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:
maximum =6
amt boys = 3
amt girls =2
therefore expression =3+2=6
Domain ; <span><span>[7,∞)</span>,<span>{x|x≥7<span>}
Range : </span></span></span><span><span>[0,∞)</span>,<span>{y|y≥0<span>}</span></span></span>
<h3>
Answer: d. (4,3)</h3>
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Explanation:
Point C is located at (-6,3). The x coordinate is x = -6.
Going from x = -6 to x = -1 is 5 units to the right. We move another 5 units to the right to go from x = -1 to x = 4 when arriving at its reflected location (4,3). The y coordinate stays the same the entire time.
It might help to refer to the diagram below.