Answer:
<u>12.5% ≈ 13%</u>
Step-by-step explanation:
The rest of the question is: There are 75 pills in the batch.
The current batch of pills is the first of the day and our goal is to produce a total of 600.
If there are 75 pills in the batch so percentage of goal completed will be
[(number of pills in the batch)/(Total pills)] * 100

≈ 13% to the nearest whole number
Answer:

Step-by-step explanation:
Given information:
- Total number of pupils = 20
- 3 pupils play the flute
- 6 pupils play the piano only
- 6 play both instruments
We can assume that 3 pupils play the flute <u>only</u> as we have also been told that 6 pupils play both instruments.
To calculate the probability that a randomly chosen pupil plays neither instrument, first determine how many pupils do not play an instrument by subtracting the number of pupils who do play an instrument from the total number of pupils:
⇒ total number of pupils - pupils who play instruments
⇒ 20 - (3 + 6 + 6)
⇒ 20 - 15
⇒ 5
Therefore, 5 pupils do not play the piano and/or flute.
To calculate probability:

Therefore:

Given system of inequalities: y > 2/5(x) - 2 and y < -3x + 1.
We are given greater than (>) inequality symbol in first inequality.
For "only greater than inequality symbol" we would draw only a "dotted line".
Also in second inequality we have less than (<) inequality symbol.
For "only greater than inequality symbol" we would draw only a "dotted line".
In the shared graph, both the lines plotted are solid lines.
Therefore, correct option is B) The lines graphed should be dashed.
Let x = blank
25 + 5x = 35
5x = 35 - 25
5x = 10
x = 10/5
x = 2
Done!
Answer:
Check the explanation
Step-by-step explanation:
1) Algorithm for finding the new optimal flux: 1. Let E' be the edges eh E for which f(e)>O, and let G = (V,E). Find in Gi a path Pi from s to u and a path
, from v to t.
2) [Special case: If
, and
have some edge e in common, then Piu[(u,v)}uPx has a directed cycle containing (u,v). In this instance, the flow along this cycle can be reduced by a single unit without any need to change the size of the overall flow. Return the resulting flow.]
3) Reduce flow by one unit along 
4) Run Ford-Fulkerson with this sterling flow.
Justification and running time: Say the original flow has see F. Lees ignore the special case (4 After step (3) Of the elgorithuk we have a legal flaw that satisfies the new capacity constraint and has see F-1. Step (4). FOrd-Fueerson, then gives us the optimal flow under the new cePacie co mint. However. we know this flow is at most F, end thus Ford-Fulkerson runs for just one iteration. Since each of the steps is linear, the total running time is linear, that is, O(lVl + lEl).