Answer:
Step-by-step explanation:
You are to make 5 assemblies.
Each assembly requires the use of 1 Type A bolt.
To make the 5 assemblies, you need 5 Type A bolts.
The container of bolts has a total of 60 bolts.
The focus - Type A bolts - is 20 out of this 60.
The probability of obtaining a Type A bolt at all, is 20/60, which is = 1/3
(A) What is the probability of taking the exact number of Type A bolts you need for your 5 assemblies, if you randomly take 10 bolts from the container?
- The exact number of Type A bolts you need for the 5 assemblies is 5
1/3 × 5/10 = 5/30 = 1/6 = 0.167
(B) What is the probability of taking/having less than 5 Type A bolts out of the randomly selected 10 bolts? The solution is to sum up the following:
1/3 × 4/10 = 0.133
1/3 × 3/10 = 0.1
1/3 × 2/10 = 0.067
1/3 × 1/10 = 0.033
1/3 × 0/10 = 0
TOTAL = 0.333
<span>−1 1/3 - 1/6
= -1 2/6 - 1/6
= (-1 2/6) + (-1/6)
= -1 3/6
= -1 1/2</span>
Answer:
An example would be
(x+3)-(y+12)
The two constants are 3 and 12 (I'm assuming you meant "constant" as in the number that stands by itself and not "constatant" lol)
The two variables are x and y
The two operations are addition and subtraction
In words you could say "the quantity of x plus three minus the quantity of y plus twelve"
