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Your (a) is not correct. Although the answer is correct. $p$ is dependent variable, $x$ is independent variable. The slope $m$ should be $$\frac{58-78}{50-30}$$ To find $b$, you plugged the wrong numbers: $58=-50+b$ The maximum profit is $P(40)$. You can find it by plugging $40$ into the profit function. Price per unit: plug $40$ into $p(x)$. Total ...

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Here is a good beginners graph theory note (see https://homepages.warwick.ac.uk/~masgax/Graph-Theory-notes.pdf). It was designed for a fourth year undergraduate (equivalent to a Masters) module in Mathematics at the University of Warwick. It gives a good insight to graph theory that will be important in giving you a very good exposure. There are also a ...

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Rewrite the thesis as $$E[u(X)] = \int u(x) dP^X \ge \int u(x + y) dP^X \otimes dP^Y = E[u(X + Y)]$$ Now let's concentrate on the right hand side; in particular we can use fubini tonelli to write $$\int u(x + y) dP^X \otimes dP^Y = \int \left(\int u(x+y)dP^Y\right) dP^X$$ The inner integral is simply $E[u(x + Y)]$ for some constant $x$. Since $u$ is ...

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$$\int_0^1 d(uF^k(u)) = \left[uF^k(u)\right]_0^1 = \int_0^1 u dF^k(u) + \int_0^1 F(u)^k du$$ this is leads to $$\left[uF^k(u)\right]_0^1 = 1\cdot F^k(1) - 0\cdot F^k(0) = 1 = \int_0^1 u dF^k(u) + \int_0^1 F(u)^k du$$ re-arrange $$\int_0^1 u dF^k(u) = 1 - \int_0^1 F(u)^k du$$

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0) I think you have some typos in the question. Let us make things more clear by assuming $Y$ has standard deviation $\left(\sqrt{\lambda}\right) \sigma$. 1) Draw a picture and compare the midpoints of the two line segments. 2) Define "centered densities": \begin{align} \tilde{f}_X(x) &= f_X(x+\mu)\\ \tilde{f}_Y(x) &= f_Y(x + \mu) \end{align} ...

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An equilibrium will be one where everyone bids 11.11% more than their valuation. That is, a bidder with private valuation $p$ reports $p/0.9$. Everyone is bidding more than private valuation, but the information revealed is same as in a regular 2nd-price auction.

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Since $y=\frac{X}{N}$, it follows that $\frac{\dot{y}}{y} = \frac{\dot{X}}{X} - \frac{\dot{N}}{N}$ (Take logs of both sides then differentiate wrt time). Call this equation (1). Also, since $X=N^{\sigma}$, it also follows that $\frac{\dot{X}}{X} = \sigma\frac{\dot{N}}{N}$ If we substiute this into equation (1) we get \$\frac{\dot{y}}{y} = ...

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