# Tag Info

## Hot answers tagged average

4

We have the following system of equations: $$x_1+\cdots+x_{11}=50\cdot11 \tag{1}$$ $$x_1+\cdots+x_{6}=48\cdot6 \tag{2}$$ $$x_6+\cdots+x_{11}=51\cdot6 \tag{3}$$ Adding (2) and (3), we obtain: $$x_1+\cdots+2x_6+\cdots +x_{11}=6\cdot(48+51)=6\cdot99.$$ Substituting in (1), $$x_6+50\cdot11=6\cdot99 \implies x_6=44.$$

3

The first one is correct from the perspective of interval arithmetic. Meaning that if you are absolutely certain that each volume is within $0.05$ mL of the true value, then you are absolutely certain that the average volume is within $0.05$ mL of the true value as well. This works out for the reason you described. The second one is never correct. Dividing ...

1

There are two possibilities here. Which one is correct depends on the context: Add up the ratios as you've said, then divide by the count of numbers you've added together (from your statement, it sounds like you have more than 1 data item per row - you need to count data items summed, not rows). You add up all the $\omega_f$ and add up all the $\omega_i$ ...

1

Suppose you have $1$ pound of super-lean beef. You then have $0.03$ pounds of fat in there. If you add $x$ pounds of lean beef, you then have $1+x$ pounds of beef, of which $0.03+0.1x$ pounds are fat. The condition you want to satisfy is that $0.03+0.1x$ is $8$ percent—that is, $0.08$—of $1+x$. Symbolically: $$0.03+0.1x = 0.08(1+x) = 0.08+0.08x$$ ...

1

I'd approach it in this way: Let $L$ be the total amount of lean beef, $L_f$ its net fat content. Same for $S$ (super lean) and $S_f$. And let $M$ be the mixture (and $M_f$ its fat). Then you know that $$L_f =0.1\, L$$ $$S_f=0.03 \, S$$ $$M_f=0.08 \,M$$ Also: $M=L+S$ and $M_f = L_f+S_f$ Then $$0.1 L + 0.03 S = 0.08 \, (L+S)$$ $$0.02\, L = 0.05 \, S ... 1 You say you take the average of all the averages, but I notice that you have a sample count column. Are these averages over different sample sizes? If so, then you would probably want a weighted average for your aggregate average:$$\text{Aggregate Average} =\frac{\sum_i (\text{sample size})_i(\text{average})_i}{\sum_i (\text{sample size})_i} But without ...

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