Answer:
1) The probability that ten students in a class have different birthdays is 0.883.
2) The probability that among ten students in a class, at least two of them share a birthday is 0.002.
Step-by-step explanation:
Given : Assume there are 365 days in a year.
To find : 1) What is the probability that ten students in a class have different birthdays?
2) What is the probability that among ten students in a class, at least two of them share a birthday?
Solution :

Total outcome = 365
1) Probability that ten students in a class have different birthdays is
The first student can have the birthday on any of the 365 days, the second one only 364/365 and so on...

The probability that ten students in a class have different birthdays is 0.883.
2) The probability that among ten students in a class, at least two of them share a birthday
P(2 born on same day) = 1- P( 2 not born on same day)
![\text{P(2 born on same day) }=1-[\frac{365}{365}\times \frac{364}{365}]](https://tex.z-dn.net/?f=%5Ctext%7BP%282%20born%20on%20same%20day%29%20%7D%3D1-%5B%5Cfrac%7B365%7D%7B365%7D%5Ctimes%20%5Cfrac%7B364%7D%7B365%7D%5D)
![\text{P(2 born on same day) }=1-[\frac{364}{365}]](https://tex.z-dn.net/?f=%5Ctext%7BP%282%20born%20on%20same%20day%29%20%7D%3D1-%5B%5Cfrac%7B364%7D%7B365%7D%5D)

The probability that among ten students in a class, at least two of them share a birthday is 0.002.
Answer:
The graph is possible for 
Step-by-step explanation:
we know that
The discriminant of a quadratic equation of the form
is equal to

If D=0 the quadratic equation has only one real solution
If D>0 the quadratic equation has two real solutions
If D<0 the quadratic equation has no real solution (complex solutions)
In this problem , looking at the graph, the quadratic equation has two real solutions (the solutions are the x-intercepts)
so

therefore
The graph is possible for 
Answer:
Here, BRX and NJY are two triangles in which,
BR = 30 cm, RX = 40 cm, BX = 60 cm, NJ = 15 cm, JY = 20 cm and NY = 30 cm,
Also, m∠B = m∠N, m∠R = m∠J and m∠X = m∠Y,
By the property of congruence,
,
and 
Thus, By AAA similarity postulate,

Hence, proved.
For the answer to the question above, we'll have to use the formula for velocity which is v =<u> d
</u> t and d = vt
Since the total distance is given, therefore we can get the time and it will look like this.
220t + 250t = 1880
<u>470t</u> = <u>1880
</u>470 470
the problem indicates that the train runs 250kph
Let's substitute:
d = 250(4)
the distance traveled by Lian is 1000km
<u>
</u>