Answer:
She payed $269 for the chainsaw :)
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:

V= 735.62
volume formula for sphere is v= 4/3 π r3
the radius is half of the diameter so v= 4/3 π 5.6 3
i think?
There are many factors, it is all a matter of preference, normally, you want to try to solve for the easiest one to get to
example
if y ou had
(x-3)^2+y=9
you would solve for y becuase it is less tricky
it is all a matter of preference
Answer:4b+5=1+5b
Step-by-step explanation: b= 14b I HOPE THIS HELPS AND HAVE A GOOD DAY!!!! And i really do hope this did help.