 0.6445).

Explanation:

To compute the margin of error for the proportion of companies likely to require higher employee contributions for health care coverage, we can use the formula:

Margin of error = Z * sqrt((p * (1-p)) / n)

where Z is the Z-score corresponding to the desired confidence level (95% in this case), p is the proportion of companies likely to require higher employee contributions, and n is the sample size. Substituting the given values into the formula, we have:

Margin of error = 1.96 * sqrt((0.62 * (1-0.62)) / 800)

Calculating this value gives us a margin of error of approximately 0.0245.

To compute the 95% confidence interval for the proportion of companies likely to require higher employee contributions, we can use the formula:

Confidence interval = p ± margin of error

Substituting the given values into the formula, we have:

Confidence interval = 0.62 ± 0.0245

Calculating this value gives us a confidence interval of (0.5955, 0.6445).

Learn more about Margin of error for a proportion here:

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${x}^(2) + √(x) + \sqrt[5]{x}$
what is f'(3) of this equation?

Answers

Answer:

$3 + (1)/(2√(3) ) + \frac{1}{5\sqrt[5]{81} }$

Step-by-step explanation:

Just to make it easier to see, $√(x) = x^{(1)/(2) }$ and $\sqrt[5]{x} = x^{(1)/(5) }$ This way we could more easily use the power rule of derivatives.

So if f(x) = $x^(2) +x^{(1)/(2) } +x^{(1)/(5) }$ then f'(x) will be as follows.

f'(x) = $x^(1) +(1)/(2) x^{-(1)/(2) } +(1)/(5) x^{-(4)/(5) } = x +\frac{1}{2x^{(1)/(2) }} +\frac{1}{ 5x^{(4)/(5) }} = x +(1)/(2√(x)) +\frac{1}{ 5\sqrt[5]{x^4} }$