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非参数统计代写|Nonparametric Statistics代考|高质保分代写非参数统计作业

问题 1.

M10 (Robert Beezer) Each sentence below has at least two meanings. Identify the source of the double meaning, and rewrite the sentence (at least twice) to clearly convey each meaning.
1. They are baking potatoes.
2. He bought many ripe pears and apricots.
3. She likes his sculpture.
4. I decided on the bus.

 

1. Does “baking” describe the potato or what is happening to the potato? Those are potatoes that are used for baking. The potatoes are being baked.
2. Are the apricots ripe, or just the pears? Parentheses could indicate just what the adjective “ripe” is meant to modify. Were there many apricots as well, or just many pears?
He bought many pears and many ripe apricots.
He bought apricots and many ripe pears.3. Is “sculpture” a single physical object, or the sculptor’s style expressed over many pieces and many years?
She likes his sculpture of the girl.
She likes his sculptural style.
4. Was a decision made while in the bus, or was the outcome of a decision to choose the bus. Would the sentence “I decided on the car,” have a similar double meaning?
I made my decision while on the bus.
I decided to ride the bus.

 

 

 
问题 2.

M70 (Robert Beezer) We have seen in this section that systems of linear equations have limited possibilities for solution sets, and we will shortly prove Theorem PSSLS that describes these possibilities exactly. This exercise will show that if we relax the requirement that our equations be linear, then the possibilities expand greatly. Consider a system of two equations in the two variables $x$ and $y$, where the departure from linearity involves simply squaring the variables.
$$
\begin{aligned}
& x^2-y^2=1 \\
& x^2+y^2=4
\end{aligned}
$$
After solving this system of non-linear equations, replace the second equation in turn by $x^2+2 x+y^2=3$, $x^2+y^2=1, x^2-4 x+y^2=-3,-x^2+y^2=1$ and solve each resulting system of two equations in two variables. (This exercise includes suggestions from Don Kreher.)

 

Solution (Robert Beezer) The equation $x^2-y^2=1$ has a solution set by itself that has the shape of a hyperbola when plotted. Four of the five different second equations have solution sets that are circles when plotted individually (the last is another hyperbola). Where the hyperbola and circles intersect are the solutions to the system of two equations. As the size and location of the circles vary, the number of intersections varies from four to one (in the order given). The last equation is a hyperbola that “opens” in the other direction. Sketching the relevant equations would be instructive, as was discussed in Example STNE.
The exact solution sets are (according to the choice of the second equation),
$$
\begin{aligned}
x^2+y^2 & =4: & & \left\{\left(\sqrt{\frac{5}{2}}, \sqrt{\frac{3}{2}}\right),\left(-\sqrt{\frac{5}{2}}, \sqrt{\frac{3}{2}}\right),\left(\sqrt{\frac{5}{2}},-\sqrt{\frac{3}{2}}\right),\left(-\sqrt{\frac{5}{2}},-\sqrt{\frac{3}{2}}\right)\right\} \\
x^2+2 x+y^2 & =3: & & \{(1,0),(-2, \sqrt{3}),(-2,-\sqrt{3})\} \\
x^2+y^2 & =1: & & \{(1,0),(-1,0)\} \\
x^2-4 x+y^2 & =-3: & & \{(1,0)\} \\
-x^2+y^2 & =1: & & \{\}
\end{aligned}
$$

.

 

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