CPhill

d. The doubling time, t,  is independent of any amount

So

2A  =  A (1.062)^t      divide both sides by  A

2 = 1.062^t      take the log of both sides

log 2  =  log (1.062)^t     and we can write

log (2)  =  t * log (1.062)     divide both sides by log (1.062)

log (2) / log (1.062)  = t    ≈ 11.52 yrs  ≈ 12 years   .i.e., it takes any starting amount about 12 years to double

In order we have

86 , 88, X, 94, 96, 97

The median  would be    [ 94 + X ]  / 2  = 92   where X is the 6th attempt

94 + X  =  184      subtract 94 from both sides

90  = 6th attempt time

5 * 8^(4x)  =  375      divide both sides by 5

8^(4x)  = 75     take the log of both sides

log 8 ^(4x)  [ log (75)    and we can write

(4x) log (8)  = log(75)       divide both sides by  4 log (8)

x =    log (75)  /  [ 4 log (8) ]

 0.8 ⁴  =  0.4096

(x - 1) (x -2) (x - 3) (x + 3)  =

[ x^2 - 3x + 2] [ x^2 - 9 ] =

x^4  -  3x^3 + 2x^2

+                 - 9x^2 +  27x - 18

________________________

x^4   -  3x^3  -  7x^2 + 27x - 18   

 4x^3  -  20x^2  +  24x   =  0       divide through by  4

x^3  -  5x^2  + 6x  = 0      factor

x (x^2 - 5x + 6)  = 0

x( (x - 3) (x -2)  = 0

The zeroes  are    0, 2 , 3

x^3  -  216  =  0        factor as a difference of cubes

(x - 6) (x^2  + 6x  +  36)

6 is a root

And using the quadratic formula  the other  roots will be complex

So we will have

[ -6 ±  √ [ 36 - 144] ] / 2

[ -6 ±  √ [  - 108] ] / 2

[ -6 ±  √ [  - 36 * 3 ] ] / 2

[ -6 ±  6i√ [  3] ] / 2

-3  + 3i√ 3       and  -3  - 3i√ 3

Area  = product of the diagonals  / 2

120  =  D * 10 / 2

120  =  D * 5

24 =   D  = the other diagonal length

And  the diagonals bisect each other...so 1/2 of each of their lenghts will form legs of a right triangle  with the side asa hypotenuse.....so we have that

5^2 + 12^2  =  side^2

169  = side ^2          take the square root of each side

13  = side

And the perimeter is 4 times this  = 52

I end up with the same result!!!...however...if we make a slight change such that Kenny has twice as many as Penny...it becomes do-able

J + K + P  =  50

K  = 2P  = 2(12) =   24

So...doing some substitution

J + 24 + 12  = 50

J  +  36  = 50

J  =  14

The table is correct...!!!

This formula is more concise than any complete sentences.....

N (A or C)  =  N(A)  + N(C)  -  N(A and C)  =

29  +  11   -   5  =

35

Just Chemistry....6/40   are enrolled in just Chemistry so we would expect the same proportion out of 2000....so    (6/40) * 2000  =   (3/20)*2000  =  3 (2000/20)  =  3 * 100   =  300

If the permeter of PQRS  is 24....remember that PQ  = SR   and   PS  = QR

So...... PQ  + SR  +  PS + QR  = 24

5  +  5    + PS + QR   =   24       sub  PS for QR

10  +  PS  + PS  =  24       subtract 10 on each side

2PS  =   14     divide both sides by 2

PS  = 7  = QR

And the area of PQRS   =  PS  * height of PQRS (=ST)  ......so we have that

28  = 7 * ST    divide both sides by 7

4  = ST

And since RST is a right triangle.... using the Pythagorean Theorem...

 RS^2 - ST^2  =  RT^2

5^2  - 4^2  = RT^2

25  - 16  = RT^2

9  = RT^2      take the square root of both sides

3  = RT

So........QT  = QR + RT  =    PS + RT  =    7 + 3   = 10

So...the perimeter of  QOST  =  2 [ QT + ST]  = 2 [ 10 +  4]  =  2 * 14  =  28

LOL!!!!....not really a genius....in fact, these probability problems give many of us trouble....including me....!!!

Answered here , NSS :  http://web2.0calc.com/questions/help-please_52349#r1