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- After day 3: 4,500,000 × 0.75 = 3,375,000
- #### 3,375,000**Question: A herpetologist studying reptile population dynamics models the growth of a lizard population with a quadratic polynomial \( h(x) \) such that \( h(0) = 3 \), \( h(1) = 8 \), and \( h(2) = 15 \). Find \( h(x^2 + 2) \).
- To find the quadratic polynomial \( h(x) = ax^2 + bx + c \), we use the given conditions:
- \( h(1) = 8 \) gives us \( a(1)^2 + b(1) + 3 = 8 \) which simplifies to \( a + b + 3 = 8 \). Thus, \( a + b = 5 \).
- \( h(2) = 15 \) gives us \( a(2)^2 + b(2) + 3 = 15 \) which simplifies to \( 4a + 2b + 3 = 15 \). Thus, \( 4a + 2b = 12 \).
- Now, solve the system of equations:
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Each corresponds to exactly one ordered triple with $ a < b < c $ satisfying the condition.
Question: A topologist labels 6 distinct regions of a cell membrane using 3 red and 3 blue topological markers, assigning one marker per region uniformly at random. What is the probability that no two adjacent regions (in a fixed linear chain) are both colored red?
This is a classic non-adjacent independent set problem.
Let $ f(n,k) $ be the number of ways to place $ k $ non-adjacent red markers in $ n $ positions in a line.
We need $ f(6,3) $ with no two reds adjacent.
Check manually: list all binary strings of length 6 with exactly three 1s (red), no two adjacent.
Such configurations must have at least one blue between reds. So min spacing: R-B-R-B-R, requires 5 positions. We have 6, so one extra red to place.
Start with the base: R-B-R-B-R, and have one extra R to place in the 4 possible gaps: before first, between, or after last.
Actually, the standard transformation: place 3 reds with at least one blue between, using stars and bars with spacing.
Thus, number of favorable outcomes: 4
