Even prime number = 2
∴ Probability = \(\frac{1}{2}\).

The height of the bar corresponding to conveyance is the maximum.

3 \(\times\)3 \(\times\) 3 = 27.

\(\frac{3}{7}+\left(\frac{-6}{11}\right)+\left(\frac{-8}{21}\right)+\left(\frac{5}{22}\right)\) 
\(=\frac{198}{462}+\left(\frac{-252}{462}\right)+\left(\frac{-176}{462}\right)+\left(\frac{105}{462}\right)\) (Note that 462 is the LCM of 7, 11, 21 and 22)
\(=\frac{198-252-176+105}{462}=\frac{-125}{462}\) 
We can also solve it as.
\(\frac{3}{7}+\left(\frac{-6}{11}\right)+\left(\frac{-8}{21}\right)+\frac{5}{22}\) 
\(=\left[\frac{3}{7}+\left(\frac{-8}{21}\right)\right]+\left[\frac{-6}{11}+\frac{5}{22}\right]\) (by using commutativity and associativity)
\(=\left[\frac{9+(-8)}{21}\right]+\left[\frac{-12+5}{22}\right]\) (LCM of 7 and 21 is 21; LCM of 11 and 22 is 22)
\(=\frac{1}{21}+\left(\frac{-7}{22}\right)=\frac{22-147}{462}=\frac{-125}{462}\) 

30

The given number is 17576.
Step 1 Form groups of three starting from the rightmost digit of 17576.
17 576. In this case one group i.e., 576 has three digits whereas 17 has only two digits.
Step 2 Take 576.
The digit 6 is at its one’s place.
We take the one’s place of the required cube root as 6.
Step 3 Take the other group, i.e., 17.
Cube of 2 is 8 and cube of 3 is 27. 17 lies between 8 and 27.
The smaller number among 2 and 3 is 2.
The one’s place of 2 is 2 itself. Take 2 as ten’s place of the cube root of 17576.
Thus, \(\sqrt[3]{17576}\) = 26

Here, 324 -1 = 323,323 - 3 = 320
320 - 5 = 315,315 -7 = 308
308 - 9 = 299, 299 - 11 = 288
288 -13 = 275,275 -15 = 260
260 - 17 = 243, 243 - 19 = 224
224 - 21 = 203,203 - 23 =180
180 - 25 =155, 155 - 27 =128
128 - 29 = 99, 99 - 31 = 68
68 - 33 = 35, 35 - 35 = 0
So, to get 0, we use 18 steps.
Hence, square root of 324 is 18.

To find the square root of 169, we subtract successive odd number starting from 1 as follows:
169 -1 = 168, 168 - 3 = 165,165 - 5 = 160
160 - 7 = 153, 153 - 9 = 144, 144 -11 = 133
133 -13 = 120, 120 -15 = 105, 105 -17 = 88
88 - 19 = 69, 69 - 21 = 48,48 - 23 = 25
25 - 25 = 0
We observe that the number 100 reduced to zero after subtracting first 13 odd numbers.
So, 169 is a perfect square.
\(\therefore \ \sqrt { 169 } =13\)
Hence, the square root of 169 is 13.

\(-{11\over8}\)

(a) The lower limit of the second class interval 20 - 30 is 20.
(b) The upper limit of the last class interval.ls 60.
(c) The frequency of the third class is 14.
(d) The interval 10 - 20 has a frequency of 15.
(e) The interval which has the lowest frequency is 20 - 30.
(f) The class size is 20 - 10 = 10.

(i) Here, maximum height of the bar is 32 for the interval 4 - 5, so the maximum number of students watching TV is 32 and they watch TV for 4 to 5 h.
(ii) It is clear from the given graph that, students watch TV for 1 - 2, 2 - 3, 3 - 4, 4 - 5, 5 - 6 and 6 - 7 h. So, students who watch TV for less than 4 h lie in the intervals 1 - 2, 2 - 3 and 3 - 4. Also, their corresponding frequencies are 4,8 and 22.
\(\therefore\) Number of students watching TV for less than 4 h
= 4 +8 + 22 =34
(iii) Students who watch TV for more than 5 h lie in the intervals 5 - 6 and 6 - 7. Also, their corresponding frequencies are 8 and 6. 
\(\therefore\) Number of students watching TV for more than 5 h
= 8 + 6 = 14.

We find the mean of the given rational numbers.
As given in the above example, the mean is \(\frac{3}{8} \text { and } \frac{1}{4}<\frac{3}{8}<\frac{1}{2} \text {. }\)
We now find another rational number between \(\frac{1}{4} \text { and } \frac{3}{8}\) For this, we again find the mean of \(\frac{1}{4} \text { and } \frac{3}{8}\). That is, \(\left(\frac{1}{4}+\frac{3}{8}\right) \div 2=\frac{5}{8} \times \frac{1}{2}=\frac{5}{16}\)
Now find the mean of \(\frac{3}{8} \text { and } \frac{1}{2}\). We have, \(\left(\frac{3}{8}+\frac{1}{2}\right) \div 2=\frac{7}{8} \times \frac{1}{2}=\frac{7}{16}\) 
Thus we get \(\frac{1}{4}<\frac{5}{16}<\frac{3}{8}<\frac{7}{16}<\frac{1}{2}\)
Thus, \(\frac{5}{16}, \frac{3}{8}, \frac{7}{16}\) are the three rational numbers between \(\frac{1}{4} \text { and } \frac{1}{2}\)

Yes; \(5 \times [(-7)\times (-8)]\)=\([5 \times (-7)]\times (-8)\) = 280.

1188 = 2 x 2 x 3 x 3 x 3 x 11
The primes 2 and 11 do not appear in groups of three. So, 1188 is not a perfect cube. In the factorisation of 1188 the prime 2 appears only two times and the prime 11 appears once. So, if we divide 1188 by 2 x 2 x 11 = 44, then the prime factorisation of the quotient will not contain 2 and 11.
Hence the smallest natural number by which 1188 should be divided to make it a perfect cube is 44.

By prime factorisation,
we have
175616 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 7x 7 x 7 = 2 x 2 x 2 x 7 = 56

Thus, the cube root of 175616 is 56.

71 = 70 + 1
Therefore, 71² = (70 + 1)²
= 70 (70 + 1) + 1 (70 + 1)
= 4900 + 70 + 70 + 1
= 5041

3^-2=\(\frac{1}{3^{2}}=\frac{1}{9}\)    [∵ \(a^{-m}=\frac{1}{a^{m}}\)]

We know that, if a number has 1 or 9 in the unit's place, then its square ends with digit 1. Therefore,161² and 109² would end with digit 1.

We know that, 289 = 17 x 17
\(\therefore\) Equivalent rational number of
\({-13\over17}={-13\times17\over17\times 17}={-221\over289}\)

There are ten separate slips having 1 to 10 numbers (one number on one slip) out of which one slip can be chosen in 10ways.
So, total number of outcomes = 10
Here, a slip containing the number 6 can be chosen in one way only.
So, favourable outcomes = 1
\(\therefore\) Probability of getting the number 6 = \(\frac{Favourable\ outcomes}{Total\ number\ of\ outcomes}=\frac{1}{10}\)

Here, number of balls is 5, which are of different colours. If we pull out a ball, then it may be R, W, B, G or Y. So, possible outcomes are R, W, B, G or Y.