How to eliminate this using multiplication?
1 answer:
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6 - a = 
(9a - 18)
First, simplify
to 
/ Your problem should look like: 
Second, multiply both sides by 3. / Your problem should look like:
Third, expand. / Your problem should look like:
Fourth, add 3a to both sides. / Your problem should look like:
Fifth, simplify 9a - 18 + 3a to 12a - 18. / Your problem should look like:
Sixth, add 18 to both sides. / Your problem should look like:
Seventh, add 18 + 18 to get 36. / Your problem should look like:
Eighth, divide both sides by 12. / Your problem should look like:
Ninth, since 3 goes into 12 to get 36, simplify
to 3. / Your problem should look like: 
Tenth, switch sides. / Your problem should look like:
Answer: a = 3 (one solution)
First, simplify
Second, multiply both sides by 3. / Your problem should look like:
Third, expand. / Your problem should look like:
Fourth, add 3a to both sides. / Your problem should look like:
Fifth, simplify 9a - 18 + 3a to 12a - 18. / Your problem should look like:
Sixth, add 18 to both sides. / Your problem should look like:
Seventh, add 18 + 18 to get 36. / Your problem should look like:
Eighth, divide both sides by 12. / Your problem should look like:
Ninth, since 3 goes into 12 to get 36, simplify
Tenth, switch sides. / Your problem should look like:
Answer: a = 3 (one solution)
Answer:
Step-by-step explanation:
For each component, there are only two possible outcomes. Either it fails, or it does not. The components are independent. We want to know how many outcomes until r failures. The expected value is given by
In which r is the number of failures we want and p is the probability of a failure.
In this problem, we have that:
r = 1 because we want the first failed unit.
The expected number of systems inspected until the first failed unit is 2.5