Desmos II: Systems, Tables, Regressions and Sliders

Typed as they stand, the lines cross at $(3,5)$. By hand: $y=2x-1$, so $4x+3(2x-1)=27$ gives $10x=30$, $x=3$ and $y=5$. $\boxed{(3,5)}$

The crossing reads $(3.818,\,2.273)$. Those are rounded. Doubling the first equation gives $6x+4y=32$; adding the second gives $11x=42$, so $x=\tfrac{42}{11}$ and $y=\tfrac{16-3x}{2}=\tfrac{25}{11}$. Check: $\tfrac{42}{11}=3.8\overline{18}$ and $\tfrac{25}{11}=2.\overline{27}$, matching the display. $\boxed{\left(\tfrac{42}{11},\tfrac{25}{11}\right)}$

Substituting: $x^2-4=3x-8$, so $x^2-3x+4=0$ with discriminant $9-16=-7<0$: no real solution. The graph shows the line passing entirely below the parabola. $\boxed{0}$

The differences are $4.2$ each time, so the model is $y=7+4.2x$, and $y(10)=7+42=\boxed{49}$. In Desmos, enter the table and then a new line y_1 ~ mx_1+b; the panel reports $m=4.2$, $b=7$, $R^2=1$.

Consecutive ratios are all $1.15$, so $P(t)=1200(1.15)^t$ and $P(6)=1200(1.15)^6=1200(2.313\ldots)=2775.67\ldots\approx\boxed{2776}$. An exponential regression returns $a=1200$, $b=1.15$, $R^2=1$ — the same model, obtained without spotting the ratio.

Here $\bar x=3$ and $\bar y=5.1$. The slope is $\dfrac{\sum(x-\bar x)(y-\bar y)}{\sum(x-\bar x)^2}=\dfrac{9.4}{10}=0.94$, and the intercept is $5.1-0.94(3)=2.28$. So $\hat y=2.28+0.94(8)=\boxed{9.8}$. Desmos returns exactly these two numbers in one step; the arithmetic above is shown only so that the output is recognisable rather than magic.

At $x=0$: (A) gives $3$, (B) gives $3$, (C) gives $0$, (D) gives $2$. Choices (C) and (D) are gone after one substitution. At $x=2$: (A) gives $9$, (B) gives $12$. The table says $12$. $\boxed{\text{(B)}}$ — two substitutions, no typing.

The slope is the predicted change in $y$ for a one-unit increase in $x$, so membership is predicted to rise by about $\boxed{4.2\text{ members per year}}$. The intercept $180$ is the predicted membership in $2010$. Answer choices in this family almost always include “total members” (that is $\hat y$) and “$4.2$ years” (units swapped).

The prediction is $\hat y=4.2(5)+180=201$, and the actual value is $208$, so the residual is $208-201=\boxed{7}$. A positive residual means the point sits above the line, so the model underestimates there.

Setting them equal: $x^2+k=4x-1$, so $x^2-4x+(k+1)=0$. One solution means discriminant zero: $16-4(k+1)=0$, giving $k+1=4$ and $\boxed{k=3}$. Dragging a slider on $k$ shows the parabola rising until it just touches the line at $k=3$.

Parallel lines need $\dfrac{3}{6}=\dfrac{k}{8}$, so $k=4$. Then the first equation doubles to $6x+8y=24$, which contradicts $6x+8y=5$, so the lines really are distinct and there is indeed no solution. $\boxed{k=4}$

The total is $127$, so the mean is $\tfrac{127}{9}=14.11\ldots$; the fifth of nine ordered values is the median, $12$. The mean is larger. $\boxed{\text{mean }14.11,\ \text{median }12}$ — the single value $40$ drags the mean up and leaves the median untouched, which is exactly what “skewed to the right” means.

$Q$ spreads from $4$ to $24$ while $P$ spreads from $10$ to $18$, so $Q$ has the greater standard deviation — $\texttt{stdev}$ returns about $3.16$ for $P$ and about $7.91$ for $Q$. Adding $5$ to every value slides both samples along the number line without changing any distance between values, so both standard deviations are unchanged while both means become $19$. $\boxed{Q\ \text{is more spread out; the standard deviations do not change}}$