# Thread: Math B Regents Help?

1. Originally Posted by Jhevon
$\displaystyle P(k) = C(n,k) p^k q^{n-k}$ ........This is just a formula you should memorize. Topsquark told you how to calculate C(n,r), i will show how to break down his formula
question: so this formula whats the name of it? the beroulli theroum? i never heard of that name before...
is that also for when you need to calculate things when they are exactly?

2. 32)

$\displaystyle y = -2x^2 + 38x + 10$

where $\displaystyle y$ is the height, in feet, and $\displaystyle x$ is the time in seconds.

we want $\displaystyle x$ so that $\displaystyle y$ is at least 125. remember, i said "at least" means "greater than or equal to" so what we want to find is what $\displaystyle x$'s solve the follwoing inequality:

$\displaystyle -2x^2 + 38x + 10 \geq 125$

$\displaystyle \Rightarrow -2x^2 + 38x + 10 - 125 \geq 0$

$\displaystyle \Rightarrow -2x^2 + 38x - 115 \geq 0$

By the quadratic formula, we have

$\displaystyle x \geq \frac {-38 \pm \sqrt {38^2 - 920}}{-4}$

$\displaystyle \Rightarrow x \geq \frac {-38 + \sqrt {524}}{-4} \mbox { or } x \geq \frac {-38 - \sqrt {524}}{-4}$

that is $\displaystyle x \geq 3.8 \mbox { or } x \geq 15.2$

Now check the intervals on a number line or something. that is, test values that fall in all the regions separated by the x-values. so you would test say, $\displaystyle x = 3, x = 5, x = 16$

you can go through the hassle and plug these into the inequality, or just realize it is a downward opening parabola, so the part above zero would be between the two values given. either way, your conclusion should be:

$\displaystyle 3.8 \leq x \leq 15.2$

3. Originally Posted by Jhevon
well, you need to get good with trig. there are a lot of problems that use trig, as you can see. for this problem you need to know two things:

(1) How to solve a quadratic equation

(2) $\displaystyle \cos 2 \theta = \cos^2 \theta - \sin^2 \theta = 1 - 2 \sin^2 \theta$

$\displaystyle 3 \cos 2 \theta + \sin \theta - 1 = 0$

$\displaystyle \Rightarrow 3 \left( 1 - 2 \sin^2 \theta \right) + \sin \theta - 1 = 0$

$\displaystyle \Rightarrow 3 - 6 \sin^2 \theta + \sin \theta - 1 = 0$

$\displaystyle \Rightarrow -6 \sin^2 \theta + \sin \theta + 2 = 0$

$\displaystyle \Rightarrow 6 \sin^2 \theta - \sin \theta -2 = 0$
i was able to break it down only this far,im not sure how to factor these numbers with sin and stuff

4. Originally Posted by shanegoeswapow
question: so this formula whats the name of it? the beroulli theroum? i never heard of that name before...
is that also for when you need to calculate things when they are exactly?
it is called "Bernoulli Trial"

we use it when:

- we have only two outcomes, one can be interpreted as a success and the other a failure

- we are performing a number of trials, and each trial is independent, that is, the outcome of one trial does not affect the other

- and i think there is one more condition that i cant recall at the moment

when we have exactly, we just find P(k). there is no need to add several of them, we just find one.

note that you may have to be cunning with these. don't forget that all possibilities add up to 1.

so if they said, we want to do something 17 times, what is the probability that it will succeed at least 2 times.

i don't expect you to find P(2) + P(3) + ... + P(17). that would take forever, and is not a smart move.

realize that since all possibilties add up to 1, all you need to find is the probability of it succeeding LESS THAN 2 times, and subtract that from 1, or

1 - (P(0) + P(1)) = 1 - P(0) - P(1)

5. Originally Posted by shanegoeswapow
i was able to break it down only this far,im not sure how to factor these numbers with sin and stuff
treat it the sine like any regular variable: what if it was $\displaystyle 6x^2 - x - 2 = 0$? how would you deal with that? can we factor that? no! use the quadratic formula:

we get: $\displaystyle \sin x = \frac {1 \pm \sqrt{1 + 48}}{12} = \frac {1 \pm 7}{12}$

Can you take it from here?

6. Hello, shanegoeswapow!

27. Two tracking stations, A and B, are on an east-west line 110 miles apart.
A forest fire is located at F, on a bearing of 42° northeast of station A
and 15° northeast of station B.
How far, to the nearest mile, is the fire from station A?
In the diagram, we have triangle $\displaystyle AFB$ with: $\displaystyle \angle FAB = 48^o$ and $\displaystyle \angle ABF = 105^o$
. . Hence: .$\displaystyle \angle AFB \:=\:180^o - 48^o - 105^o \:=\:27^o$

Using the Law of Sines: .$\displaystyle \frac{AF}{\sin105^o} \:=\:\frac{110}{\sin27^o}$

Hence: .$\displaystyle AF \:=\:\frac{110\sin105^o}{\sin27^o} \:=\:234.0397843$

Therefore: .$\displaystyle AF \:\approx\:\boxed{234\text{ miles}}$

29. The probability that a planted watermelon will spout is $\displaystyle \frac{3}{4}$.
If Peyton plants seven seeds, find, to the nearest ten-thousandths,
the probabiity at least five will sprout.
You're expected to know the Binomial Probability formula.

We have: .$\displaystyle P(\text{sprout}) = 0.75,\;P(\text{not}) = 0.25$

"At least five" means: "five spout" or "six sprout" or "seven spout".
We must find these three probabilities and add them.

$\displaystyle P(\text{5 sprout}) \;=\;{7\choose5}(0.75)^5(0.25)^2\;=\;0.311462402$

$\displaystyle P(\text{6 sprout}) \;=\;{7\choose6}(0.75)^6(0.25^1 \;=\; 0.311462402$

$\displaystyle P(\text{7 sprout}) \;=\;{7\choose7}(0.75)^7(0.25)^0 \;=\;0.133483887$

Therefore: .$\displaystyle P(\text{at least 5}) \;=\;0.311462402\,+\,0.311462402\,+\,0.133483887 \;=\;0.756408691$ $\displaystyle \;\approx\;\boxed{0.7564}$

7. Originally Posted by Jhevon
treat it the sine like any regular variable: what if it was $\displaystyle 6x^2 - x - 2 = 0$? how would you deal with that? can we factor that? no! use the quadratic formula:

we get: $\displaystyle \sin x = \frac {1 \pm \sqrt{1 + 48}}{12} = \frac {1 \pm 7}{12}$

Can you take it from here?
math is so confusing!

8. I need to finish my other homework being how it is now around 12:05am over here
I need to wake up like at 6am so i can go to math extra help and try to understand more
i really want to thank you guys for all your help, it helped me alot, im planning on checking over the rest of them and asking more questions possibly tommorow if i have time (i have to study for 3 other regents coming up this week)
i think there were a few more questions i had if not, thanks everyone for your help
so ill be back here tommorow with more questions and hopefully i can get some more great help?
fun fun fun

9. Originally Posted by shanegoeswapow
math is so confusing!
Picking up where we left off

$\displaystyle \sin x = \frac {1 \pm 7}{12}$

$\displaystyle \Rightarrow \sin x = \frac {8}{12} = \frac {2}{3} \mbox { or } \sin x = - \frac {6}{12} = - \frac {1}{2}$

if $\displaystyle \sin x = \frac {2}{3}$

$\displaystyle \Rightarrow x = \arcsin \left( \frac {2}{3} \right)$

$\displaystyle \Rightarrow x \approx 42^{\circ}, 138^{\circ}$

if $\displaystyle \sin x = - \frac {1}{2}$

$\displaystyle \Rightarrow x = \arcsin \left( \frac {1}{2} \right)$

$\displaystyle \Rightarrow x = 210^{\circ}, 330^{\circ}$

So there are four solutions for x between 0 and 360. (x should be $\displaystyle \theta$)

To get all the angles you have to be able to work with reference angles

10. back again
on the same regents can i get help with the proof number 33
and also on 34

11. Originally Posted by shanegoeswapow
back again
on the same regents can i get help with the proof number 33
and also on 34
Did you see Soroban's solutions?

12. i found his confusing

what did you mean when you said

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:

can you go into that external angeles of a traingle and like how i would say that for this kind of proof?

13. Originally Posted by shanegoeswapow
i found his confusing

what did you mean when you said

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:

can you go into that external angeles of a traingle and like how i would say that for this kind of proof?
this is what i meant. see the diagram below.

the theorem says, that in this setup, angle C = angle A + angle B

why? well, think about it. all angles in a triangle add up to 180.

so A + B + D = 180
=> A + B = 180 - D

also, angles on a straight line add up to 180
so C + D = 180
=> C = 180 - D

so you see that A + B = C

that was the theorem i used. Soroban used a different one though

and i still don't recall what this theorem is called, i think it had its name has something to do with "external angle of a triangle", since C is outside of the triangle

14. yeah i liek that one better

how do you guys suggest to study for my regents? (which is like 2 days away)

15. Originally Posted by shanegoeswapow
yeah i liek that one better

how do you guys suggest to study for my regents? (which is like 2 days away)
hopefully you have been doing problems on your own and have not just been copying answers from us, otherwise you won't make it.

i would also recommend that you begin memorizing formulas. i dont know if you get a formula sheet for your exam, if you don't you will have to know every formula we used by heart

you can also give yourself an exam. do a past paper and time yourself. it will get you used to the time constraints

then practice, practice, practice!

the night before the exam get a lot of rest. you should not be killing yourself with heavy studying. by that time you should be comfortably browsing through problems to see if you can do them or not. if you can't, you can pop in here and ask a quick question or two

EDIT: Oh wait, you guys do get some of the formulas. Page 19 has a lot of the trig formulas you will need. But not formulas for things like binomial distribution or Bernoulli trials

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