A fair six – faced die is rolled. Deduce the probability of obtaining: (a) 4 (b) 1 or 2 (c ) an odd number (d) an even number (e) a 5 or 6

(f) 3 or more ( g) less than 5 (h) a prime number (i) more than 6 (j)
2 ½

substitute,

circle 1 = 5

circle 2 = 20

circle 3 = 44

in this case no matter how you add, you will always get a perfect square..

hence proved

happy to

pls mark as brainliest

the answer is 0052

0 0 5 2

5 0 2 6 3 5

− 0

2 6

− 0

2 6 3

− 2 5 0

1 3 5

− 1 0 0

3 5

look at the picture.

1.

ef : ab = 2 : 3

gf : cb = 2 : 3

gh : cd = 2 : 3

he : da = 2 : 3

2.

∠a ≅ ∠e

∠b ≅ ∠f

∠c ≅ ∠g

∠d ≅ ∠h

3.

use pythagorean theorem:

the length of the abcd trapezoid sides:

|ab| = 3 cm

|bc| = 10 cm

|cd| = 3(3) cm = 9 cm

|da| = 8 cm

the perimeter of abcd = 3 + 10 + 9 + 8 = 30 cm

the ratio of perimeters of the tarapezoids is equal 2 : 3.

substitute:

      cross multiply

      divide both sides by 3