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A zero-coupon bond sells for $ 250. At maturity in 16 years, it will be redeemed for $ 1,000. What approximate annual rate of growth does this represent?
Using the future value formula: \(FV = PV \times (1+r)^n\). So \(1000 = 250\times(1+r)^{16}\). \((1+r)^{16} = 4\). \(1+r = 4^{1/16} = 4^{0.0625}\). \(\ln(1+r) = \frac{\ln 4}{16} = \frac{1.386}{16} = 0.0866\). \(1+r = e^{0.0866} \approx 1.0905\). So \(r \approx 9\%\). Wait: \(4^{1/16}\): \(\log(4)/16 = 0.602/16=0.0376\), antilog \(=1.089\), so \(r\approx 9\%\) = option (B). Re-examining: \(4^{0.0625}=1.0905\Rightarrow r\approx9\%\). Answer: option (B) = 9 percent.
Use: $P(\text{exactly one of }X\text{ or }Y) = P(X)+P(Y)-2P(X\cap Y)$.
Adding all three equations:
$2[P(A)+P(B)+P(C)] - 2[P(A\cap B)+P(B\cap C)+P(C\cap A)] = \dfrac{3}{4}$
$\Rightarrow P(A)+P(B)+P(C) - [P(A\cap B)+P(B\cap C)+P(C\cap A)] = \dfrac{3}{8}$ ...(i)
Using inclusion-exclusion:
$P(A\cup B\cup C) = P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C)$
$= \dfrac{3}{8} + \dfrac{1}{16} = \dfrac{6}{16}+\dfrac{1}{16} = \dfrac{7}{16}$
Bulk modulus $K = \rho \frac{dP}{d\rho}$.
- Rearranging for change in density $d\rho$: $d\rho = \frac{\rho dP}{K}$
- Given $dP = P$, the increase is $\frac{\rho P}{K}$.
- Total isotopes = 3 (Protium, Deuterium, and Tritium).
- Radioactive isotopes = 1 (Only Tritium is radioactive, emitting low-energy $\beta^{-}$ particles).
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