In detail: Based on the central limit theorem, the distribution appears normal due to the large sample size. The confidence interval is presented in the format: (Sample mean - margin of error, sample mean + margin of error). The sample mean, denoted as x, serves as the point estimate for the population mean. The confidence interval is computed as: mean ± z × σ/√n, where σ represents the population standard deviation. The formula transforms into confidence interval = x ± z × σ/√n, with specific values: x = $75, σ = $24. To find the z score, we subtract the confidence level from 100% which gives α as 1 - 0.96 = 0.04; halving this results in α/2 = 0.02, signifying the tail areas. To ensure we account for the center area, we have 1 - 0.02 = 0.98, corresponding to a z score of 2.05 for the 96% confidence level. The confidence interval becomes 75 ± 2.05 × 24/√64 = 75 ± 2.05 × 3 = 75 ± 6.15. The lower limit is 75 - 6.15 = 68.85, while the upper limit stands at 75 + 6.15 = 81.15. For n = 400, with x = $75 and σ = $24, the z score remains 2.05, resulting in the confidence interval calculated as 75 ± 2.05 × 24/√400 = 75 ± 2.05 × 1.2 = 75 ± 2.46. Subsequently, the lower bound becomes 75 - 2.46 = 72.54, and the upper limit adds up to 75 + 2.46 = 77.46. Lastly, when n = 400, x = $200, and σ = $80, the z score tied to a 94% confidence level is 1.88. Thus, the confidence interval is expressed as 200 ± 1.88 × 80/√400 = 200 ± 1.88 × 4 = 200 ± 7.52, giving us a margin of error of 7.52.
The diagrams for parts A and C are included here. For part B, we have circle O. We begin by drawing two radii OA and OC, connecting points A and C to create chord AC. The radius intersects chord AC at point B, bisecting AC into equal segments AB and BC. This gives us two triangles, ΔOBA and ΔOBC, where OA equals OC (since they're radii), OB equals OB (by the reflexive property), and AB is equal to BC (as stated in the question). By applying the SSS triangle congruence criterion, we conclude that ΔOBA is congruent to ΔOBC, allowing us to deduce that ∡OBA equals ∡OBC, both measuring 90°. Thus, OB is perpendicular to AC. Moving on to part D, we again work with circle O and draw the two radii OA and OC, joining points A and C to create chord AC. The radius intersects AC at point B, where AB is perpendicular to AC, meaning ∡B equals 90°. We then consider the right triangles ΔOBA and ΔOBC, and given OA equals OC (the radii), and OB equals OB (reflexive property), we conclude through the HL triangle congruence that ΔOBA is congruent to ΔOBC. Consequently, we find BA equal to BC, thus OB bisects AC.
