Calculate the probability that a head will show

[有問題的人]

A kind of the trigonometry question...

Let θ be an angle in standard position such that cotθ=-4/3 and sinθ<0.
Determine the exact value of sec θ




Determine the number of solutions for (asinx-b)(acosx-a)(bsinx+a)=0
where 0≦x＜2\pi, if 0＜a＜b.




A fair coin is tossed 60 times. Using the normal approximation to the binomial distribution, calculate the probability that a head will show between 32 and 36 times.




Please give me as many explanations as possible~ thanks

 

[BoundByAxioms]

A kind of the trigonometry question...

Let θ be an angle in standard position such that cotθ=-4/3 and sinθ<0.
Determine the exact value of sec θ

Determine the number of solutions for (asinx-b)(acosx-a)(bsinx+a)=0
where 0≦x＜2\pi, if 0＜a＜b.

A fair coin is tossed 60 times. Using the normal approximation to the binomial distribution, calculate the probability that a head will show between 32 and 36 times.

Please give me as many explanations as possible~ thanks

Well, where are sine and cot negative at the same time? Quadrant IV. So, draw a triangle whose adjacent side is 4, and whose opposite side is (negative) three. By doing this, you are in essence drawing cot(-4/3). Solve for the hypotenuse of this triangle, and find out what secant is.

For problem two:
sinx=b/a, cosx=1, sinx=-a/b. Since a<b, sinx=b/a does not exist, because |b/a|>1, which is not in the range of the arcsin function. Cosx=1 at one place (x=0), and sinx=b/a in two places: where sinx is positive. I'm not sure on this, but I think you can say that there are only two solutions, because the coefficient of the x is only 1. So, on the interval [0,2\Pi], there are three solutions.

For problem three:
I'm not sure on these...haven't done probability in a very long time. Here's my guess though. Type this on your calculator: binomcdf(60,0.5,36)-binomcdf(60,0.5,32). Another guess would be normalcdf(32,36) (also on your calculator).

 

[Defennder]

Interesting username.

 

[rootX]

Interesting username.

and the thread title

 

[rootX]

A kind of the trigonometry question...

Let θ be an angle in standard position such that cotθ=-4/3 and sinθ<0.
Determine the exact value of sec θ

Determine the number of solutions for (asinx-b)(acosx-a)(bsinx+a)=0
where 0≦x＜2\pi, if 0＜a＜b.

A fair coin is tossed 60 times. Using the normal approximation to the binomial distribution, calculate the probability that a head will show between 32 and 36 times.

Please give me as many explanations as possible~ thanks

Please show your own efforts (in future) before asking questions (see rules).

 

[有問題的人]

Interesting username.

It's ok ==~~ not a creative name...

 

[有問題的人]

and the thread title

how's the title interesting @@

You mean the form?

 

[有問題的人]

Well, where are sine and cot negative at the same time? Quadrant IV. So, draw a triangle whose adjacent side is 4, and whose opposite side is (negative) three. By doing this, you are in essence drawing cot(-4/3). Solve for the hypotenuse of this triangle, and find out what secant is.

For problem two:
sinx=b/a, cosx=1, sinx=-a/b. Since a<b, sinx=b/a does not exist, because |b/a|>1, which is not in the range of the arcsin function. Cosx=1 at one place (x=0), and sinx=b/a in two places: where sinx is positive. I'm not sure on this, but I think you can say that there are only two solutions, because the coefficient of the x is only 1. So, on the interval [0,2\Pi], there are three solutions.

For problem three:
I'm not sure on these...haven't done probability in a very long time. Here's my guess though. Type this on your calculator: binomcdf(60,0.5,36)-binomcdf(60,0.5,32). Another guess would be normalcdf(32,36) (also on your calculator).

I got the second one ~~

however... I still don't know how to do the first one and the third one

 

[有問題的人]

Well, where are sine and cot negative at the same time? Quadrant IV. So, draw a triangle whose adjacent side is 4, and whose opposite side is (negative) three. By doing this, you are in essence drawing cot(-4/3). Solve for the hypotenuse of this triangle, and find out what secant is.

For problem two:
sinx=b/a, cosx=1, sinx=-a/b. Since a<b, sinx=b/a does not exist, because |b/a|>1, which is not in the range of the arcsin function. Cosx=1 at one place (x=0), and sinx=b/a in two places: where sinx is positive. I'm not sure on this, but I think you can say that there are only two solutions, because the coefficient of the x is only 1. So, on the interval [0,2\Pi], there are three solutions.

For problem three:
I'm not sure on these...haven't done probability in a very long time. Here's my guess though. Type this on your calculator: binomcdf(60,0.5,36)-binomcdf(60,0.5,32). Another guess would be normalcdf(32,36) (also on your calculator).

Please show your own efforts (in future) before asking questions (see rules).

so many rules =o=

 

[rootX]

1) draw x-y axis.

plot cot (t) = -4/3 (it can be at two positions)
cot(t) = x/y
so you know what x and y would give that

once you have two vectors(or lines) find where the second condition is met,
you would left with one vector (or line)

using that line find your sec t3. no clue (can't remember how to do those questions)

remove [solved] if possible from the thread ...

------------------------

@thread title:
It's font. I know when you use chinese font and write in english it looks really cool to me!

 

[有問題的人]

1) Draw X-y Axis.

Plot Cot (t) = -4/3 (it Can Be At Two Positions)
Cot(t) = X/y
So You Know What X And Y Would Give That

Once You Have Two Vectors(or Lines) Find Where The Second Condition Is Met,
You Would Left With One Vector (or Line)

Using That Line Find Your Sec T


3. No Clue (can't Remember How To Do Those Questions)

Remove [solved] If Possible From The Thread ...

------------------------

@thread Title:
It's Font. I Know When You Use Chinese Font And Write In English It Looks Really Cool To Me!

　Ｏｈ～～　Ｉ　ｄｉｄ　ｇｅｔ　ｔｈｅ　ｆｉｒｓｔ　ｏｎｅ。。。　

　Ｔｈａｎｋｓ　ｆｏｒ　ｔｈｅ　ｉｎｓｐｉｒａｔｉｏｎ

　ｔｈｅ　ｔｈｉｒｄ　ｏｎｅ　ｉｓ　。。。ｓｔｉｌｌ　ｃｏｎｆｕｓｅｄ。。。

　ｈａｈａ。。。　Ｉ　ｇｕｅｓｓ　ｉｔ’ｓ　ｃｕｔｅｒ。。　ｓｏ　ｔｈａｔ’ｓ　ｗｈｙ　ｉ　ｕｓｅ　ｉｔ　ａｌｌ　ｔｈｅ　ｔｉｍｅ