CPhill

To make this easier, let us suppose that the total games to be played = 200

So far,  we have played  .75 (200) =  150 games

And we  have won  .42 (150) =  63 games

And we  must  win 100 to have the same number of wins as  losses

So....we need 100 - 63 =  37 more wins to  get to 100

And we have 50 games  left to play

So....we have to  win 37 out of 50 games  remaining to get to 100 wins

So

37/50  = 74/ 100  =  .74

So....we must  win 74% of the remaining games  to  have the same number of wins as losses

For the circumradius we have

( product of the sides)  / (4 * Area)

You should be able to  finish this one......

Inradius =   Area /  semi-perimeter

semiperimeter  (s)  =   ( 7 + 5 + 4) / 2  =  16 / 2  =  8  

Using Heron's Formula to  find the area =   sqrt [ s ( s -a) (s - b) (s -c) ]  where a,b,c  are the sides

So

Area = sqrt  [ 8 (8 -7) (8 -5) ( 8-4) ]  = sqrt [ 8 * 1 * 3 * 4 ] =  sqrt [ 96] =  sqrt [ 16 * 6] = 4sqrt (6)

So

inradius=   4sqrt (6) / 8  =   sqrt (6) / 2  

Juan =   c cheesecakes

Clarence = 3c - 6     cheesecakes

Darlene =   (3c - 6)/2 + 4 =   (3c - 6 + 8) / 2  =   (3c + 2) / 2 =   (3/2)c + 1    cheesecakes

Total number of  cheesecakes = 

c + 3c -6 + (3/2)c + 1  =     

4c + (3/2)c - 5  =

(8/2)c + (3/2)c - 5 =

(11/2)c - 5

Thank you !!!

Before the transfer from B to A  let the number of shoes in each rack = A, B  and C

We have these equations

A = .40(B + C)    →  C = A /.40 - B    (1)

.75B = (A + C)    →   C = .75B - A     (2)

Equate (1) and (2)

A/.40 - B = .75B - A

A/.40 + A = .75B + B

2.5A + A = 1.75B

3.5A = 1.75B    (1)

After the transfer

A + 42 = B - 42   simplify

A = B - 84      (2)

Sub (2) into (1)

3.5 ( B - 84)  = 1.75B

3.5B - 294 =  1.75B

3.5B - 1.75B = 294

1.75B = 294

B = 294/1.75  =  168

A = B - 84  =  168 - 84 =  84

C = .75B - A  =   .75 (168)  - 84 =   42

So the total number of shoes =    168 + 84 + 42 =   294

Correct  !!!

The 67 - 70   range is 3 standard deviations from  the  mean

2.1% of the scores will fall into  this range

So

.021 * 170  =  3.57 ≈  4 scores will fall into  this range 

(I cheated a little on this one !!! )

                  C

          E

A     9        F   3     D          x             B

In triangle  ABC,   DE is paralllel to  BC

So....AE / AC = AD / AB     (1)

In triangle  ACD,  EF is parallel to CD

So.... AE/ AC = AF / AD      (2)

So  using (1) and (2)

AD/ AB  = AF / AD

So

AD^2 = AB * AF

(9+ 3)^2  = (12 + x) (9)

12^2  = 108 + 9x

144 = 108 + 9x

144-108 = 9x

36  = 9x

36/9  = x  = 4 =    BD

Red balls = 48 / 3   = 16

White balls = 30 / 5 =  6

Total balls =  30 + 48 + 16 + 6 =   100

If 7/10  are taken away  3/10  must  remain......can you solve it ???