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There are 4 suits in a standard deck 52 cards.Number of ways of selecting 4 cards out of 13 cards( i.e. a suit) are

13!/(4!9!)Note that we can choose 4 cards out of any 4 possible suits.So total ways become:

P = 4 * ( 13!/9!4! )

Also, no of ways to select 4 cards out of 52 cards are:

Q = 52!/(4!48!)

Hence The probability is:

P/Q

= (4*13!/4!9!)/(52!/4!48!)

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in suits total 13 cardsnow we have to draw 4 different cardsso probability =(13*12*11*10)/(52*51*50*49)=(12*11)/(4*51*5*49)=(11)/(17*5*49)=11/4165

Solution:c) 10N

Explanation:Upthrust = Weight of water displaced= mass of water * g= (density*Volume) of water * g= 1000 *(1/1000)* 9.8= 9.8 NHere I have used Kg/m^3 for density and m^3 for volume.

745=45×6+25; 45/5=9,

cards are of same suit so(13C4)/(52C4)=(13×12×11×10)/(52×51×50×49)=(12×11)/(4×51×5×49)=(11/(17×5×49)=11/4165

Total No. Of cards =52

No.Of cards for each suit = 13

Total types Of French suit = 4(hearts,spades, diamonds,clubs)

Prob. = Conditional case /total case

For this conditional case of selecting all four same suit,it can be either hearts or spades or clubs or diamonds

So Total no. Of conditional case= no. of case of selecting all four cards of hearts suit +no.of case of selecting all four cards of spades suit +no.of case of selecting all four cards of clubs suit + no. of case of selecting all four cards of diamonds suit.

= 13C4+13C4+13C4+ 13C4

=4*(13C4)=2860

Total no. of case= 52C4=270725

So, probability of getting all four cards of the same suit=2860/270725

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A= lost card is spade

B= not spade

P(A)=13/52=1/4

P(B)=39/52=3/4

Let E be three cards drawn which is spade

P(E/A)=12/51*11/50*10/49=1320/124950=0.0105

P(E/B)=13/51*12/50*11/49=1716/124950=0.013

P(A/E)=P(E/A)P(A)/[P(A)*P(E/A)+P(E/B)*P(B)]

=0.0105*1/4/[1/4*0.0105+0.013*3/4]=0.0729

the average and.cloud not show

Let a and b are two adjacent sides,

h , H are their corresponding heights

of the parallelogram .

a = 15 cm , h = 6 cm

b = 12 cm , H = ?

Therefore ,

bH = ah

12 × H = 15 × 6

H = ( 15 × 6 ) /12

H = 7.5 cm

thank you

1 - quadrilateral2 - pentagon3 - heptagon4 - octagon5 - nonagon 6 - n-gon7 - decagon 8 - hexagon

(1)10sides.....,,,,,,,,,,,,,,,,,,,,,,,

Ans :- Thelaw of reflection statesthat the incident ray, thereflectedray, and the normal to the surface of the mirror all lie in the same plane. Furthermore, the angle of reflectionis equal to the angle of incidence .

THE cube of even number is even and cube of odd number is odd

AS we know that 3 x 3 x3 x 3=81so x=4is the answer

answers IS D CORRECT

D the cube of every even even every odd odd

D is correct answers

answer is d plz follow me or mark on best answer

D is correct answer I am also study 9th my exam is complete on Monday

the cube of every even number is even, and odd number is odd option D

Let the number be x

Its fifth part is = x/5

Its fourth part is= x/4

A.T.Q

(X/5) +30 = (X/4) - 30

(X/5) -(X/4) = -30-30

(4X-5X)/20 = -60

-X = -60 *20

-X= - 1200

X= 1200------------------------------------------------------------------------------------------------------The number is 1200

----------------------------------------------------------------------------------------------------

1200 is correct answer.

number = a; fourth part = a/4, fifth part = a/5, (a/5) + 30 = ( a/4) - 30; ( a/5)( a/4)=-30-30; (4a - 5a/20)=-60; (4a - 5a) = -60 × 20; -a = -1200,; a = 1200

1200 is the right answer

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the number is 1200 according to the above conditions.

1200 is the correct answer of the following question

best answer is the 1200

x = 1200 is the right answer

Let the required number be X this number fifth part increased by 30= X/5 +30and this number fourth part decreased by 30= X/4 -30=X/5 +30 = X/4 - 30=X/5 - X/4 = -30-30LCM of 5& 4 is 20=4X-5X/20 = -60=4X-5X = -60×20= -X = -1200= X = 1200So the required number is 1,200.

1200 is the answer of the following

1200 is the correct answer. please like as best

what happend bro

Let a be any odd positive integer.

We apply the division lemma with

a and b = 3 .

Since 0 ≤ r < 3 ,

the possible remainders are 0 , 1

and 2 .

That is , a can be 3q , or 3q + 1 , or

3q + 2 , where q is the quotient .

Now ,

a² = ( 3q )² = 9q²

Which can be written in the form

= 3 ( 3q² )

= 3m , [ since m = 3q² ]

It is divisible by 3 .

Again ,

a² = ( 3q + 1 )²

= 9q² + 6q + 1

= 3( 3q² + 2q ) + 1

= 3m + 1 [ since m = 3q² + 2q ,

3( 3q² + 2q ) is divisible by 3 ]

Lastly ,

a² = ( 3q + 2 )²

= 9q² + 12q + 4

= ( 9q² + 12q + 3 ) + 1

= 3( 3q² + 4q + 1 ) + 1

Which can be written in the form

3m + 1 , since

3( 3q² + 4q + 1 ) is divisible by 3.

Therefore ,

The square of any positive integer

is either of the form 3m or 3m + 1

for some integer m .

For knowing how many small boxes did a big box contain we should first get the volume of both boxesVolume of big box=length × breadth × height=(1×100)cm×60cm×40xm=240000cubic cmNow,volume of small boxes=5×8×10=400 cubic cmNo. of small boxes in the big box = volume of big box/volume of small box=240000/400=600 boxes

let ' a' be any positive integer and b = 3.

we know, a = bq + r , 0< r< b.

now, a = 3q + r , 0<r < 3.

the possibilities of remainder = 0,1 or 2

Case I - a = 3q

a²= 9q²

= 3 x ( 3q²)

= 3m (where m = 3q²)

Case II - a = 3q +1

a²= ( 3q +1 )²

= 9q²+ 6q +1

= 3 (3q²+2q ) + 1

= 3m +1 (where m = 3q²+ 2q )

Case III - a = 3q + 2

a²= (3q +2 )²

= 9q²+ 12q + 4

= 9q²+12q + 3 + 1

= 3 ( 3q²+ 4q + 1 ) + 1

= 3m + 1 where m = 3q²+ 4q + 1)

From all the above cases it is clear that square of any positive integer ( as in this case a²) is either of the form 3m or 3m +1.

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By Euclid’s division algorithm a = bq + r, where 0 ≤ r ≤ b Put b = 3 a = 3q + r, where 0 ≤ r ≤ 3 If r = 0, then a = 3q If r = 1, then a = 3q + 1 If r = 2, then a = 3q + 2 Now, (3q)^2 = 9q^2 = 3 × 3q^2 = 3m, where m is some integer (3q + 1)^2 = (3q)^2 + 2(3q)(1) + (1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 = 3m + 1, where m is some integer (3q + 2)^2 = (3q)^2 + 2(3q)(2) + (2)^2 = 9q^2 + 12q + 4 = 9q^2 + 12q + 4 = 3(3q^2 + 4q + 1) + 1 = 3m + 1, where m is some integer Hence the square of any positive integer is of the form 3m, or 3m +1 But not of the form 3m + 2

16856 is the correct answer.

the correct answer is 16856

16856 is the right answer

16856 is the right answer

c is the correct answer

16856 is correct answer. by the no of terms...

1 i sthe right answer

For knowing how many small boxes did a big box contain we should first get the volume of both boxesVolume of big box=length × breadth × height=(1×100)cm×60cm×40xm=240000cubic cmNow,volume of small boxes=5×8×10=400 cubic cmNo. of small boxes in the big box = volume of big box/volume of small box=240000/400=600 boxes

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Let the nos be x and (x+2) becuz they are consecutive even nos .

so:(x+2)2-x2=36

(x+2+x)(x+2-x)=36

(2x+2)(2)=36

2x+2=18

2x=16

x=8

how is is solved

u can search it in Google , Wikipedia and other search engines there u better know about this

Let the smaller natural number be x and larger natural number be yHence x2= 4y → (1)Given y2– x2= 45⇒ y2– 4y = 45⇒y2– 4y – 45 = 0⇒y2– 9y + 5y – 45 = 0⇒y(y – 9) + 5(y – 9) = 0⇒(y – 9)(y + 5) = 0⇒ (y – 9)= 0 or (y + 5) = 0∴ y = 9 or y = -5But y is natural number, hence y ≠ - 5Therefore, y = 9Equation (1) becomes,x2= 4(9) = 36∴ x = 6Thus the two natural numbers are 6 and 9.

Meets has more speed

solve the sum

Meera solves 28 sums in 4 hours.Shabnam solves 36 sums in 8 hours.

Shabnam solves 18 sums in 4 hours. ( Making it half).Merra solves 28 sums in 4 hours while shabnam solves 18 sums in 4 hours. Therefore Meera has more speed than shabnam because meera solves more sums than shabnam in 4 hours.

Distance travelled = perimeter of the park

Given,length = 140 mbreadth = 19 mPerimeter= 2 ( l + b )= 2 ( 140 + 19 ) m= 2 ( 159 ) m= 318 m

Now, number of rounds = 5Total distance covered = 5 × 318 = 1590 m

which one we have to do?

Let Charus age be xSo, Karan = (x+3) 6 years ago,(x+3-6)=4(x-6) x-3=4x-24 21=3xx=7 So Charu's age is 7 years and Karans age is 10 years.

Length (l) of cardboard = 8 cmBreadth (b) of cardboard = 5 cm

Area of cardboard including margin = l × b = 8 × 5 = 40 cm2

From the figure, it can be observed that the new length and breadth of the cardboard, when margin is not included, are 5 cm and 2 cm respectively.

Area of the cardboard not including the margin = 5 × 2 = 10 cm^2

Area of the margin = Area of cardboard including the margin − Area of cardboard notincluding the margin= 40 − 10 = 30 cm^2