Aybalance:
Pls I need help with this:differential equation

A particle mass m is attached to one end of a light spring of modulus (lambda), the other end being fixed and the spring vertical.prove that the velocity of the particle when it has transverses a distance 'a' is...

Please help me solve this question

trinity0088:


Please, anyone in Auburn or has a friend currently in Auburn should reach me on 08130528870 or chukwuebukajude22@gmail.com.

I just got admitted to a PhD program. I need all the information I can get

I'm an undergraduate, didn't know anyone in Auburn

thedeedray:
Please help me solve this question

X=1
Or 1.4339

Mrshape:

X=1
Or 1.4339

Good morning boss
Please I knew how to solve this before but have forgotten
I'd be grateful if u solve them. All

Martinez39s:
Use sum formula for tangent and the fact that tan 45° = 1.

Substitute, expand, and solve, you will get tan² θ = 1/3. The solutions of this equation are 30°, 210°, 150°, and 330°.

I need help, please. All

Orestino:



From the rule of surds
Sqr(ab) = sqr(a) × sqr(b)
sqr(1225) = sqr(100 × 12.25) = sqr(100) × sqr(12.25)
= 10 × 3.5
=35
hence sqr(1225) = 35

Help me solve then please. All

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Among a group of 100 random students, what is the expected number of students who don't share birthdays with others in the group?

Note: by birthday, it means same month and day, and not necessarily year.


Assumptions: none of the students were born on February 29 and that number of days in a year is 365.

Mrshape, dejt4u, martinez39s, Richiez, mathefaro, naturalwaves, other mathematicians please help

jackpot:
Among a group of 100 random students, what is the expected number of students who don't share birthdays with others in the group?

Note: by birthday, it means same month and day, and not necessarily year.


Assumptions: none of the students were born on February 29 and that number of days in a year is 365.

Mrshape, dejt4u, martinez39s, Richiez, mathefaro, naturalwaves, other mathematicians please help

Is require critical thinking o

jackpot:
Among a group of 100 random students, what is the expected number of students who don't share birthdays with others in the group?

Note: by birthday, it means same month and day, and not necessarily year.


Assumptions: none of the students were born on February 29 and that number of days in a year is 365.

Mrshape, dejt4u, martinez39s, Richiez, mathefaro, naturalwaves, other mathematicians please help

N =100
P = 1 - (1/365) = 364/365

Expected value = N×P = 100 × (364/365) = 99.73

jackpot:
Among a group of 100 random students, what is the expected number of students who don't share birthdays with others in the group?

Note: by birthday, it means same month and day, and not necessarily year.


Assumptions: none of the students were born on February 29 and that number of days in a year is 365.

Mrshape, dejt4u, martinez39s, Richiez, mathefaro, naturalwaves, other mathematicians please help

Do you actually mean the number of possible ways in which the students can have their birthdays without any two sharing the same birthday? If that is what you mean, your answer is

³⁶⁵P₁₀₀ = 365!/265!

dejt4u:


N =100
P = 1 - (1/365) = 364/365

Expected value = N×P = 100 × (364/365) = 99.73

The guru at work again. I greet you sir

dejt4u:


N =100
P = 1 - (1/365) = 364/365

Expected value = N×P = 100 × (364/365) = 99.73

That is exactly my first line of thought when I first saw the problem. I think that your answer is from the perspective of a particular member of the group. When you throw in 100 random birthdays, one expects some couple of shared birthdays and a fairly larger proportion of unshared birthdays.

Read up birthday paradox: it says that among a group of 23 persons there's an approx. 50-50 chance that there's at least a shared birthday within the group. For example, in the 2014 world cup, 16 teams out of the 32 teams had at least 2 team members sharing a birthday. The paradox is strange but true.

Martinez39s:
Do you actually mean the number of possible ways in which the students can have their birthdays without any two sharing the same birthday? If that is what you mean, your answer is

No. That is not what I mean.

If one throws in 100 random birthdays, one expects some birthdays to match and some not to match. If this random experiment is conducted continuously (so to say), on the average, how many birthdays do not match with the others in a group of 100 persons?

ifada123:

Is require critical thinking o

yes bro.

Get to work. I need to see your solution.

jackpot:
yes bro.

Get to work. I need to see your solution.

Let's go.

Probability of first person number having birth day with another is 364/365
Second 363/365
Third 362/365
......... 265/365
So
Prob of not have same of 100 persons will be =

(First ) and (second ) and (third)and....×(100th no)
=(First not) x (second not) x (third not)x.....(100th not)
(364/365)×(363/365)×(362/365).......x(265/360)
=(1/365)^100 x 364 P100
=

Almost everyone will share same birth day

ifada123:

Let's go.

Probability of first person number having birth day with another is 364/365
Second 363/365
Third 362/365
......... 265/365
So
Prob of not have same of 100 persons will be =

(First ) and (second ) and (third)and....×(100th no)
=(First not) x (second not) x (third not)x.....(100th not)
(364/365)×(363/365)×(362/365).......x(265/360)
=(1/365) x 364 P100
=

Almost everyone will share same birth day

the answer then would be totally strange. Or don't you think?

I thought in same direction but I don't think E=NP should apply.

jackpot:
Among a group of 100 random students, what is the expected number of students who don't share birthdays with others in the group?

Note: by birthday, it means same month and day, and not necessarily year.


Assumptions: none of the students were born on February 29 and that number of days in a year is 365.

Mrshape, dejt4u, martinez39s, Richiez, mathefaro, naturalwaves, other mathematicians please help

SOLUTION
We need to take a step by step approach in order to get this.
Since 29th of February is not included, this means that we are dealing with just 365 days.
Having said that, what is the expected probability of the 1st student? That is going to be 365/365 days
What is the expected probability of the 2nd student? 364/365
Third? 363/365
Fourth....362/365
Fifth..................361/365
This is how it will go on till we get to the 100th student which will be 266/365 but we cannot do this as it will be too cumbersome.
Now, let us sort the denominator. Every component in the denominator is 365 i.e 365/365 * 364/365 * 363/365................*266/365.
This means that the denominator will be 365¹⁰⁰.

How about the numerator? We have 365! but we are not going to be needing the whole of it. We only need the area from 365 to 266. How do we get this?
All we need do is to divide 365! by 265!.

Thus, the desired solution to the problem is;
(365!) / (265! * 365¹⁰⁰).

Modified:
So, I have gotten a calculator,
(365!) / (265! * 365¹⁰⁰). = 3.0725 * 10^-7 approximately.
That is around 0.00000030725 approx.
As you can see, the probability is very slim as it is almost unlikely that out of 100 students, none will share a birthday which makes a lot of sense.

cc: Jackpot, Dejt4u, Martinez39s, Mrshape, Richiez, mathefaro,ifada123, Goalnaldo

This is also one of the best advanced calculators available: https://keisan.casio.com/calculator

naturalwaves:


SOLUTION
We need to take a step by step approach in order to get this.
Since 29th of February is not included, this means that we are dealing with just 365 days.
Having said that, what is the expected probability of the 1st student? That is going to be 365/365 days
What is the expected probability of the 2nd student? 364/365
Third? 363/365
Fourth....362/365
Fifth..................361/365
This is how it will go on till we get to the 100th student which will be 266/365 but we cannot do this as it will be too cumbersome.
Now, let us sort the denominator. Every component in the denominator is 365 i.e 365/365 * 364/365 * 363/365................*266/365.
This means that the denominator will be 365¹⁰⁰.

How about the numerator? We have 365! but we are not going to be needing the whole of it. We only need the area from 365 to 266. How do we get this?
All we need do is to divide 365! by 265!.

Thus, the desired solution to the problem is;
(365!) / (265! * 365¹⁰⁰.

I think I like this

dejt4u:
I think I like this

Question is quite tricky and requires a high thought process. Will get a calculator to compute final answer soon.

Maybe we should try to simulate it using real data. This is the [url]https://img.fifa.com/image/upload/hzfqyndmnqazczvc5xdb.pdf [/url]for 2018 fifa world cup data with the players birthdays.

We can decide to pick any consecutive 100 players. Either from top, middle or bottom.

Let's go!

jackpot:
the answer then would be totally strange. Or don't you think?

I thought in same direction but I don't think E=NP should apply.

To get number of people they did not ask us for the probability
So that's why I said almost everyone will have same birth date with someone.

Or

It is unlikely for someone not to have same birth date with another.


Does figures are there for proof.


But humans are discrete data so you make it discrete

That is 0 no person no birth date mate

jackpot:
Maybe we should try to simulate it using real data. This is the [url]https://img.fifa.com/image/upload/hzfqyndmnqazczvc5xdb.pdf [/url]for 2018 fifa world cup data with the players birthdays.

We can decide to pick any consecutive 100 players. Either from top, middle or bottom.

Let's go!

Did you see the solution I posted?

naturalwaves:


SOLUTION
We need to take a step by step approach in order to get this.
Since 29th of February is not included, this means that we are dealing with just 365 days.
Having said that, what is the expected probability of the 1st student? That is going to be 365/365 days
What is the expected probability of the 2nd student? 364/365
Third? 363/365
Fourth....362/365
Fifth..................361/365
This is how it will go on till we get to the 100th student which will be 266/365 but we cannot do this as it will be too cumbersome.
Now, let us sort the denominator. Every component in the denominator is 365 i.e 365/365 * 364/365 * 363/365................*266/365.
This means that the denominator will be 365¹⁰⁰.

How about the numerator? We have 365! but we are not going to be needing the whole of it. We only need the area from 365 to 266. How do we get this?
All we need do is to divide 365! by 265!.

Thus, the desired solution to the problem is;
(365!) / (265! * 365¹⁰⁰).

Modified:
So, I have gotten a calculator,
(365!) / (265! * 365¹⁰⁰). = 3.0725 * 10^-7 approximately.
That is around 0.00000030725 approx.
As you can see, the probability is very slim as it is almost unlikely that out of 100 students, none will share a birthday which makes a lot of sense.

cc: Jackpot, Dejt4u, Martinez39s, Mrshape, Richiez, mathefaro,ifada123

This is also one of the best advanced calculators available: https://keisan.casio.com/calculator

Yes I agree with this

naturalwaves:


SOLUTION
We need to take a step by step approach in order to get this.
Since 29th of February is not included, this means that we are dealing with just 365 days.
Having said that, what is the expected probability of the 1st student? That is going to be 365/365 days
What is the expected probability of the 2nd student? 364/365
Third? 363/365
Fourth....362/365
Fifth..................361/365
This is how it will go on till we get to the 100th student which will be 266/365 but we cannot do this as it will be too cumbersome.
Now, let us sort the denominator. Every component in the denominator is 365 i.e 365/365 * 364/365 * 363/365................*266/365.
This means that the denominator will be 365¹⁰⁰.

How about the numerator? We have 365! but we are not going to be needing the whole of it. We only need the area from 365 to 266. How do we get this?
All we need do is to divide 365! by 265!.

Thus, the desired solution to the problem is;
(365!) / (265! * 365¹⁰⁰).

Modified:
So, I have gotten a calculator,
(365!) / (265! * 365¹⁰⁰). = 3.0725 * 10^-7 approximately.
That is around 0.00000030725 approx.
As you can see, the probability is very slim as it is almost unlikely that out of 100 students, none will share a birthday which makes a lot of sense.

cc: Jackpot, Dejt4u, Martinez39s, Mrshape, Richiez, mathefaro,ifada123

This is also one of the best advanced calculators available: https://keisan.casio.com/calculator

but in reality, this is not valid

jackpot:

Read up birthday paradox: it says that among a group of 23 persons there's an approx. 50-50 chance that there's at least a shared birthday within the group. For example, in the 2014 world cup, 16 teams out of the 32 teams had at least 2 team members sharing a birthday. The paradox is strange but true.

Yes..

dejt4u:
but in reality, this is not valid

How is it not valid bro in reality?? If you have a group of 100 people, the probability that none will share same birthday is almost non existent. The probability that at least 2 people will share the same birthday will be quite high. Therefore, it is very valid in reality.

naturalwaves:

Question is quite tricky and requires a high thought process. Will get a calculator to compute final answer soon.

so the answer is not complete yet?

Goalnaldo:
so the answer is not complete yet?

Look up now. The solution is there. Let me tag you on it so you can get it.

naturalwaves:


SOLUTION
We need to take a step by step approach in order to get this.
Since 29th of February is not included, this means that we are dealing with just 365 days.
Having said that, what is the expected PROBABILITY of the 1st student? That is going to be 365/365 days
:
:
:
Modified:
So, I have gotten a calculator,
(365!) / (265! * 365¹⁰⁰). = 3.0725 * 10^-7 approximately.
That is around 0.00000030725 approx.
As you can see, the PROBABILITY is very slim as it is almost unlikely that out of 100 students, none will share a birthday which makes a lot of sense.

cc: Jackpõt, Dèjt4u, Martiñez39s, Mrshãpe, Rićhiez, mathêfaro, ifadã123, Goalñaldo

This is also one of the best advanced calculators available: https://keisan.casio.com/calculator

Jackpot never mentioned anything about probability. How did probability get involved in the solution?

Martinez39s:
Jackpot never mentioned anything about probability. How did probability get involved in the solution?

you can get your expected values from probability.. It is even simpler in this case since our N is 100