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A biotech company’s stock value grows continuously at 8% per year. If an investor buys $10,000 worth, what will the value be after 5 years?
Use \( A = Pe^{rt} = 10000 \cdot e^{0.08 \cdot 5} = 10000 \cdot e^{0.4} \approx 10000 \cdot 1.4918 = 14918 \).
#### 14918**Question:** A museum curator is organizing a display of 7 unique historical scientific instruments, but the virtual cataloging system can only showcase 4 at a time. How many different sets of 4 instruments can be chosen from the 7?
Solution:** To solve this problem, we use the concept of combinations, as the order in which the instruments are showcased does not matter. The number of ways to choose 4 instruments from 7 is given by the binomial coefficient \(\binom{7}{4}\). The formula for combinations is:
Substituting \(n = 7\) and \(r = 4\), we get:
\binom{7}{4} = \frac{7!}{4! \times (7-4)!} = \frac{7 \times 6 \times 5 \times 4!}{4! \times 3 \times 2 \times 1} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35
Thus, there are \(\boxed{35}\) different sets of 4 instruments that can be chosen from the 7.
Question:** An agricultural systems officer in California is planning a community program and needs to select 3 types of sustainable crops from a list of 10. In how many ways can they choose 3 crops if the order does not matter?
Solution:** This is another combinations problem, where we need to choose 3 crops from a list of 10. The number of combinations is given by:
\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8 \times 7!}{3! \times 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120