Answer:
<em>Most likely time, </em>according to PERT (Program evaluation and review technique).
Step-by-step explanation:
PERT is "a statistical tool used in <em>project management" (Program evaluation and review technique (2020), </em>in Wikipedia), and it is commonly used with CPM <em>(Critical Path Method)</em> to manage projects.
Inside PERT, there are different defined times to accomplished an activity in a project, that is:
- An <em>optimistic time</em> or minimum time required to accomplished an activity, i.e., if everything goes better than normal, the activity is accomplished before expected.
- A <em>pessimistic time, </em>a time quite the opposite to optimistic time.
- A <em>most likely time</em>, or a time required to accomplished an activity if everything goes as expected or normally.
- An <em>expected time</em>, an statistical estimation.
Considering the question, we have that the <em>time</em> when "the first module of the project could be completed":
- "[...] in as few as 15 days" is the <em>optimistic time</em>.
- "[...] or could take as many as 25 days" is the <em>pessimistic time</em>.
- "[...] but most likely will require 20 days" is the <em>most likely time</em>.
As a result, the <em>20-day estimate</em> is called the <em>most likely time</em> in the context of the PERT/CPM techniques.
3x - 10 < 2
3x < 2 + 10
3x < 12 Divide through by 3.
x < 12/3
x < 4
Answer: log₂
+ 1 = n
<u>Step-by-step explanation:</u>
h(n) = 6 * 2ⁿ⁻¹
<u> ÷6 </u> <u>÷6 </u>
= 2ⁿ⁻¹
log₂
= log₂(2ⁿ⁻¹) →→<em>log₂2 cancels out</em>
log₂
= n - 1
log₂
= n
2 x 3 = 6
6 x 3 = 18
18 x 3 = 54
The answer is 54
I'm not sure what you're really asking but 6 and 9's LCM is 18