Rom

\(P[\text{getting scholarship A}] = 0.05 \\ P[\text{getting scholarship B}] = 0.01\\ E[\text{Value of scholarship A }]= 0.05(2000-10) + 0.95(-10) = $90\)

similarly

\(E[\text{Value of scholarship B}] = 0.01(4000-25) + 0.99(-25) = \$15\)

\(x+2 = \sqrt{3x+10} \\ (x+2)^2 = 3x + 10 \\ x^2 + 4x+4 = 3x + 10 \\ x^2 + x -6 = 0 \\ (x+3)(x-2) = 0 \\ x = -3, 2 \\ \text{however }\dots \text{when we plug these back in to check the solutions} \)

\(-3+2 \overset{?}{=} \sqrt{3(-3) + 10} \\ -1 \overset{?}{=} \sqrt{\text{of anything}} \\ \text{over the real numbers... no} \\ \\ 2+2 \overset{?}{=} \sqrt{3(2) + 10} \\ 4 \overset{?}{=} \sqrt{16} \\ \text{yes}\)

\(\text{So we see that x=-3 is an extraneous solution}\\ \text{arising because we squared things in order to solve the original equation} \\ \\ \text{the range of the square root function is only non-negative reals} \\ \text{and thus we need to be careful and check solutions whenever we square things to solve an equation}\)

\(62 = \dfrac {62}{1} = \dfrac{124}{2} = \dots \\ \mathbb{Z} \subset \mathbb{Q} \\ \text{so yes 62 is a "fraction", more accurately a rational number}\\ \text{it also happens to be a completely wrong answer in this context but that's another story}\)

any probability is between 0 and 1 so 62 is a non-starter

there are 32 total marbles in the bag

a) \(P[\text{drawing a green marble}] = \dfrac{6}{32}=\dfrac{3}{16}\)

b) \(P[\text{drawing a red marble}] = \dfrac{10}{32}=\dfrac{5}{16}\)

c) The easiest way to do this is find the probability of drawing a yellow marble and subtracting that from 1

\(P[\text{drawing a !yellow marble}] = 1-P[\text{drawing a yellow marble}] = 1 - \dfrac{12}{32} = \dfrac{20}{32} = \dfrac{5}{8}\)

no good answer... must have been late.

\(\dfrac{p + 5n + 10d}{p+n+d}=7 \\ \dfrac{p+5 + 5(n-1) + 10d}{p+5+(n-1)+d} = \dfrac{p+5n+10d}{p+n+d+4}=6 \\ 7(p+n+d) = 6(p+n+d+4) \\ p+n+d = 24\)

\(p+5n+10d = (24)(7) = 168 \\ \text{clearly }p=3 \text{ so }\\ 5n+10d = 165,~n+d = 21 \\ 5(21-d)+10d = 165 \\ 105 + 5d = 165 \\ 5d = 60 \\ d=12 \\ n=21-12 = 9\)

\(p=3,~n=9,~d=12\)

The first flip must be heads which takes her to floor 7.

The second flip must be heads, which takes her to floor 8.

Now for the 3rd flip she can flip either heads or tails going to either floor 9 or floor 7

At floor 7 for the 4th flip she must flip heads to get to floor 8

At floor 9 for the 4th flip she can flip either heads or tails going to either Floor 10 or Floor 8

For the 5th flip, as she is on either floor 8 or floor 10, she can flip either heads or tails

so we end up with the following valid flip sequences

\(\left( \begin{array}{ccccc} H & H & T & H & T \\ H & H & T & H & H \\ H & H & H & T & T \\ H & H & H & T & H \\ H & H & H & H & T \\ H & H & H & H & H \\ \end{array} \right)\)

You can see there are 6 of them.

The total number of flip possibilities is \(2^5 = 32\)

So  \(p=\dfrac{6}{32}=\dfrac{3}{16}\)

This isn't quite correct as it doesn't take into account words with repeated letters.

Suppose the first letter is a vowel, there are 26 valid choices for the second letter.  There are 6 vowels.

Suppose the first letter is a consonant.  There are 6 valid choices for the second letter.  There are 20 consonants.

So the total number of valid words is

6 x 26 + 20 x 6 = 276

Is there some diagram that goes with this problem?  I don't see any way to solve it w/o knowing something about the geometry of the ladder.