ilovepizza547

So, we have that there were 21-4=17 students who either passed the test or completed HW. So, there were 11+14-17 = 8 students that passed the test and compelted the HW. So, the probability is $\frac{8}{14} = \frac47.$

Yup, I think that's right...

$750 - 750*0.544 - 750*0.36 = 72.$

Well, let the common length of the altitude be $h$. Then, the length of each side is $\frac{2K}{h}$ where K is the area, so it is equilateral.

Hint: Use the parametric form of the circle $(x,y) = (42000 \cos \theta, 42000 \sin \theta)$

Guest would be right if the groups were distinct, but if they are indistiguishable it would be half that or $\frac{\binom{26}{13}}{2}$

Just use RRT, you get the possible roots are $\pm \frac12, \pm 1$ so there are 4 possible rational roots.

What does this mean, where is the problem?

We can see that if we put in all of the odd terms, we get $2^{99}$ , however, the odd terms are missing. 

But, we see that for each even term $\binom{99}{2x}$ there is a odd term $\binom{99}{99-2x}$ which is equal to it.

So, the sum is $2^{98}$.

Mean is sum divided by number of elements so it is $\frac{52}{7}$.

Median is simply 6

Mode is most often occuring so it is also 6

Range is 15-2=13