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Seven sticks with lengths 2, 3, 5, 7, 11, 13 and 17 inches are placed in a box. Three of the sticks are randomly selected. What is the probability that a triangle can be formed by joining the endpoints of the sticks? Express your answer as a common fraction.

17,13,        and 5 or 7 or 11         (3)

17, 11,       and 7                          (1)

13, 11,       and  3,5,7                  (3)

13, 7,         and    none

11, 7           and 5                        (1)

7, 5,           and 3                        (1)

5, 3,           and none

equals  9  favourable outcomes   (you need to check)

Can you see how I have done this?  What needs to happen for three lengths be able to form a triangle?

7C3=35

Probability = 9/35

5 what?

I should not have used the = sign.

tan90 does not equal undefined it just IS undefined   

$\text{Let's say we have an arrangement of lamps such that}\\ \text{the leftmost lamp is blue}\\ \text{and the rightmost lamp is red}\\ \text{Now we select 3 lamps at random and turn them on}\\ \text{There are a total of }\dbinom{6}{3}=20 \text{ ways they can be chosen}\\ \text{If we insist on the left lamp being off and the right one being on}\\ \text{we have }\dbinom{4}{2}=6 \text{ ways of selecting the remaining lamps that will go on}\\ \text{So given an arrangement as described above}\\ P[\text{choosing random 3 lamps to turn on results in left off and right on}]=\dfrac{6}{20}=\dfrac{3}{10}$

$\text{Now we need to determine the probability of}\\ \text{having the left lamp blue and right lamp red}\\ \text{We fix those two and for the remaining slots we have}\\ \dbinom{4}{2} = 6 \text{ way to select slots for say the red lamps}\\ \text{and having done that the blue lamps are determined}\\ \text{so there are 6 total arrangements satisfying the requirement}\\ \text{out of a possible }\dbinom{6}{3}=20 \text{ ways to arrange the lamps, so}\\ P[\text{leftmost lamp is blue, rightmost lamp is red}]=\dfrac{6}{20}=\dfrac{3}{10}$

$\text{The probability we're after is just the product of these two}\\ P[\text{left blue lamp off and right red lamp on}]=\left(\dfrac{3}{10}\right)^2=\dfrac{9}{100}$

Here is the graph of y=tanx

You can see that as x tends to 90 degrees from below   the tan of it approaches +infinity.

AND

as x tends to 90 from above x tends to - infinity.

So it is correct to say this

$\displaystyle\lim_{x\rightarrow \infty^-}\;\;tanx=+\infty\\~\\ \displaystyle\lim_{x\rightarrow \infty^+}\;\;tanx=-\infty\\~\\ but\\ \displaystyle\lim_{x\rightarrow \infty}\;\;tanx=\text{undefined}$

yes perfect, thank you again for teaching me a new method 

OK

Note  that the "n" in the polynomial represents the nth term in the summation

So.....the first term is  "6"

So....we know that

a(n)^3 + b(n)^2 + c(n) + d = 6

So....in the first equation where we are trying to find the values for a, b, c and d we have

a(1)^3 + b(1)^2 + c(1) + d   =  6

a + b + c + d =  6

But....we have already found  a,b and c    as   1, - 6 and 11

So

1 - 6 + 11 + d = 6

-5 + 11 + d = 6

6 + d =  6

So...    d  =  0

Does that make sense????

the one in the equation to solve the system 

a+b+c+d = 6

Which "6"  are you referring to   ????

ah i see thank you!

the method makes perfect sense and is fascinating but can u please explain where the 6 came from? just to clear up the little uncertainty thanks so much

as an alternative approach we can use induction

$P_1=3(-1)(-2) = (0)(-1)(-2)+6 = True\\ \text{show }P_n \Rightarrow P_{n+1}\\ 3\sum \limits_{r=1}^{n+1} ~(r-2)(r-3) = \\ \left(3\sum \limits_{r=1}^n(r-2)(r-3)\right)+3(n-1)(n-2) = \\ \left((n-1)(n-2)(n-3)+6\right) + 3(n-1)(n-2) = \\ (n-1)(n-2)((n-3)+3)+6 = \\ (n-1)(n-2)(n)+6 = \\ ((n+1)-1)((n+1)-2)((n+1)-3)+6 \Rightarrow P_{n+1}=True$

Yeah....I see that you might be able to use some type of induction to prove this, as well

Here's how to solve the system....it's lengthy but not that difficult

a + b + c + d  = 6

8a + 4b + 2c + d = 6

27a + 9b + 3c + d = 6

64a + 16b + 4c + d = 12

Multiply the first equation by - 1   and add it to the second equation

7a + 3b + c  = 0     (4)

Multiply the second equation by - 1 and add it to the third equation

19a + 5b + c  = 0       (5)

Multiply the third equation by -1 and add it to the fourth equation

37a + 7b + c  =  6       (6)

Multiply  (4)  by  - 1   and add it to (5)

12a + 2b  =  0      (7)

Multiply (5) by -1 and add it to (6)

18a + 2b =  6        (8)

Multiply (7) by -1 and add it to (8)

6a =  6

a = 1

Sub this into  (7) to find b

12(1) + 2b = 0

2b = - 12

b = -6

Sub a, b into (4) to find c

7(1) + 3(-6) + c = 0

7 - 18 + c = 0

-11 + c = 0

c = 11

Sub a,b, c into  the first equation to find d

1 - 6 + 11 + d  = 6

-5 + 11  + d = 6

6 + d = 6

d = 0

So

a = 1   b =  -6   c = 11    d  = 0

Q#1 Suppose a researcher goes to a small college of 200 faculty, 12 of which have blood type O-negative. She obtains a simple random sample of size n = 20 of the faculty. Let the random variable X represent the number of faculty in the sample of size n = 20 that have blood type O-negative. a)What is the probability that 3 of the faculty have blood type O-negative? b) What is the probability that at least one of the faculty has blood type O-negative? c)Find the probability distribution of the aculty have blood type O-negative d) also verify your result using Formula

$X \text{ is binomially distributed with parameters}\\ n=20,~p=\dfrac{12}{200}=\dfrac{3}{50}$ 

$P[X=3] = \dbinom{20}{3}\left(\dfrac{3}{50}\right)^3\left(\dfrac{47}{50}\right)^{17} = \\ \dfrac{41011174284214815703116854545293}{476837158203125000000000000000000} \approx 0.086$

$P[1 \leq X] = 1-P[X=0] = \\ 1-\left(\dfrac{47}{50}\right)^{20} = \\ \dfrac{6770074452100164190549251988657599}{9536743164062500000000000000000000} \approx 0.7099$

I don't understand questions (c) and (d)

thank you and

Please do, i think this is a different approach to the question than the one im taught but i really want to learn it 

The "3"  is just a sum multiplier

We can use the sum  of differences to find a polynomial

The frist 7 terms produced by the summation are :

      6            6         6         12         30        66      126

             0          0         6        18        36       60

                   0          6        12        18      24

                         6          6        6          6

We have 3 rows of non-zero differences.....so we have a third power polynomial of the form  

an^3 + bn^2  + cn + d

And we have this system

a(1)^3 + b(1)^2 + c(1) + d =  6

a(2)^3 + b(2)^2 + c(2) + d = 6

a(3)^3 + b(3)^2 + c(3) + d = 6

a(4)^2 + b(4)^2 + c(4) + d = 12

a + b + c + d  = 6

8a + 4b + 2c + d = 6

27a + 9b + 3c + d = 6

64a + 16b + 4c + d = 12

a = 1  b = -6   c = 11   d  = 0

So.......we have the polynomial

n^3 - 6n^2 + 11n         

We can show that this is equal to  (n - 1)(n -2)(n -3) + 6 

 (n - 1)(n - 2) (n -3) + 6   =

(n^2 - 3n + 2) (n - 3) + 6  =

 (n^3 - 3n^2 + 2n - 3n^2 + 9n - 6) +  6   =

(n^3 - 6n^2 + 11n - 6) + 6 =

n^3 - 6n^2 + 11n      

PS, YEEEEEET......I can show you how to solve the system of equations if you want me to....

Note that this is just a permute of all the possible sets of choosing  any 2 things from 12

Note that  each set will be   (Satrap, Vizier)....and for any two people we choose, we have two possible arrangements for each set...

So....the number of possible sets =   2 * (12C2) =  132  possibilities

Equate the "y's"

2x^2 - 10x - 10 = x^2 - 4x + 6   rearrange as

x^2 - 6x - 16  =  0        factor

(x - 8) ( x + 2) =  0

Set both factors to 0 and solve for x  and we have that  x = -2  and x = 8

So....when x = - 2, y = (-2)^2 - 4(-2) + 6 =  18

And when x = 8, y =  (8)^2 -4(8) + 6 =  38

So

(-2, 18)   ;  (8, 38)

OK, CS.....!!!!

[ BTW....this is kind of a neat problem..... ]

1/3, 2/9, 1/9, 4/81, 5/243, 6/729.......etc
a(n) =3^(-n) * n - closed form
∑[[3^(-n) * n], n, 1,∞] = 3/4

Mine is to get into the life of church, I am Christian and I think I should be going to church more! :P

Thank you so much!