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(I'm just saying I'm an 11 year old that doesn't understand a lot of what is posted)

I literally substituted for your old version from back then!

PartialMathematician,

I do understand that you want lots of points but you are being a little too obvious about it.

It is also rude to give yourself points and not to give points to the other answerers, who answered properly before you added your 1 cent worth.

Hi Red,

Welcome back  

I certainly do not mind if you throw a daily riddle our way.

I miss the fun that we used to have here.     

If you can reinvent some of it I think that would be nice :))

There are 210 students in a twelfth grade high school class. 90 of these students have at least one sister and 105 have at least one brother. Out of these, there are 45 who have at least one sister and one brother.

Let A be the event that a randomly selected student in the class has a sister and B be the event that the student has a brother. Based on this information, answer the following questions.

What is P(A), the probability that the student has a sister? _____    \(\frac{90}{210}=\frac{3}{7}\)

What is P(B), the probability that the student has a brother?_____   \(\frac{105}{210}=\frac{1}{2}\)

What is P(A and B), the probability that the student has a sister and a brother? _____  \(\frac{45}{210}=\frac{3}{14}\)

What is P(B | A), the conditional probability that the student has a brother given that he or she has a sister?   \(\frac{45}{90}=\frac{1}{2}\)

Is P(B | A)=P(B)?   YES   Are the events A and B independent?   Yes I think they are independent.

Here is the corresponding Venn diagram.

The town of Hamlet has  3 people for each horse, 4 sheep for each cow, and  3 ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?

Answer Choices:

A. 41

B. 47

C. 59

D. 61

E. 66

1Horse, 3People, 9ducks = some multiple of 13

1Cow, 4 sheep = some multiple of 5

So just like Rom said the answer will be such some multiple of     13H+5C

Now the multiples of 13 are   0, 13, 26, 39, 42, 55, 68 and 68 is bigger than any choice so I do not need to go further

5C will end in 0 or 5 so I can add  0 or 5 to the last digit of the multiples of 13 above

41=26+15

47= ???  to get an answer of 47 the multiple of 13 would have to end in 7 or 2 and none do so it is not a possible answer.

No, this is not the answer either.

I got the same answer,

Here is my take.

A line segment is broken at two random points along its length. What is the probability that the three new segments can be arranged to form a triangle?

Let the length of the line be 1 unit.

Let the lengths of the segments be  x, 1-y, and y-x   units.      (if you add these they equal 1)

Now all these segments must be between 0 and 1 unit

So I will graph this and the area will give the full population of possible cuts.

Represented on this contour map in dark green

Now lets see how much of this is included if the peices can form a triangle.

To make a triangle the sum of the smaller peices has to be bigger than the length of the longest one.

so all three of the following conditions must be met.

\(x+1-y>y-x, \qquad x+y-x>1-y, \qquad 1-y+y-x>x\\ \text{These simplify to }\\ y 0.5, \qquad x<0.5\)   

So now I can plot these over the population and see what happens.

I can see that a quarter of the population triangle is included so

The probablility that the peices will be able to form a triangle is 1/4

(I keep editing this because some of what I write keeps getting deleted. There must be more elements here than web2.0 can happily cope with. I will see what happens this time but if it is not complete you will have to join the dots for youselves :/)

you just plug in h=1,10, c=1,10

and sort them

I guess what I'd do is solve 13h+5c = (41,47,59,61,66)

and remember that h and c have to be non-negative integers

\(13h+5c = 41\\ c(h) = \dfrac{41-13h}{5}\\ c(2) = 3\\ \text{so 41 is a legit choice}\)

\(13h+5c=47\\ c(h)=\dfrac{47-13h}{5}\\ \text{there are no integer solutions to this equation}\\ \text{so this must the the non legit total}\)

checking for thoroughness

\(13\cdot 3 + 5\cdot 4 = 59\)

\(13\cdot 2 + 5 \cdot 7 = 61\)

\(13 \cdot 2 + 5 \cdot 8 = 66\)

For example, if I were taking a test, what would be an easier way to get my answer choice, which is B, instead of writing out all the possible numbers?

How do you find the first 100 possible totals from the equation?

you don't give any choices...

total = p + h + s + c + d = 3h + h + 4c + c + 3(3h) = 13h + 5c

here's the first 100 possible totals

18, 23, 28, 31, 33, 36, 38, 41, 43, 44, 46, 48, 49, 51, 53, 54, 56,
57, 58, 59, 61, 62, 63, 64, 66, 67, 69, 70, 71, 72, 74, 75, 76, 77,
79, 80, 82, 83, 84, 85, 87, 88, 89, 90, 92, 93, 95, 96, 97, 98, 100,
101, 102, 103, 105, 106, 108, 109, 110, 111, 113, 114, 115, 116, 118,
119, 121, 122, 123, 124, 126, 127, 128, 129, 131, 132, 134, 135, 136,
137, 139, 140, 141, 142, 144, 145, 147, 149, 150, 152, 154, 155, 157, 
160, 162, 165, 167, 170, 175, 180

Don't know, I've answered this question before, more then once, but I never find it  simple....

mmmmm  I could just chase down an old answer .... would that be cheating?  That is rhetorical,    LOL.

P(B | A), the conditional probability that the student... Out of 210 students, 90 have at least one sister. There are 45 who have at least one sister and one brother. \(\dfrac{45}{90}\) simplifies to \(\dfrac{1}{2}\). 

- PM

this is correct then?

Across Quadrants I and II. 

You are very welcome!

:P

According to desmos, the best fit line is \(y = 12.7288x + 24.678\), which you can round to \(y = 12.73x + 24.68\) or \(y = 13x + 25\) .

Hope this helps,

- PM

Why do you want another set of eyes Rom?  

Here is a graph:

https://www.desmos.com/calculator/vgvpzwtqbr

- PM

Why are they duplicates of the same point in the question?

Exponential Growth. When X increases, Y increases. When X decreases, Y decreases.

https://support.desmos.com/hc/en-us/articles/202532159-Regressions