# Math Help - Need Help With Math B Regents June 2005

1. ## 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.

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)

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

.......................... $= 3 + 1 - 3 + 1 + 3 + 1 - 3 + 1$

.......................... $= 4$

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

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

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

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

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

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

................................. $= 0.3115 + 0.3115 + 0.1335$

.................................
$= 0.7565$

I hope you know how to calculate $C(n,k)$

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 $s$ is given by the formula:

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

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

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

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

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

$\Rightarrow \theta \approx 94^{ \circ}$

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:

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

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

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

since $\angle BAC$ is positive, $\angle 1 > \angle 2$

Jhevon did a fantastic job!

I had a slightly different approach to #33.

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

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

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

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

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

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

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

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

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

Draw radii $OB = OE = r$

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

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

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

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