Answer:
Part A:
The probability that all of the balls selected are white:

Part B:
The conditional probability that the die landed on 3 if all the balls selected are white:

Step-by-step explanation:
A is the event all balls are white.
D_i is the dice outcome.
Sine the die is fair:
for i∈{1,2,3,4,5,6}
In case of 10 black and 5 white balls:






Part A:
The probability that all of the balls selected are white:


Part B:
The conditional probability that the die landed on 3 if all the balls selected are white:
We have to find 
The data required is calculated above:

Answer:
x=3
Step-by-step explanation:
f(x) = 2^x
Let f(x) =8
8 = 2^x
Rewriting 8 as 2^3
2^3 = 2^x
The bases are the same so the exponents are the same
3=x
Answer:
i can help
Step-by-step explanation:
Answer:
Option B. The value of x is 20
Step-by-step explanation:
we know that
<u>The intersecting chords theorem</u> states that the products of the lengths of the line segments on each chord are equal.
so
In this problem
