Assume that an IR, system returns a ranked list of 10 total documents for a given query. Assume that according to a gold-standard labelling there are 5 relevant documents for this query, and that the only relevant documents in the ranked list are in the 2nd, 3rd, 4th and 8th positions in the ranked results. Calculate and clearly show the interpolated precision value for each of the following standard recall levels: (0.0, 0.1 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0) for this individual query.

Step-by-step explanation:

Precision-Recall values corresponding to relevant documents positions

Doc. No    Recall         Precision

2nd          1/5 = 0.2      1/2 = 0.5

3rd           2/5 = 0.4     2/3 = 0.67

4th           3/5 = 0.6     3/4 = 0.75

8th           4/5 = 0.8     4/8 = 0.5

Going by the above results, the interpolation of Precision and Recall will be:

Interpolated Precision-Recall values

Recall    Precision

0.0.           0.75

0.1             0.75

0.2             0.75

0.3              0.75

0.4              0.75

0.5               0.75

0.6               0.75

0.7                 0.5

0.8                 0.5

0.9                   0

1.0                     0

50 min

step-by-step explanation:

they will be decreasing both their times in half

60 / 2 = 30

40/2 = 20

30 + 20 = 50

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841