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