1 |
2 |
3 |
4 |
3 |
8 |
5 |
16 |
3 |
4 |
Then, E | = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} |
∴ n(E) = 27.
∴ P(E) = | n(E) | = | 27 | = | 3 | . |
n(S) | 36 | 4 |
n(S) = 6, n(E) = (4, 6) = 2
? P(E) = 2/6 = 1/3
Here, A = ? 8800, T =2 yr, R = 5%
We know
SI = ART/(100 + RT) = (8800 x 5 x 2) / (100 + 5 x 2)
= (8800 x 10) / 110
= ? 800
3rd term = (2nd term) x 3 - 4 = 26 x 3 - 4 = 74.
4th term = (3th term) x 3 - 4 = 74 x 3 - 4 = 218.
5th term = (4th term) x 3 - 4 = 218 x 3 - 4 = 650.
∴ 5th term must be 650 instead of 654.
Other side = ?5 2 - 42
= ? 9
=3 m
So The area of the rectangular field = 4 x 3
= 12 m2
0.27 = (27 - 2)/90 = 25/90 = 5/18
For 50 students, food is sufficient for 45 days
? For 1 student, food is sufficient for 45 x 50 days
and for 75 students, food is sufficient for (45 x 50)/75 days. i,e., for 30 days.
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