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antoniya [11.8K]
3 years ago
12

Which of the following statements is CORRECT? a. A graph of the SML as applied to individual stocks would show required rates of

return on the vertical axis and standard deviations of returns on the horizontal axis. b. An increase in expected inflation, combined with a constant real risk-free rate and a constant market risk premium, would lead to identical increases in the required returns on a riskless asset and on an average stock, other things held constant. c. If two "normal" or "typical" stocks were combined to form a 2-stock portfolio, the portfolio's expected return would be a weighted average of the stocks' expected returns, but the portfolio's standard deviation would probably be greater than the average of the stocks' standard deviations. d. If investors become more risk averse, then (1) the slope of the SML would increase and (2) the required rate of return on low-beta stocks would increase by more than the required return on high-beta stocks. e. The CAPM has been thoroughly tested, and the theory has been confirmed beyond any reasonable doubt.
Mathematics
1 answer:
SIZIF [17.4K]3 years ago
5 0

b. An increase in expected inflation, combined with a constant real risk-free rate and a constant market risk premium, would lead to identical increases in the required returns on a riskless asset and on an average stock, other things held constant.

Hope this helps :)

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The half-life is a substance is 375 years. If 70 grams is present now, how much will be present in 500 years?
Lynna [10]

Answer:

27.76 grams will be present in 500 years

Step-by-step explanation:

The given formula is A=A_{o}e^{kt} , where A is the value of the substance in t years, and A_{o} is the initial value

∵ The half-life is a substance is 375 years

- Substitute A by \frac{1}{2}A_{o} and t by 375 to find the value of k

∴ \frac{1}{2}A_{o}=A_{o}e^{375k}

- Divide both sides by A_{o}

∴ \frac{1}{2}=e^{375k}

- Insert ㏑ in both sides

∴ ㏑( \frac{1}{2} ) = ㏑ ( e^{375k} )

- Remember ㏑ ( e^{n} ) = n

∵ ㏑ ( e^{375k} ) = 375 k

∴ ㏑( \frac{1}{2} ) = 375 k

- Divide both sides by 375

∴ k ≈ -0.00185

∴  A=A_{o}e^{-0.00185t}

∵ 70 grams is present now

- That means the initial value is 70 grams

∴ A_{o} = 70

∵ The time is 500 years

∴ t = 500

- Substitute the values of A_{o} and t in the formula

∵ A=70e^{-0.00185(500)}

∴ A = 27.76

∴ 27.76 grams will be present in 500 years

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Which set of values are equivalent. ,
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<span>the answer is C. x ≥ 8.00 and x < 9.50
proof 
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