Rom

\(f(x)=3(x-2)^2-5\\ \text{This is the function}\\ y = x^2\\ \text{that has been shifted down 5 units, shifted to the right 2 units, and scaled by 3}\)

\(\text{The domain is all real numbers}\\ \text{The range is }[-5,\infty)\\ f(x) \text{ is a functional relation as each element in the domain maps to only a single element in the range}\)

\(\text{do you mean}\\ \sqrt{2^{i7\pi/4}}\)

\(-2-2i = 2\sqrt{2} e^{i5\pi/4}\\ z^3 = 2\sqrt{2} e^{i5\pi/4}\\ z = \sqrt{2} \exp\left(i\dfrac{\frac{5\pi}{4}+2k\pi}{3}\right), ~k=0,1,2\\ z =\sqrt{2}\exp\left(i\dfrac{5\pi}{12}\right),~ \sqrt{2}\exp\left(i\dfrac{13\pi}{12}\right),~\sqrt{2}\exp\left(i\dfrac{7\pi}{4}\right)\\ \text{and of these }\\ z = \sqrt{2}\exp\left(i\dfrac{7\pi}{4}\right) \text{ lies in the fourth quadrant}\)

\(\text{There are 7 valid digits for the first number, and 8 valid digits for numbers 2 and 3}\\ \text{Thus there are }\\ n = 7\cdot 8^2 = 448 \text{ 3 digit numbers with no 7's or 9's}\)

the answer is take the hint I gave you and do your own #$)(*# work

\(\text{If you have a function }g(x) \text{ which is neither even nor odd (such as }2^x)\\ g_e(x)= \dfrac{g(x)+g(-x)}{2} \text{ is even and }\\ g_o(x) = \dfrac{g(x)-g(-x)}{2} \text{ is odd and }\\ g_e(x) + g_o(x) = g(x) \)

\(\text{Apply this to the two terms of the expression that are neither even nor odd}\)

It seems that the dishonest coin principle is the fact that the Central Limit theorem states that

given enough samples the outcome a set of Bernoulli trials as described in the problem

is approximately normally distributed with parameters

\(\mu = n p\\ \sigma = n p (1-p)\\ \text{Where }p \text{ is the probability of the success of a single event}\)

\(\text{In this problem }n=400,~p=\dfrac{1}{10}\\ \mu = 40\\ \sigma = \sqrt{(400)(0.1)(0.9)} =6\)

\(\text{Denoting the CDF of the standard normal as }\Phi(x) \text{ we have}\\ P[n\leq 40] = \Phi\left(\dfrac{40-40}{6}\right) = \Phi(0) = \dfrac 1 2\)

Now just apply this same principle to (B)

see edited first post.  I had reversed the number of juniors and seniors in deriving the distribution.

\(\text{First choose the driver. There are }\dbinom{2}{1}=2 \text{ ways to do this.}\\ \text{Now arrange the remaining 4 students. As the seats are distinct there are }4!=24 \text{ ways}\\ \text{Thus we have }2\cdot 24 = 48 \text{ ways off arranging the students}\)

oops we sure do