A95% confidence interval for the population proportion of townspeople that are in favor of the proposal is a (.86, .92). if there had been 1035 townspeople present at the meeting and 921 of those 1035 had been in favor of the proposal, what would happen to this interval?

921 of those 1035 people in the meeting had been in favor of the proposal confirms the 95% confidence interval (.86, .92)

Step-by-step explanation:

95% confidence interval for the population proportion of townspeople that are in favor of the proposal is a (.86, .92) means that 95% of the time population proportion of townspeople that are in favor of the proposal falls between .86 and .92.

If 921 of those 1035 had been in favor of the proposal, then the proportion in the meeting who are in favor of the proposal is ≈ 0.89.

Since this value falls  within the confidence interval, this evidence confirms the confidence interval.

yes the total will be same because in both of them the expression.is rearranged but their sign remains the same hence it hss just been shuffled