amygdaleon305

Post the diagram along with the question..

This is a really question so I'll give you some hints... 

There're 14 females and 13 males, so total = 27

Commitee of 9 can be chosen in total  = $^{27} C_9$ ways

And, 3 females and 6 males can be chosen = $^{14} C_3 ×{ ^{13}C_6}$ ways

Hope you got it :) 

This is blatantly a homework question so I'll just give a few hints...

First find the no. of routes to (5,7) and (4,3) and then subtract the latter from the former got it?

So, to get to (5,7) Marvin must take 5 steps right and 7 up in any order.

Similarly to get to (4,3) he does 4 right and 3 up in any order.

Now, no. of routes to (5,7)$={12!\over 5!7!}$

Similarly you can find the no. of routes to (4,3). 

There you go, I've given enough hints for this one..😏

Let  $f(a) = a^3+2a^2-9a-18$

⇒ (a + 2) is a factor of f(a) 

So dividing f(a) by (a+2) we get 

$f(a) = (a+2)(a^2-9)$

         $=(a+2)(a+3)(a-3)$

∴ Factors = -2, 3, -3

First group them into a frequency-distribution table

                x                                   f                                   c.f                            c.f (without outliers)

               20                                  1                                   1                                       

               60                                  1                                   2                                             1

               65                                  2                                   4                                             3

               70                                  2                                   6                                             5

               75                                  4                                  10                                            9

               80                                  1                                   11                                           10

               85                                  2                                   13                                           12

               90                                  1                                   14                                           13

               95                                  1                                   15                                           14

              229                                 1                                   16

                                                ∑f = 16

a) Mean $={20+60+65(2)+70(2)+75(4)+80+85(2)+90+95+229\over 16}$

    Mode = 75 

    Median is at (16/2)^th and (16/2+1)^th positions i.e 8^th and 9^th positions. 

    Median $={75+75\over 2}$ $=75$

b) The outliers are 20 and 229. 

c) Without outliers, 

    Mean $={60+65(2)+70(2)+75(4)+80+85(2)+90+95\over 14}$

    Mode = 75

    Median is at 7th and and 8th positions

    Median $={75+75\over 2} = 75$

   How is mean affected? Calculate the difference between both the means. 

d) Now you might as well think about that one... its easy, would you rather pay less or pay more? 😏 

For product 10 = {(1,10), (2,5)}                           = 2 pairs                   

For product 20 = {(1,20), (2,10), (4,5)}               = 3 pairs                    

For product 30 = {(1,30), (2,15), (3,10), (5,6)}    = 4 pairs                   

For product 40 = {(1,40), (2,20), (4,10), (5,8)}    = 4 pairs                    

For product 50 = {(1,50), (2,25), (5,10)}              =3 pairs                     

For product 60 = {(1,60), (2,30), (3,20), (4,15), (5,12), (6,10)}   = 6 pairs

For product 70 = {(1,70), (2,35), (5,14), (7,10)}                          = 4 pairs

For product 80 = {(1,80), (2,40), (4,20), (5,16), (8,10)}              = 5 pairs

For product 90 = {(1,90), (2,45), (3,30), (5,18), (6,15), (9,10)}   = 6 pairs

For product 100 = {(1,100), (2,50), (4,25), (5,20), (10,10)}        = 5 pairs

⇒ No. of ways = 2 × (2 + 3 + 4 +  4  + 3 + 6 + 4 + 5 + 6 + 5) 

                        = 2 × 42 

                        = 84 

Thanks!! 

Check the original post, I've posted the answer there.

a) We have eight slips of 2 and two slips of 9, lets make a probability distribution

E(x) = (2 × 4/5) + (9 × 1/5) = 17/5 = 3.4

b) one additional 9 is added so, 

E(x) = (2 × 8/11) + (9 × 3/11) = 43/11 = 3.9

c) another 9 is added to the bag so, 

E(x) = (2 × 2/3) + (9 × 1/3) = 13/3 = 4.3

d) In order to make the expected value 6,

     Let probability of 2 be x then, probability of 9 is (1-x)

    ⇒ 2x + 9(1-x) = 6

         2x + 9 - 9x = 6

         7x = 3   ⇒x = 3/7

     Probability of 9 = 1- 3/7 = 4/7 

   Let no. of 9 cards added be n then

    ${4+n\over 12+n} = {4\over 7}$

    28 +7n = 48 + 4n 

     3n  =  20  ⇒ n = 20/3 

e) In order to make the expected value 8,

     Let probability of 2 be x then, probability of 9 is (1-x)

    ⇒ 2x + 9(1-x) = 8

         2x + 9 - 9x = 8

         7x = 1   ⇒x = 1/7

     Probability of 9 = 1- 1/7 = 6/7 

   Let no. of 9 cards added be n then

    ${4+n\over 12+n} = {6\over 7}$

    24 + 6n = 48 + 4n 

     2n  =  24   ⇒n = 12

∴ '12' 9s  should be added. 

Perimeter of rhombus = 4 × side = 148                        A                        B

⇒ side = 37 

Consider rhombus ABCD                                                     O

      AB = BC = CD = DA  &  AC ⊥ BD

 AC & BD intersect at O.                                       D                       C

BD = 20  so, OB = OD = 10

Since ∠COD = 90             [diagonals of rhombus intersect at right angles]

⇒ OC = 35.6

Diagonal AC =  71.2 units