Rom

looks to be 144

1 4 9 16 36 144

\(\text{Binomial distribution with }n=5,~p=0.01\\ a) P[0]=(1-p)^5 = (0.99)^5 = 0.951\\ b) P[>2] = 1 - P[0]-P[1]-P[2] = \\ 1 - (0.99)^5 - \dbinom{5}{1}(0.01)(0.99)^4 - \dbinom{5}{2}(0.01)^2(0.99)^3 = 9.8506\times 10^{-6}\)

If 12 and 15 are factors then LCM(12,15)=60 is a factor

Thus 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

are the 12 factors of x=60

1) for each possible roll of 2 dice determine the absolute difference

2) for each value generated in (1) count the number of times it occurs.  Divide this count by 36.

This is the probability that this absolute difference occurs.

3) What you now have is a distribution on the absolute difference

4) multiply each value by it's probability and sum up the results.  This is the mean value.

5) multiply each value squared by it's probability and sum the results.  The is the expected value of of the absolute difference squared.

6) Find the variance as The number you found in (5) minus the number you found in (4) squared.

i.e. var = E[|roll diff|²] - (E[|roll diff|])² 

7) standard deviation is the square root of variance

\(\text{The largest angle is opposite the longest side}\\ \text{Use the law of cosines}\\ c^2 = a^2 + b^2 - 2 a b \cos(C)\\ \text{Where }c \text{ is the length of the longest side, and }\\ a,b \text{ are the lengths of the other two sides}\)

No.  0 is just an extraneous root of the quadratic you synthesized.

Just use the quadratic formula.

Use a bit of algebra to get all the terms on one side with 0 on the other.

\(g=2b\\ g+b = 30\\ b = ?\\ \dfrac{b}{30}=?\)

\(-3x-1 = x^2 - 3x + 4\\ x^2 +5=0\\ \text{This should pretty clearly be seen to have no real solutions}\)

incorrect answer deleted

Go Melody!

\(\text{Consider a line from the right bottom corner to the center of the circle}\\ \text{Convince yourself that it bisects the angle the triangle makes with the right side}\\ \text{and thus it is a }15^\circ \text{ angle}\)

\(\tan(15^\circ) = \dfrac{r}{s-r}\\ s = \dfrac{r(1+\tan(15^\circ))}{\tan(15^\circ)}\)

\(\text{I'll let you plug and chug to see that it's }s=18\)