![\[\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{9}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{1}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c30dd6bfab0df4e91e525631f2ba2f4a_l3.png "Rendered by QuickLaTeX.com")
Given: \[\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{9}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{1}\] \[\frac{{{x}^{2}}}{\frac{1}{4}}+\frac{{{y}^{2}}}{\frac{1}{9}}=1\]….(i) Since, ...
, find the values of tan 2xAnswer: Replacing x by 2x, we get tan 2x = sin 2x / cos 2x Putting the values of sin 2x and cos 2x, we get
, find the values of cos 2xAnswer: We know that, cos 2x = 2cos2 x – 1 Putting the value, we get
, find the values of sin 2xAnswer: We know that, sin2x = 2 sinx cosx …(i) Here, we don’t have the value of sin x. So, firstly we have to find the value of sinx We know that, cos2 x + sin2 x = 1 Putting the values, we get...
, find the values of tan 2xAnswer: We know that: Replacing x by 2x, we get tan 2x = sin 2x / cos 2x Putting the values of sin 2x and cos 2x, we get
, find the values of cos 2xAnswer: We know that, cos 2x = 1 – 2sin2 x Putting the value, we get
, find the values of sin 2xAnswer: Given: $\sin x=\frac{\sqrt{5}}{3}$ To find: sin2x We know that, sin2x = 2 sinx cosx …(i) Here, we don’t have the value of cos x. So, firstly we have to find the value of cosx We know that,...
![\[\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{16}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{144}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-862342dfce29b8b9c9392c90bd44304d_l3.png "Rendered by QuickLaTeX.com")
Given: \[\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{16}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{144}\] Divide by \[144\] to both the sides, we get...
is 
Solution: Option(C) is correct. To Find: The Principle value of $\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$ Let the principle value be given by $\mathrm{x}$ Now, let $x=\cos...
Given: \[\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{16}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{144}\] Divide by \[144\] to both the sides, we get...
Given: \[\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{16}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{144}\] Divide by \[144\] to both the sides, we get...
![\[{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b1c4d22d5b39a8859feb9ee71c371bb0_l3.png "Rendered by QuickLaTeX.com")
Given \[{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...
Given \[{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...
Given \[{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...
![\[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{25}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{400}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-aed3022a6a6b09b7823a798e1b61f3a4_l3.png "Rendered by QuickLaTeX.com")
Given: \[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{25}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{400}\] Divide by \[400\] to both the sides, we get...
Given: \[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{25}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{400}\] Divide by \[400\] to both the sides, we get...
Given: \[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{25}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{400}\] Divide by \[400\] to both the sides, we get...
![\[\frac{{{x}^{2}}}{49}+\frac{{{y}^{2}}}{36}=1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-07b8fa9753bba0be6e51668e79506671_l3.png "Rendered by QuickLaTeX.com")
Given: \[\frac{{{x}^{2}}}{49}+\frac{{{y}^{2}}}{36}=1\]…(i) Since, \[49\text{ }>\text{ }36\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] …(ii)...
Given: \[\frac{{{x}^{2}}}{49}+\frac{{{y}^{2}}}{36}=1\]…(i) Since, \[49\text{ }>\text{ }36\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] …(ii)...
![\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d32e615a6517b76e65b04fd25ae8bf0a_l3.png "Rendered by QuickLaTeX.com")
Given: \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]…(i) Since, \[25>9\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]…(ii)...
Given: \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]…(i) Since, \[25>9\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]…(ii)...
Answer:
Answer: Taking LHS
Answer:
Answer: Taking LHS
, prove that 
Answer: cosx + cosy = 1/3--------- i sinx + siny = 1/3----------ii dividing ii by i we get Using the formula, sinA + sinB = 2sin(A+B)/2 . cos(A-B)/2 cosA + cosB = 2cos(A+B)/2 ....
Answer; Taking LHS = 3/16
= 3/16
Answer: = 1/16
Answer: = cot 5x Using the formulas, 2cosAsinB = sin (A + B) – sin (A - B) 2sinAsinB = cos (A - B) – cos (A + B)
Answer: = tan 2x Using the formulas, 2cosAsinB = sin (A + B) – sin (A - B) 2cosAcosB = cos (A + B) + cos (A - B) 2sinAsinB = cos (A - B) – cos (A + B)
Answer: =Sin3x+sin2x-sinx = (sin3x- sinx)+sin2x = ( 2cos(3x+x)/2 . sin(3x-x)/2 ) + sin2x = 2cos2xsinx +sin2x = 2cos2xsinx + 2sinxcosx = 2sinx (cos2x + cosx ) = 2sinx (2cos(2x+x)/2 . cos(2x-x)/2 ) =...
Answer: Using the formula, sinA + sinB = 2sin(A+B)/2 . cos(A-B)/2 sinA - sinB = 2cos(A+B)/2 . sin(A-B)/2
Answer: Using the formula, cosA - cosB = -2sin(A+B)/2 . sin(A-B)/2 cosA + cosB = 2cos(A+B)/2 . cos(A-B)/2
Answer: Using the formula, cosA - cosB = -2sin(A+B)/2 . sin(A-B)/2 sinA - sinB = 2cos(A+B)/2 . sin(A-B)/2
Using the formula, cosA - cosB = -2sin(A+B)/2 . sin(A-B)/2 sinA - sinB = 2cos(A+B)/2 . sin(A-B)/2
Answer: = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = (2sin(3x+x)/2 . cos(3x-x)/2) sin x + (-2sin(3x+x)/2 . sin(3x-x)/2) cosx = (2sin2x cosx) sinx-(2sin2x sinx) cosx = 0. Using the formula,...
Answer: L.H.S cot 4x (sin 5x + sin3x) = cot 4x (2sin(5x+3x)/2 . cos (5x-3x)/2 ) = cot 4x (2 sin4x cosx) = (cos4x / sin 4x).(2 sin4x cosx) = 2cos4xcosx R.H.S cot x (sin 5x - sin3x) = cot x...
Answer: = sin 6x / cos 6x = tan 6x Using the formula, sinA + sinB = 2sin(A+B)/2 . cos(A-B)/2 cosA + cosB = 2cos(A+B)/2 . cos(A-B)/2
Answer: Using the formula, cosA - cosB = -2sin(A+B)/2 . sin(A-B)/2 sinA - sinB = 2cos(A+B)/2 . sin(A-B)/2
Answer: = tan4x Using the formula, sinA + sinB = 2sin(A+B)/2 . cos(A-B)/2 cosA + cosB = 2cos(A+B)/2 . cos(A-B)/2
Answer: = sin x / cos x = tanx Using the formula, sinA - sinB = 2cos(A+B)/2 . sin(A-B)/2 cosA + cosB = 2cos(A+B)/2 . cos(A-B)/2
Answer: = cos x / sin x = cot x Using the formula, sinA + sinB = 2sin (A+B)/2 . cos (A-B)/2 cosA - cosB = -2sin (A+B)/2 . sin (A-B)/2
Answer: iii) 2cos7xcos3x = cos (7x+3x) + cos (7x – 3x) =cos10x + cos 4x Using, 2cosAcosB = cos (A+ B) + cos (A - B) iv)2sin8xsin2x = cos (8x - 2x) – cos (8x + 2x) = cos6x – cos10x Using, 2sinAsinB =...
Answer: i) 2sin 6x cos 4x = sin (6x+4x) + sin (6x-4x) = sin 10x + sin 2x Using, 2sinAcosB = sin (A+ B) + sin (A - B) ii) 2cos 5x sin3x = sin (5x + 3x) – sin (5x – 3x) = sin8x – sin2x Using,...
Answer:
Answer:
Answer: In this question the following formulas will be used:
Answer: In this question the following formulas will be used: sin (A - B) = sinA cos B - cosA sinB cos (A - B) = cosAcosB+ sinAsinB
Answer: In this question the following formulas will be used: cos (A + B) = cosAcosB – sinAsinB cos (A - B) = cosAcosB+ sinAsinB = cos x + cos 120 cosx – sin120sinx + cos 120 cosx+sin120sinx = cosx...
Answer: In this question the following formula will be used: Sin( A +B)= sinA cos B + cosA sinB Sin( A - B)= sinA cos B - cosA sinB = sin150 cosx + cos 150 sinx + sin150 cosx – cos150 sinx =2sin150...
Sol: Steps of Construction with base AB’ = 8 cm. Step 1: Draw a line segment AB = 6cm. Step 2: At A, draw ÐXAB = 30°. Step 3: At B, draw ÐYBA = 60°. Suppose AX and BY intersect at C. Thus, DABC is...
Answer : Given: We have 4 letters and 5 letter boxes To Find: Number of ways of posting letters. One letter can be posted in any of 5 letter boxes. We have to assume that all the letters are...
of men and 
of women have grey hair. A grey-haired person is selected at random. What is the probability of this person being male? Assume that there is an equal number of males and females.Let MG : Men having grey hair WG: Women having grey hair G: Having grey hair Given an equal number of males and females. So let's assume both the probability be $\frac{1}{2}$ We want to find $P(M G...
of the boys and 
of the girls have an IQ of more than 
In this class, 
of the students are boys. If a student is selected at random and found to have and IQ of more than 150, find the probability that the student is a boy.Let, I : students having IQ more than 150 B : Boys in the class G: Girls in the class We want to find $P(B \mid I)$ i.e. probability that selected student having IQ greater than 150 is a boy...
of boys and 
of girls are taller than 
meters. Furthermore, of the students are girls. If a student is selected at random and is taller than meters, what is the probability that the selected student is a girl?Let, T :students taller than $1.75$ B: Boys in class G: Girls in class We want to find $P(G \mid T)$, i.e. probability that selected taller is a girl $\begin{array}{l} \mathrm{P}(\mathrm{G} \mid...
and B. plant A produces 
and plant B produces 
of the total product. 
of the scooters produced at pant and 
of the scooters produced at plant 
are of standard quality. A scooter produced by the company is selected at random, and it is found to be of standard quality. What is the probability that it was manufactured at plant A?Let S : Standard quality We want to find $\mathrm{P}(\mathrm{A} \mid \mathrm{S})$, i.e. probability that selected standard scooter is from plant A $\mathrm{P}(\mathrm{A} \mid...
, manufacture 
and 
of the total production respectively. Of their respective outputs, 
and are defective. A bulb is drawn at random from the total product, and it is found to be defective. Find the probability that it was manufactured by machine 
.Let D : Bulb is defective We want to find $P(C \mid D)$, i.e. probability that the selected defective bulb is manufactured by $C$ $\mathrm{P}(\mathrm{C} \mid \mathrm{D})=\frac{\mathrm{P}(\mathrm{C})...
Answer: Given, A(3, 2, 0) B(5, 3, 2) C(-9, 6, -3)
Answer: The vertices of the parallelogram be A(3, 4, -3), B(7, 10, -3) and C(5, -2, 7), and the fourth coordinate be D(a,b,c). The property of parallelogram is the diagonal bisect each other. The...
Answer: The centroid of a triangle $ = \left( {\frac{{{x_2} + {x_1} + {x_1}}}{3},\frac{{{y_2} + {y_1} + {y_2}}}{3},\frac{{{z_2} + {z_1} + {z_3}}}{3}} \right)$ The points are A(2, -4, 3) and B(3, -1,...
Answer: The centroid of a triangle $ = \left( {\frac{{{x_2} + {x_1} + {x_1}}}{3},\frac{{{y_2} + {y_1} + {y_2}}}{3},\frac{{{z_2} + {z_1} + {z_3}}}{3}} \right)$ The points are A(a,1,3) and B(-2, b,...
Answer: The midpoints of the sides of a triangle are (1, 5, -1), (0, 4, -2) and (2, 3, 4). Let its vertices be A(x1,y1,z1), B(x2,y2,z2), C(x3,y3,z3) . The mid point of AB is (1,5,-1). $\frac{{{x_2}...
Answer : To Find: Value of n Given: (n+1)! = 12× [(n-1)!] Formula Used: n! = (n) × (n-1) × (n-2) × (n-3)............. 3 × 2 × 1 Now, (n+1)! = 12× [(n-1)!] ⇒ (n+1) × (n) × [(n-1)!] = 12 × [(n-1)!] ⇒...
Given : let $\mathrm{H}$ be head, and $\mathrm{T}$ be tails where as $1,2,3,4,5,6$ be the numbers on the dice which are thrown when a head comes up or else coin is tossed again if its tail....
Given : let $\mathrm{H}$ be head, and $\mathrm{T}$ be tails where as $1,2,3,4,5,6$ be the numbers on the dice which are thrown when a head comes up or else coin is tossed again if its tail....
and 
be two the switches and let their probabilities of working be given by 
and 
Find the probability that the current flows from terminal A to terminal , when 
and 
are installed in parallel, as shown below:Solution: Given: $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ are two swiches whose probabilities of working be given by $\mathrm{P}\left(\mathrm{S}_{1}\right)=\frac{2}{3}$ and...
and be the two switches and let their probabilities of working be given by 
and 
. Find the probability that the current flows from the terminal A to terminal B when and are installed in series, shown as follows:Solution: Given: $S_{1}$ and $S_{2}$ are two swiches whose probabilities of working be given by $\mathrm{P}\left(\mathrm{S}_{1}\right)=\frac{4}{5} \text { and }...
and 
respectively. What is the probability that at least one shot hits the plane?Given:Let $A, B, C$ and $D b e$ first second third and fourth shots whose probability of hitting the plane is given i.e, $\mathrm{P}(\mathrm{A})=0.4, \mathrm{P}(\mathrm{B})=0.3,...
Answer: The coordinates of point R that divides the line segment joining points P (x1, y1, z1) and Q (x2, y2, z2) in the ratio m: n are point A( 3, 2, 5 ) and B( -4, 2, -2 ), m and n are 4 and 3....
Answer: The coordinates of point R that divides the line segment joining points P (x1, y1, z1) and Q (x2, y2, z2) in the ratio m: n are point A( 2, 1, -3 ) and B( 5, -8, 3 ), m and n are 2 and 1....
Answer: The coordinates of point R that divides the line segment joining points P (x1, y1, z1) and Q (x2, y2, z2) externally in the ratio m: n are point A( -2, 4, 7 ) and B( 3, -5, 8 ), m and n are...
Answer: Let the ratio be k:1 in which point R divides point P and point Q, where m and n are k and 1. The point which this formula gives is already given. R(5,4,-6) and the joining points are P(3,...
and 
, respectively. What is the probability that the machine will fail?Given: let $A, B$ and $C$ be the three components of a machine which works only if all its three compenents function.the probabilities of the failures of $A, B$ and $C$ respectively is given i.e,...
Answer: Let the ratio be k:1 in which point R divides point P and point Q, where m and n are k and 1. The point which this formula gives is already given. R(5,9,-14) and the joining points are P(2,...
Answer: Let the plane XZ divides the points A(-1, -3, 4) and B(4, 2, -1) in ratio k:1. On XZ plane, Y co- ordinate of every point be zero. Formula - Using the above formula, $\begin{array}{l} =...
Answer: Let the plane XY divides the points A(3,4,1) and B(5, 1, 6) in ratio k:1. On XY plane, Z co- ordinate of every point be zero. Formula - Using the above formula, $\begin{array}{l} = \left(...
Answer: The plane x – 2y + 3z = 5 divides the join of A(3, -5, 4) and B(2, 3, -7) in ratio k:1. The point which will come by section formula will be in the plane. Putting that in the plane equation...
What is the probability thatGiven: Let $A$ and $B$ be two fire extinguishing engines. The probability of availability of each of the two fire extinguishing engines is given i.e., $\mathrm{P}(\mathrm{A})=0.95$ and...
and 
. In the process of manufacture of part X. 8 out of 100 parts may be defective. Similarly, 5 out of 100 parts of may be defective. Calculate the probability that the assembled product will not be defective.Given: $X$ and $Y$ are the two parts of a company that manufactures an article. Here the probability of the parts being defective is given i.e, $\mathrm{P}(\mathrm{X})=\frac{8}{100}$ and...
in these courses are 
and 
respectively. Find the probabilities that Neelam receives exactly 2 A grades.Given : let $A, B$ and $C$ represent the subjects physics,chemistry and mathematics respectively ,the probability of neelam getting $A$ grade in these three subjects is given i.e, $P(A)=0.2,...
in these courses are and respectively. Find the probabilities that Neelam receivesGiven : let $A, B$ and $C$ represent the subjects physics,chemistry and mathematics respectively ,the probability of neelam getting $A$ grade in these three subjects is given i.e, $P(A)=0.2,...
can hit 2 times in 3 shots. Calculate the probability that
and 
all hit the target
and hit and 
does not hit the target.Given : let $A, B$ and $C$ chances of hitting a target is given i.e, $P(A)=\frac{4}{5}, P(B)=\frac{3}{4}$ and $P(C)=\frac{2}{3}$ $\Rightarrow \mathrm{P}(\bar{A})=\frac{1}{5},...
solving a problem are 
and 
, respectively. If all the three try to solve the problem simultaneously, find the probability that exactly one of them will solve it.Given : let $A, B$ and $C$ be three students whose chances of solving a problem is given i.e, $P(A)=\frac{1}{3}, P(B)=\frac{1}{4}$ and $P(C)=\frac{1}{6}$ $\Rightarrow...
Answer : It is also in the form of a circle, So we need to arrange 16flowers in Circle 16 flowers can be arranged by 15! Now each flower have the same neighbour in the clockwise and anticlockwise...
and , respectively. Find the probability that the problem is solved.Given : let $A, B$ and $C$ be three students whose chances of solving a problem is given i.e, $P(A)=\frac{1}{4}, P(B)=\frac{1}{5}$ and $P(C)=\frac{1}{6}$ $\Rightarrow...
, and the probability that B can solve the same problem is \%, find the probability that and will solve the problemGiven : Here probability of $A$ and $B$ that can solve the same problem is given, i.e., $P(A)=\frac{2}{3}$ and $P(B)=\frac{3}{5} \Rightarrow P($ $\bar{A})=\frac{1}{3}$ and...
Answer : We know that necklace in the form of a circle, So we need to arrange 20 pearls in Circle 20 pearls can be arranged by 19! Now each pearl have the same neighbour in the clockwise and...
Answer : By using the formula (n-1)! (Mention in Solution-1) So 8 persons can be arranged by 7! Now each person have the same neighbour in the clockwise and anticlockwise arrangement Total number of...
and that of B’s selection is 
Find the probability thatGiven : $A$ and $B$ appear for an interview, then $P(A)=\frac{1}{6}$ and $P(B)=\frac{1}{4} \Rightarrow P(\bar{A})=\frac{5}{6}$ and $P(\bar{B})=\frac{3}{4}$ Also, $A$ and $B$ are independent. A and...
and that of B’s selection is Find the probability thatGiven : $A$ and $B$ appear for an interview, then $P(A)=\frac{1}{6}$ and $P(B)=\frac{1}{4} \Rightarrow P(\bar{A})=\frac{5}{6}$ and $P(\bar{B})=\frac{3}{4}$ Also, $A$ and $B$ are independent. A and...
, and that of Ved’s rejection is 
Find the probability that at least one of them will be selected.Given : let A denote the event 'Arun is selected' and let B denote the event 'ved is selected'. Therefore, $\mathrm{P}(\mathrm{A})=\frac{1}{4}$ and $\mathrm{P}(\bar{B})=\frac{2}{3} \Rightarrow...
Answer : Let suppose 4 candidates be C1, C2, C3, C4 and 5 men be M1, M2, M3, M4, M5 Now M1 choose any one candidates from four (C1, C2, C3, C4) and give the vote to him by any 4 ways Similarly, M2...
, and that of Vimal’s selection is 
Find the probability that only one of them will be selected.event 'vimal is selected'. Therefore, $\mathrm{P}(\mathrm{A})=\frac{1}{3}$ and $\mathrm{P}(\mathrm{B})=\frac{1}{5}$ Also, $A$ and $B$ are independent .A and not $B$ are independent, not $A$ and $B$...
Answer : Let suppose 4 prizes be P1, P2, P3, P4 and 3 boys be B1, B2, B3 Now P1 can be distributed to 3 boys(B1, B2, B3) by 3 ways, Similarly, P2 can be distributed to 3 boys(B1, B2, B3) by 3 ways,...
Answer: (x1,y1,z1) = (1, 2, 3) (x2,y2,z2) = (-1, -2, -1) (x3,y3,z3) = (2, 3, 2) (x4,y4,z4) = (4, 7, 6) $\begin{array}{l} Length AB = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2}...
Answer: (x1,y1,z1) = (2, 3, 5) (x2,y2,z2) = (-4, 7, -7) (x3,y3,z3) = (-2, 1, -10) (x4,y4,z4) = (4, -3, 2) $\begin{array}{l} Length PQ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} +...
and be the events such that 
and 
or 
.Given: $A$ and $B$ are the events such that $P(A)=\frac{1}{2}$ and $P(B)=\frac{7}{12}$ and $P(\operatorname{not} A$ or $\operatorname{not} B)=\frac{1}{4}$ To Find: i)if A and B are mutually...
Answer: (x1,y1,z1) = (1, 3, 4) (x2,y2,z2) = (-1, 6, 10) (x3,y3,z3) = (-7, 4, 7) (x4,y4,z4) = (-5, 1, 1) $\begin{array}{l} Length PQ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2}...
Answer: (x1,y1,z1) = (3, 2, -5) (x2,y2,z2) = (-3, 8, -5) (x3,y3,z3) = (-3, 2, 1) (x4,y4,z4) = (-1, 4, -3) $\begin{array}{l} Length AD = \sqrt {{{({x_4} - {x_1})}^2} + {{({y_4} - {y_1})}^2} +...
Answer: (x1,y1,z1) = (-2, 3, 5) (x2,y2,z2) = (1, 2, 3) (x3,y3,z3) = (7, 0, -1) $\begin{array}{l} Length AB = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \\...
Answer: (x1,y1,z1) = (3, -5, 1) (x2,y2,z2) = (-1, 0, 8) (x3,y3,z3) = (7, -10, -6) $\begin{array}{l} Length AB = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}}...
Answer: (x1,y1,z1) = (3, -2, 4) (x2,y2,z2) = (1, 1, 1) (x3,y3,z3) = (-1, 4, -2) $\begin{array}{l} Length PQ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \\...
Answer: Let us take, C(x,y,z) point equidistant from points A(-1, 2, 3) and B(3, 2, 1). ∴ AC = BC $\sqrt {{{(x + 1)}^2} + {{(y - 2)}^2} + {{(z - 3)}^2}} = \sqrt {{{(x - 3)}^2} + {{(y - 2)}^2} +...
Answer: Let us take, C(0,y,0) point which lies on y axis and is equidistant from points A(3, 1, 2) and B(5, 5, 2). ∴ AC = BC $\sqrt {{{(0 - 3)}^2} + {{(y - 1)}^2} + {{(0 - 2)}^2}} = \sqrt {{{(0 -...
Answer: Let us take, C(0,0,z) point which lies on z axis and is equidistant from points A(1, 5, 7) and B(5, 1, -4). ∴ AC = BC $\sqrt {{{(0 - 1)}^2} + {{(0 - 5)}^2} + {{(z - 7)}^2}} = \sqrt {{{(0 -...
Answer : Let Suppose Letterbox be B1, B2 and letters are L1, L2, L3 So L1 can be posted in any 2 letterboxes (B1, B2) by 2 ways Similarly, L2 can be posted in any 2 letterbox (B1, B2) by 2 ways...
Answer: Let us take, D(x,y,z) point equidistant from points A(a, 0, 0), B(0, b, 0), C(0, 0, c) and O(0, 0, 0). ∴ AD = OD $\sqrt {{{(x - a)}^2} + {{(y - 0)}^2} + {{(z - 0)}^2}} = \sqrt {{{(x -...
Answer: The general point on yz plane is D(0, y, z). Consider this point is equidistant to the points A(3, 2, -1), B(1, -1, 0) and C(2, 1, 2). ∴ AD = BD $\sqrt {{{(0 - 3)}^2} + {{(y - 2)}^2} + {{(z...
Answer: The general point on xy plane is D(x, y, 0). Consider this point is equidistant to the points A(2, 0, 3), B(0, 3, 2) and C(0, 0, 1). ∴ AD = BD $\sqrt {{{(x - 2)}^2} + {{(y - 0)}^2} + {{(0 -...
Answer : As there is 5 banana, So suppose it as B1, B2, B3, B4, B5 And Let the Boy be A1, A2, A3 So B1 can Be distributed to 3 Boys (A1, A2, A3) by 3 ways, Similarly, B2, B3, B4, B5 Can be...
Answer : The first marble can be put into the pockets in 6 ways, i.e. Choose 1 Pocket From 6 by 6C1=6 Similarly second, third, Fourth, fifth & Sixth marble. Thus, the number of ways in which the...
Answer : To find: Number of 5 - digit numbers that can be formed 2 numbers are of 1 kind, and 2 are of another kind Total number of permutations = 30 number can be formed
Answer : To find: number of ways letters can be arranged such that all S’s are together Let all S’s be represented by a single letter Z New word is AAINATIONZ Number of arrangements = Letters can be...
Answer : To find: number of words such that C and T are never together Number of words where C and T are never the together = Total numbers of words - Number of words where C and T are together...
Answer : To find: number of words where L do not come together Let the three L’s be treated as a single letter say Z Number of words with L not the together = Total number of words - Words with L’s...
Answer : To find: Number of distinct signals possible Total number of fags = 7 2 are of 1 kind, 3 are of another kind, and 2 are of the 3rd kind ⇒ Number of distinct signals Hence 210 different...
Answer : To find: number of 3 digit numbers he can make If all were distinct, he could have made 3! = 6 numbers But 2 number are the same So the number of possibilities He can make 3 three - digit...
Answer : (i) two of them, Rajan and Tanvy, are always together Consider Rajan and Tanvy as a group which can be arranged in 2! = 2 ways. The 3 children with this 1 group can be arranged in 4! = 24...
Answer : (i) There is no restriction on letters The word VOWELS contain 6 letters. The permutation of letters of the word will be 6! = 720 words. Each word begins with Here the position of letter E...
Answer : To find: the number of permutations of 10 objects, taken 4 at a time. Formula Used: Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is...
Answer : (i)To distribute 3 prizes among 4 girls where no girl gets more than one prize the possible number of permutation possible are:4P3=24 To distribute 3 prizes among 4 girls where a girl may...
Answer : Given: code consists of digits 3, 5, 6, 9. To find: the largest possible number of trials necessary to obtain the correct code. The customer remembers that this 4 digit code consists of...