All Answers 355923 Answers

16 possible outcomes

P         N           D           Q                     Value  of Heads

H        H           H           H                        .41

T        H           H           H                        .40

T        T           H            H                        .35

T        T           T            H                        .25

T        T           T           T                            0

H       T            T           T                         .01

H       H           T            T                        .06

H      H            H            T                        .16

T       H           T            T                         .05

T      T            H            T                         .10

H      T            T            H                         .26

T      H           H             T                         .15

T      H           T             H                         .30

H     T            H            T                          .11

H     T            H            H                          .36

H     H           T             H                          .31

P ( 15 cents or more of heads)  = 10/16 =  5/8

Ok!

CPhil answered the question as    672    if Bob and Yogi are NOT together.

THEN, as a bonus,  he showed the answer IF Bob and Yogi could play on the same team

Thanx to BOTH of you !  

There are 4 ways to choose which 3 days it rains. It would be 

((3/4)^3)(1/4) = 27/256 because each day has a 3/4 chance of rain and 1/4 chance of sunshine.

For each of the 4 ways this can happen, that would be:

4 * (27/256) 

27/64

You can position all of the wombats in 7! ways BUT you have over counted the 3 hairy wombats. So the answer is:

7!/3! = 840 

@CPhill, you are wrong. Everything was correct up to this(From your post:)

Note.....the only remaining team is the one where both are on it and we can pick any 3 of the other 10 players  = C (10,3)   =  120

BUT...If you look at the question, it says that Bob and Yogi refuse to play together.

The Screamers are coached by Coach Yellsalot. The Screamers have 12 players, but two of them, Bob and Yogi, refuse to play together. How many starting lineups (of 5 players) can Coach Yellsalot make, if the starting lineup can't contain both Bob and Yogi? (The order of the 5 players in the lineup does not matter; that is, two lineups are the same if they consist of the same 5 players.)

I get a little different answer from Ninja

Possible teams :

Team 1 : Bob is on the team and we can pick any 4 of the other 10 players [not counting Yogi ]  = C(10,4) = 210 possible teams

Team 2 :  Same as team 1 except that Yogi is on the team and we can pick any 4 of the other 10 players [not counting Bob ]  = C(10,4) = 210 possible teams

Team 3 :   Neither is on the team and we can pick any 5 of the other 10 players   = C(10,5) = 252 possible teams

So....the total teams, if the starting lineup can't contain both/either Bob and Yogi  = 2 * 210  + 252   = 672

Note.....the only remaining team is the one where both are on it and we can pick any 3 of the other 10 players  = C (10,3)   =  120

So all possible .teams = .....672  + 120  = 792   =  C(12,5)

There are 6^2=36 possibilities of rolls.

There are only three of them that sum less than or equal to 3. (1,1), (1,2), (2,1). That means 3/36 = 1/12 don't work. The ones that ARE greater than 3 are 1 - 1/12 = 11/12

There are 16 cards in a deck of 52 which is an ace or a diamond or both. So that means the probability of NOT choosing an ace or a diamond or both on both turns is:

((52-16)/52)^2 

(36/52)^2

(9/13)^2

81/169

This means that the probability of choosing an ace or a diamond or both on both turns is:

1 - 81/169

88/169

Another approach

We have 15 ways to choose a goalie.....and for each of these, we can pick any 6 of the other 14 remaining people....so......the total  ways of choosing a team  =   

15 * C(14,6)   = 15 * 3003 =   45,045 ways

This would be much easier to visualize if I could upload a diagram, but unfortunately I can't. So I will try and give a brief description of what actually happens.

Draw a semi circle with a diameter of 200 m, and the label the centre "A". Draw a perpendicular line from from centre "A" to a point on the circumference, which will call "B", which will be the radius =100m. The dog starts swimming from the centre A, within the 1st quadrant, in an arc from A to B. This arc, AB, will form a smaller circle within the 1st quadrant, whose diameter is the radius of semi circle, which is = 100 meters.

The circumference of this smaller circle within the semi circle will be=100π. By the time the dog reaches the duck at point B, he/she has already swum 1/2 of circumference of the smaller circle, or

100π/2 =157.08 meters! And that is how much the dog will have to swimm to meet the duck!.

And that is my take on it.                                                                                                                            Now, you guys with fancy graphing implements would see it clearly if you drew a picture of it.

There are 15 people for goalie, and 1 less for each of the other positions. So it is 15 * 14 * 13 * 12 * 11 * 10 * 9.

But because the order of the other 6 guys(or girls) don't matter, you divide by 6!. So it is:

(15 * 14 * 13 * 12 * 11 * 10 * 9)/6!

32432400/720

45045

x = 3*(2/3)

x=2

There are 3 different cases for the starting lineup.

Case 1: Bob starts (and Yogi doesn't). In this case, the coach must choose 4 more players from the 10 remaining players (remember that Yogi won't play, so there are only 10 players left to select from). Thus there are 10C2 lineups that the coach can choose.

Case 2: Yogi starts (and Bob doesn't). As in Case 1, the coach must choose 4 more players from the 10 remaining players. So there are 10C4 lineups in this case.

Case 3: Neither Bob nor Yogi starts. In this case, the coach must choose all 5 players in the lineup from the 10 remaining players. Hence there are 10C4 lineups in this case. To get the total number of starting lineups, we add the number of lineups in each of the cases:

Case 1 + Case 2 + Case 3 = 210 + 210 + 252 = 672

Bogus Solution:(This is what I did the first time)

Case 1(and 2): Bob is in the team but Yogi isn't. There are 11 people that can be on the team. 11 * 10 * 9 * 8 * 7 * 2(for Yogi's case). That is 110880.

Case 3: neither are on the team. 10 * 9 * 8 * 7 * 6 = 30240

Sum is 141120

Way off 

The series closer to the theoretical 50/50 cannot be determined unless the total number of rolls for both series is given. 

What do you want to do?  Convert it to decimal degree?

60 minutes in each degree    60 seconds in each minute  (this 3600 seconds in each degree)

Example    Convet   35 degrees 18' 24 "   do decimal degrees   ( ' symbol is minute symbol   " is seconds symbol)

  35 +  18/60  + 24/3600 = 35.3066 degrees  

Multiply this by pi/180 to get  . 6162 Radians

Since there is a total of  9313  and  1676  are business majors,  9313 - 1676  =  7637  are not business majors.

The probability of not being a business major is:  7637 / 9313  =  0.82  (approximately)

Average 72 inches per day?

1. 249 - 40%

249 - 0.4

248.6

2.(if you mean 40 percent OF 249)

249 - 249(40%)

249 - 99.6

149.4

5...

Tooooo funny !    Apology accepted, Melody !!    Not me (THIS time )     LOL

OK.....what's the question....???

pH is calculated as the negative log of the hydrogen ion concentration, H+  or  H3O

- log[1.4 x 10^(-5) ]  = 4.85

Sorry...that was just a typo (I got a '2' in the way)....but the calculation was made using the correct numbers.....

Sharp eye !     Thanx !

SORRY MAX! EVEN MUCH LONGER THAN YOURS.

Take the integral:
 integral sqrt(tan(x)) dx
For the integrand sqrt(tan(x)), substitute u = tan(x) and du = sec^2(x) dx:
 = integral sqrt(u)/(u^2 + 1) du
For the integrand sqrt(u)/(u^2 + 1), substitute s = sqrt(u) and ds = 1/(2 sqrt(u)) du:
 = 2 integral s^2/(s^4 + 1) ds
For the integrand s^2/(s^4 + 1), use partial fractions:
 = 2 integral(-s/(2 sqrt(2) (-s^2 + sqrt(2) s - 1)) - s/(2 sqrt(2) (s^2 + sqrt(2) s + 1))) ds
Integrate the sum term by term and factor out constants:
 = -1/sqrt(2) integral s/(s^2 + sqrt(2) s + 1) ds - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
Rewrite the integrand s/(s^2 + sqrt(2) s + 1) as (2 s + sqrt(2))/(2 (s^2 + sqrt(2) s + 1)) - 1/(sqrt(2) (s^2 + sqrt(2) s + 1)):
 = -1/sqrt(2) integral((2 s + sqrt(2))/(2 (s^2 + sqrt(2) s + 1)) - 1/(sqrt(2) (s^2 + sqrt(2) s + 1))) ds - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
Integrate the sum term by term and factor out constants:
 = -1/(2 sqrt(2)) integral(2 s + sqrt(2))/(s^2 + sqrt(2) s + 1) ds + 1/2 integral1/(s^2 + sqrt(2) s + 1) ds - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
For the integrand (2 s + sqrt(2))/(s^2 + sqrt(2) s + 1), substitute p = s^2 + sqrt(2) s + 1 and dp = (2 s + sqrt(2)) ds:
 = -1/(2 sqrt(2)) integral1/p dp + 1/2 integral1/(s^2 + sqrt(2) s + 1) ds - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
The integral of 1/p is log(p):
 = -(log(p))/(2 sqrt(2)) + 1/2 integral1/(s^2 + sqrt(2) s + 1) ds - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
For the integrand 1/(s^2 + sqrt(2) s + 1), complete the square:
 = -(log(p))/(2 sqrt(2)) + 1/2 integral1/((s + 1/sqrt(2))^2 + 1/2) ds - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
For the integrand 1/((s + 1/sqrt(2))^2 + 1/2), substitute w = s + 1/sqrt(2) and dw = ds:
 = -(log(p))/(2 sqrt(2)) + 1/2 integral1/(w^2 + 1/2) dw - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
Factor 1/2 from the denominator:
 = -(log(p))/(2 sqrt(2)) + 1/2 integral2/(2 w^2 + 1) dw - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
Factor out constants:
 = -(log(p))/(2 sqrt(2)) + integral1/(2 w^2 + 1) dw - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
For the integrand 1/(2 w^2 + 1), substitute v = sqrt(2) w and dv = sqrt(2) dw:
 = -(log(p))/(2 sqrt(2)) + 1/sqrt(2) integral1/(v^2 + 1) dv - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
The integral of 1/(v^2 + 1) is tan^(-1)(v):
 = (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) - 1/sqrt(2) integral s/(-s^2 + sqrt(2) s - 1) ds
Rewrite the integrand s/(-s^2 + sqrt(2) s - 1) as 1/(sqrt(2) (-s^2 + sqrt(2) s - 1)) - (sqrt(2) - 2 s)/(2 (-s^2 + sqrt(2) s - 1)):
 = (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) - 1/sqrt(2) integral(1/(sqrt(2) (-s^2 + sqrt(2) s - 1)) - (sqrt(2) - 2 s)/(2 (-s^2 + sqrt(2) s - 1))) ds
Integrate the sum term by term and factor out constants:
 = (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + 1/(2 sqrt(2)) integral(sqrt(2) - 2 s)/(-s^2 + sqrt(2) s - 1) ds - 1/2 integral1/(-s^2 + sqrt(2) s - 1) ds
For the integrand (sqrt(2) - 2 s)/(-s^2 + sqrt(2) s - 1), substitute z_1 = -s^2 + sqrt(2) s - 1 and dz_1 = (sqrt(2) - 2 s) ds:
 = (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + 1/(2 sqrt(2)) integral1/z_1 dz_1 - 1/2 integral1/(-s^2 + sqrt(2) s - 1) ds
The integral of 1/z_1 is log(z_1):
 = (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + (log(z_1))/(2 sqrt(2)) - 1/2 integral1/(-s^2 + sqrt(2) s - 1) ds
For the integrand 1/(-s^2 + sqrt(2) s - 1), complete the square:
 = (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + (log(z_1))/(2 sqrt(2)) - 1/2 integral1/(-(s - 1/sqrt(2))^2 - 1/2) ds
For the integrand 1/(-(s - 1/sqrt(2))^2 - 1/2), substitute z_2 = s - 1/sqrt(2) and dz_2 = ds:
 = (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + (log(z_1))/(2 sqrt(2)) - 1/2 integral1/(-z_2^2 - 1/2) dz_2
Factor -1/2 from the denominator:
 = (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + (log(z_1))/(2 sqrt(2)) - 1/2 integral2/(-2 z_2^2 - 1) dz_2
Factor out constants:
 = (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + (log(z_1))/(2 sqrt(2)) - integral1/(-2 z_2^2 - 1) dz_2
Factor -1 from the denominator:
 = (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + (log(z_1))/(2 sqrt(2)) + integral1/(2 z_2^2 + 1) dz_2
For the integrand 1/(2 z_2^2 + 1), substitute z_3 = sqrt(2) z_2 and dz_3 = sqrt(2) dz_2:
 = (tan^(-1)(v))/sqrt(2) - (log(p))/(2 sqrt(2)) + (log(z_1))/(2 sqrt(2)) + 1/sqrt(2) integral1/(z_3^2 + 1) dz_3
The integral of 1/(z_3^2 + 1) is tan^(-1)(z_3):
 = -(log(p))/(2 sqrt(2)) + (tan^(-1)(v))/sqrt(2) + (log(z_1))/(2 sqrt(2)) + (tan^(-1)(z_3))/sqrt(2) + constant
Substitute back for z_3 = sqrt(2) z_2:
 = -(log(p))/(2 sqrt(2)) + (tan^(-1)(v))/sqrt(2) + (log(z_1))/(2 sqrt(2)) + (tan^(-1)(sqrt(2) z_2))/sqrt(2) + constant
Substitute back for z_2 = s - 1/sqrt(2):
 = -(log(p))/(2 sqrt(2)) - (tan^(-1)(1 - sqrt(2) s))/sqrt(2) + (tan^(-1)(v))/sqrt(2) + (log(z_1))/(2 sqrt(2)) + constant
Substitute back for z_1 = -s^2 + sqrt(2) s - 1:
 = -(log(p))/(2 sqrt(2)) + (log(-s^2 + sqrt(2) s - 1))/(2 sqrt(2)) - (tan^(-1)(1 - sqrt(2) s))/sqrt(2) + (tan^(-1)(v))/sqrt(2) + constant
Substitute back for v = sqrt(2) w:
 = -(log(p))/(2 sqrt(2)) + (log(-s^2 + sqrt(2) s - 1))/(2 sqrt(2)) - (tan^(-1)(1 - sqrt(2) s))/sqrt(2) + (tan^(-1)(sqrt(2) w))/sqrt(2) + constant
Substitute back for w = s + 1/sqrt(2):
 = -(log(p))/(2 sqrt(2)) + (log(-s^2 + sqrt(2) s - 1))/(2 sqrt(2)) - (tan^(-1)(1 - sqrt(2) s))/sqrt(2) + (tan^(-1)(sqrt(2) s + 1))/sqrt(2) + constant
Substitute back for p = s^2 + sqrt(2) s + 1:
 = (log(-s^2 + sqrt(2) s - 1))/(2 sqrt(2)) - (log(s^2 + sqrt(2) s + 1))/(2 sqrt(2)) - (tan^(-1)(1 - sqrt(2) s))/sqrt(2) + (tan^(-1)(sqrt(2) s + 1))/sqrt(2) + constant
Substitute back for s = sqrt(u):
 = (log(-u + sqrt(2) sqrt(u) - 1))/(2 sqrt(2)) - (log(u + sqrt(2) sqrt(u) + 1))/(2 sqrt(2)) - (tan^(-1)(1 - sqrt(2) sqrt(u)))/sqrt(2) + (tan^(-1)(sqrt(2) sqrt(u) + 1))/sqrt(2) + constant
Substitute back for u = tan(x):
 = -(tan^(-1)(1 - sqrt(2) sqrt(tan(x))))/sqrt(2) + (tan^(-1)(sqrt(2) sqrt(tan(x)) + 1))/sqrt(2) + (log(-tan(x) + sqrt(2) sqrt(tan(x)) - 1))/(2 sqrt(2)) - (log(tan(x) + sqrt(2) sqrt(tan(x)) + 1))/(2 sqrt(2)) + constant
Factor the answer a different way:
 = (-2 tan^(-1)(1 - sqrt(2) sqrt(tan(x))) + 2 tan^(-1)(sqrt(2) sqrt(tan(x)) + 1) + log(-tan(x) + sqrt(2) sqrt(tan(x)) - 1) - log(tan(x) + sqrt(2) sqrt(tan(x)) + 1))/(2 sqrt(2)) + constant
An alternative form of the integral is:
 = (-2 tan^(-1)(1 - sqrt(2) sqrt(tan(x))) + 2 tan^(-1)(sqrt(2) sqrt(tan(x)) + 1) + log(-(tan(x) - sqrt(2) sqrt(tan(x)) + 1)/(tan(x) + sqrt(2) sqrt(tan(x)) + 1)))/(2 sqrt(2)) + constant
Which is equivalent for restricted x values to:
Answer: |= (-2 tan^(-1)(1 - sqrt(2) sqrt(tan(x))) + 2 tan^(-1)(sqrt(2) sqrt(tan(x)) + 1) + log(tan(x) - sqrt(2) sqrt(tan(x)) + 1) - log(tan(x) + sqrt(2) sqrt(tan(x)) + 1))/(2 sqrt(2)) + constant