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Ira Lisetskai [31]
3 years ago
13

A sample survey contacted an SRS of 2854 registered voters shortly before the 2012 presidential election and asked respondents w

hom they planned to vote for. Election results show that 51% of registered voters voted for Barack Obama. We will see later that in this situation the proportion of the sample who planned to vote for Barack Obama (call this proportion V) has approximately the Normal distribution with mean μ 0.52 and standard deviation σ = 0.009.
(a) If the respondents answer truthfully, what is P (0.5くV < 0.54)? This is the probability that the sample proportion v estimates the population proportion 0.52 within plus or minus 0.02.
P (0.5<= V <=0.54) (±0.0001)=

(b) In fact, 50% of the respondents said they planned to vote for Barack Obama V = 0.5. If respondents answer truthfully, What is P(V <=0.5)?
P (V <=0.5) (±0.0001) =

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
3 0

Answer:

a) 97.37%

b) 1.31%

Step-by-step explanation:

a)  

Here we want to calculate the area under the Normal curve with mean 0.52 and standard deviation 0.009 between 0.5 and 0.54

This can be easily done with a spreadsheet and we get

<em>P (0.5くV < 0.54) = 0.9737 or 97.37% </em>

(See picture 1)

b)

Here we want the area under the Normal curve with mean 0.52 and standard deviation 0.009 to the left of 0.5.

<em>P(V ≤ 0.5) = 0.0131 or 1.31% </em>

(See picture 2)

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Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.
Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

\mathbf{ 5e^{-x^2} cos (4x)  }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

Step-by-step explanation:

GIven that:

f(x) = 5e^{-x^2} cos (4x)

The Maclaurin series of cos x can be expressed as :

\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...  \ \ \ (1)}

\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0}  \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!}  -\dfrac{x^6}{3!}+... \ \ \  (2)}

From equation(1), substituting x with (4x), Then:

\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}

The first three terms of cos (4x) is:

\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}

\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}

\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}

Multiplying equation (2) with (3); we have :

\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }

\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }

\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }

\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

Finally , multiplying 5 with \mathtt{ e^{-x^2} cos (4x) } ; we have:

The first three nonzero terms in the Maclaurin series is

\mathbf{ 5e^{-x^2} cos (4x)  }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

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3 years ago
Vanessa draws one side of equilateral ΔABC on the coordinate plane at points A(–2, 1) and B(4,1). Select all the possible coordi
Radda [10]

(1, \sqrt{27})  and (1, - \sqrt{27})

Step-by-step explanation:

Step 1 :

The co ordinates of the given equilateral triangle are A(-2,1) and B(4,1)

The distance between these 2 points is the length of the given triangle

Distance between the 2 points is  \sqrt{(x2-x1)^{2}+ (y2-y1)^{2}} = sqrt (sq(4-(-2)) + sq(1-1)) = 6

Hence the given triangle has 3 equal side of length 6 unit.

Step 2:

The length of the other side should be 6. Let (x,y) be the co-ordinate of the point C

We have then x = 1 (because the perpendicular from C to AB bisects AB we have the point C to have the x co-ordinate as 1)

Also we have the distance between the point B(4,1) and C(1,y) to be 6 as this is an equilateral triangle

Hence \sqrt{(4-1)^{2} + (1-y)^{2 } = 6

=> 9 + 1 + y^{2} -2 y = 6

=> y^{2}-2 y - 26 = 0

=> y = 2± sqrt(4+104) / 2 = 1 ± sqrt(27)

Hence the possible co ordinates of C are (1, \sqrt{27})  and (1, - \sqrt{27})

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Answer:

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Answer:

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