Answer:
Step-by-step explanation:
3/5/7=3/35
Let me use Python to find the answers
Program.

![\tt Prices=[75,120,120,145,150,150,150,175,175,200,225,275]](https://tex.z-dn.net/?f=%5Ctt%20Prices%3D%5B75%2C120%2C120%2C145%2C150%2C150%2C150%2C175%2C175%2C200%2C225%2C275%5D)



Output:;
- Mean=163.333
- Median=150.0
- Mode=150
Console attached
Well, you would istribute
using
a(b+c)=ab+ac
so
3(g+5)=3(g)+3(5)=3g+15
2(3g-6)=2(3g)+2(-6)=6g-12
now we gots
3g+15+6g-12
using commutative, we can arrange and group like terms because
a+b=b+a
so
3g+6g+15-12
add
9g+3
Answer:

Possible values of x: Any from 0 to 5.






Step-by-step explanation:
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.
In this question:

So

Possible values of x: 5 trials, so any value from 0 to 5.
For each value of x calculate p(☓ =x)






