G Suppose we take a survey and use the sample proportion to calculate a confidence interval. Which level of confidence gives the confidence interval with the largest margin of error? 90%, 95%, and 99%?

The 99% confidence interval has the higher value of z, so it has the largest margin of error

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

The margin of error is:

So the higher the value of Z, the higher the margin of error.

90% confidence level

So , z is the value of Z that has a pvalue of , so .

95% confidence level

So , z is the value of Z that has a pvalue of , so .

99% confidence level

So , z is the value of Z that has a pvalue of , so .

The 99% confidence interval has the higher value of z, so it has the largest margin of error

- 2 + 10x³ - 21x² + 19x - 4

step-by-step explanation:

each term in the second factor is multiplied by each term in the first factor

x²(- 2x² + 4x - 1) - 3x(- 2x² + 4x - 1) + 4(- 2x² + 4x - 1) ← distribute

= - 2 + 4x³ - x² + 6x³ - 12x² + 3x -8x² + 16x - 4

collect like terms

= - 2 + 10x³ - 21x² + 19x - 4 ← in standard form

c

step-by-step explanation: