A city will soon be voting on a bond issue which will provide the necessary funds for a new middle school. The school board wants to know how likely it is that the bond issue will pass. They randomly sample three groups of 200 people and ask each person whether they are for or against the bond issue. The results are recorded in the table below.
For Against
Sample #1 111 89
Sample #2 97 103
Sample #3 104 96

Based on the collected data, do you think the bond issue will pass or fail? How confident are you? Explain your reasoning using complete sentences.
Using the mean of the samples, approximately how many of the 15,000 voters would you expect to vote against the bond issue?

No as the average would be 58% and the decrease group of 42% could be biased across a category group like age, gender etc... Endorsement rates are required within data, this can distinguish other things like missing category groups to avoid bias. Other test scores are useful to ensure within test scores for the organizers that further data can be distinguished that can rule out whether the group membership has weight and significance by itself.  The general patterns that show different forms of bias are split into 2 one is predictive bias (category and influence) and the other is test based and shows more detail in the differentiating samples that appear to be similar but ultimately could affect the products success.

Step-by-step explanation:

111 97 104 89 103 96  + bias gn gp but cna make gp void by tests and not test predictions. We then total back each sum 111a-gn  89b-gn  97c-gn 103d-gn  104e-gn  96f-gn and we still have 3 groups divided by 2 divided by 2. Hereafter, we can assume when we multiply back we can only multiply to find the mean. lets say the gn =18. We can assume it would weigh heavy on the positive groups and becomes 40% average. which is 120/300 = 1 / 0.4   but with your data it is 158/300 =1 / 0.58 mean. For influenced bias 15,000.00 voters the unfair vote would be 0.58 x 15000 = 8700 / 15000 but with gn tests etc and void predictive bias. 0.18 x 8700 =  1566  8700-1566 = 7134 /15000 positive vote, where the negative vote under my predictive facts making predictive bias void and averaging text bias test data averaging 18% higher for positive votes, we can prove that the negative would be of significance as 15000 - 7134 = 7866 when percentages are made without negative and void data as 0.18 x 8700 can be used first instead to be correct- you will see that  (18% - 58% = 40% of 600 =240% we need )  0.40 x 15000 = 6000 and therefore 6000/15000 votes to me seems more a likely event. Based on prediction that everyone that considers voting actually votes and voted within a reasonable voting heavily marketed duration.

2x^2+5x+17

your welcome. : )

- both graphs increase on their domain.

- the graphs of both functions have the same x-intercept.

- neither graph has a y-intercept.