Answer:
6. To find the distance between Springfield and Bloomington, we can use the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
where (x1, y1) = (-74, -21) and (x2, y2) = (-29, 27). Substituting these values into the formula, we get:
d = sqrt((-29 - (-74))^2 + (27 - (-21))^2) = sqrt(45^2 + 48^2) = sqrt(4059) ≈ 63.7
Therefore, the distance between Springfield and Bloomington is approximately 63.7 miles.
7. With a confidence level of 95% and a margin of error of plus or minus 3.5 percentage points, we can interpret the results of the poll as follows: if the same poll were conducted many times, 95% of the time the percentage of Americans who support the proposed infrastructure plan would fall within 3.5 percentage points of the reported value of 29%. In other words, the true percentage of Americans who support the plan is likely to be between 25.5% and 32.5%, with a high degree of confidence.
8. To calculate the margin of error with a 95% confidence level, we can use the formula:
margin of error = zsqrt(p(1-p)/n)
where z is the z-score associated with a 95% confidence level (which is approximately 1.96), p is the proportion of students with significant credit card debt (which is 0.22), and n is the sample size (which is 200). Substituting these values into the formula, we get:
margin of error = 1.96sqrt(0.22(1-0.22)/200) ≈ 0.07
Therefore, the margin of error for this survey with a 95% confidence level is approximately 0.07, or 7%.
9. To find the sample size needed to achieve a margin of error of 4%, we can use the formula:
n = (zsqrt(p(1-p))/e)^2
where z is the z-score associated with a 95% confidence level (which is approximately 1.96), p is the proportion of students with significant credit card debt (which is 0.22), and e is the desired margin of error (which is 0.04). Substituting these values into the formula, we get:
n = (1.96sqrt(0.22(1-0.22))/0.04)^2 ≈ 1385
Therefore, the researcher would need to survey approximately 1385 students to achieve a margin of error of 4%.
Explanation: