Given: \(n=200\), reported mean \(=40\) ⇒ reported total $$S_{\text{reported}}=200\times 40=8000.$$
Two scores were misread: true \((43,35)\) but used as \((34,53)\).
Correct the total: $$S_{\text{correct}}=S_{\text{reported}}-\big(34+53\big)+\big(43+35\big) =8000-87+78=7991.$$
Corrected mean: $$\bar{x}_{\text{correct}}=\frac{S_{\text{correct}}}{n} =\frac{7991}{200}=39.955\;\approx\;39.96.$$
Answer: \(39.955\) (≈ \(39.96\)).
AP: \(a,\,a+d,\,a+2d,\,\dots,\,a+2nd\) has \(2n+1\) terms. The mean is the middle term \(a+nd\).
Mean deviation from the mean (MD): average of absolute deviations from \(a+nd\).
Symmetry gives pairs at distances \(kd\) for \(k=1,\dots,n\). Hence
$$\text{MD}=\frac{1}{2n+1}\Big[2\sum_{k=1}^{n} kd\Big] =\frac{2d}{2n+1}\cdot\frac{n(n+1)}{2} =\frac{d\,n(n+1)}{2n+1}.$$Answer: \(\displaystyle \boxed{\frac{d\,n(n+1)}{2n+1}}\).
Sets: \(A=\{(x,y)\mid y=\tfrac{1}{x},\ x\in\mathbb{R}\setminus\{0\}\}\), \(B=\{(x,y)\mid y=-x,\ x\in\mathbb{R}\}\).
Intersection: Solve \( \frac{1}{x} = -x \) with \(x\neq 0\). \[ \frac{1}{x} = -x \;\Longrightarrow\; 1 = -x^2 \;\Longrightarrow\; x^2 = -1, \] which has no real solution.
Conclusion: \(A \cap B = \varnothing\) (they are disjoint in \(\mathbb{R}^2\)).
Note: Over complex numbers, the intersection would be at \(x=\pm i\), but for real \(x\), there is none.
Given averages: Boys = 52, Girls = 42, Combined = 50.
Let the proportion of boys be \(p\) (so girls = \(1-p\)). Weighted mean gives: $$52p + 42(1-p) = 50$$ $$52p + 42 - 42p = 50 \Rightarrow 10p = 8 \Rightarrow p = 0.8.$$
Percentage of boys \(= 0.8 \times 100\% = \boxed{80\%}\).
The number is 4-digit and must be between 4000 and 7000.
So the thousands digit can be: 4, 5, or 6.
The number must be even, so the units digit must be one of {0, 2, 4, 6, 8}.
Total = 224 + 280 + 224 = 728.
There are 728 such even integers.
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