Answer:
466mEq
Step-by-step explanation:
First, we need to know the concentration of NaCl in a normal saline solution, this is by definition 0.9%, meaning we have 0.9g of NaCl per 100ml of solution, we want to know how much NaCl we have in 3L (3000ml):

So, we have 27000mg in 3L of normal saline solution.
Now, acording to our milliequivalent (mEq) equation (
) where pE is de molecular mass of NaCl divided by their charges, in this case 1:

Finally we substitute in the mEq formula:

I hope you find this information useful! Good luck!
Answer:
11 will be shared out and there will be 2 rubbers left.
Step-by-step explanation:
the last number that 3 goes into before 35 is 33. 33 ÷ 3 = 11. So each friend gets 11 rubbers. then from 33, there will be 2 rubbers left.
Answer: it going to be 10.2% percent
Step-by-step explanation: that how you get your answer as your -
10.2 %
Actually it's
y =

it's a line. so 2 points is enough
choose x so that x+4 would be divisible by 3 :)
1) x = 2, then y = (2+6) / 3 = 2 - sketch point (2,2)
2) x = 5, then y = (5+4) / 3 = 3 - sketch point (5,3)
3) line goes though those points. use ruler :)
Answer:
C. Median
Step-by-step explanation:
If both sides are <em>symmetrical,</em> then there is no need to try and find the mean or anything because you already have the middle point and no outliers. <u>If there were outliers it'd be different</u>, unless the outliers were symmetrical with the rest of it, then the median would still be a good option.