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The solution is z = 2/5.

Let S be the number of students who studied and NS be the number of students who did not study

We can represent the total number of students (T) with the following equation

T = S + N

We are given the average scores for each group and the overall class average:

Average score for those who studied (AS): AS = 78

Average score for those who did not study (ANS): ANS = 54

Overall class average (Avg): Avg = 70

We can express the total score for all students using the average scores and the number of students in each group:

Total score for those who studied (TS): TS = AS * S

Total score for those who did not study (TNS): TNS = ANS * NS

Total score for the entire class (Total): Total = Avg * T (since Avg represents the average score per student)

Since the total score for the entire class is the sum of the scores from each group:

Total = TS + TNS

We can substitute the expressions for each score:

Avg * T = AS * S + ANS * NS

Now we can substitute the values we are given:

70 * (S + NS) = 78 * S + 54 * NS

We can rearrange the equation to isolate NS (the number of students who did not study):

20 * (S + NS) = 24 * S

20S + 20NS = 24S

-4S = 20NS

NS = (-4/20) * S (we can divide both sides by -4 as long as S is not 0, which we can safely assume since there must be at least one student who studied)

NS = -(1/5) * S

Since NS represents a number of students, it cannot be negative. Therefore, we can flip the signs to get a positive value for the fraction representing the portion of the class that did not study:

NS/T = (1/5) * S/T

We are interested in the portion of the class that did not study, so we can express this as:

Fraction of class who did not study = NS/T = 1/5.

You can use Stirling's formula:

n! = sqrt(2*pi*n)*(n/e)^n

You can use Stirling's formula:

n! = sqrt(2*pi*n)*(n/e)^n

Let's analyze the scenario step-by-step:

Initial Flips: We are given that after 2 flips, Erica has more heads than Alan. This can happen in 3 ways:

Erica gets HH and Alan gets TT (2 heads for Erica, 0 for Alan).

Erica gets HT and Alan gets TT (1 head for Erica, 0 for Alan).

Erica gets TH and Alan gets TT (1 head for Erica, 0 for Alan).

Remaining Flips: There are 3 remaining flips for each person. Since the coin is fair (each flip has a 50% chance of heads or tails), there are 23=8 possible outcomes for each person's remaining flips.

Favorable Outcomes: We only care about the outcomes where Erica maintains more heads than Alan after 5 flips. Consider the 3 initial scenarios:

HH vs TT: In this case, Erica needs any combination of heads and tails in her remaining 3 flips. There are 8 possibilities for her flips. For Alan, he cannot have any heads, so he needs all tails. There is only 1 possibility for his flips. So, there are 8⋅1=8 favorable outcomes.

HT vs TT: Here, Erica needs at least 2 heads in her remaining flips. There are (23​)+(33​)=3+1=4 ways for her to achieve this (2 heads and 1 tail, or 3 heads). Again, Alan needs all tails, so there is only 1 possibility for his flips. This gives us 4⋅1=4 favorable outcomes.

TH vs TT: Similar to HT vs TT, Erica needs at least 2 heads. There are 4 favorable outcomes. Alan still needs all tails, resulting in 1 possibility. So, there are 4⋅1=4 favorable outcomes.

Total Favorable Outcomes: Summing up the favorable outcomes from each initial scenario, we get a total of 8+4+4=16 successful outcomes for Erica.

Total Outcomes: There are 23=8 possibilities for Erica's remaining flips and 23=8 possibilities for Alan's remaining flips, leading to a total of 8⋅8=64 total outcomes.

Probability: The probability that Erica has more heads than Alan after 5 flips is the number of favorable outcomes divided by the total number of outcomes: $ \frac{16}{64} = \frac{1}{4}$.

Converting the fraction to relatively prime integers, we have m=1 and n=4. Therefore, m+n=5​.

By Vieta's formulas, the sum of the roots of f(x) is −5, so the sum of the roots of x3+3x−3=0 is −(r1​+r2​+⋯+r7​)=5.

By the Factor Theorem, x−5 is a factor of x3+3x−3. We can use polynomial division (or inspection) to find that x3+3x−3=(x−5)(x2+8x+3). Then

\begin{align*} g(r_1) g(r_2) \dotsm g(r_7) &= (r_1 - 5)(r_1^2 + 8r_1 + 3) (r_2 - 5)(r_2^2 + 8r_2 + 3) \dotsm (r_7 - 5)(r_7^2 + 8r_7 + 3) \ &= (r_1 - 5)(r_2 - 5) \dotsm (r_7 - 5) (r_1^2 + r_2^2 + \dots + r_7^2 + 8(r_1 + r_2 + \dots + r_7) + 21). \end{align*}

By Vieta's formulas again, r12​+r22​+⋯+r72​=49−2⋅4+2⋅5−9+1=43, and r1​+r2​+⋯+r7​=−5.

Hence, \begin{align*} g(r_1) g(r_2) \dotsm g(r_7) &= (-5)(-5) \dotsm (-5) (43 + 8 \cdot -5 + 21) \ &= \boxed{125a + 3125}. \end{align*}

By the integer root theorem, the an integer root must be a factor of 72, like 1, 2, 3, 4, 6, ...  Checking these, we find that 18 is an integer root.

The answer is x = 985.

The area of triangle ABD is 40*sqrt(2) + 12.

BX = 20*sqrt(7)