BuilderBoi

Thank you!

Is it supposed to be \(y - \lfloor y \rfloor \) or \(y + \lfloor y \rfloor \)?

Alright. Notice that the red-shaded region is a quarter circle with a radius of 1. This means that its area is \((1^2 \pi) \div 4 = {\pi \over 4}\)

The area of the square is just 1 x 1 = 1, so the probability is just \(\color{brown}\boxed{ \pi \over 4}\)

First, move everything to the left-hand side: \(4t^2 - 22t + 7 \leq 0\)

Now, let's assume that we want to solve \(4t^2 - 22t + 7 = 0\). Using the quadratic formula, this would be \(t = {22 \pm \sqrt{22^2-4 \times 7 \times 4} \over 2 \times 4} ={ 22 \pm 2\sqrt{93} \over 8} = {11 \pm \sqrt{93} \over 4}\)

But, we haven't solved the problem yet! We need to find the values that satisfy the inequality. Note that all the points in between the interval of our zeros (\({11 - \sqrt {93} \over 4}\) and \({11 +\sqrt {93} \over 4}\)) will either be positive of negative. 

So, let's plug in \(t = 1\) because it's in the interval. This gives us \(4(1)^2 -22 (1) + 7 = -11\), which is less than 0. This means that the value of quadratic must be less than 0 within this interval. 

In interval notation, this would be \(\color{brown}\boxed{{11 - \sqrt {93} \over 4}, {11 + \sqrt{93} \over 4}}\)

Note: There's probably a better, easier way of solving this that I can't think of.

First, simplify the expression to \(-2t^2 + 20t - 10\)

Now, recall that the vertex occurs at \(t = -{b \over 2a} = -{20 \over -4} = 5\)

This means that if we substitute \(t = 5\) into the expression, we get the vertex. 

Want to try it?

Note that the equation is a circle centered at (0, 0) with a radius of 1. 

Using this, we can draw a diagram representing the outcomes:

Basically, the shaded region (red) is the area of the success and the black square (side length 1) is the area of the total region

Can you take it from here?

Consider the equation as two separate equations: \(5x + 3 = 4x - 1\) and \(-(5x + 3) = 4x - 1\)

Now, solve both equations for x as you normally would. Be careful of the negative in the second equation. 

The only way for 3 squares to be colored red is if the first, third, and last squares are red. 

This means that the two remaining squares (second and fourth) must be either red or blue. 

So, there are \(2 \times 2 = \color{brown}\boxed{4}\) ways to color the five squares. 

The area of the shaded region is the sum of the circle and triangle minus the white region. 

The white region is a 60-degree sector, so its area is \({10 \times {60 \over 360}} = {5 \over 3}\)

This means that the area of the shaded region is \(20 - {5 \over 3 } = {\color{brown}\boxed{55 \over 3}}\)

I think that the probability \(WX > YZ\) equals the probability of \(WX < YZ\).

This means that each has a probability of \(\color{brown}\boxed{1 \over 2}\)