Can a monkey outrun a bullet and still save her litter?

[JaredJames]

is Zero(0) an even number, odd or prime??

It's none of those.

 

[JaredJames]

do women get bald? is it normal? explain it, please.

Of course women can go bald.

There are a number of causes but I'd say the main cause (not including voluntary shaving) would be cancer treatment (Chemotherapy).

 

[evra]

jarednjames; then tell me where can you catagorise zero(0) under? and for baldness in women is it strictly abnnormal, genetically?

 

[JaredJames]

jarednjames; then tell me where can you catagorise zero(0) under? and for baldness in women is it strictly abnnormal, genetically?

Scrap that, apparently 0 is even, although I don't agree with the logic on it I'm in no position to argue it: http://en.wikipedia.org/wiki/0_(number)#As_a_number

Female Baldness: http://www.bbc.co.uk/health/physical_health/conditions/hair_loss_women.shtml

 

[Darken-Sol]

can you use the word "and" 5 times in a row in a sentence that is grammatically correct?

 

[Darken-Sol]

can you use the word "and" 5 times in a row in a sentence that is grammatically correct?

if i was painting a sign for ben and jerries i'd want equal spacing between ben and and and and and jerries.

 

[DaveC426913]

Can you use the word had eight times in a row in a grammatically correct and meaningful (if fictional) sentence?

 

[Darken-Sol]

Can you use the word had eight times in a row in a grammatically correct and meaningful (if fictional) sentence?

i cannot.

 

[Darken-Sol]

c'mon dave you got to tell us. i will own up to being a slacker. i just keep saying "had" after two hads i get nowhere.

 

[regor60]

Can you use the word had eight times in a row in a grammatically correct and meaningful (if fictional) sentence?

eight times with no intervening words ?

 

[DaveC426913]

eight times with no intervening words ?

Yes.

Spoiler

In teaching the class about tenses, the teacher made them write out several sentences like "If I'd had the time I'd have gotten gas too."

When she checked their papers afterward, she saw that Jimmy had had 'had', while Karen had had 'had had'. 'Had had' had had a better effect on the teacher.

 

[BOYALSAHAB]

NOt right;
Can anyone answer this?
2) A murderer is condemned to death. He has to choose between three rooms: The first is full of raging fires, the second is full of assassins with loaded guns, and the third is full of lions that haven't eaten in 3 years. Which room is safest for him?

Third Room Because Lions Are Not Live Because Of lions that haven't eaten in 3 years.


Third Room Because Lions Are Not Live Because Of lions that haven't eaten in 3 years.

ANSWERED BY BOYALSAHAB

 

[evra]

i have a bus that takes 100 passengers and i need only 100dollars id the bus is full. the fare for;
1 man is 10dollars
5 children is 1dollar
4 women is 5dollars
i need the bus to be full and i need 100dollars only out of it..

how many children, men and women should I've in the bus??

 

[DaveC426913]

i have a bus that takes 100 passengers and i need only 100dollars id the bus is full. the fare for;
1 man is 10dollars
5 children is 1dollar
4 women is 5dollars
i need the bus to be full and i need 100dollars only out of it..

how many children, men and women should I've in the bus??

Women have a fee different from men? Do they have to sit at the back too?

 

[evra]

Women have a fee different from men? Do they have to sit at the back too?

that is not important. anyone can sit anywhere.

 

[DavidSnider]

So the constraint is exactly 100 people and exactly 100 dollars?

I don't think there is a solution.

 

[evra]

So the constraint is exactly 100 people and exactly 100 dollars?

I don't think there is a solution.

there is! just do your best and see. if you can't, that does not mean that there is no solution.

 

[DavidSnider]

Well, I brute forced it... Maybe my math is wrong, take a look:

var answers = [];
for(var i = 0; i<=100; i++) {
	for(var j = 0; j<=100; j++) {
		for(var k = 0; k<=100; k++) {
		    if(((i*10.0) + (j*0.2) + (k*1.25)) == 100.0 && (i + j + k == 100.0)) {
		        answers.push({ men: i, children: j, women: k, total: (i * 10.0 + j * 0.2 + k * 1.25) });
			}
		}
	}
}
answers;

 

[evra]

check it again.. u can do better than this from wat i saw.

 

[Soca fo so]

This might be a stupid question, but is the driver included in the 100 passengers?

 

[Soca fo so]

There seems to be no solution when you treat the three groups, men, women, children as distinct but if you consider that each child is also male or female then consider a female child as being half a woman so should be charged half the fare for a woman as well as the fare for a child so 5/8 + 1/5 dollars in total and similarly for a male child with the fare in this case being 5 + 1/5 dollars, then 4 male children plus 96 female children would give you 100 passengers and 100 dollars.

 

[DaveC426913]

there is! just do your best and see. if you can't, that does not mean that there is no solution.

Clearly there is s trick to it, as

10x + .2x + 1.25x = 100 does not have an integer solution.

 

[Soca fo so]

If you were to charge pregnant women the fare for a woman plus the charge for a child i.e. 5/4 + 1/5 dollars that would give a solution of 15 pregnant women, 21 non pregnant women, 60 children, and 4 men. The issue would be whether to count a pregnant woman as two passengers, but since they would be taking up one seat I would think they would be counted as one passenger.

Another thing that comes to mind is that perhaps there is a difference in the fare for a male child and a female child in the same way as for adults i.e. an 8 to 1 ratio male to female then let b be the number of boys and g be the number of girls, both natural numbers or zero, and let 1/f be the fare for girls, then the following two equations could be used to fix the two different fares for girls and boys:

b + g = 5

b*(8/f) + g*(1/f) = 1

these can be solved for f by noting that b must be a natural number or zero and must be less than or equal to 5 the first constraint means f must be equivalent to 12 mod 7 and the second constraint in addition to the first means that f must in fact be 12.

Then that leads to the following two simultaneous equations:

m + w + b + g = 100

10*m + (5/4)*w + (2/3)*b + (1/12)*g = 100

where m is the number of men and w is the number of women both natural numbers or zero.

If a solution exists it can be found by brute force I expect, I look forward to seeing the solution! (if it exists of course).

 

[DaveC426913]

If you were to charge pregnant women the fare

I would consider that a cheating answer. Charging a pregnant woman for her unborn child is unheard of.

 

[DaveC426913]

i have a bus that takes 100 passengers and i need only 100dollars id the bus is full. the fare for;
1 man is 10dollars
5 children is 1dollar
4 women is 5dollars
i need the bus to be full and i need 100dollars only out of it..

how many children, men and women should I've in the bus??

Perhaps we are overthinking the answer.

I can answer the question quite simply, really:

Q: how many children, men and women should I've in the bus??
A: 100.

 

[Soca fo so]

Another thing that comes to mind is that perhaps there is a difference in the fare for a male child and a female child in the same way as for adults i.e. an 8 to 1 ratio male to female then let b be the number of boys and g be the number of girls, both natural numbers or zero, and let 1/f be the fare for girls, then the following two equations could be used to fix the two different fares for girls and boys:

b + g = 5

b*(8/f) + g*(1/f) = 1

these can be solved for f by noting that b must be a natural number or zero and must be less than or equal to 5 the first constraint means f must be equivalent to 12 mod 7 and the second constraint in addition to the first means that f must in fact be 12.

I've made a bit of a mistake in the above; there are 6 possible values of f from those equations of which 12 is one.

Perhaps we are overthinking the answer.

I can answer the question quite simply, really:

Q: how many children, men and women should I've in the bus??
A: 100.

Yes of course!

 

[evra]

This might be a stupid question, but is the driver included in the 100 passengers?

no he's not!

 

[evra]

Clearly there is s trick to it, as

10x + .2x + 1.25x = 100 does not have an integer solution.

there is! very correct infact! try!

 

[evra]

Perhaps we are overthinking the answer.

I can answer the question quite simply, really:

Q: how many children, men and women should I've in the bus??
A: 100.

lol! nooo that's not it.

 

[I like Serena]

there is! very correct infact! try!

You skipped over the pregnant-woman-solution.
So is that it?