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Thread: Need Help With Math B Regents June 2005

  1. #1
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    Need Help With Math B Regents June 2005

    I'm currently reviewing for the Math B regents and I have some trouble with 5 questions I looke up the answers online but i still need help showing the work to arrive to the answers. Here is the link to the the answers:http://www.nysedregents.org/testing/mathre/b-key605.pdf
    And here is the link to the questions:http://www.nysedregents.org/testing/mathre/b605.pdf

    These are the question i need help showing work for. 23,29,31,33, and 34.
    In 34 i got the second part i just need help figuring out the diameter of the circle.
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    Quote Originally Posted by badandy328 View Post
    I'm currently reviewing for the Math B regents and I have some trouble with 5 questions I looke up the answers online but i still need help showing the work to arrive to the answers. Here is the link to the the answers:http://www.nysedregents.org/testing/mathre/b-key605.pdf
    And here is the link to the questions:http://www.nysedregents.org/testing/mathre/b605.pdf

    These are the question i need help showing work for. 23,29,31,33, and 34.
    In 34 i got the second part i just need help figuring out the diameter of the circle.
    23)

    $\displaystyle \sum_{k=0}^{3} \left( 3 \cos k \pi + 1 \right) = (3 \cos(0) + 1) + (3 \cos( \pi) + 1) + (3 \cos(2 \pi) + 1) + (3 \cos(3 \pi) + 1) $

    ..........................$\displaystyle = 3 + 1 - 3 + 1 + 3 + 1 - 3 + 1$

    ..........................$\displaystyle = 4$
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by badandy328 View Post
    I'm currently reviewing for the Math B regents and I have some trouble with 5 questions I looke up the answers online but i still need help showing the work to arrive to the answers. Here is the link to the the answers:http://www.nysedregents.org/testing/mathre/b-key605.pdf
    And here is the link to the questions:http://www.nysedregents.org/testing/mathre/b605.pdf

    These are the question i need help showing work for. 23,29,31,33, and 34.
    In 34 i got the second part i just need help figuring out the diameter of the circle.


    29)
    We proceed by Bernoulli Trials:

    Recall, the probability of $\displaystyle k$ successes from $\displaystyle n$ trials, where the probability of success is $\displaystyle p$ and the probability of failure is $\displaystyle q$ is given by the following:

    $\displaystyle P(k) = C(n,k) p^k q^{n-k}$

    We want the probability of at least 5 successes from 7 trials

    The at least means 5 successes or more, that means we want $\displaystyle P(5) + P(6) + P(7)$

    we are told that $\displaystyle p = \frac {3}{4}$, that means $\displaystyle q = 1 - p = \frac {1}{4}$

    So $\displaystyle P(5) + P(6) + P(6) = C(7,5) \left( \frac {3}{4} \right)^5 \left( \frac {1}{4} \right)^2 + C(7,6) \left( \frac {3}{4} \right)^6 \left( \frac {1}{4} \right)^1 + C(7,7) \left( \frac {3}{4} \right)^7 \left( \frac {1}{4} \right)^0$

    .................................$\displaystyle = 0.3115 + 0.3115 + 0.1335 $

    .................................
    $\displaystyle = 0.7565$

    I hope you know how to calculate $\displaystyle C(n,k)$
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by badandy328 View Post
    I'm currently reviewing for the Math B regents and I have some trouble with 5 questions I looke up the answers online but i still need help showing the work to arrive to the answers. Here is the link to the the answers:http://www.nysedregents.org/testing/mathre/b-key605.pdf
    And here is the link to the questions:http://www.nysedregents.org/testing/mathre/b605.pdf

    These are the question i need help showing work for. 23,29,31,33, and 34.
    In 34 i got the second part i just need help figuring out the diameter of the circle.
    31)

    The formula for the length of an arc $\displaystyle s$ is given by the formula:

    $\displaystyle s = \frac { \theta}{360} \cdot 2 \pi r$
    where $\displaystyle \theta$ is the angle that subtends the arc in degrees, and $\displaystyle r$ is the radius

    we are given $\displaystyle s = 247$ and $\displaystyle r = 150$

    $\displaystyle \Rightarrow 247 = \frac { \theta}{360} 2 \pi (150)$

    $\displaystyle \Rightarrow \frac { \theta}{360} = \frac {247}{300 \pi}$

    $\displaystyle \Rightarrow \theta = \frac {360 \cdot 247}{300 \pi}$

    $\displaystyle \Rightarrow \theta \approx 94^{ \circ}$
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by badandy328 View Post
    I'm currently reviewing for the Math B regents and I have some trouble with 5 questions I looke up the answers online but i still need help showing the work to arrive to the answers. Here is the link to the the answers:http://www.nysedregents.org/testing/mathre/b-key605.pdf
    And here is the link to the questions:http://www.nysedregents.org/testing/mathre/b605.pdf

    These are the question i need help showing work for. 23,29,31,33, and 34.
    In 34 i got the second part i just need help figuring out the diameter of the circle.
    33)

    Now i don't recall what this theorem is called but it has something to do with "external angles of a triangle" or something like that. basically, according to this theorem:

    $\displaystyle \angle 1 = \angle ACB + \angle BAC$

    But $\displaystyle \angle ACB = \angle 2$ since they are alternate interior angles, so we have:

    $\displaystyle \angle 1 = \angle 2 + \angle BAC$

    since $\displaystyle \angle BAC$ is positive, $\displaystyle \angle 1 > \angle 2$
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    Hello, badandy!

    Jhevon did a fantastic job!


    I had a slightly different approach to #33.

    Since $\displaystyle AD \parallel BC:\:\angle1 \,=\,\angle DAB$ .(corresponding angles)

    So we have: .$\displaystyle \angle1\:=\:\angle2 + \angle CAB$

    Since $\displaystyle \angle CAB$ is positive: .$\displaystyle \angle1 \:>\:\angle 2$



    34) We are given: .$\displaystyle m(\overline{ADB}):m(\overline{AEB}) \:=\:3:2$

    Let $\displaystyle m(\overline{ADB}) = 3k,\;m(\overline{AEB}) = 2k$

    . . Then: .$\displaystyle 3k + 2k\:=\:360^o\quad\Rightarrow\quad k = 72^o$

    Hence: .$\displaystyle m(\overline{ADB}) = 216^o,\;m(\overline{AEB}) = 144^o$


    Theorem: The angle between two tangents to a circle is measured
    . . . . . . . .by one-half the difference of the intercepted arcs.

    Therefore: .$\displaystyle \angle ACB \:=\:\frac{1}{2}\left(216^o - 144^o\right) \;=\;\boxed{36^o}$


    Draw radii $\displaystyle OB = OE = r$

    Since $\displaystyle \angle OBC = 90^o$, we have: .$\displaystyle OB^2 + BC^2 \:=\:OC^2$

    . . Hence: .$\displaystyle r^2 + 60^2\:=\:(r + 43.6)^2\quad\Rightarrow\quad r^2 + 3600\:=\:r^2 + 78.2r + 1900.96$

    . . Then:. . $\displaystyle 87.2r \:=\:1609.04\quad\Rightarrow\quad r \:= \:\frac{1609.04}{87.2}\:=\:19.48440367$

    Therefore, the diameter is: .$\displaystyle 2r \:=\:38.96880734 \:\approx\:\boxed{ 39\text{ m}}$

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