RiemannIntegralzzz

Listing out the divisors of 720, we have:

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 30, 36, 40, 45, 48, 60, 72, 80, 90, 120, 144, 180, 240, 360, and 720.

The divisors of 720 that are a multiple of 4 are:

4, 8, 12, 16, 20, 24, 36, 40, 48, 60, 72, 80, 120, 144, 180, 240, 360, 720

There are 30 divisors of 720, and 18 of them are a multiple of 4, so the probability of picking a divisor of 720 that is a multiple of 4 is $\frac{18}{30}=\boxed{\frac{3}{5}}$

However, this method is very time consuming, and any algebraic solution would be much appreciated.

There is a total of $3+1+4+1+1+5=15$ total students, and getting the total number of days missed is $0*3+1*1+2*4+3*1+4*1+5*5=0+1+8+3+4+25=41$ total days missed. This means the average days missed is $\frac{41}{15}=2$ $\frac{11}{15}$. The mode number of days missed is 5, because the most common number of days missed was 5. What I don't understand is that the mean is LESS than the mode, not GREATER than the mode. So the absolute difference between them would be $|5-2$ $\frac{11}{15}|=2$ $\frac{4}{15}$, or, as a common fraction, $\boxed{\frac{34}{15}}$

The next multiple of 3 after 62 is 63...and 273 is the nearest multiple of 3 that is less than 275... the number of multiplyes is given by:

$\frac{273-63}{3}+1=\frac{210}{3}+1=70+1=\boxed{71}$

*Credit, CPhill's answer to this problem: https://web2.0calc.com/questions/how-many-multiples-of-3-are-between-62-and-215*

Please show all your work for these answers! I will not be using them unless they have work.

Actually, I figured it out myself:

You can use complementary counting/combinations here:

The number of ways to choose a committee of all boys or all girls is $2*C(10,4)$(2 for the girls and boys)$=420$. The total number of committees is $C(20, 4)=4845$. So the answer is $1-\frac{420}{4845}=\frac{4425}{4845}=\boxed{\frac{295}{323}}$

That was incorrect, but I have one more try.

Thank you, it helped!

We let the two numbers be x and y. The set of all possible pairs (x,y) satisfies the inequalities 0

We want to find the area of the set of points which also satisfies x^2+y^2<=4. The set of ordered pairs that satisfy this inequality is a circle with radius 2 centered at the origin. The overlap between the circle and the square is the quarter-circle in the first quadrant with radius 2. This region has area 1/4(2)^2$\pi$=$\pi$

Thus the desired probability is $\boxed{\frac{\pi}{4}}$

Sorry, my image isn't displaying.. I'll try to type it out by hand