You plan to construct a confidence interval for the mean \muμ of a Normal population with (known) standard deviation \sigmaσ. Which of the following will reduce the size of the margin of error? Group of answer choices Use a lower level of confidence. Increase the sample size. Reduce \sigma σ . All of the answers are correct.

increase the sample size

Step-by-step explanation: the margin of error for any confidence interval is given by the formulae below.

Margin of error = critical value × standard deviation/√n

From what we can see the critical value and standard deviation are constants, the only variables here are the margin of error and sample size which are inversely proportional to each other, that is the margin of error is inversely proportional to the square of the sample size.

Hence, reducing the sample size will increase the margin of error while increasing the sample size will reduce the margin of error.

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0