Aresearcher compares the effectiveness of two different instructional methods for teaching physiology. a sample of 181 students using method 1 produces a testing average of 57.5. a sample of 227 students using method 2 produces a testing average of 65. assume the standard deviation is known to be 14.26 for method 1 and 7.48 for method 2. determine the 98% confidence interval for the true difference between testing averages for students using method 1 and students using method 2.

98% confidence interval using Method 1 is (55.01, 59.99) and Method 2 is (63.84, 66.16)

Step-by-step explanation:

Confidence Interval (CI) = mean + or - (t×sd)/√n

Method 1

Mean = 57.5, sd = 14.26, n = 181, degree of freedom = n - 1 = 181 - 1 = 180

t-value corresponding to 180 degrees of freedom and 98% confidence level is 2.3474

Lower bound = 57.5 - (2.3474 × 14.26)/√181 = 57.5 - 2.49 = 55.01

Upper bound = 57.5 + (2.3474 × 14.26)/√181 = 57.5 + 2.49 = 59.99

98% confidence interval using Method 1 is (55.01, 59.99)

Method 2

Mean = 65, sd = 7.48, n = 227, degree of freedom = n - 1 = 227 - 1 = 226

t-value corresponding to 226 degrees of freedom and 98% confidence level is 2.3434

Lower bound = 65 - (2.3434 × 7.48)/√227 = 65 - 1.16 = 63.84

Upper bound = 65 - (2.3434 × 7.48)/√227 = 65 - 1.16 = 6384

98% confidence interval using Method 2 is (63.84, 66.16)

geometry: triangles

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call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

25.45584

i believe the answer is c.