Rom

there are 2 0 point tiles out of 100

thus the probability of choosing a 0 point tile is 2/100 = 1/50

similarly there are 10 4 point tiles so the probability of choosing one is 10/100 = 1/10

there are 2 tiles worth 10 points and again the probability is 2/100=1/50 of choosing a 10 point tile

the probability of choosing a tile worth at least 4 point is that of choosing either a 4, 5, 8, or 10 point tile.

I leave you to figure out what that probability is.  (Hint: sum the number of tiles 4 points and greater)

I'm seeing it's 425.

I don't have a suggestion on how to find that though.

I guess you could start at 400 and add 1 for every square or cube you encounter in the list

There are 20 squares <= 400, 1-20,

There are 7 cubes <= 400, 1-7

But you seem to be excluding 1 from the series and that's counted here twice so we only have 25 squares and cubes.

Thus the 400th element is 425.

there are 5 ways for 1 drawer to be open

there are 3+2+1=6 ways for 2 drawers to be open

there is 1 way for 3 drawers to be open

5+6+1 = 12 ways total

pretty simple to list these out

(0,0,25)

(0,1,20)

(0,2,15)

(0,3,10)

(0,4,5)

(0,5,0)

(1,0,15)

(1,1,10)

(1,2,5)

(1,3,0)

(2,0,5)

(2,1,0)

12 ways

[-4,4] x [-3,3] = 9 x 7 = 63 integer coordinate points are strictly in the interior

no 2x2 squares have 4 black squares

all 3x3 squares have at least 4 black squares, as do all 4x4, 5x5, 6x6, 7x7, and 8x8 squares.

there are 6x6=36 3x3 squares

there are 5x5=25 4x4 squares

there are 4x4=16 5x5 squares

there are 3x3=9  6x6 squares

there are 2x2=4  7x7 squares

there is   1          8x8 square

Summing we get 91 squares

were you able to solve that w/o trial and error?

\(\dfrac{p}{y} = \dfrac{3}{5} \Rightarrow p=\dfrac{3}{3+5}72,~y = \dfrac{5}{3+5}72\\ p=3\cdot 9 = 27,~y = 5\cdot 9 = 45\)

\(4^x = (2^2)^x = 2^{2x} = (2^x)^2 = y^2\)

\(\text{what you've written isn't correct on it's own}\\ y=2^x \Rightarrow y^2 = (2^x)^2 = 2^{2x} = 4^x \\ \text{did you mean to write express }4^x \text{ in terms of }y?\)