# Need Help With Math B Regents June 2005

• May 27th 2007, 04:15 PM
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.
• May 27th 2007, 04:27 PM
Jhevon
Quote:

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)

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

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

..........................$\displaystyle = 4$
• May 27th 2007, 04:49 PM
Jhevon
Quote:

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 $\displaystyle C(n,k)$
• May 27th 2007, 05:00 PM
Jhevon
Quote:

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 $\displaystyle r = 150$

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

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

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

$\displaystyle \Rightarrow \theta \approx 94^{ \circ}$
• May 27th 2007, 05:13 PM
Jhevon
Quote:

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, $\displaystyle \angle 1 > \angle 2$
• May 27th 2007, 06:05 PM
Soroban

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: .$\displaystyle 2r \:=\:38.96880734 \:\approx\:\boxed{ 39\text{ m}}$