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So \( \frac{2}{5} h^{5/2} = 199.726 - 3.648 = 196.078 \)
\( h^{5/2} = \frac{5}{2} \cdot 196.078 = 490.195 \)
Now \( h = (490.195)^{2/5} \)
\( \ln(490.195) \approx 6.192 \), \( \frac{2}{5} \times 6.192 = 2.477 \), \( e^{2.477} \approx 11.94 \)
So \( h \approx 11.94 \) cm.
#### 11.94
A satellite orbits Earth in an elliptical path where the Earth is at one focus. The semi-major axis is 15,000 km and the eccentricity is 0.2. What is the distance from the satellite to the Earth’s center at the closest approach (perigee)?
Perigee distance = \( a(1 - e) = 15000 \times (1 - 0.2) = 15000 \times 0.8 = 12,000 \, \text{km} \).
A factory produces widgets with a defect rate of 3%. A quality inspector selects a random sample of 150 widgets. What is the probability that exactly 5 are defective?
Use binomial distribution: \( n = 150 \), \( p = 0.03 \), \( k = 5 \).