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The study of explanatory or independent differences based on the dependent variable is a simple statistical method. The correlation between two variables is researched in which only the explanatory variable of them would be present, and the further discussion can be defined as follows:

For point (i):

The log(distances) seem to be optimistic, and their valuation is 0.312, which also means that the cost of a house will rise even as the distance increases.
In this, the certainly is what everybody expects the better the house from the sewage treatment is from a hygienic viewpoint.

For point (ii):

In this regard, we cannot use straightforward regression, even as non-biased price estimation might not even be provided.
The price-distance relationship here is not so much linear.

For point (iii):

Several factors influence the house's price, the biggest contributor is house length, market nearness or public areas, number of rooms, etc.
Thus the only influence is not the mere distance from the incinerator. It's one of the factors affecting housing pricing.
The proximity of markets and places can be connected to the distance from the crematorium.
However, factors such as house size and the number of stories aren't related to the distance from the incineration plant.

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brainly.com/question/19729277

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3 years ago

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Orlov [11]

angle QRO is 126 degrees

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3 years ago

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Prove that H c G is a normal subgroup if and only if every left coset is a right coset, i.e., aH = Ha for all a e G

Kaylis [27]

$\Rightarrow$

Suppose first that $H\subset G$ is a normal subgroup. Then by definition we must have for all $a\in H$ , $xax^{-1} \in H$ for every $x\in G$ . Let $a\in G$ and choose $(ab)\in aH$ ( $b\in H$ ). By hypothesis we have $aba^{-1} =abbb^{-1}a^{-1}=(ab)b(ab)^{-1} \in H$ , i.e. $aba^{-1}=c$ for some $c\in H$ , thus $ab=ca \in Ha$ . So we have $aH\subset Ha$ . You can prove $Ha\subset aH$ in the same way.

$\Leftarrow$

Suppose $aH=Ha$ for all $a\in G$ . Let $h\in H$ , we have to prove $aha^{-1} \in H$ for every $a\in G$ . So, let $a\in G$ . We have that $ha^{-1} =a^{-1}h'$ for some $h'\in H$ (by the hypothesis). hence we have $aha^{-1}=h' \in H$ . Because $a$ was chosen arbitrarily we have the desired .

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3 years ago