Answer:
0.625 = 62.5% probability that part B works for one year, given that part A works for one year.
Step-by-step explanation:
We use the conditional probability formula to solve this question. It is

In which
P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
P(A) is the probability of A happening.
The probability that part A works for one year is 0.8 and the probability that part B works for one year is 0.6.
This means that 
The probability that at least one part works for one year is 0.9.
This means that: 
We also have that:

So


Calculate the probability that part B works for one year, given that part A works for one year.

0.625 = 62.5% probability that part B works for one year, given that part A works for one year.
Answer:
m=-2/3 b=-3
Step-by-step explanation:
Answer:
$9.50h, $(9.50h +2.20rh)
Step-by-step explanation:
Every hour ----- $9.50
h hours ----- h ×$9.50
Amount of money Mr Smith earns in h hours
= $9.50h
If he fixes a watch per hour, additional income= $2.20h
If he fixes r watches per hour, additional income= $2.20rh
Total income
= $(9.50h +2.20rh)
A factor of 30 is chosen at random. What is the probability, as a decimal, that it is a 2-digit number?
The positive whole-number factors of 30 are:
1, 2, 3, 5, 6, 10, 15 and 30.
So, there are 8 of them. Of these, 3 have two digits. Writing each factor on a slip of paper, then putting the slips into a hat, and finally choosing one without looking, get that
P(factor of 30 chosen is a 2-digit number) = number of two-digit factors ÷ number of factors
=38=3×.125=.375