Rom

\(P[\text{2 red plates}]=\dfrac 5 9 \dfrac 4 8 = \dfrac{5}{18}\\ P[\text{2 blue plates}] = \dfrac 4 9 \dfrac 3 8 = \dfrac{1}{6}\\ P[\text{plates are same color}]=\dfrac{5}{18}+\dfrac 1 6 = \dfrac 4 9\)

Jeeze where are you getting these problems?

First off the two spins have the following distribution on the resulting movement.

Movement to the right is positive, movement to the left is negative.

\(P[(-2,~0,~2)] = \left(\dfrac 1 9,~\dfrac 4 9,~\dfrac 4 9\right)\)

Second we note the possible multiples of 3 which are (0, 3, 6, 9, 12)

So with probability 1/9 of ending up on a multiple of 3 we can choose (2,5,8)

With probability 4/9 of ending up on a multiple of 3 we can choose (3, 6, 9)

With probability 4/9 of ending up on a multiple of 3 we can choose (1, 4, 7, 10)

All cards have probability 1/10 of being chosen and we end up with the following probabilities

\(P[\text{ending up on multiple of 3}] = \dfrac{3}{10} \dfrac 1 9 + \dfrac{3}{10}\dfrac 4 9 + \dfrac{4}{10}\dfrac 4 9 = \dfrac{31}{90}\)

Look at the following sequence of probability tables

We see the answer is 36, choice B.

not a terribly well worded question.  I'm going to assume the stated speeds are with respect to a stationary object on the ground.

Given that the resultant speed of the two cars approach is

s=70+80=150mph

t = d/s = 750miles/150mph = 5hr

choice B

in general the formula for the volume of a cylinder is

\(V = \pi r^2 h\)

looking at #1 we have \(V = \pi (12^2) (8) \approx 3619.115 \)

it looks like you squared the diameter instead of the radius

Looking at #2 \(V = \pi (7.5^2)(4) = 706.858\)

again you used the diameter instead of the radius

You got the 3rd one correct because they gave you the radius, not the diameter, in the diagram.

A gets reflected to B=(3,-4) which then gets reflected to C=(-3,-4)

That gets us a right triangle with legs of length 6 and 8.

Can you figure out the area from that?

This means the limit of f(x) as x approaches 2 from below.

In the function graphed below we see that 

\(\lim \limits_{x\to 2^-} = 1 \\ \lim \limits_{x \to 2^+} = 5 \\ \lim \limits_{x\to 2} \text{ does not exist.}\)

we k**l and eat Gilligan

I don't see any really clever way of doing this.  You can expand it using the binomial theorem to get

\(\left(2+\sqrt{3}\right)^9 = \sum \limits_{k=0}^9~\dbinom{9}{k}2^k~3^{(9-k)/2}\)

if you then expand all that out and combine things you'll have an expression that looks like

\(a + b \sqrt{3}\)

b above turns out to be 40545 so you'll need to take the square root of 3 out to 5 decimal places

there are 3 rolls whose product totals 180

(5,6,6), (6,5,6), (6,6,5)

There are 216 possible rolls of 3 six sided dice.

\(P[\text{product of 3 dice = 180}] =\dfrac{3}{216}=\dfrac{1}{72}\)