1. Originally Posted by shanegoeswapow
so i just need to factor it out then?
yes, factoring will work. try it and tell me what answer you get

2. 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) $\cos 2 \theta = \cos^2 \theta - \sin^2 \theta = 1 - 2 \sin^2 \theta$

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

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

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

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

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

Note that this is quadratic in $\sin \theta$, you can see this explicity if you were to say let $\sin \theta = x$ then you would get $6x^2 - x - 2 = 0$. But don't bother with that, it's usually a waste of time. Just solve this like a regular quadratic and tell me the answers you get for $\theta$
when you say solve liek a regular quadratic you mean like factor it out or use the quadratic formula thing (x= -b + or - the square root of b^2 - 4ac all over 2a?

3. ps can jhe can you check post #27 from me? i dont get those questions

4. Originally Posted by shanegoeswapow
more help if you dont mind? im not good with these problemos ><
(29)

By the method of Bernoulli Trials we have that, the probability of $k$ successes in $n$ trials is given by:

$P(k) = {n \choose k}p^kq^{n-k}$

where $p$ is the probability of success, and $q = 1 - p$ is the probability of failure.

In this question, $n = 7$ since we planted seven plants. we want the probability of at least 5 will sprout, that is, 5 or more (meaning 5 OR 6 OR 7). which is

$P(k \geq 5) = P(5) + P(6) + P(7)$

$\Rightarrow P(k \geq 5) = {7 \choose 5}\left( \frac {3}{4} \right)^5 \left( \frac {1}{4} \right)^2 + {7 \choose 6} \left( \frac {3}{4} \right)^6 \left( \frac {1}{4} \right)^1 + {7 \choose 7} \left( \frac {3}{4} \right)^7 \left( \frac {1}{4} \right)^0$

Note, ${7 \choose 5}$ is the same as $_7C_5$ and so on.

Now can you take it from here?

Originally Posted by shanegoeswapow
ps can jhe can you check post #27 from me? i dont get those questions

5. Originally Posted by Jhevon
the post number 27 i think i asked for number 25 and i think 26

6. there are full solutions to numbers 23, 29, 31, 33 and 34 here

i'd appreciate it if you thanked Soroban for the hard work
------
25)

See the diagram below:

in the diagram, $\theta = 70^{ \circ}$, $a = 13$ and $b = 11$

use the formula:

$A = \frac {1}{2}ab \sin \theta$

7. okie i got 25
26: how do i do this?
32: is just calculator?
29: can you go over that? i dont get understand how you broke it down for badandy
31: that formula i need to just remember?

8. nvm 26 i got it

9. 26)

Do you know how to form composite functions? it is very important that you do! Here not so much, we don't have to deal with composites if we don't want to, but you should know it. if you have any doubts about what i am talking about, please say so!

we have $h(x) = 70 + 0.2x$ and $g(t) = 300(0.8)^t$

$\Rightarrow g(4) = 300(0.8)^4 = 122.88$

Finally, $h \left(g(4) \right) = 70 + 0.2(g(4)) = 70 + 0.2(122.88) 94.576 \approx 95$ to the nearest whole number

10. Originally Posted by Jhevon
26)

Do you know how to form composite functions? it is very important that you do! Here not so much, we don't have to deal with composites if we don't want to, but you should know it. if you have any doubts about what i am talking about, please say so!

we have $h(x) = 70 + 0.2x$ and $g(t) = 300(0.8)^t$

$\Rightarrow g(4) = 300(0.8)^4 = 122.88$

Finally, $h \left(g(4) \right) = 70 + 0.2(g(4)) = 70 + 0.2(122.88) 94.576 \approx 95$ to the nearest whole number
i thought it was calculator work, but when i read the question over i got it thanks

11. Originally Posted by shanegoeswapow
okie i got 25
32: is just calculator?
no

29: can you go over that? i dont get understand how you broke it down for badandy
will do, soon

31: that formula i need to just remember?
all formulas i give you are ones you need to remember, all of them. also any formula topsquark gave you, remembe rthem as well. it seems rare for them to ask you to derive formulas on this exam, so any formula you see us use, its a formula you should know by heart

12. uh so you think you can go over 32 and 29 then thanks

13. Originally Posted by Jhevon
all formulas i give you are ones you need to remember, all of them. also any formula topsquark gave you, remembe rthem as well. it seems rare for them to ask you to derive formulas on this exam, so any formula you see us use, its a formula you should know by heart
so far i wrote them all down from before just in case i did need to remember them

(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:

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

We want the probability of at least 5 successes from 7 trials ....i already explained what "at least" means

The at least means 5 successes or more, that means we want $P(5) + P(6) + P(7)$ ......with probability, "or" means "add"

we are told that $p = \frac {3}{4}$, that means $q = 1 - p = \frac {1}{4}$ ......do i have to explain this? the probability that a plant will sprout is 3/4, that is what we can a success. a failure is when a plant doesn't sprout, which is 1 - 3/4. why? the probability of all possible outcomes add up to 1. there are only two outcomes for the plant, it will sprout or it won't. so P(sprout) + P(does not sprout) = 1. therefore, P(does not sprout) = 1 - P(sprout)

So $P(5) + P(6) + P(6) = C(7,5) \left( \frac {3}{4} \right)^5 \left( \frac {1}{4} \right)^2 + C(7,6) \left( \frac {3}{4} \right)^6 \left( \frac {1}{4} \right)^1 + C(7,7) \left( \frac {3}{4} \right)^7 \left( \frac {1}{4} \right)^0$ ......we simply plug the values into the formula. make sure you understand what n and k means. n is the total number of trials, k is the number of successes we are looking for

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

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

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

now to calculate C(n,r) we do the follwoing. You can plug it into your calculator as i directed you in a previous post, or you can use topsquark's formula.

C(n,k) = n!/(k!(n - k)!)

now ! means factorial, and it is defined as the product of all integers from 1 up to the number you want to find the factorial of.

example, 5! = 1*2*3*4*5 = 120

so C(n,k) = [n*(n - 1)*(n - 2)*...*1]/{k(k - 1)(k - 2)*...*1[(n - k)(n - k - 1)(n - k - 2)*...*1]}

now what (n - k)! in the deminator does, is knock off all the numbers that are k and below in the n! above. and then we cancel accordingly to find the answer. so for instance:

C(5,3) = 5!/(3!(5 - 3)!) = 5!/(3!2!) ..........this 2! cancels with the 2*1 contained in the 5! so,

C(5,3) = (5*4*3)/(3*2*1) = 5*2 = 10 since the 3's cancel, and 2 goes into 4 2 times.

it is also good to know that C(n,k) = C(n, n - k)

so C(5,3) = C(5,5-3) = C(5,2) = (5*4)/(1*2) = 10

so in general, if passing the test is your only concern, just memorize the formula and plug in the numbers. all you need to know is what the variables represent

15. Originally Posted by Jhevon

so in general, if passing the test is your only concern, just memorize the formula and plug in the numbers. all you need to know is what the variables represent
lol sadly i cant just pass the test. i pretty much need an 85 or an above (mastery) to stay in my math advanced class. im okay in math its just that i dont like my teacher i have and the way she explains things
just from today on these forums i feel like i learned more here than in that class lol

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