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2 years ago

If the population density for the city is 10,000 people per square mile, what is the population of the city?

Mathematics

1 answer:

ololo11 [35]2 years ago

6 0

Answer:

10,000 * (square miles in the city)

Step-by-step explanation:

You've been given a population of people for every square mile (10,000) and you're trying to figure out how many people live in the city.

To convert people per a square mile into people, you'll need to know how many square miles you have. I assume you've been given this as the size of the city or how many square miles the city takes up. Just multiply the amount of people per a square mile by the amount of square miles in the city, and it'll give you the population of people in every square mile of the city (the population).

Note:

This is people per a square mile, or in other words, for every square mile of city, 10,000 people live there. If you have 2 square miles of city, then 2 * 10,000 = 20,000 people live there.

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4. Dean Pelton wants to perform calculations to impress the accreditation consultants, but upon asking for information about GPA

Leya [2.2K]

Answer:

the minimum records to be retrieved by using Chebysher - one sided inequality is 17.

Step-by-step explanation:

Let assume that n should represent the number of the students

SO, $\bar x$ can now be the sample mean of number of students in GPA's

To obtain n such that $P( \bar x \leq 2.3 ) \leq .04$

⇒ $P( \bar x \geq 2.3 ) \geq .96$

However ;

$E(x) = \int\limits^4_2 Dx (2+e^{-x} ) 4x = D \\ \\ = D(e^{-x} (e^xx^2 - x-1 ) ) ^D_2 = 12.314 D$

$E(x^2) = D\int\limits^4_2 (2+e^{-x})dx \\ \\ = \dfrac{D}{3}[e^{-4} (2e^x x^3 -3x^2 -6x -6)]^4__2}}= 38.21 \ D$

Similarly;

$D\int\limits^4_2(2+ e^{-x}) dx = 1$

⇒ $D*(2x-e^{-x} ) |^4_2 = 1$

⇒ $D*4.117 = 1$

⇒ $D= \dfrac{1}{4.117}$

$\mu = E(x) = 2.991013 ; \\ \\ E(x^2) = 9.28103$

∴ $Var (x) = E(x^2) - E^2(x) \\ \\ = .3348711$

Now; $P(\bar \geq 2.3) = P( \bar x - 2.991013 \geq 2.3 - 2.991013) \\ \\ = P( \omega \geq .691013) \ \ \ \ \ \ \ \ \ \ (x = E(\bar x ) - \mu)$

Using Chebysher one sided inequality ; we have:

$P(\omega \geq -.691013) \geq \dfrac{(.691013)^2}{Var ( \omega) +(.691013)^2}$

So; $(\omega = \bar x - \mu)$

⇒ $E(\omega ) = 0 \\ \\ Var (\omega ) = \dfrac{Var (x_i)}{n}$

∴ $P(\omega \geq .691013) \geq \dfrac{(.691013)^2}{\frac{.3348711}{n}+(691013)^2}$

To determine n; such that ;

$\dfrac{(.691013)^2}{\frac{.3348711}{n}+(691013)^2} \geq 0.96 \\ \\ \\ (.691013)^2(1-.96) \geq \dfrac{-3348711*.96}{n}$

⇒ $n \geq \dfrac{.3348711*.96}{.04*(.691013)^2}$

$n \geq 16.83125$

Thus; we can conclude that; the minimum records to be retrieved by using Chebysher - one sided inequality is 17.

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