Maths

So, radius, r = $\frac{40}{2}$ cm = 20 cm

Length of chord (AB) = 20cm

In OAB

OA = OB=AB=20 cm

Hence, AOAB is equilateral triangle and end of the angle is 60°

:. Ø =60° = 60 *
$\frac{\mathrm{}\mathrm{\pi }}{180}$ radian =$\frac{\mathrm{\pi }}{3}$ radian

Hence, length of minor are of the chord, l=rØ.

l = 20 * $\frac{\mathrm{\pi }}{3}$ cm

l = $\frac{2\mathrm{0\pi }}{3}$ cm.

4. Here l = 22cm.

r =100cm.

Ø =?

Hence by r = $\frac{1}{\mathrm{Ø}}$

= Ø = $\frac{l}{r}$ = $\frac{22}{100}$ radian

= $\frac{22}{100}$ * $\mathrm{180°}$/π

$\frac{\mathrm{}\mathrm{}}{}$$\frac{}{}$= $\frac{22}{100}$ * 180° * $\frac{7}{22}$

= ${\left(\frac{63}{5}\right)}^0$

=12 $\frac{3°}{5}$ = 12° $\frac{3*60\text{'}}{5}=12°36\text{'}$

3. Given that a wheel makes 360 revolutions in one minute

Then, number of revolutions in one second = $\frac{360}{60}$ =6.

In 1 complete revolution the wheel turns 360°= 2π radian.

So, In 6 revolution, the wheel will turns 6*2π radian = 12π radian.

Hence, in one second the wheel will turn an angle of 12π radian.

1. (i) 25°

Solution:We know that 180° = π radian.

Hence, 25° = $\frac{\pi }{180}$ 25 radian= $\frac{\mathrm{5\pi }}{36}$ radians.

(ii) 47°30′

Solution: We know that 180° = π radian,

Hence, -47°30′= -47 * $\frac{1}{2}$ degree= $\frac{-47}{2}$ * $\frac{\pi }{\mathrm{180°}}$ radians.

= $-\frac{1\mathrm{9\pi }}{72}$ radians

(iii) 240°

Solution:We know that, 180°= radian.

Hence, 240°= 240* $\frac{\pi }{\mathrm{180 }}$ radian.

= $\frac{\mathrm{4\pi }}{3}$ radian.

(iv) 520°

Solution: We know that, 180= radian.

Hence, 520°= 520°* $\frac{\pi }{\mathrm{180 }}$radian.

= $\frac{2\mathrm{6\pi }}{9}$ radian.

90. There are four entries in a determinant of 2 x 2 order. Each entry may be filled up in two ways with 0 or 1.

 Number of determinants that can be formed = (2)⁴ = 16

The value of determinants is positive in the following cases:

$\left|\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right|\left|\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right|\left|\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right|$

Therefore, the probability that the determinant is positive = 3/16

88. Let R be the event of drawing the red marble.

Let EA, EB, and EC respectively denote the events of selecting the box A, B, and C.

Total number of marbles = 40

Number of red marbles = 15

P (R) = 15/40 = 3/8

Probability of drawing the red marble from box A is given by P (EA|R).

$?P\left(E_A|R\right)=\frac{P\left(E_A?R\right)}{P\left(R\right)}=\frac{\frac{1}{40}}{\frac{3}{8}}=\frac{1}{15}$

Probability that the red marble is from box B is P (EB|R).

$?P\left(E_B|R\right)=\frac{P\left(E_B?R\right)}{P\left(R\right)}=\frac{\frac{6}{40}}{\frac{3}{8}}=\frac{2}{5}$

Probability that the red marble is from box C is P (EC|R).

$?P\left(E_C|R\right)=\frac{P\left(E_C?R\right)}{P\left(R\right)}=\frac{\frac{8}{40}}{\frac{3}{8}}=\frac{8}{15}$

86. Let the man toss the coin n times. The n tosses are n Bernoulli trials.

Probability (p) of getting a head at the toss of a coin is 1/2.

∴ p = 1/2 ⇒ q = 1/2

$\therefore P\left(X=x\right)={\mathrm{nC}}_x$
$p^{n-x}{q}^x={\mathrm{nC}}_x$
${\left(\frac{1}{2}\right)}^{n-x}{\left(\frac{1}{2}\right)}^x={\mathrm{nC}}_x$
${\left(\frac{1}{2}\right)}^n$

It is given that,

P (getting at least one head) > 90/100

P (x ≥ 1) > 0.9

⇒ 1 − P (x = 0) > 0.9

$1-{\mathrm{nC}}_0$
$.\frac{1}{2^n}>0.9\\ {\mathrm{nC}}_0$
$\begin{array}{l}.\frac{1}{2^n}<0.1\\ \frac{1}{2^n}<0.1\\ 2^n>\frac{1}{0.1}\\ 2^n>10......(1)\end{array}$

The minimum value of n that satisfies the given inequality is 4.

Thus, the man should toss the coin 4 or more than 4 times.