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dimaraw [331]
3 years ago
10

Suppose that a particular candidate for public office is in fact favored by 48%of all registered voters in the district. A polli

ng organization will take a random sample of 500voters and will use p, the sample proportion, to estimate p. What is the approximate probability that p will be greater than 0.5, causing the polling organization to incorrectly predict the result of the upcoming election?
Mathematics
1 answer:
Mumz [18]3 years ago
7 0

Answer: 0.1854

Step-by-step explanation:

Given : Suppose that a particular candidate for public office is in fact favored by 48% of all registered voters in the district.

Let \hat{p} be the sample proportion of voters in the district favored a particular candidate for public office .

A polling organization will take a random sample of n=500 voters .

Then, the probability that p will be greater than 0.5, causing the polling organization to incorrectly predict the result of the upcoming election :

P(\hat{p}>0.5)=P(\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}>\dfrac{0.5-0.48}{\sqrt{\dfrac{0.48(0.52)}{500}}})\\\\=P(z>0.8951)\ \ [\because\ z=(\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}]\\\\=1-P(z\leq0.8951)\ \ [\because\ P(Z>z)=1-P(Z\leq z)]\\\\ = 1-0.8146=0.1854

∴ Required probability = 0.1854

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x = peach trees

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3 years ago
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In calculating the interest on a car we have a formula

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7 0
3 years ago
Thor invests $750 in an account, with interest compounded continuously. If his investment doubles in value after 9 years, how mu
elixir [45]
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3 years ago
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A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo
djverab [1.8K]

Answer:

a) y(t)=50000-49990e^{\frac{-2t}{25}}

b) 31690.7 g/L

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is \frac{dy}{dt}=R_{i}-R_{o}, where R_{i} is the rate of salt entering and R_{o} is the rate of salt going outside.

Then we have, R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}, and

R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}

So we obtain.  \frac{dy}{dt}=4000-\frac{2y}{25}, then

\frac{dy}{dt}+\frac{2y}{25}=4000, and using the integrating factor e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}, therefore  (\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}, we get   \frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}, after integrating both sides y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C, therefore y(t)= 50000 +Ce^{\frac{-2t}{25}}, to find C we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions y(0)=10, so 10= 50000+Ce^{\frac{-0*2}{25}}

10=50000+C\\C=10-50000=-49990

Finally we can write an expression for the amount of salt in the tank at any time t, it is y(t)=50000-49990e^{\frac{-2t}{25}}

b) The tank will overflow due Rin>Rout, at a rate of 80 L/min-40L/min=40L/min, due we have 500 L to overflow \frac{500L}{40L/min} =\frac{25}{2} min=t, so we can evualuate the expression of a) y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7, is the salt concentration when the tank overflows

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77julia77 [94]
A must be parallel to b.

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Then, a must be parallel to b (transverse line d cuts parallel lines producing converse alternate angles that are equal (ie </span>∠15 = <span>∠3))
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