Calculate the probability that a head will show

  • #1
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[tex]\pi[/tex], 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
 

Answers and Replies

  • #2


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[tex]\pi[/tex], 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[tex]\Pi[/tex]], 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).
 
Last edited:
  • #3
Defennder
Homework Helper
2,591
5


Interesting username.
 
  • #5
412
4


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[tex]\pi[/tex], 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).
 
  • #8


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[tex]\Pi[/tex]], 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
 
  • #9


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[tex]\Pi[/tex]], 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=
 
  • #10
412
4


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!
 
  • #11


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
 

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