CPhill

Not sure about this, but I believe it will be the squares of all primes from 2 to 100

So

4, 9,25, 49, 121 , 169, 289, 361, 529, 841,961,1369, 1681, 1849, 2209, 2809 , 3481, 3721, 4489, 5041, 

5329, 6241, 6889, 7921, 9409

I count 25....but....as I did it hurriedly, I might have missed one or two.....better double-check 

P(0 girls)  = (.52)^2  ≈ .27

P(1 girl )   = 2C1 * (.52)(.48) ≈ .50

P(2 girls)  = (.48)^2   ≈ .23

Answer here :

https://web2.0calc.com/questions/helps-please-also-ty-cphill-for-helping-me-on-the-first-one

First one

Number of possible new plates  =  (10)^5  =  100,000

Number of possible old plates = (10 * 9 *8 * 7* 6)  =  30,240 

Just subtract these to get the answer

Second one.....considering that all rotations are the same

The two boys can be seated in 2 ways

And the girls can be seated in 3! ways

So   2 * 3! =  2 * 6  = 12 ways

Just sum the frequencies.....then....divide each frequency by this sum to get each %

Perimeter  = 2 ( L + W)

18 = 2(L + W)    divide through by 2

9 = L + W

9 - L  = W

Area = L  * W

20 = L (9 - L)

20 = 9L - L^2

L^2  - 9L + 20  = 0      factor

(L - 5) (L - 4) = 0

Set both factors to 0  and solve for L

L = 5     and L  = 4

And these are the dimensions....5 x 4

Here you go, GM

5 + 10 + 2 + 9 + 33 + 12 + 15 + 3 + 1  = 90

0 times  =  5/90 = .06

1 time  = 10/90 = .11

2 times = 2/90 = .02

3 times = 9/90 =  .1

4 times =  33/90 = .37

5 times = .12/90 = .13

6 times = 15/90 = .17

7 times = 3/90 = .03

8 times = 1/90 = .01

a₁ = 2, a_n = 3a_n−1

So  we have

2, 3(2), (3)(3)(2), (3)(3)(3)(2)   =

2, 6, 18, 36

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