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P(Moon Hit)  =  Area of Moon  / Area  of Cover

{ 2 1/2   = 2.5  }

Area of Moon  = pi (diameter/ 2)^2  =  pi (2.5/2)^2  = pi (1.25)^2 =  4.9 in ^2

Area of Cover  = (5 1/2) ( 8 1/2)  = (5.5) (8.5)  = 46.75 in^2

So

4.9  / 46.75  ≈ .1048 ≈   10.5%

P(Romance novel)  = 36 / 91  = .396

P(Romance Novel Given a Paperback)  = 28/ 62  = .452

Hint:

You're trying to find \(z^{53}+z^{52}+z^{51}+z^{50}+z^{49}\), so factor out \(z^{49}\) to get \(z^{49}(z^4+z^3+z^2+z+1)\). We know what \(z^2+z+1\) is, so substitute it in. You now have\(z^{49}(z^4+z^3)\), which can be written as \(z^{51}(z^2+z)\). Now, we know that \(z^2 + z + 1=0\), so \(z^2+z=-1\). Substituting, we get \(-z^{51}\).

Continue from here.

-24

-1, -3, -5, -7

The common difference is - 2

14th term 

-1  - 2 (14 - 1)  =

-1 - 2(13)  =

-1 - 26  =

-27

#1) Since the surface area is 100x of the Earth, the volume should be \(100^{3/2}=1000\) times of the Earth.

I would discourage asking problems from AoPS, as it is in their ToS to not ask for third-party help.

Question 2

4 feet

tan x  =  4/20

arctan (4/20) ≈  11.3°

6 feet

tan x  = (6/20)

arctan (6/20)  ≈  16.7°

11.3° ≤ x ≤ 16.7°

Question 1

Here you go.....

We have right triangle ABC

Tan A  = 4.8 / 2  = 2.4

Area  =  = product of legs / 2   =  (4.8)(2) / 2  =  4.8  units^2 

Exponential Growth rate equation

y = 2000(1.5)^t

Ohhhh okay makes sense! Thank you!

2C1   means " two choose one"

In this case....we could have a girl either born first or second...choose either

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 

2C1 * (.52)(.48) ≈ .50

is that a C before the 2 and after the 1?

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

Thank you so much! I've been having a lot of trouble with this concept.

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

When the weight of an object equals the weight of the water displaced (or pushed away) by the object, the object is neutrally buoyant and easy to lift to the surface.

A tanker made of 2.5\times10^3\text{ m}^32.5×103 m32, point, 5, times, 10, start superscript, 3, end superscript, space, m, start superscript, 3, end superscript of steel sank to the bottom of the ocean. The density of steel is approximately 8\times10^3\text{ kg/m}^38×103 kg/m38, times, 10, start superscript, 3, end superscript, space, k, g, slash, m, start superscript, 3, end superscript. The density of seawater is approximately 1\times10^3\text{ kg/m}^31×103 kg/m31, times, 10, start superscript, 3, end superscript, space, k, g, slash, m, start superscript, 3, end superscript. Because gravity is pretty much the same everywhere on earth, the weight of the tanker can be assumed to be approximately equal to its mass.

In kilograms, what total weight of seawater needs to be displaced in order to lift the tanker to the surface?

It said it was wrong, I'm doing MathXL homework. 

a₁ = 2, a_n = 3a_n−1

So  we have

2, 3(2), (3)(3)(2), (3)(3)(3)(2)   =

2, 6, 18, 36